Your SlideShare is downloading. ×
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies

3,876

Published on

This four-day course gives a solid practical and intuitive understanding of the fundamental concepts of discrete and continuous probability. It emphasizes visual aspects by using many graphical tools …

This four-day course gives a solid practical and intuitive understanding of the fundamental concepts of discrete and continuous probability. It emphasizes visual aspects by using many graphical tools such as Venn diagrams, descriptive tables, trees, and a unique 3-dimensional plot to illustrate the behavior of probability densities under coordinate transformations. Many relevant engineering applications are used to crystallize crucial probability concepts that commonly arise in aerospace CONOPS and tradeoffs

Published in: Technology, Education
0 Comments
1 Like
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total Views
3,876
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
15
Comments
0
Likes
1
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. Course Sampler From ATI Professional Development Short Course Fundamentals of Engineering Probability Visualization Techniques & MatLab Case Studies Instructor: Dr. Ralph E. MorgansternATI Course Schedule: http://www.ATIcourses.com/schedule.htmATIs Engineering Probability: http://www.aticourses.com/Fundamentals_of_Engineering_Probability.htm
  • 2. www.ATIcourses.comBoost Your Skills 349 Berkshire Drive Riva, Maryland 21140with On-Site Courses Telephone 1-888-501-2100 / (410) 965-8805Tailored to Your Needs Fax (410) 956-5785 Email: ATI@ATIcourses.comThe Applied Technology Institute specializes in training programs for technical professionals. Our courses keep youcurrent in the state-of-the-art technology that is essential to keep your company on the cutting edge in today’s highlycompetitive marketplace. Since 1984, ATI has earned the trust of training departments nationwide, and has presentedon-site training at the major Navy, Air Force and NASA centers, and for a large number of contractors. Our trainingincreases effectiveness and productivity. Learn from the proven best.For a Free On-Site Quote Visit Us At: http://www.ATIcourses.com/free_onsite_quote.aspFor Our Current Public Course Schedule Go To: http://www.ATIcourses.com/schedule.htm
  • 3. Fundamental Probability Concepts • Probabilistic Interpretation of Random Experiments (P) – Outcomes: sample space – Events: collection of outcomes (set theoretic) – Probability Measure: assign number “probability” P ε [0,1] to event • Dfn#1-Sample Space (S): Fine-grained enumeration (atomic - parameters) – List all possible outcomes of a random experiment – ME - Mutually exclusive - Disjoint “atomic” – CE - Collectively exhaustive - Covers all outcomes • Dfn#2- Event Space (E): Coarse-grained enumeration (re-group into sets) – ME & CE List of Events S (all outcomes) Atomic Outcomes Events: A,B,C ME but not CE A D (Disjoint by dfn) Events: A,B,C ,D both ME & CE C B 14 INDEXDiscrete parameters uniquely define the coordinates of the Sample Space (S) and the collection of allparameter coordinate values defines all the atomic outcomes. As such atomic outcomes are mutuallyexclusive (ME) and collectively exhaustive (CE) and constitute a fundamental representation of the SampleSpace S.By taking ranges of the parameters such as A, B, C, and D, one can define a more useful Event Space whichshould consist of ME and CE events which cover all outcomes in S without overlap as shown in the figure. 14
  • 4. Fair Dice Event Space Representations d2 • Coordinate Representation: 6 – Pair 6-sided dice 5 A: d1=3, d2 =arb. 4 – S={(d1,d2): d1,d2 = 1,2,…,6} 3 2 C: d1=d2 – 36 Outcomes Ordered pairs 1 d1 1 2 3 4 5 6 B: d1+d =7 • Matrix Representation: 1  [1 2 3 4 5 6]  (1,1) (1,2) (1,3) (1,4) (1,5) (1,2 )  6   (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) – Cartesian Product: 2   3   = (3,1)  (3,2) (3,3) (3,4) (3,5) (3,6)  – {d1} x {d2} = d1 d2T 4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)   (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) 5     (6,1)  (6,2) (6,3) (6,4) (6,5) (6,6)  6 • Tree Representation: d2 d1 (1,1) (1,2) 1 (1,3) 36 Outcomes (1,4) Ordered Pairs 2 (1,5) 3 (1,6) • Polynomial Generator for Sum Start 4 2 Dice 5 (6,1) (6,2) ( x1 + x 2 + x3 + x 4 + x5 + x 6 ) 2 = 1x 2 + 2 x3 + 3 x 4 + 4 x5 + 5 x 6 + 6 x 7 6 (6,3) (6,4) Exponents represent + 5 x8 + 4 x9 + 3 x10 + 2 x11 + 1x12 (6,5) (6,6) 6-sided die face numbers Exponents represent pair sums Coefficients represent #ways 16It is helpful to have simple visual representations of Sample and Event SpacesFor a pair of 6-sided dice, coordinate, matrix, and tree representations are all useful representations. Alsothe polynomial generator for the sum of a pair of 6-sided dice immediately gives probabilities for each sum.Squaring the polynomial (x1+x2+x3+x4+x5 +x6)2 yields a generator polynomial whose exponents representall possible sums for a pair of 6-sided dice S={2,3,4,5,6,7,8,9,10,11,12}and whose coefficients C={1,2,3,4,5,6,5,4,3,2,1} represent the number of ways each sum can occur. Dividing by the coefficients C bythe total #outcomes 62 = 36 yields the probability “distribution” for the pair of dice.Venn diagrams for two or three events are useful; for example, the coordinate representation in the topfigure can be used to visualize the following events A: {d1 = 3 and d2 = arbitrary, B= {d1 + d2 = 7}, and C= {d1 = d2}Once we display these two events on the coordinate diagram their intersection properties are obvious, viz.,both A & B and A & C intersect, albeit at different points, while B & C do not intersect (no pointcorresponding to sum=7 and equal dice values). More than three intersecting sets, become problematic forVenn diagrams as the advantage of visualization is muddled somewhat by the increasing number ofoverlapping regions in theses cases (see next two slides). 16
  • 5. Venn Diagram for 4 Sets 4C = (4C1 4-Singles) – (4C2 6-Pairs) + (4C3 4-Triples ) - ( 4C4 1-Quadruple) 0 A B AB BD AC ABD ABC AD ABCD BC ACD BCD CD D C 17As we go to Venn diagrams with more than 3 sets the labeling of regions becomes a practical limitation totheir use. In this case of 4 sets A,B,C, D, the labeling is still pretty straightforward and usable.The 4 singles A,B,C,D are labeled in an obvious manner at the edge of each circle.The 6 pairs AB,AC,AD,BC,BD,CD are labeled at the intersection of two circles. The 4 triples ABC, ABD,BCD, ACD are labeled within “curved triangular areas” corresponding to the intersections of three circles.The 1 quadruple ABCD is labeled within the unique “curved quadrilateral area” corresponding to theintersection of all four circles. 17
  • 6. Trivial Computation of Probabilities of Events sum = d1 + d2 d2 Ex#1 Pair of Dice E1 S={(d1,d2): d1,d2 = 1,2,…,6} 6 12 5 11 E2 10 E1={(d1,d2): d1+d2 ¥ 10} 4 9 8 P(E1)=6/36=1/6 3 7 6 2 5 E2={(d1,d2): d1+ d2 = 7} 4 P(E2)=6/36=1/6 1 3 2 d1 1 2 3 4 5 6 Ex#2 Two Spins on Calibrated Wheel S={(s1,s2): s1,s2 ε [0,1]} s2 E1={(s1,s2): s1+s2 ¥ 1.5}--> P(E1) = ----- =.52/2=1/8 1 1 E1 0.5 E3 E2={(s1,s2): s2 § .25} --> P(E2)=1(.25)/1=.25 E2 0 s1 E3={(s1,s2): s1= .85; s2= .35}--> P(E3)=0/1=0 0 0.5 1 20For equally likely atomic events the probability of any outcome Event is easily computed as the (#atomicoutcomes in Event)/(total # outcomes). For a pair of dice, the total # of outcomes is 6*6=36 and hencesimple counting of the # points in E /36 yields P(E), etc.Two spins on a calibrated wheel [0, 1) can be represented by the unit square in the (s1 , s2)-plane and ananalogous calculation can be performed to obtain the probability for the event E by dividing the areacovered by the event by the area of the event space (“1”): P(E)= area(E)/ 1. 20
  • 7. DeMorgans’ Formulas - Finite Unions and Intersections i) Compl(Union) = Intersec(Compls): ( E1 ∪ E2 ∪ c ∪ En ) c = E1 ∩ E2 ∩ c ∩ En c c c c ii) Compl(Intersec) = Union(Compls): ( E1 ∩ E2 ∩ ∩ En ) c = E1 ∪ E2 ∪ ∪ En Useful Forms: A∪ B i’) Union expressed ( A ∪ B) c = Ac B c Visualization Compl(Union) Intersec(Compl) ( A ∪ B)c as an Intersection (( A ∪ B) ) c c = A ∪ B = ( Ac B c ) c A Ac Intersect grey areas B Bc Ac & B c ii’) Intersection ( AB) c = Ac ∪ B c Ac B c Yields one Union(Compl) grey area Ac B c expressed as a Union Compl(Intersec) with A and B excluded (( AB) ) c c ( = AB = Ac ∪ B c )c Taking its complement ( Ac B c )c yields white area, i.e., A ∪ B 24 INDEXDeMorgan’s Laws for the complement of finite unions and intersections states thati) The complement of unions equals the intersections of the complements, andii) The complement of intersections equals the union of complementsThe alternate forms obtained by taking the complements of the original equations are often more useful because they give a direct decomposition of the union and the intersection of two or more setsi’) The union equals the complement of the (intersection of complements)ii’) The intersection equals the complement of the (union of complements)A graphical construction of A U B = (Ac Bc)c is also shown in the figure..Ac and Bc are the two shaded areas in the middle planes which exclude A and B respectively (white) ovalsIntersecting these two shaded areas and taking the complement leaves the white oval areas which is A U B 24
  • 8. Set Algebra Summary Graphic Union A ∪ B = A ∪ Ac B = B ∪ Bc A Union AUB “A-B” “B-A” Intersection A ∩ B = A ⋅ B = AB A Bc A AB B Ac B x ∈ AB iff x ∈ A & x ∈ B Intersection Difference A − B ≡ A ∩ B c = AB c x ∈ A − B iff x ∈ A and x ∉ B Differences DeMorgans A ∪ B = ( Ac B c )c ( A ∪ B )c = Ac B c means ( ) c AB = Ac ∪ B c complement of (At least one) = (not any) 27This summary graphic illustrates the set algebra for two sets A , B and their union intersection anddifference.DeMorgans Law can be interpreted as saying “the complement of (“at least one”) is “not any”Associativity and commutivity of the two operations allows extension to more than two sets. 27
  • 9. Basic Counting Principles Principle #0: Take Case n=3-4; generalize to n Binomial Expansion: (a+b)3 (a+b)n Repetitions Allowed Principle #1: Product Rule for Sub-experiments: 6- Bins = 263 ⋅103 m Num Suit Licenses ⋅ nm = ∏ nk 26 26 26 10 10 10 n = n1 ⋅ n2 H 1 D S C H 5 16- Bins k =1 Start 2 D Binary S 2 216 = 65,536 C H 13 D 2 2 2 2 ... 2 Generate “tree” of outcomes S C Digits #ways: 13 * 4 = 52 No Repetitions Principle #2: Perm n distinguish-obj take k k=n Arrange 11 Travel 5 Cooking 4 Garden All Books n! “Fill k-bins” 11! 5! 4! n Pk = (n) k = 3! Permute Groups (n − k )! k<n 11 Travel Books in 5 bins 11| 10| 9 |8 |7 Principle #3:Perm n-obj take n with r - Arrange 4! groups of indistinguishable objects Letters “TOOL” = 12 2!⋅1!⋅1! hable  # Distinguis n! 10!  Sequences  = n !⋅n !⋅ ⋅ n !   r − groups {4”r”, 3”s”, 2”o”, 1 “t”} 4!⋅3!⋅2!⋅1! = 12,600   1 2 r Principle #4: Combination of n-objects take k Committee of 4 22! 22! C4 = 22 = = 7315 from 22 people (22 − 4)!4! 18!4! n n! n Ck =   =  k  k ! ( n − k )! k ≤ n Order not   Committee of 3 {2M, 1F} 6⋅5 important! from {6M, 3F} 6 C2 ⋅3 C1 = ⋅ 3 = 45 2! = Principle #3 with {taken , not taken} not counted 28 INDEXOutcomes must be distinguished by labels. They are characterized by either i) distinct orderings or ii)distinct groupings. A grouping consists of objects with distinct labels; changing order within a group is nota new group, but is a new permutation. The four basic counting principles for groups of distinguishableobjects are summarized and examples of each are displayed in the table.Principle#0: This is practical advice to solve a problem with n= 2,3,4 objects first and then generalize the“solution pattern” to general n.Principle#1: This product rule is best understood in terms of the multiplicative nature of outcomes as we“branch out” on a tree. For a a single draw from a deck of cards there are 13 “number” branches and, inturn, each of these has 4 “suit” branches yielding 13*4 =52 distinguishable cards or outcomes.Principle#2: Permutation (ordering) of n objects take k at a time is best understood by setting up “k-containers” putting one of “n” in the first, one of “n-1” , ... and finally one of “n-k+1” in the kth container.The total #ways is obtained by the product rule as n*(n-1)*...*(n-k+1) = n!/(n-k)!Principle#3: Permutation of all ”n” objects consisting of “r “ groups of indistinguishable objects {3 t , 4s 5 u}. If all objects were distinguishable then the result would be n! permutations; however permutationswithin the r groups does not create new outcomes and therefore we divide by factorials of the numbers ineach group to obtain n!/(n1! n2! ... nr!)Principle#4: Combination of n objects take k is related to Principles#2, #3. There are n! permutations;ignoring permutations within r= 2 groups {“taken” , “not taken”} yields n!/(n! (n-k)!) 28
  • 10. Counting with Replacement Refills Drop Down Select “B” from Alphabet and Replace A A B B ... Y Y Z Z Always have 26 letters to choose from A A B B Y Y Z Z 23 =8 distinct 4 distinct Permutation of “n” obj with (# drws) orderings groupings replacement taken “k” at a time n Pk = # replaceable objects = nk A {AAA} 3 “A” B {AAB} 2 “A”& 1”B” A n n n n n…n A {ABA} 2 “A”& 1”B” A B B {ABB} 2 “B”& 1”A” Bin# 1 2 3 …k S A A {BAA} 2 “A”& 1”B” B n=2 , k=3 B B {BAB} 2 “B”& 1”A” A {BBA} 2 “B”& 1”A” B {BBB} 3 “B” Combination of “n” obj with replacement taken “k” at a time effective # objects  n + k − 1  n + k − 1 n Ck = / n + (k-1) = n + k −1 Ck =  =  Note: “k” can be larger than “n” (draw k)  k   n −1  Example: From 2 objects {A, B} choose 3 with replacement (Only Way!) After each draw of an A or B “drop 4 Outcomes down a replacement” add 1 after each A B A/B A/B {AAA},{BBB} draw except last 4! {ABB},{AAB} (effective # objects) = 2 +(3-1)=4 2 C3 = 2+3−1C3 = 4C3 = / =4 3! 1! 41 INDEXCounting permutations and combinations with replacement is analogous to a candy machine purchase inwhich a new object drops down to replace the one that has been drawn, thus giving the same number ofchoices in each draw.Permutation of n obj taken k at a time with replacement: Each of the k draws has the same number ofoutcomes n because of replacement, the result is n*n*n... *n = nk and is written nPk with an “over-slash” onthe permutation symbol. The case n=2, k=3 of 3 draws with 2 replaceable objects {A,B} shows the slash-2 P3 =23 = 8 permutations that result.Combination of n obj taken k at a time with replacement: For n=2, k=3, 2 take 3 does not make anysense. However, with replacement, it does since each draw except the last drops down an identical item andhence the number of items to choose from becomes n +(k-1) and slash-nCk = n+(k-1)Ck. The tree verifies thisformula and explicitly shows that there are 4 distinct groupings {3A, 3B, 2A1B, 1A2B} exactly the numberof combinations with replacement given by the general formula slash-2C3 = 2+(3-1)C3 = 4C3 =4 41
  • 11. II) Fundamentals of Probability 1. Axioms 2. Formulations: Classical, Frequentist, Bayesian, Ad Hoc 3. Adding Probabilities: Inclusion / Exclusion, CE & ME 4. Application of Venn Diagrams & Trees 5. Conditional Probability & Bayes’ “Inverse Probability” 6. Independent versus Disjoint Events 7. System Reliability Analysis 47As a theory, Probability is based on a small set of axioms which set forth fundamental properties ofconstruction.In practice, probability may be formulated theoretically, experimentally, or subjectively, but must alwaysobey the basic Axioms.Evaluating probabilities for events, is naturally developed in terms of their unions and intersections usingVenn Diagrams, Trees and Inclusion/Exclusion techniques.Conditional probabilities, their inverses (Bayes’ theorem), and the dependence between two or more eventsflow naturally from the basic axioms of probability.System reliability analysis utilizes all these fundamental concepts 47
  • 12. Inclusion / Exclusion Ideas ME Events A,B - Disjoint AB= φ A B P(A∩B) = P(A) + P(B) No intersections ”Add Prob” No intersections Intersect: “CE, not ME” “Recast” as Disjoint Union “CE & ME” Not Disjoint AB∫φ A A B-A B ∫ AB P(A∩B) = P(A) + P(B-A) = P(A) + P(BAc) Intersection “AB” Counted Twice!! P(A∩B) ∫ P(A) + P(B) B = B ⋅ S = B ⋅ ( A ∪ Ac ) = BA ∪ BAc Subtract “P(AB)” from sum; count only once A BAC B P ( A ∪ B ) = P ( A) + P ( B ) − P ( AB ) AB P( BAc ) = P( B) − P( AB) Generalization by Induction: let D = B ∪ C P ( A ∪ B ∪ C ) = P ( A ∪ D ) = P ( A) + P ( D) − P ( AD ) = P ( A) + P ( B ∪ C ) − P( A ⋅ ( B ∪ C )) = P ( A) + {P ( B ) + P (C ) − P ( BC )} − {P ( AB ) + P ( AC ) − P ( ABAC )} Inclusion / P ( A ∪ B ∪ C ) = P ( A) + P ( B ) + P (C ) − P ( AB ) − P ( AC ) − P ( BC ) + P ( ABC ) Exclusion add singles subtract pairs add triples 54 INDEXIt is important to realize that although probabilities are simply numbers that add, the probability of theunion of two events P(A U B) is not equal to the sum of individual probabilities for the two events P(A) +P(B).This is because points in this overlap region AB are counted twice; to correct for this one needs to subtractout “once” the double counted points in the overlap yielding P(A U B) = P(A) + P(B)-P(AB).Only in the case of non-intersection AB = φ does the simple sum of probabilities hold.The generalization for a union of three or more sets alternates inclusion and exclusion; for A,B,C theprobability P(AUBUC) adds the singles, subtracts the doubles and adds the triple as shown. 54
  • 13. Venn Diagram Application: Inclusion/Exclusion Given following information find how many club members play at least one sport T or S or B T (36) TS (22) S (28) Club: 36 T , 28 S, 18 B TSB (4) SB (9) Let N= Total # members (unknown) TB (12) 36 28 18 B (18) Write Probabilities as P(T) = ; P(S) = ; P(B) = ; etc. N N N CLUB Method 1: Subs into Formula for Union P ( T ∪ S ∪ B) = P (T ) + P( S ) + P( B ) − P (TS ) − P(TB ) − P ( BS ) + P (TBS ) 36 28 18 22 12 9 4 TS (22) = + + − − − + T (36) STc (6) N N N N N N N 43 18 1 = Thus 43 of “N” Club Members play 6 TSB N at least one sport. (N is irrelevant) (4) 5 8 SB (9) Method 2: Disjoint Union - Graphical TB (12) 1 T ∪ S ∪ B = T ∪ ST ∪ BT Sc c c BTcSc (1) CLUB P(T ∪ S ∪ B) = P(T ) + P( ST c ) + P( BT c S c ) 36 6 1 43 = + + = N N N N 68 INDEXThis example illustrates the ease by which a Venn diagram can display the probabilities associated with thevarious intersections of 3 sets T, S, and B.The number of elements in each of the 7 distinct regions is easily read off the figure; they are required toestablish the total number in their union T U S U B via the inclusion/exclusion formula.Another method of finding P(T U S U B ) is to decompose the union T U S U B into a union of disjoint setsT* U S* U B* for which the probability is additive, i.e., P(T* U S* U B* ) = P(T*) + P(U*) + P(B*). 68
  • 14. Matching Problem – 1 “N” men throw hats onto floor; Each man in turn randomly draws a hat a) No Matches - Find Probability None draw own hat. Let Event Ei = ith man chooses his own hat ; compute: P(0 − matches) = 1 − P( E1 ∪ E2 ∪ ∪ EN ) 1|2|3|… | k | k+1 | … |N Hats i1 |i2 | i3 | … | in in+1 | in+2 | in+3 | … | iN Men Probability that M1 & M2 &...&Mn irrespective of what n “Ei s” choose own hats (N-n) Does not Matter draw own hats other men draw (Matched or Not Matched ) Total # of“n-tuple” N # perms ( N − n)!   P( Ei1 Ei2 Ein ) = = selections from N n Total# perms N!    N  ( N − n)! N! ( N − n)! 1 Sum Joint Probabilities ∑ P( Ei1 Ei2 Ein ) =   ⋅ = n !( N − n)! N ! = over all “n-tuples” n −tuples All n-tuples Eq. Likely n N! n!   P (0 − Matches ) = 1 − P ( E1 ∪ E2 ∪ E3 ) = 1 −  ∑ P ( Ei1 ) − ∑ P ( Ei1 Ei2 ) + ∑ P( Ei1 Ei2 Ei3 ) = 1 − {1 − 2! + 3!} = 1 1 1 3 1− tuples pairs triples  P(0 − matches) = 1 − P( E1 ∪ E2 ∪ ∪ EN ) = 1 − 1 + 1 − 2! 3! 4! 5! 1 + + ( −1) N 1 N!  e−1 N →∞ → b) k- Matches Poisson with success rate λ=1/N & “time  k! ⋅ e−1 →1 1 1 1 N −k 1   − + + + ( −1)  ( N − k )!  P(k matches) =  2! 3! 4! N→∞ intvl” t = N samples; a=λ *t =(1/N)*N =1 k! 69 INDEXHere is an example that requires the inclusion/exclusion expansion for a large number of intersecting sets.Since it becomes increasingly difficult to use Venn diagrams for a large number of intersecting sets, wemust use the set theoretic expansion to compute the probability. We shall spend some time on this problemas it is very rich in probability concepts.The problem statement is simple enough: “N men throw their hats onto the floor; each man in turnrandomly draws a hat. “a) What is the probability that no man draws his own hat?b) What is the probability of exactly k-matches?Key ideas: define Event Ei = ith man selects his own hat then take union of N sets E1 U E2 U ... U EN and P(no-matches)=1- P(E1 U E2 U ... U EN)The expansion of the P(E1 U E2 U ... U EN) involves addition and subtraction of P(singles), P(pairs),P(triples), etc. ( The events Ei are CE but not ME so you cannot simply sum up the P(Ei ) for k singles toobtain an answer to part b)) .This slide shows a key part of the proof which establishes the very simple result that the sum over singles,P(singles) = 1/(1!); sum over pairs is P(pairs)= 1/(2!) ; sum over triples is P(triples)=1/(3!); sum over 4-tuples, P(4-tuples) = 1/(4!); ... sum over N-tuples, P(N-tuple) = 1/(N!).Limit as N large approaches a Poisson Distribution with success rate for each draw λ=1/N and data lengtht =N i.e., parameter a =λ t =1 69
  • 15. Man Hat Problem n =3 Tree/Table Counting M#1 M#2 M#3 M.E. Match Tree#1 Drw#1 Drw#2 Drw#3 Outcomes Outcomes M#1 M#2 M#3 #Matches E2 1 2 3 E3 {E1 E2 E3 } triple 1 2 3 3 1/2 Br#1 1 E1 1/2 3 2 c {E1 E2 E3 } c single 1 3 2 1 1/3 1 1/2 1 1 3 E3 {E1c E2 c E3 } single 2 1 3 1 E1C Br#2 Start 1/3 2 1/2 1 1 c c {E1 E2 E3 } c No-match 2 3 1 0 3 1/3 1/2 3 1 1 2 c c {E1 E2 E3 } c No-match 3 1 2 0 Br#3 E1C 1/2 2 E2 1 1 c {E1 E2 E3 } c single 3 2 1 1 P(Ei) = 1/3 2/6 2/6 From Table: From Tree: Connection: Matches & Events Prob[0-matches]=2/6 Prob[0-matches]=1-Pr[E1 U E2 U E3] Prob[1-matches]=3/6 Prob[Sgls]=P[E1]=P[E2]=P[E3]=1/3 =1-{Sum[Sngls]-Sum[Dbls]+Sum[Trpls]} Prob[2-matches]=0/6=0 Prob[Dbls] = P[E1E2]=(1/3)(1/2)=1/6 =1-{3(1/3) -3(1/6)+1(1/6)}=2/6 Prob[3-matches]=1/6 Prob[Trpls] = P[E1E2E3]=(1/3)(1/2)=1/6 Alternate Trees Yield: P[E1E3]= P[E2E3]=1/6 75This slide shows the complete the tree and associated table for the Man - Hat problem in which n=3 menthrow their hats in the center of a room and then randomly select a hat. The drawing order is fixed asMan#1, Man#2, Man #3, and the 1st column of nodes labeled as circled 1, 2, 3 shows the event E1 in whichthe Man#1draws his own hat, and the complementary event E1c i.e., Man#1 does not draw his own hat . The2nd column of nodes corresponds to the remaining two hats in each branch shows the event E2 in which theMan#2 draws his own hat; note that E2 has two contributions of 1/6 summing to 1/3. Similarly, the 3rd drawresults in the event E3 in two positions shown again summing to 1/3.The tree yields ME & CE outcomes expressed as composite states such as {E1E2E3}, {E1E2cE3c, etc., orequivalently in terms of the number of matches in the next column. The nodal sequence in the tree can betranslated into the table on the right which is analogous to the table we used on the previous slide. Thenumber of matches can be counted directly from the table as shown.The lower half of the slide compares the “ # of matches” events with the “compound events” formed fromthe “Ei”s{ no-matches, singles, pairs, and triples }. The connection between these two types of events isbased on the common event “no-matches,” i.e., the inclusion/exclusion expansion of the expression [1-P(E1U E2U E3) ] in terms of singles doubles and triples yields P(0-matches). 75
  • 16. Conditional Probability - Definition & Properties ˆ P ( AS )  2 • Definition of Conditional Probability ˆ P( A | S ) ≡ =  ˆ P( S )  3 • In terms of atomic events si we can formally write ˆ ˆ P( ∪ si S ) ∑ P( s S ) ˆ i (# pts in Sˆ & A) A = ∪ si ˆ ) = P ( A S ) = si ∈ A = si ∈ A = si ∈ A P( A | S ˆ P( S ) ˆ P( S ) ˆ P( S ) (# pts in Sˆ ) ˆ • Note in case S = S it reduces to P(A) as it must A B •Asymmetry of Conditional Probability BA P(BA) P ( BA)  fraction  BA P ( B | A) = = = P ( A)  BA over A    A Given A Not Symmetrical! P( BA)  fraction  BA P( A | B) = = = P( B)  BA over B    Given B B 82 INDEXThe formal definition of conditional probability follows directly from the renormalization concept discussedon the previous slide. It is simply the joint probability defined on the intersection of the set A and S-cap,P(AS-cap) divided by the normalizing probability P(S-cap).It can also be written explicitly in terms of a sum over atomic events given in the second equation.Conditional probability is not symmetric because the joint probability on the intersection of A and B isdivided by probability of the conditioning set which is P(A) in one case and P(B) in the other. This is alsoeasily visualized using Venn diagrams where the “shape division” are obviously different in the two cases. 82
  • 17. Examples - Coin Flips, 3-Sided Dice nH > nT Flip#3 Example#1: Three Coin Flips Flip#2 H {HHH} Given the first flip is H, Find Flip#1 H T {HHT} ˆ S Prob #H > #T H {HTH} H T T {HTT} #H > #T S 4 1 1 1 ˆ P ( S ) = ; P( HHH ) = ; P ( HHT ) = ; P( HTH ) = S H H {THH} 8 8 8 8 T T T {THT} 3 P ( HHH ) + P ( HHT ) + P ( HTH ) 3 = 8= H {TTH} P (nH > nT | H ) = ˆ) P( S 4 4 T 8 {TTT} Example#2: 4-Sided Dice Given the first “die” d1= 4” d1 d2 Find Prob of Event A: “d2= 4” 1 P(d2=4| d1= 4)=? 2 S S 3 (4,1) ˆ 4 1 4 (4,2) ˆ S P ( S ) = P( d1 = 4) = ; P( 4,4) = (4,3) 16 16 d2 (4,4) A 1 4 P(4,4) 1 P (d 2 = 4 | d1 = 4) = = 16 = ˆ P( S ) 4 4 3 ˆ S Reduced 16 2 Sample space 1 d1 1 2 3 4 83Here are two examples illustrating conditional probability.The first involves a series of three coin flips and a tree shows all possible outcomes for the original space S.The reduced set of outcomes conditions on the statement “ 1st draw is a head (red circle)” and S-cap onlytakes the upper branch of the tree and leads to a reduced set of outcomes. The conditional probability iscomputed either by considering outcomes in this conditioning space S-cap or by computing the probabilityfor S (the whole tree) and then renormalizing by the probability for S-cap ( upper branch). The second example involves the throw of a pair 4-sided dice and asks for the probability that d2 =4 giventhat d1=4, P(d2 =4 | d1 =4 ). The answer is obtained directly from the definition of conditional probabilityand is illustrated using a tree and a coordinate representation of the dice sample space with a Venn diagramoverlay for the event (d1, d2) = (4,4) (green) and the subspace S-cap {d1=4} (red rectangle). 83
  • 18. Probability of Winning in the “Game of Craps” Rules for the “Game of Craps” First Throw - dice sum=(d1+d2) Subsequent Throws - dice sum=(d1+d2) 2, 3, 12 - “Lose” (L) “Point” - “Win” (W) 7, 11 - “Win” (W) 7 “Lose” (L) Other (O) - first time defines your “Point” = “5” say Other (O) “Throw Again” Thr#1 2 L Thr#2 Thr#3 Thr#4 4 S=d1+d2 #Ways #Prob 3 L 36 5 2, 12 1 1/36 4 W 4 6 5 Point L 3, 11 2 2/36 36 7 36 5 W 6 26 4 P Start O 6 o 4, 10 3 3/36 7 W 7 L 36 36 36 5 W i 8 26 6 n 5, 9 4 4/36 O 9 7 L t 36 36 s 6, 8 5 5/36 10 26 O 11 7 6 6/36 W 36 12 L   4  1  2 2 3 4 4  26  4  26  4  26  P (W | 5) = +  +   +   + =  = 36 36  36  36  36  36  36  36  1 − 26  5    36  P(W ) = P(7) + P(11) + ∑ P(W | Point )P(Point ) Points 6 2   = + + 2  P(W | 4) P (4) + P(W | 5) P (5) + P (W | 6) P(6) = .4929 36 36  1/ 3  3 / 36 2/5 4 / 36 5 / 11 5 / 36   85 INDEXHere we compute the probability of winning the game of craps previously described by the rules for the 1stand subsequent throws given in the box and illustrated by the tree. Since there are 36 equally likelyoutcomes the #ways for the two dice summing to either 2 or 12 is obviously 1/36, for 3 or 11 it is 2/36, andthe remaining sums of two dice can be read directly off the sum axis coordinate representation and aredisplayed in the table on the right.We have labeled the partial tree “given the point 5” by their conditional probabilities derived from the table.The probability for the three outcomes W(“5”), L (“7”), “Other (not “5 or 7”) can be read off the table asP(5)= 4/36, P(7)=6/36, P(Other)= 1-(4+6)/36 =26/36. Note that these are actually conditional probabilities;but since the throws are independent the conditionals are the same as the a prioris as taken from the table.The P(W|5) is obtained by summing all paths that lead to a win on this “infinite tree”. Thus the 2nd throwyields W with probability 4/36 and the 3rd throw yields W with probability P(5|Other)P(5)=(26/36)(4/36),and the 4th throw yields W with probability P(5|Other,Other)P(5)=(26/36)2 (4/36), ... leading to an infinitegeometric series which sums to (4/36)*1/(1-26/36)=2/5.The total probability of winning is the sum of winning on the 1st throw (“7” or “11”) plus winning on thesubsequent throws for each possible “point.” The infinite sum for the other points is obtained in a similarmanner to that for “5” and (taking points by pairs in the table leads to the factor of two) the final result isshown to be .4929, i.e., a 49.3% chance of winning! 85
  • 19. Visualization of Joint, Conditional, & Total Probability Binary Comm Signal - 2 Levels {0,1} Binary Decision - {R0, R1}={(“0” rcvd , “1” rcvd} x = 0,1 Joint Probability (Symmetric) 0 1 sent P(0,R0) = P(R0,0) ovly R1 “0” sent & R0 (“0” rcvd ) & y =R0 ,R1 R0 rcvd R0 (“0” rcvd ) “0” sent Conditional Probability 0R1 (Non-Symmetric) R0 ,R1 1R1 Joint P(0|R0) ∫ P(R0|0) 0R0 1R0 “0” sent given R0 (“0” rcvd ) x = 0 ,1 P(0) = P(0, R0 ) + P(0, R1 ) P(R0 ) = P(R0 ,0) + P(R0 ,1) R0 (“0” rcvd ) given “0” sent Total Probability P(0) Total Probability P(R0) sum up joint on R0,R1 sum across joint on 0,1 Conditional Probability P( R0 ,0) P( R0 ,0) P ( R0 | 0) ≡ = Requires Total Probability P ( 0) P( R0 ,0) + P( R0 ,1) Re-normalize Joint Probability P(0), P(R0), etc. P( R0 ,0) P ( R0 ,0) P (0 | R0 ) ≡ = P ( R0 ) P ( R0 ,0) + P ( R0 ,1) 88 INDEXAnother way to visualize the communication channel is in terms of an overlay of a Signal Plane divided(equally) into “0”s and “1”s and a Detection Plane which characterizes how the “0”s and “1”s are detectedand is structured as shown so that when we overlay the two planes we obtain an Outcome Plane with fourdistinct regions whose areas represent probabilities of the four product (joint) states { 0R0, 0R1, 1R0, 1R1}(similar to the tree outputs).In this representation the total probability of a “0” P(0) can be thought of as decomposed into two partssummed vertically over the “0”-half of the bottom plane shown by the break arrow P(0) = P(0,R0) + P(0,R1)[Note: summing on the “1”-half of the bottom plane yields P(1) = P(1,R0) + P(1,R1).]Similarly the total probability P(R0) can be thought of as decomposed into two parts summed horizontallyover the “R0”-portion of the bottom plane shown by the break arrow P(R0) = P(R0,0) + P(R0,1); similarlywe have P(R1) = P(R1,0) + P(R1,1).The Total Probability of a given state is obtained by performing such sums over all joint states. 88
  • 20. Log-Odds Ratio - Add & Subtract Measurement Information Note: Revisit Binary Comm Channel P( R0 | 0) = .95 P ( R1 | 1) = .90 P(0)=.5 E = “1” P( R1 | 0) = .05 P ( R0 | 1) = .10 P(1)=.5 Ec = “0” Relation between  P (1 | R1 )  P (1 | R1 ) e L1 L1 ≡ ln 1 − P(1 | R )  ⇒ e = 1 − P(1 | R ) ⇒  L1 P(1 | R1 ) = L1 and P(1|R1)  1  1 1 + e L1  P(1 | R1 )   P (1)   P ( R1 | 1)   P(1)   P( R1 | 1)  L1 ≡ ln 1 − P(1 | R )  = ln 1 − P(1)  + ln 1 − P( R | 1)  = ln P(0)  + ln P ( R | 0)            1     1     1  ≡ L0 ≡ ∆L1  P( R1 | 1)  Additive Meas Updates for L Lnew = Lold + ∆LR1  P (1)   P(0)  ; ∆LR1 = ln P( R | 0)  Lold = ln       1  Updates Meas#1: R1 Meas#2: R0 Alternate Meas#2: R1  .5   P( R0 | 1)   .10   P( R1 |1)  Lold = ln  = 0 ∆LR0 = ln  .90   .5   P( R | 0)  = ln .95     ∆LR1 = ln   = ln    0   P( R1 | 0)   .05   .9  = −2.25129 ∆LR1 = ln  = +2.8903  .05  Lnew = Lold + ∆LR0 Lnew = Lold + ∆LR1 = 2.8903 = 2.8903 + (−2.25129) = .63901 = 2.8903 + 2.8903 = 5.7806 Lnew = 0 + 2.8903 e 2.8903 e.63901 e 5.7806 P(1 | R1 ) = = .947 P(1 | R1 R0 ) = = .655 P (1 | R1 R0 ) = = .997 1 + e 2.8903 1 + e.63901 1 + e 5.7806 96 INDEXRevisiting the binary communication channel we now compute updates using the log odds ratio which areadditive updates. The update equation simply starts from the initial log odds ratio which isLold=ln[P(1)/P(1c)] =ln(.5/.5)=0 for the communication channel. There are two measurement types R1 andR0 and each adds an increment ∆L determined by its measurement statistics, viz.,R1: ∆LR1 =ln[(P(R1|1)/P(R1|1c)]=ln(.90/.05) = +2.8903 (positive “confirming”)R0: ∆LR0 = ln[(P(R0|1)/P(R0|1c)]=ln(.10/.95)= -2.25129. (negative “refuting”)The table illustrates how easy it is to accumulate the results of two measurements R1 followed by R0 by justadding the two ∆Ls to obtainLnew= 0+2.8903-2.25129=.63901,or alternately R1 followed by R1 to obtainLnew=0+2.8903+2.8903=5.7806.These log odds ratios are converted to actual probabilities by computing P= eLnew / (1+ eLnew ) yielding .655and .997 for the above two cases.If we want to find the number of R1 measurements needed to give .99999 probability of “1” we need onlyconvert .99999 to an L =ln[(.99999)/(1-.99999)] =11.51 and divide the result by 2.8903 to find 3.98 so that4 R1 measurements are sufficient. 96
  • 21. Discrete Random Variables (RV) –Key Concepts • Discrete RVs: A series of measurements of random events • Characteristics: “Moments:” Mean and Std Deviation • Prob Mass Fcn: (PMF), Joint, Marginal, Conditional PMFs • Cumulative Distr Fcn: (CDF) i) Btwn 0 and 1, ii) Non-decreasing • Independence of two RVs • Transformations - Derived RVs • Expected Values (for given PMF) • Relationships Btwn two RVs: Correlations • Common PMFs Table • Applications of Common PMFs • Sums & Convolution: Polynomial Multiplication • Generating Function: Concept & Examples 122 INDEXThis slide gives a glossary of some of the key concepts involving random variables (RVs) which we shalldiscuss in detail in this section. Physical phenomena are always subject to some random components sothat RVs must appear in any realistic model and hence their statistical properties provide a framework foranalysis of multiple experiments using the same model. These concepts provide the rich environment thatallows analysis of complex random systems with several RVs by defining the distributions associated withtheir sums and transformations of these distributions inherent in the mathematical equations that are used tomodel the system.At any instant, a RV takes on a single random value and represents one sample from the underlying RVdistribution defined by its probability mass function (PMF). Often we need to know the probability for somerange of values of a RV and this is found by summing the individual probability values of the PMF; thus acumulative distribution function (CDF) is defined to handle such sums. The CDF formally characterizes thediscrete RV in terms of a quasi-continuous function that ranges between [0,1] and which has a uniqueinverse.Distributions can also be characterized by single numbers rather than PMFs or CDFs and this leads toconcepts of mean values, standard deviations, correlations between pairs of RVs and expected values.There are a number of fundamental PMFs used to describe physical phenomena and these common PMFswill be compared and illustrated through examples. Finally, the relationship between the sum of two RVsand the concept of convolution and the generating function for RVs will be discussed. 122
  • 22. Transformation of Sample Space: Sum & Difference - 4-Sided Dice Fair 4-sided dice thrown twice: RVs: Sum= “S” & Absolute Difference “D” Uniform PMF pD1D2 (d1,d2) = 1/16 Find New PMF pDS(d,s) = ? Labels: D/S=3/5 d pS(6) Collapse on s- d2 S=d1+d2 Rotated to D, S “missing” axis points D=|d2-d1| Coordinates 4 3/5 2/6 1/7 0/8 4 3 2/16 Collapse on d-axis pD(3) 2/4 1/5 0/6 1/7 2/16 2/16 d2 2 3 D 2/16 2/16 2/16 Collapse on 1 4 1/3 d-axis pD(1) 0/4 1/5 2/6 2 Fold over s 1/16 1/16 1/16 1/16 3 0 3/ 5 0/2 1/3 2/4 3/5 D/S=3/5 S-Axis 2 2/ 0 1 2 3 4 5 6 7 8 2/ 1 6 4 1 1/ 1/ d1 7 1/ 5 3 0/ d 0/ 8 6 0/ 1 2 3 4 0/ S 2 4 pSD ( s, d ) 4 1/ 7 1/ 1/ 5 3 1 pD1D2(d1,d2) d 2/ 2/ 3 6 4 2 2 3/ 2 4 5 3 1 3 D /S 1/16 =3 0 /5 2 4 d1 2/16 2/16 2/16 1 Absolute Difference Doubles 1/ 2/ 2/ 2/ 6 1 1/16 0 6 1 6 1 6 1 Values above S-Axis 1 2/16 2/16 2 1/ 6 1 2/ 6 1 2/ 6 1 3 1/16 1 4 2/ 2/16 2 5 1/ 6 1 6 1 3 6 1/16 4 7 1/ 6 1 8 1/16 d 1 s 125 INDEXIn the game with 4-sided dice, we are interested in the distribution of the sum random variable S = D1 + D2 ,pS(s) and not the joint distribution pD1,D2(d1d2). This slide and several to follow illustrate the procedure forobtaining the desired “marginal” (or collapsed ) distribution pS(s). In the process, we shall develop therelationship between distributions under transformation of coordinates, and define conditional, andmarginal, distributions involving a pair of RVs {D1,D2}.We start with the 2- and 3-dimensional dice representations of equally likely outcomes of 1/16 as shown onthe left. Recall that the points (d1, d2) for dice outcomes may alternately be expressed by points (s,d) theirsum and difference coordinates, where s = d1+ d2 and d = d2 - d1 . These coordinate axes are shown in thetop left figure where the sum and difference each take on 7 values: s={2,3,4,5,6,7,8} and d={-3,-2,-1,0,1,2,3}We consider a slightly different transformation s = d1+ d2 and |d| = |d2 - d1| and now the absolute difference|d| takes on only 4 values {0,1,2,3}; this has the effect of doubling the probability values of {1,2,3} byfolding over the negative difference values onto and doubling them. If we label each point in this figure bythe “|d |/ s” values we see for example that the points (d1d2) =(1,4) and (d1d2) =(4,1) at opposite corners ofthe grid are both now labeled with |d| / s = 3 / 5 . Labeling all points in this manner and rotating the figureclockwise 90o so D is up and S is to the right (central figure) we have found the new joint distributionpSD(s,|d|) as illustrated in the two right figures where points are now labeled by (s,|d|) values. Note that thenew distribution has doubled the positive d values to 2/16 each and that certain coordinate points(s,|d|)=(3,0) are not occupied (green). The marginal distribution pS(s) defined as the sum of the jointdistribution pSD(s,|d|) over all |d| values and is easily picked off the upper right figure by collapsing valuesdown along the s-axis. Similarly, the distribution pD(|d|) defined as the sum of the joint distributionpSD(s,|d|) over all s-values. The table shows the results. 125
  • 23. Common PMFs and Properties -1 RV Name PMF Mean Variance E[ X ] = ∑ x⋅ p x = 0 ,1 X ( x) var( X ) = E[ X 2 ] − E[ X ]2  p X = 1 (success) Bernoulli p X ( x) =  1 − p = q X = 0 (failure) E [ X 2 ] = 0 2 ⋅ (1 − p ) + 12 ⋅ p 1-Trial E [ X ] = 0 ⋅ (1 − p ) + 1 ⋅ p = p X=x succ. = p var( X ) = p − p 2 = p (1 − p ) “0” or “1” x “Atomic” RV successes = pq 0 1 p X (x) Binomial  n p X ( x) =   p x q n − x    x n n n - Trials E[ X ] = ∑ x  p x q n − x 6/16 5/16   var( X ) = npq x = 0,1, n x=0  x  X=x Succ. 4/16 3/16 = np How many Independent 2/1 6 1/16 succ “x” in Bernoulli Trials 0 x “n” trials ? 0 12 3 4 p X (x) Geometric p X ( x) =  pq x −1 x = 1,2, ∞ d ∞ x  1/2 E[ X ] = ∑ x ⋅ pq x −1 = p ∑q var( X ) = q X=x Trials  0 (otherwise) 7/16 dq x =1 6/16 x =1 p2 1- Success 5/16 d  1  +p 1 =p  = = dq  1 − q  (1 − q) 2 p 4/16 How many One Sequence 3/16   trials “x” 2/16 1/16 As p decr. Expected num. trials for “1” succ 0 x “x” for 1-succ must incr. 0 1 2 3 4 5 ... ∞  x − 1 r x − r Negative  x − 1 r −1 x − r E[ X ] = ∑ x ⋅   r − 1 p q  q  r − 1 p q ⋅ p p X ( x) =     var( X ) = r ⋅ Binomial   x=r p2 succ. on Geom RV = Neg Binom r next trial = X=x Trials ( r −1) succ. in ( x −1) trials for r=1 succ. p x = r , (r + 1), ( r + 2), ∞ As p decr. Expected num. trials r- Successes Many Sequences “x” for r-succ must incr. 137 INDEXThis table and one to follow compare some common probability distributions and explore theirfundamental properties and how they relate to one another. A brief description is given under the “RVName” column followed by the PMF formula and figure in col#2; formulas for the mean and variance areshown in the last two columns.The Bernoulli RV X answers the question “what is the result of a single Bernoulli trial?” It takes ononly two values, namely “1”=Success with probability p and “0”=Fail with probability q=1-p.The Binomial RV “X” answers the question “how many successes X in n Bernoulli trials?” It takes onvalues corresponding to the number of successes “X” in “n” independent Bernoulli trials; the sum RVX=X1+ X2+ ...+Xn of n Bernoulli RVs has nCx tree paths for X=x successes yielding a pmf nCx px qn-x asshown.The Geometric RV X answers the question “how many Bernoulli trials X for 1 success?” It takes onvalues from 1 to infinity and is the sum of n-1 failed Bernoulli trials followed by one successful trial; thesum RV X=X1+ X2+ ...+Xn of n Bernoulli RVs has only one tree path with X= x trials yielding 1-successand so has a pmf qx-1 p1 as shown.The Negative Binomial RV X answers the question “how many Bernoulli trials X for r- successes?” Ittakes on values from r to infinity and is the sum of n Geometric random variables; the sum RV X=G1+G2+ ...+Gr of “r” Geometric RVs with probability pr-1 qx-r p1 and has x-1Cr-1 tree paths for X=x-1 trialsyielding (r-1)-successes followed by one final success and so has a pmf x-1Cr-1 pr-1 qx-r p1 with x = r, r+1,... inf, as shown 137
  • 24. Bernoulli/Binomial Tree Structures RV Name PMF  p X = 1 (success) Bernoulli p X ( x) =  (q+p) x Prob 1-Trial 1 − p = q X = 0 (failure) F q 0 q X=x succ. START p 1 p “0” or “1” x “Atomic” RV S successes 0 1 Prob Binomial  2 p X (x) x p X ( x) =   p x q 2 − x 2 - Trials  x (q+p)2 F {FF} 0 q2 2C 1/2 q 0 x = 0,1, 2 X=x Succ. q F S {FS} 1 qp p 2C 1/4 How many Independent START q F {SF} 1 pq 1 succ “x” in Bernoulli Trials x p S {SS} 2 p2 2C “2” trials ? p S 2 0 1 2 (q+p)2 = q2 + 2pq + p2 = 2C0 p0 q2 + 2C1 p1 q1 + 2C2 p2 q0 138 INDEXThe RVs of the last slide are grouped in pairs {Bernoulli,Binomial} and {Geometric, Negative Binomial}for a reason. The sum of many independent Bernoulli trials generates a Binomial distribution and similarlythe a sum of many independent Geometric trials generates the Negative Binomial distribution. This slideand the next give a graphical construction of these trees for these two groups of paired distributions byrepeatedly applying the basic tree structure of the underlying Bernoulli or Geometric tree structure asappropriate.In the first panel we show the PMF properties for Bernoulli on the left and on the right we displayBernoulli tree structure where the upper branch q=Pr{Fail] goes to the state X= 0 and the lower branch p =Pr[Success] goes to the state X= 1.In the second panel we show the PMF properties for a simple n=2 trial Binomial. The corresponding treestructure for this Binomial is obtained by appending a second Bernoulli tree to each output node of the firsttrial, thus yielding the 4 output states {{FF}, {FS}, {SF}, {SS}}. We see that there is 2C0 tree paths leadingto {FF} p0q2 , 2C1 tree paths leading to{FS} p1q1 , and 2C2 tree paths leading to {SS} p2q0 , which isprecisely as expected from the Binomial PMF for n=2.This can be continued for n=3, 4, ... by repeatedly appending a Bernoulli tree to each new node. Further wesee that this structure for n=2 is represented algebraically by (q+p)2 inasmuch as the direct expansion gives1=q2 + 2q1p1 +p2 ; expanding an expression corresponding to n Bernoulli trials (q+p)n obviously yields theappropriate Binomial expansion for general exponent n.Thus the Binomial is represented by the repetitive tree structure or by the repeated multiplication of thealgebraic structure 1=(q+p) by itself n-times to obtain 1n=(q+p)n . 138
  • 25. Geometric/NegBinomial Tree Structures RV Name PMF p X (x) Geometric  pq x −1 x = 1,2, 1/2 [(1-q)-1 p] X=x Trials p X ( x) =  7/16 q F  0 (otherwise) 6/16 p 1- Success 5/16 F S 4/16 q How many 3/16 START p trials “x” for One Infinite 2/16 S “1” succ Sequence 1/16 p 0 x 0 1 2 3 4 5 ... S Negative  x − 1 2−1 x − 2 F Binomial p X ( x) =  p q ⋅ p [(1-q)-1 p ]2 q  2 − 1 succ. on q F X=x Trials (2 −1)succ. in ( x −1) trials next trial S p S 2- Successes x = 2,3, 4, ∞ q F p p F S S p X (x) START q p q F 1/4 S q F p S p 3/16 S S p Many Infinite 1/8 Sequences S 1/16 F q 0 x S q F 0 1 2 3 4 5 ... p S p2 (1-q)-2 = p {1+(-2)1-3(-q)1 +[(-2)(-3)/2] 1-4(-q)2 +[(-2)(-3)(-4)/(2)(3)] 1-5(-q)3 + ...} p p ...} S ={ 1C p 1 + 2C 1 pq1 + 3C 1 p1 q2 + 4C1 p1 q3 + p 139 INDEXThis slide first gives a graphical construction of a Geometric tree from an infinite number of Bernoullitrials and then shows how the Negative Binomial tree is the result of appending a Geometric tree toitself in a manner similar to that of the last slide. In the first panel we repeat the PMF properties forGeometric RV. On the right side of this panel we display Geometric tree structure whose branches endin a single success. This tree has a Bernoulli trial appended to each failure node and is constructed froman infinite number of Bernoulli trials. The 1st Bernoulli trial yields X=1 with p=Pr[Success] and thisends the lower branch; its upper branch yields X=0 with q=Pr{Fail]; this failure node spawns a 2ndBernoulli trial which again leads to X=1 or X=0; this process continues indefinitely. It accuratelydescribes the probabilities for a single success in 1, 2, 3,... inf number of trials and is algebraicallyrepresented by the expression 1=[(1-q)-1 p] which expands to [1 + q1 + q2 + q3 +....]*p corresponding toexactly 0, 1, 2, 3,... “failures before a single success”In the second panel we show the PMF properties for an r=2 Negative Binomial; on the right we displaythe Negative Binomial tree structure obtained by applying the basic Geometric tree to each node(infinite number) corresponding to a 1st success. This leads to a doubly infinite tree structure for the r=2Negative Binomial which gives the number of trials X =x required for r=2 successes. We can verify thefirst few terms in the Negative binomial expansion given under PMF in the lower panel using the tree.This process may be extended to r=3, 4, ... successes by repeatedly applying the Geometric tree to eachsuccess node. For n=2, direct expansion of the algebraic identity 12=[(1-q)-1 p]2 yields { 1C1 p + 2C1 pq1+ 3C1 p1 q2 + 4C1 p1 q3 + ...}p in agreement with the n=2 Negative Binomial terms in the table. In ananalogous fashion expansion of 1r=[(1-q)-1 p]r yields results for the r-success Negative Binomial. Notethat the “Negative” modifier to Binomial is a natural designation in view of the (1-q)-1 term in thealgebraic structure. 139
  • 26. Bernoulli, Geometric, Binomial & Negative Binomial PMFs • Bernoulli RV as Probability “Indicator” for Outcomes of a Series of Experiments representing a two different Event types, namely, E1: “Success in 1 trial” X = Bernoulli RV Binomial b(k;n,p) n = # trials , k = # successes E2: “ N1 is #Trials for 1stsuccess“ N1 = Geometric RV K=# Succ for n- trials n  n K = ∑ Xi p K (k ) =   p k q n − k k i =1   n K = ∑ Xi Bernoulli Bernoulli Process Sum n Indep. i =1 Single RV , Two Outcomes 1 Bernoulli trial for Bernoulli RVs “X” E ( K ) = µ K = np Event E1 var( K ) = σ K = npq 2  p X = 1 (success) p X ( x) =  p X ( x) = p 1 − p = q X = 0 (failure) Neg. Binomial bn(nr;r,p) Sum r Indep. 1 = # trials , 0,1 = # successes Geometric RVs Geometric Process nr = #trials for r successes ”N1” E ( X ) = µ X = p ; var( X ) = σ X = 0 2 n1 Bernoulli trials for  n − 1 pNr (nr ) =  r  p r q nr − r Event E2  r −1  r pN1 (n1 ) = p1q n1 −1 N r = ∑ ( N1 )i r i =1 N r = ∑ ( N1 )i i =1 1 E[ N r ] = µ N r = rE[ N1 ] = r Nr =# Trials p for r-Succ. q var( N r ) = σ N r 2 = r var( N1 ) = r p2 140The Bernoulli RV “X” is the basic building block for other RVs ( “atomic” RV ) and has a PMFdistribution with only two outcomes X=1 with probability p and X=0 with probability q=1-p . We have seenthat n such Bernoulli variables when added yield a Binomial PMF {b(x;n,p), x=0,1,2,...,n} which gives the“#successes “x” for “n” trials.We have also seen that this Binomial PMF can be understood by repeatedly appending the Bernoulli treegraph to each of its nodes (repeated independent trials) thereby constructing a tree with 2n outcomescorresponding to the n Bernoulli trials, each with two possible outcomes.Alternately, the Geometric PMF can be constructed by repeatedly appending a Bernoulli tree graph, but thistime only to the failure node, an infinite number of times, thereby constructing a tree with an infinitenumber of outcomes all of which correspond to “x-1” failures and exactly 1 success for x=1,2, ...., inf.Just as the Bernoulli tree graph is a building block for the Binomial tree graph, the infinite Geometric PMFtree graph is a building block for the Negative Binomial. The Negative Binomial tree graph for r=2successes is constructed by appending a Geometric tree graph to itself, but this time only to the successnodes, resulting in a doubly infinite tree graph corresponding to exactly “x-1” failures and exactly 2successes for x= 2,3 ...., inf. Repeating this process r-times yields the r-fold infinite tree graphcorresponding to exactly “x-1” failures and exactly r successes for x= r,r+1, ...., inf.The mathematical transformations relating Bernoulli, Binomial,Geometric and Negative Binomial areshown in this slide. 140
  • 27. Common PMFs and Properties-2 RV Name PMF Mean Variance E[ X ] = ∑ x⋅ p x = 0 ,1 X ( x) var( X ) = E[ X 2 ] − E[ X ]2  "m-marked" "(N-m) = unmarked" x from (n-x) from Hyper-  m ( N − n) m ( N − m)  m  N − m E[ X ] = n ⋅ = n ⋅ p var( X ) = n ⋅ ( N − 1) ⋅ N ⋅ N geometric   x   N     n−x  ; x ≤ x ≤ x X=x -succ pX ( x) =  where p = m / N is the  N min max ( N − n) N= fixed pop    "initial" probability of var( X ) = ⋅n⋅ p⋅q   n ( N − 1) m= tagged  0 ; Otherwise drawing a marked item  n=test sampl m ∈ [1, N ] ; n ∈ [1, N ] ; ( N − m − n) ≤ x ≤ min(m, n) w/o rplcemt PMF Derives from  N   m + ( N − m)   m   N − m   m   N − m   m N − m  m  N − m  Binomial Identity  =  =    +   + +   + +    n  n   0   n   1   n −1   x  n − x   n  0  n≤m≤ N Poisson  ( a x / x !)  x = 0,1, 2, ∞ Trials p X ( x) =  ea E[ X ] = a var( X ) = a 0 Otherwise X=x Succ  Limit of Binomial a = lim(n ⋅ p) = λ ⋅ t = (aver. arrival rate)*time n →∞ p →0 Zeta(Zipf)   ( )  1 xs p X ( x; s ) =  ζ ( s) = "ζ − term " x = 1, 2, ; s >1 ( ∞ E[ X ; s ] = ζ 1s ) ⋅ ∑ x⋅ 1s x ( ∞ Var ( X ; s ) = ζ 1s ) ⋅ ∑ x2 ⋅ 1s − E[ X ; s]2 x =1 x n - Trials ζ (s) x =1 X=x Succ.    0 Otherwise ( ∞ = ζ 1s ) ⋅ ∑    1    = ζζ( s( −1) s) = ζ ζ( s(−)2) − s ( ζ (s) ) ζ ( s −1) 2  x s −1  −1 x =1   ( ) ∞ ∞ ζ (1.5) ζ (2.5) 2 ∑ x =1 = 1 ⇒ C =  ∑  1s   = ζ 1s ) C       xs       x =1  x   ( E[ X ; s = 3.5] = ζζ( s( −1) = 1.191 Var ( X ; s = 3.5) = ζ (3.5) − ζ (3.5) s) = .856 Riemann Zeta Fcn ζ (s) 141 INDEXThis second part of the Common PMFs table shows the Hyper-geometric, Poisson and Riemman Zeta (orZipf ) PMFsThe Hyper-geometric RV “X” answers the question “how many successes (defectives) X are obtainedwith n test samples (trials without replacement) from a production run (sample space) that contains mdefective and N-m working items?” X takes on values corresponding to the number of successes(defectives) “X” in “n” dependent Bernoulli trials; the distribution is best understood in terms of theBinomial identity NCn = mC0 N-m Cn + ...+ mCx N-m Cn-x +... + mCm N-m Cn-m which when divided by NCnyields the distribution mCx N-m Cn-x where X takes on values x=[xmin, xmax] where xmin=N-n-m and xmax=min(n,m).as allowed by the combinations w/o replacementThe Poisson RV “X” answers the question “how many successes X in n Bernoulli trials with n verylarge?” We shall discuss this in more detail in the second part of the course where we pair it with acontinuous distribution. For now it is sufficient to know that it represents a limiting behavior of theBinomial PMF in the limit that n-> inf and its terms represent single terms in the expansion of ea where a=λ∗ t is called the Poisson parameter, where λ is a “rate” and t is a time interval for the data run. The PMFis therefore the ratio of the single term in the expansion to ea over ea which ispX(x)={ ax/ x!} / ea for x=0,1,2,3,... The Poisson RV has many applications in physics and engineering.The Riemman Zeta RV “X” has applications to Language processing and prime number theory and itsproperties are given in the table. Note that the exponent must satisfy α >0 in order to avoid the harmonicseries which will does not converge and therefore cannot satisfy the sum to unity condition on the PMF. 141
  • 28. Chapter 5 – Continuous RVs Probability Density Function (PDF) f X (x) Event E = {x : a ≤ x ≤ b} : b Pr[ x ∈ E ] = ∫ f X ( x)dx = ∫ f X ( x)dx Pr[a ≤ x ≤ b] E a a x 2.0 b Pr[ x = 2.0] = ∫f x = 2.0 X ( x)dx = 0 Prob at a point = 0 Except for δ-fcn at a point αδ ( x − x0 ) uniform Mixed Continuous & Discrete Outcomes – Dirac δ-fcn f X (x) β (b − a ) β f X ( x) = αδ ( x − x0 ) + (b − a ) b x0 + ε x ∫ αδ ( x − x )dx = ∫ ε αδ ( x − x )dx =α a 0 x0 − 0 a x0 b Sampled Continuous Fcn g(x) f X (x) α k δ ( x − xk ) n g (x) f X ( x ) = ∑ α k δ ( x − xk ) k =0 b α k = ∫ g ( x)δ ( x − xk ) =g ( xk ) a x0 x1 xk xn x 2/24/2012 3In Discrete Probability a RV is characterized by its probability mass function (PMF) pX(x) whichspecifies the amount of probability associated with each point in the discrete sample space. Continuousprobability generalizes this concept to a probability density function (PDF) fX(x) defined over acontinuous sample space. Just as the sum of pX(x) over the whole sample space must be unity, theintegral of fX(x) over the whole sample space must also be unity. An event E is defined by a sum orintegral over a portion of the sample space as shown by the shaded area in the upper figure between x=aand x=b.The middle panel gives an example of a mixed distribution containing continuous uniform distributionβ/(b-a) and a Dirac δ-function at the point x0 α∗ δ(x-x0) corresponding to a discrete contribution at thatpoint. The uniform distribution is shown as a continuous horizontal line at “height” y = β between a andb and the Dirac δ-function is shown with an arrow corresponding to a probability mass “α” accumulatedat a single point x=x0.. The integral over the continuous part gives (b-a)* β/(b-a) = β and the integral ofthe Dirac δ-function α∗ δ(x-x0) over any interval containing x0 yields α. Thus, in order for thisexpression to be a valid probability density function, we require the sum of the two contributions beunity: α+ β =1 .Consider the continuous curve fX(x) = g(x) in the bottom panel and take the sum of products αk*δ(x-xk).Is this a valid discrete “PMF”? In order for this to be so the sum of the contributions αk must be unity.Does it represent a digital sampling of g(x)? No, in order to actually write down an appropriate“sampled” version of g(x), we need to develop a “sampling” transformation Yk=Yk(X) for k=0,1,2,...,n soas to transform the original continuous fX(x) to a discrete fY(yk) (See slide#26 ) 3
  • 29. Cumulative Distribution Function (CDF) x FX ( x) = Pr[ X ≤ x] = ∫ f X ( x ) dx Probability Density PDF x =−∞ integrates to yield CDF fX(x) fX(x) PDF Bdy Values : FX (−∞) = 0 ; FX (+∞) = 1 PDF 1 1 ¼ δ(x-1) 1/2 Monotone Non - decr. : FX (b) ≥ FX (a ) ; if b ≥ a 0 x x 0 0 1/2 1 3/2 0 1/2 1 3/2 Prob Interpretation : Pr[a ≤ x ≤ b] = FX (b) − FX (a) FX(x) CDF FX(x) CDF Density PDF : d dx FX ( x) = f X ( x) 1 1 ¼ 1/2 1/2 or, dFX ( x) = FX ( x + dx) − FX ( x) = f X ( x)dx 0 x 0 x 0 1/2 1 3/2 0 1/2 1 3/2 2/24/2012 7The cumulative distribution function (CDF) for a continuous probability density function fX(x) is definedin a manner similar to that for discrete distributions pX(x) except that the cumulative sum over a discreteset is replaced by an integral over all X less than or equal to a value x. This integral yields a function of“x” FX(x) = Pr[X<=x] which has the following important properties(i)FX(x) always starts at 0 and ends at 1(ii)FX(x) is continuous,(iii)FX(x) is non-decreasing,(iv)FX(x) is invertible; i.e., FX -1 (x) exists, and(v)The density fX(x)=d/dx{FX(x)} (since exact differential d FX(x) = FX(x+dx) - FX(x) = fX(x)dx )It is important to note all five properties of FX(x) as they have important consequences.The figure shows the relationship between the density fX(x) and the cumulative distribution FX(x) for twocases (i) two regions of constant density (two “boxes”) and (ii) one region of constant density plus a deltafunction (one “box” and an arrow “spike”) .In case (i) FX(x) ramps from a value of 0 to ½ in the region [0, ½ ] from the 1st constant density box, thenremains constant at ½ over the region [ ½ , 1] and finally ramps from ½ to 1 from the 2nd constantdensity box. Note that the slopes of the two ramps are both “1” in this case and that the total area underthe density curves 1* [1/2-0] + 1* [3/2-1] = 1.In case (ii) FX(x) ramps from a value of 0 to ½ in the region [0, 1] by virtue of the constant “½” densitybox, then jumps by “1/4” because of the delta function, and finally continues its ramp from the value ¾ to1. Note that this is simply the superposition of a constant density of “ ½“ plus a delta function ¼∗ δ(x-1), and again the total area under the density curves ½ * [3/2-0] + ¼ = 1 7
  • 30. Transformations of Continuous RVs • Transformation of Densities PDFs in 1 dimension • Transformation of Joint Densities PDFs in 2 or more dimensions • Two Methods: 1) CDF Method: Step#1) First find CDF FX(x) by integrating fX(x) Step#2) Invert y=g(x) transformation y = g(x) ⇒ x = g −1 ( y ) & use it to write FY ( y ) = Pr[Y ≤ y ] in terms of the known FX(x) (Note y= g(x) may not be “one-to-one” “multiplicity”) y = y Step#3) Differentiate wrt y: d d fY ( y ) = dy FY ( y ) = dy ∫f y = −∞ Y ( y )dy 2) Jacobian Method: Transform PDF fY(y) using derivatives f X ( x) fY ( y ) = Express everything in terms of variable y dy dx fY ( y )dy = f X ( x)dx ; y = g ( x) f X ( x = g −1 ( y )) = g ( x = g −1 ( y )) Note absolute value 2/24/2012 14It is very important to understand how probability densities change under a transformation of coordinatesy=g(x). We have seen several examples of such coordinate transformations for discrete variables,namely,(i) Dice: Transform from individual dice coordinates (d1, d2) to the sum and difference coordinates (s, d)corresponding to a 90 degree rotation of coordinates, and(ii) Dice: Transform from individual dice coordinates (d1, d2) to the minimum and maximum coordinates(z, w) corresponding to corner shaped surfaces of constant minimum or maximum values.There are two methods for transforming the densities of RVs, namely (i) the CDF-method and (ii) theJacobian Method. While they are both quite useful for 1-dimensional PDFs fX(x), the Jacobian method isbest for transforming joint RVs .The CDF method involves three distinct steps as indicated on the slide, namely (i) compute CDF FX(x),(ii) Relate FY(y) = Pr[Y<=y] to FX(x) and then invert the transformation x = g-1(y) and substitute to findFY(y) with a redefined y domain, and (iii) differentiate wrt “y” to obtain the transformed probabilitydensity for the RV Y: fY(y). Note that if the function is multi-valued and therefore not invertible, it mustbe broken up into intervals for which it is invertible and appropriate “fold-over” multiplicities must beaccounted for.The Jacobian Method uses derivatives of the transformation to transfer densities from the original set ofRVs to the new one; the Jacobian accounts for linear, areal, and volume changes between the coordinates.In one dimension the Jacobian is simply a derivative and is obtained by transferring the probability in theinterval x to x+dx: fX(x)dx to the probability in the interval y to y+dy: fY(y)dy Equating the twoexpressions yields fY(y) =fX(x) / |dy/dx| = fX(g-1(y) ) / |dy/dx|. Note that the absolute value is necessarysince fY(y) must always be greater than or equal to zero. 14
  • 31. Method#1 Transformation of Continuous RV - CDF Method Resistance X = R Step#1 Compute FX(x) CDF= FX(x) PDF = fX(x) 1/ 200 900 ≤ r ≤ 1100 f R (r ) =  1  0 Otherwise 1/200 r =r  0 r < 900  FR (r ) = Pr[ R ≤ r ] = ∫ f R (r )dr = (r − 900) / 200 900 ≤ r ≤ 1100 0 r =−∞  1 r > 1100 900 1100 x  Conductance Y = 1/R Step#2 Transform to FY(y) PDF = fY(y) FY ( y) = Pr[Y ≤ y] = Pr[ R ≥ 1/ y] = 1 − Pr[ R ≤ 1/ y] 6050  1− 0 = 1 1/ y < 900 CDF= FY(y)  1  ( − 900)  y 1 = 1 − FR (1/ y) = 1 − 900 ≤ 1/ y ≤ 1100  200  1 −1 = 0 1/ y > 1100 4050   Step#3 Differentiate FY(y)  0 y< 1  1100  1 0 d  1 1 fY ( y ) = FY ( y ) =  ≤ y≤ 1/1100 1/900 y dy  200 y 2 1100 900  0 1 y>   900 2/24/2012 15The Resistance X=R of a circuit has a uniform probability density function fR(r)=1/200 between 900 and1100 ohms as shown in the top panel; the corresponding CDF FR(r) is the ramp function starting at “0”for R<=900 and reaching “1” at R=1100 and beyond as shown. The detailed analytic function is given inthe slide and represents the result of Step#1 in the CDF-Method.The problem is to find the PDF for the conductance Y=1/X = 1/R. We first down the definition for FY(y)for a given value Y=y and then re-express it as a function of R =1/YFY(y) =Pr[Y<=y] = Pr[R>=(1/y)] = 1-Pr[R<=(1/y)] = 1 – FR(1/y )This last expression is now evaluated in the lower panel of the slide by substituting r=1/y into theexpression for FR(1/y ) of the upper panel. Note the resulting expression has been written down by directsubstitution and the intervals have been left in terms of 1/y. (This constitutes step#2 of the method).Finally, differentiating FY(y) wrt “y” we find (step#3) the desired PDF fY(y); we have also “flipped” the“1/y” interval specifications and reordered the resulting “y” intervals in the customary increasing order.As seen in this example, the CDF method requires careful attention to the definition of the FY(y) definedin terms of cumulative probability of the variable Y. Since Y=1/R, this leads to FY(y) = 1-FR(1/y ) and a reverse ordering of the inequalities for the intervals. 15
  • 32. Transformation of Continuous RV - Derivative (Jacobian) Method Method#2 PDF 1 / 200 900 ≤ r ≤ 1100 6050 f R (r ) =   0 Otherwise 1 fY ( y ) = fY ( y )dy = f R (r )dr ⇒ Find fY ( y ) 200 y 2 dr f (r ) 4050 fY ( y ) = f R (r ) = R dy | dy / dr | 1 1 f X ( x) = 200 dy y= fY ( y ) = f R (r ) = (1 / 200) 900 dx R | −1 / r |2 y2 hyperbola: xy = 1 dy slope = dx 1 1 1 fY ( y ) = for ≤ y≤ 200 y 2 1100 900 1100 x=R Note: fY(y) is large for small slope & vice versa. Same Differential Area (Probability) is mapped via hyperpola to yield the tall high and short fat strip areas shown for fY(y) 2/24/2012 16The Jacobian Method is much more straight forward and moreover has a very intuitive visualization inthe 3-dimensional plot shown on this slide. The uniform probability density function fR(r)=1/200 between900 and 1100 ohms is written explicitly in the first boxed equation. The Jacobian method just takes theconstant fR(r) = 1/200 and divides it by the magnitude of the derivative |dy/dr|=|-1/r2| = y2 to yield directlyfY(y)=1/(200y2) for y ε [1/1100, 1/900].The 3-dimensional plot shows exactly what is going on:i) The original uniform distribution fX(x)=1/200 displayed as a vertical rectangle in the x-z plane ii)Sample strips at either end with width “dx” have the same small probability dP= fX(x)dx as shown AtR=900, the density fX(x) is divided by the large slope |dy/dx| yielding a smaller magnitude for fY(y) asillustrated, but this is compensated by a proportionately larger “dy”and thus transfers the same smallprobability dP= fY(y)dy.iii) Conversely, the strip at R=1100 is divided by a small slope |dy/dx| and yields a larger magnitude forfY(y), which is compensated by a proportionately smaller “dy” again transferring the same dP.iv) The end point values of the transformed density fY(y) are illustrated in the figure. The strip width “dx”cuts the x-y transformation curve at two red points which have a “dy” width that is small at x =1100 andlarge at x = 900 as determined by the slope of the curve. The shape in between these end points is aresult of the smoothly varying slope of the transformation hyperbola shown in the x-y plane.Thus the slope of the transformation curve (hyperbola xy=constant in this case) in the x-y planedetermines how each “dx” strip of the uniform distribution fX(x)=1/200 in the x-z plane transfers to thenew density fY(y) shown in the z-y plane. This 3-dimensional representation de-mystifies the nature ofthe transformation of probability densities and makes it quite natural and intuitive for 1-dimensionaldensity functions. It is easily extended to two-dimensional joint distributions. 16
  • 33. Transformation of Continuous RV – Example 3 “Multiplicity Factor” Gaussian PDF : y x2 1 − f X ( x) = e − ∞ < x < +∞ 2 2π Not a 1-1 mapping Double Density Pts Find PDF for Y = X 2 ( −∞, ∞) → (0, ∞) Fold-over 1 − y density is doubled x e 2 f X (x) 2π f Y (y) = 2 =2 dy/dx 2 y 1 − y y 2π y e 2 1 − = e 2 for 0 < y < +∞ x2 2πy 1 − 2π e 2 Two Equal GeneralRule : Contributions from –x & +x f X (x) fY (y) = α ⋅ dy/dx y Double α = multiplici factor ty Density Pts " fold - over" y = x2 x 2/24/2012 18The transformation of a Gaussian PDF under the transformation Y=X2 is easily computed using theJacobian method provided one incorporates a multiplicity factor α as shown in the boxed densityequation . The multiplicity factor arises because there are two contributions to the same y-value onefrom –x and the other from +x as illustrated in the upper figure; thus folding the parabola across the x=0symmetry line yields twice the density on positive x and this corresponds to a multiplicity factor α=2 inthe boxed density transformation equation.The 3d plot shows the original Gaussian density function (grey) in the x-z plane, the transformation y=x2in the x-y plane, and the resulting distribution shown as a dashed curve in the y-z plane. The two thinvertical slices at –x and +x are mapped to the same y-value and hence doubles the density contribution tofY(y) as shown. 18
  • 34. Analog to Digital (A/D) Converter - Series of Step Functions Continuous Representation of Discrete “sampled” Distributions Y (OUT) 3 A/D converter Mapping Fcn Y = g( X ) = k +1 ; k < x ≤ k +1 -3 2 1 -2 -1 X Mapped Density fY (y) = ∑ αk ⋅ δ(y − yk ) 0 -1 1 2 3 (IN) k -2 a) Exponential b) Gaussian b) Uniform 1 −x2 / 2 ae − ax PDFX = f X ( x) = e 1 0 ≤ x ≤ 10 x≥0 2π PDFX = f X ( x) =  PDFX = f X ( x) =  0 otherwise  0 x<0 −∞ < x < ∞  k k α k = ∫ f X ( x)dx =  x =∫ −1 k  ae − ax dx = −e − ax x≥0 k 1 − x2 / 2 k k − (k − 1) k x = k −1 αk = ∫ 2π e dx = ϕ (k ) −ϕ (k − 1) αk = ∫ 1 dx = x = k −1  0 x<0 x = k −1 10 10  x=k 1 − x2 / 2 x = k −1 1 e − ak (ea − 1) x ≥ 0 ϕ (k ) ≡ ∫ 2π e dx ; k ∈ (−∞, ∞ ) = 10 ; k = 1, 2,L ,10 = ; k = 1, 2,... x =−∞  0 x<0 ∞ 10 1 fY ( y ) = ∑ e − ak (e a − 1) ⋅ δ ( y − k ) fY ( y ) = ∑ α k ⋅ δ ( y − yk ) fY ( y ) = ∑ δ ( y − k) k =1 k k =1 10 e − (0.1) k (e0.1 − 1) = .105 ⋅ e − (0.1) k fY(y) fY(y) k αk fY(y) 0.1 α kδ ( y − k ) 1 0.095 1/10 0.095 2 0.086 0.050 3 0.078 y y 0 1 5 10 y 0 10 20 11 0.035 2/24/2012 26In discussing the half-wave rectifier on the last slide we found that the effect of a “zero” slopetransformation function was to pile up all the probability in the x-interval into a single δ-function at theconstant y=“0” value associated with that part of the transformation. Here we extend that concept to a“sample & hold” type mapping function typical of an Analog to Digital (A/D) converter. The specificmapping function y=g(x) = k+1 for k < x ≤ k+1 is illustrated in the grey box as a series of horizontalsteps over the entire range of x [-3, 3]; the y-values for these steps range from y=-2 to y=+3. Eachhorizontal (zero-slope) line accumulates the integral of fX(x) from x=k to k+1 onto its associated y-valueshown as a red circle with the point of a δ-function arrow pointing up out of the page and having anamplitude given by the integral for that interval denoted by the symbol αk.The table shows several examples of a digitally sampled representation for a) Exponential, b) Gaussian,and c) Uniform distributions in the three columns. The rows of the table give the specific continuousdensities for each, the computations for the amplitudes of the discrete digital samples αk, the resultingsum of δ-functions, and finally a plot showing arrows of different lengths to represent the δ-functions ofthe sampled distributions. 26
  • 35. Order Statistics - General Case n Random Variables General Case n Variables: X1,, X2 , ... ,, Xn RVs fX (x) fX(y)dy Assume RVs are Indep and Identically Distributed (IID) FX ( y ) 1 − FX ( y ) {X1,, X2 , ... ,, Xn } fX(x) f X 1 X 2 L X n ( x1 x 2 L x n ) = f X ( x1 ) ⋅ f X ( x 2 ) ⋅ L ⋅ f X ( x n ) Reorder {X1,, X2 , ... ,, Xn } as follows: fX(y)dy Y1,= smallest {X1,, X2 , ... ,, Xn } all Yk <y y y+dy all Yk > y Y2= next smallest {X1,, X2 , ... ,, Xn } Y1 |Y2 |… |Yj-1 Yj+1 | Yj+2 | … | YN jth “order Yj= jth smallest {X1,, X2 , ... ,, Xn } (j-1) RVs statistic” (n-j) RVs Yn= largest {X1,, X2 , ... ,, Xn } Each IID: P[Yj ≤ y]= FX(y) P[Yj > y]= 1 - FX(y) Y1< Y2 < Yj <… < Yn [FX(y)]j-1 [1 - FX(y)]n-j Same PDF in variable “y” fX(y) Diff’l Prob. Find PDF for the jth “order statistic” “one sequence” = ( FX ( y ) ) j −1 ⋅ f X ( y )dy ⋅ (1 − FX ( y ) )n − j Pr[ y ≤ Y j ≤ y + dy ] = fY j ( y )dy ; j = 1, 2,L , n jth order statistic 3! [φ| X1 | X2 X3] j=1: [φ| Y1 |Y2 Y3 ] 0! 1! 2! =3 [φ| X2 | X1 X3] Case n=3 {Min, Mdl, Max};Y2 = “Mdl“statistic. Min [φ| X3 | X1 X2] Y2 could be any one of {X1,, X2 , X3 } 3! [X2,| X1 | X3], [X3,| X1 | X2] j=2: [Y1 |Y2 | Y3 ] =6 [X1,| X2 | X3], [X3,| X2 | X1] 1! 1! 1! [X1,| X3 | X2], [X2,| X3 | X1] Mdl There are 3! = 6 orderings; however, we partition into 3 [ X2 X3 |X1 |φ] 3! groups and permutations within a group is irrelevant; j=3: [Y1 Y2 | Y3 | φ] 2! 1! 0! =3 [ X1 X3 |X2 |φ] [ X1 X2 |X3 |φ] Max 48 2/24/2012Order Statistics for the general case of n IID Random Variables is detailed on this slide. The n IID RVs{X1, X2,..., Xn} are re-ordered from the smallest Y1 to the largest Yn and the jth Y in the sequence Yj iscalled the “jth order statistic”. Again we fix a value Y=y and consider the continuous range of re-orderedY-values illustrated in the figure: the small interval from y to y+dy contains the differential probabilityfor the jth order statistic Yj given by fX(y)dy; all Y-values less than this belong to the Y1 through Yj-1 andthose greater belong to Yj+1 through Yn as shown in the inset figure. Now for each of the Ys on the left wehave the probability Pr[Y1 ≤ y] = FX(y), Pr[Y2 ≤ y] = FX(y), ... Pr[Yj-1 ≤ y] = FX(y), and because they areIID the total probability of those on the left is Pr[Yleft ≤ y] = [FX(y) ]j-1; similarly on the right we findPr[Yright ≤ y] = [1-FX(y) ]n-j. So for the reordered Ys the differential probability is just the product of thesethree terms multiplied by a multiplicity factor α, viz., dP = Pr[y≤ Yj ≤ y+dy]= f Yj (y) dy = α [FX(y) ]j-1 fX(y) [1-FX(y) ]n-j dyThe multiplicity factor α results from the number of re-orderings of {X1, X2,..., Xn} for each orderstatistic Yj ; arguments for n=3 and n=4 are illustrated on this slide and the next. These arguments look(in turn) at each order statistic min, middle(s), and max and compute in each case the number of distinctarrangements of {X1, X2,..., Xn} that yield the three groups relative to the “separation point” Y=y andarrive at multinomial forms dependent upon the orderings for each statistic. The specific multiplicityfactors for the cases for n=3,4 are easily found to be α = 3C (j-1),1,(3-j) = 3! / [(j-1)! 1! (3-j)!] ; α = 4C (j-1),1,(4-j) = 4! / [(j-1)! 1! (4-j)!]and the final results for the PDF of the jth order statistic f Yj (y) in these cases are fYj (yj) = 3C (j-1),1,(3-j) [FX(yj) ]j-1 fX(yj) [1-FX(yj) ]3-j for j=1,2,3 (n=3) fYj (yj) = 4C (j-1),1,(4-j) [FX(yj) ]j-1 fX(yj) [1-FX(yj) ]4-j for j=1,2,3,4 (n=4) 48
  • 36. Random Processes – Introduction - Lec#4 • Time Series Data = Physical Measurements in time • Random Process = Sequence of random variable realizations – Geiger Counter Sequence of “detections” - Poisson Process – Communication Binary Bit Stream - Bernoulli Process “ 01001…” – E&M Propagation Phase (I-Q components) - Gaussian Process • Arrival Event: Success =“arrival” (of an event in time) • Interarrival Times for Random Processes – Not only interested in how many successes K (“ arrivals”) there are – But also interested in “specific time of arrivals,” e.g., TK = time of kth arrival – DSP Chip Interrupts: Random Number of Interarrival • Time between interrupts Process Arrivals Times • used for data processing Geiger Poisson Exponential – Waiting on Telephone: Counter • “you are 10th customer in line and … Binary Bit Stream Bernoulli Geometric • your wait will be approximately “7 minutes” 2/24/2012 61Observations of physical processes produce measurements over time which almost always havecomponents described by a random process. Some examples are Geiger counter detections (PoissonProcess), Binary bit streams (Bernoulli Process) and Electromagnetic wave I, Q Phase components(Gaussian Process).Because, these processes take place over time, the notion of a “success” is translated to an “arrival” at aspecific time. Moreover, we are not only interested in how many successes K there are, but also theirspecific arrival times, i.e., we would like to know the time of the kth arrival Tk. This has application tomany physical processes such as the timing of DSP chip interrupts relative to their “clock cycles” and thequeuing of customers in a telephone answering system. In both cases you want to make sure the systemcan handle the “load” in an appropriate manner; for the DSP chip you need to minimize the number oftimes you are near the leading or trailing “edge” of the timing pulse in order to avoid errors, while for thetelephone answering service, the 10th customer, would like to know how long he must wait in the queuebefore being served. 61
  • 37. Multi-User Digital Communication “CDMA” Arrival Slots • Two signals s1 , s2 ;Decode s1 or s2 in given time slot s1 Decoded P|s1,1]= P[1|s1] P|s1] “success” • a priori Prob: P[s1]=3/4 ; P[s2]=1/4 P[1|s1] =(2/3)(3/4) =1/2 p1=1/2 S1 2/3 • Decoding Statistics: 1/3 P[s1] 3/4 P[0|s ] P|s1,0]=1/4 decoded “1” : P[1|s1]=2/3 ; P[1|s2]=2/3 Time 1 s1 Not Slot #4 P[1|s2] P|s2,1]= P[1|s2] P|s2] Decoded not decoded “0” : P[0|s1]=1/3 ; P[0|s2]=1/3 1/4 S2 =(2/3)(1/4) =1/6 “failure” P[s2] 2/3 q1=1/2 1/3 Nr time slots (“trials”)  n − 1 r n − r P[0|s2] P|s2,0]=1/12 p N r ( n) =  p q r-Decodes of s1 p1=q1=1/2  r −1 a priori decode 4 −1 1 1 1 1 1) Pr[ 1st decode in 4th slot] Pr[ N1 = k ] = p N1 (k ) = q k −1 p1 ⇒ Pr[ N1 = 4] = p N1 (4) =     =  2   2  16 2) Pr[ 4th decode in 10th slot | 3 decodes No memory - slots 6 to 10 1 2 3 4 5 6 7 8 9 10 “1” “1” “1” in 1st 6 time slots ] 1 1 3 1 1 1 2 3 4 Pr[ N1 = 4] = p N1 (4) = q 3 p =     = 3 “1”s No Memory  2   2  16 4 3) Pr[ 2nd decode in 4th slot]  n − 1 r n − r  4 − 1 2 4 − 2 1 3 Pr[ N r = n] = p N r (n) =    p q ⇒ Pr[ N 2 = 4] = p N 2 (4) =    2 − 1 p q = 3 2  = 16   r − 1     4) Pr[ 2nd decode in 4th slot | no decodes No memory of failures in slots 3 & 4 1 2 3 4 “0”“0” 2 in 1st 2 time slots] 1 1 1 2 Pr[ N 2 = 2] = p N 2 (2) = p 2 =   =  2 4 “Renewal” { “means” N2>2 } Pr[ N 2 = 4 , N 2 > 2 ] p N 2 (4) ( 3 / 16 ) 1 Pr[ N 2 = 4 | N 2 > 2 ] = = = = Pr[ N 2 > 2 ] 1 − p N 2 ( 2 ) 1 − (1 / 4 ) 4 2/24/2012 78This example illustrates renewal properties and time slot arrivals of the Geometric and Negative Binomial RV distributions.In a multiuser environment the digital signals from multiple transmitters can occupy the same signal processing time slot solong as they can be distinguished by their modulation characteristics. Code Division Multiple Access (CDMA) uses apseudorandom code that is unique to each user to “decode” the proper signal source.Consider two signals s1 and s2 being processed in the same time slot with a priori “system usage” given by P[s1] = ¾ and P[s2]= ¼ ; further let “1” denote successful and “0” denote unsuccessful decodes respectively. Given that each signal has the same2/3 probability of a successful decode P[1|s1] = P[1|s2] = 2/3, we can use the tree to find the single trial probability of successfor decoding each signal. For signal s1 we see that the end state {s1, 1} represents a successful decode and has p1=1/2 ; all other states {s1, 0}, {s2 1},{s2, 0} represent failure to decode signal s1 with probability q1 = 1/4+1/6 + 1/12 = 1/2. Similarly for signal s2 we see that theend state {s2, 1} represents a successful decode of s2 and has p2 =1/6 ; all other states {s2, 0}, {s1 1}, {s1, 0} represent failure todecode signal s2 with probability q2 = 1/12+1/2 + 1/4 = 10/12 =5/6.We consider successive decodes of s1 as independent trials with probability of success p1=1/2 . Thus, the probability ofhaving r- successful decodings of s1 in Nr signal processing slots “trials” is given by the Negative Binomial PMF pNr(n) = n-1Cr-1p1rq1n-r with nr = r, r+1, r+2, .... with p1=q1=1/21) Pr of 1st decode (r=1) in 4th slot (N1 =4) is pN1(4) = 4-1C1-1p11q14-1 = 1(1/2)4 = 1/162) Pr of 4th decode (r=4) in 10th slot (N4 =10) given 3 previous decodes in 1st 6 slots is found by “restarting the process withslots #7 , 8, 9, 10 so we need only one decode (r =1) in 4 slots, i.e., N1 =4, which is identical to part 1) and yieldsPr[N4 = 10 | N3=6] = pN1(4) = 4-1C1-1p11q14-1 = 1(1/2)4 = 1/163) Pr of 2nd decode (r=2) in 4th slot (N2 =4) is pN2(4) = 4-1C2-1p12q14-2 = 3(1/2)4 = 3/164) Pr of 2nd decode (r=2) in 4th slot given 1st two slots were not decoded is found by “restarting the process with slots #3,4 “sowe need r=2 in the two remaining slots N2 =2 which means two successes in two trials, so we havepN2(2) = 2-1C2-1p12q12-2 = 1(1/2)2= 1/4 78
  • 38. Binary Communication with Noise Gaussian under Linear X : N ( µ X , σ X 2 )  Y : N (eµ X + f , e 2σ X 2 ) → Transformation: Y=eX+f Y = eX + f 1 24 { 4 3 ≡ µY ≡σ Y 2 Noise X: N(0,1) Y1 = N(a,1) “1” +a Y1 = a + X Threshold d1 = detect “1” Binary Modulator Channel Detector Generator -a Y0 = - a + X d0 = detect “0” “0” Y0 = N(-a,1) Threshold Threshold Detector detect “0” y=c detect “1” Y>c detect “+ a” or “1” fY|A (y|-a) fY|A (y|+a) Y≤c detect “- a” or “0” y -a 0 +a Type I Type II Prob of an Error “Missed Detection” “False Positive” for Detection a “1” P(Er “1” ) = P(Y ≤ c | +a) P(+a) + P(Y > c | -a) P(-a) Type I Error “Missed Detection” Type II Error “False Positive” Does not Exceed Threshold Exceeds Threshold But Belongs to “+a” Distrib. But Belongs to “-a” Distrib. 2/24/2012 97Consider the Binary communication channel depicted in the upper sketch: A binary sequence of “1”s and“0”s is generated and then amplitude modulated by a positive amplitude +a for “1” and –a for “0” asillustrated by the “square wave pulse train” at the modulator. Zero mean unit variance Gaussian noiseN(0,1) is added by the “channel” and the (signal + noise) outputs are two distinct Gaussian RVs : Y1= a+X ~ N(+a, 1) and Y0=–a +X ~ N(-a,1) about two different means as shown in the probability densityplot. This output is presented to a Threshold detector which attempts to detect the original sequence of“1”s and “0”s by setting a threshold Y =c (vertical dashed line) and assigning a “1” to Y-values to theright and “0” to for Y-values to the left of the threhold.Considering the detection of “1” we see that two types of error can occur as follows:Type I Missed Detection: P(Y≤c | +a) The larger hatched area on the left with Y<c which belongs to theN(+a,1) curve but is rejected because it does not exceed the threshold “c”Type II False Positive: P(Y>c | -a) The smaller hatched area on the right with Y>c which belongs to the“0” N(-a,1) curve but is falsely detected as “1” because it exceeds the threshold “c”The total probability for an error in detecting a “1” is the sum of each conditional multiplied by its apriori as shown in the bottom equation. The total probability for an error in detecting a “0” is writtendown in an analogous fashion as a sum of conditionals multiplied by their a priori s (not shown) . 97
  • 39. Common PDFs - “Continuous” and Properties RV Name PDF Generating Mean Variance ∞ Fcn ϕ ( s) = E[e Xs ] ∫ x⋅ f x = −∞ X ( x)dx var( X ) = E[ X 2 ] − E[ X ]2 f X (x)  1  f X ( x) =  b − a a≤ x≤b e sb − e sa a+b (b − a )2 Uniform  0  Otherwise s (b − a ) 2 12 x a b fT (t ) λ e − λ t t≥0 1 1 f T (t ) =  λ Exponential  0 t<0 λ λ2 λ−s “exponential wait” λ>0 t f Tr (t) Peaks at Gamma  λ e − λt (λt ) r −1  t≥0 Exponent r r fTr (t ) =  ( r − 1)! ial tmax = r −1 λ  λ    r r-Erlang   0 t<0 1 r =1 E[T1] = λ λ−s λ r = integer r =2 E[T2 ] = 2 For r=3: three λ2 λ>0 Arrival Rate λ 3 r =3 E[T3 ] = λ “exponential waits” t E[T3 ] = 1 λ + 1 λ + 1 λ 2 ( x −µ ) Normal 1 − Gaussian f X ( x) = e 2σ2 Rayleigh (σ s )2 2π ⋅ σ Peaks µs+ µ σ2 N (µ, σ ) e 2 2 at x=0 Peaks at −∞ < x < ∞ x=1/a ( s/ a )2 Rayleigh a2 x2 − 1+  a  e s − 2 π ⋅ 2−π f X ( x) = a 2 xe 2     2 1 π 2a 2 x ⋅ 1 + erf  a 2  (s/a)  x>0; a>0 0     2     2/24/2012 101This table compares some common continuous probability distributions and explores their fundamentalproperties and how they relate to one another. A brief description is given under the “RV Name” columnfollowed by the PMF formula and figure in col#2, the generating function in col#3, and formulas for themean and variance in the last two columns.The Uniform Distribution has a constant magnitude 1/(b-a) over the interval [a,b]; the mean is at thecenter of the distribution (a+b)/2 and the variance is (b-a)2/12 .The Exponential Distribution decays exponentially with time from an initial probability density λ att=0. The mean time for an arrival is E[T] = 1/ λ which equals the e-folding time of the exponential. Itsvariance is 1/ λ2 . This cumulative exponential distribution is the probability that the first arrival T1occurs outside a fixed time interval [0,t]; it equals the probability that the discrete number of Poissonarrivals K(t)=0 occurs within the interval [0,t] , that is, Pr(T1>t)= Pr(K(t)=0).The r-Erlang / Gamma Distributions for r>1, all rise from zero to reach a maximum at (r-1)/ λ and thendecay almost exponentially ~tr-1e-λt to zero. The maximum occurs after a wait of one exponential meanwait time 1/ λ for r=1, two 1/ λ waits for r=2, and r 1/ λ waits for any r. The variance is r times that ofthe exponential variance 1/ λ2 . The cumulative r-Erlang distribution is the probability that the rth arrivaltime Tr occurs outside a fixed time interval [0,t] ; this equals the probability that the discrete number ofPoisson arrivals K(t) ≤ (r-1) i.e., Pr(T1>t)= Pr(K(t) ≤ (r-1)). The Gamma density is a generalization of therth Erlang density obtained by replacing (r-1)! with Γ(r) making it valid for non-integer values of r.The Gaussian (Normal) Distribution is the most universal distribution in the sense of that the Centrallimit theorem requires that sums of many IID RVs approach the Gaussian distribution.The Rayleigh Distribution results from the product of two independent Gaussians when expressed inpolar coordinates and integrated over the angular coordinate. The probability density is zero at x=0 andpeaks at r=1/a½ before it drops towards zero with a “Gaussian-like” shape for x>0. It is compared withthe Gaussian which is symmetric for about x=0. 101
  • 40. Consequences of Central Limit Theorem Discrete Uniform PMF pX(x) / 1 11 1 p X ( x) = δ ( x − xi ) ; xi = −.5,L , 0,L ,.5 11 x -.5 -.4 -.3 -.2 -.1 0 .1 .2 .3 .4 .5 Generate uniform Sequence of N=1000 points { Xi } {Xi } .2 | .5 | -.1 | .3 | -.2 | -.1 | -.1 | .4 | -.3 | .1 | -.5 | -.1 L -.1 | .4 | -.3 | .1 | -.5 | -.1 n=2 .7 .2 -.3 .5 -.2 -.6 Sum of n Uniform Variates Xi n n=4 .9 .2 -.8 Z n = ∑ X i ; n = 2, 4,8,12 i =1 n=8 1.1 Plot Frequency of Occurrence f Zn ( z ) n = 12 .2 fZn ( z ) ≈ pZn ( z ) pX (x) = 1 11 Note: Curves give “shape” of 1.0 n = 2 freq of occur. for discrete points spaced 0.1 apart n = 4 .05 Central Limit Thm: n = 12 =>Generates a Gaussian as n=2,4,8,12, … large z 2/24/2012 -2.0 - 1.0 0 1.0 2.0 109The Discrete Uniform PMF with values at 11 discrete points ranging from x ={-.5, -.4, -.3, -.2,-.1, 0, .1,.2,.3,.4.,.5} can be expressed as a sum of 11 δ-functions with magnitude 1/11 at each of these points asshown in the figure. This can also be thought of as the result of a “sample and hold” transform (seeSlide#26) of a Continuous Uniform PMF fY(y) = 1/11 ranging along the y-axis from y=-.6 to y=+.5 ; forexample, the term 1/11*δ(x-(-.5)) is the δ-function located at x= -.5 generated by integrating thecontinuous PMF from y= -.6 to y=-.5 which gives an accumulated probability of ”.1/(.5 –(-.6)) =1/11 atthe correct x-location.Suppose that a sequence of 1000 numbers from the discrete set {-.5, -.4, -.3, -.2,-.1, 0, .1, .2,.3,.4.,.5} arerandomly generated on a computer to create the data run notionally illustrated in the 2nd panel . Now wecan create sum variables Zn consisting of the sum of n =2 or n= 4 or n= 8, or n=12 of these samplesfrom the discrete uniform PMF. According to the CLT, as we increase “n”, the resulting frequencydistribution of the sum variables “Zn“s should approach a Gaussian. The notional illustration shows whatwe should expect. The dashed rectangle shows the bounds of the original uniform discrete PMF and theother curves show the march towards a Gaussian. Note that unlike a Gaussian all these distribution arezero outside a finite interval determined by the number of variables that are summed. The triangle shapeis the sum of two RVs and obviously the min and max are [-1, 1] for Z2 ; the Z12 RV on the other hand,covers the range from [-6, 6]; the range increases as we sum more variables, but only as n-> ∞ does thesum variable fully capture the small Gaussian “tails” for large |x| as required by the CLT.This result can also be thought of in terms of an n-fold convolution of the IID RVs Xk k=1,2,...,n whichalso spreads out with each new convolution in the sequence. The next slide shows the results of a MatLabsimulation of this CLT approach to a Gaussian and a plot of the results confirming the notional sketchshown on this slide. (The MatLab script is given on the notes page of the next slide.) 109
  • 41. Examples Using Markov & Chebyshev Bounds Markov Examples: Prob “value” of RV X exceeds “r” times its Kindergarten Class mean height = 42” Find bound on mean is 1/r Prob of a student being taller than 63” 1 µ X = 42 r ⋅ 42 = 63 ⇒ r = 1.5 Pr[ X ≥ 1.5 ⋅ 42] ≤ 1 / 1.5 = 66.7% P[ X ≥ rµ X ] ≤ r or Note that for r =1 the Markov E[ X ] µ X bound is “1” or 100%; P[ X ≥ c] ≤ = Thus useful bounds c c require r >1 0 µX 2µ X 3µ X Chebyshev 1.5µ X Prob “deviation” of RV Ross Ex. 7-2a) Factory production X exceeds “r” times its std dev r σX is 1/r2 a) Given mean =50, find bound on Prob production exceeds 75, i.e., Prob[X>75] P[ X ≥ 75] ≤ E[ X ] = 50 = .667 Markov P [ X − µ X ≥ rσ X ] ≤ 1 c 75 r2 Note a upper bound: at most 66.7% or b) Given also variance = 25 , find bound on Prob production 2 between 40 and 60 σX P[ X − 50 ≥ 10] ≤ 25 P[ X − µ X ≥ k ] ≤ 10 2 = .25 Chebyshev k2 ⇒ 1 − P[ X − 50 ≥ 10] ≥ 1 − .25 = .75 Note a lower bound: at least 75% 2/24/2012 121Here are two examples of the application of the Markov and Chebyshev Bounds. The two forms for eachare stated on the LHS of the slide for reference purposes. The decision to use one or the other of thesebounds depends upon what type of information we have about the distribution. Thus if the RV X takes ononly positive values and we only know its mean, µX , then we must use the Markov bound. On the otherhand, if the RV X takes on both positive and negative values and we know the mean, µX , and variance,σX2, then we must use the Chebyshev bound. If in the latter case the RV X takes on only positive values,then we could use either Chebyshev or Markov bounds, but we would choose Chebyshev over Markovbecause it uses more of the information and hence will always be a tighter upper bound. Neither of thesebounds is very tight because the information about the distribution is very limited; knowing the actualdistribution itself always yields the best bounds.1) The mean height in a Kindergarten Class is µX = 42” and we are asked “what is the probability of astudent being taller than 63?” Short of knowing the actual distribution, the best we can do is use theMarkov inequality to find an upper bound Pr[X>63] < 42/63=.67 or 67%. This is also easily computed ifwe realize that the tail is the region beyond 63”= 1.5(42”) so r=1.5 and the answer is 1/1.5 =2/3=.67 .2) The factory production has a mean output µX = 50 units and we are asked(a) “what is the probability of a 75 unit output?” This again involves a positive quantity X the number ofunits and we choose the Markov bound for 1.5(50) = 75 units so again r=1.5 and the resulting probabilityis 67% .(b) If we are also given the variance of the production σX2 = 25 the additional information allows us touse the Chebyshev bound to find the probability in the tails on either side of the mean of 50. Thus, if wefind the probability in the 2-sigma tails (r=2) to left of 50-10 and to the right of 50+10 as Pr[Tails] ≤ 1/22= 25%. Hence the production within the bounds [40,60] is the complementary probability Pr[40 ≤ X ≤ 60] =1-Pr[Tails] ≥ 1-.25 = .75 or at least 75% 121
  • 42. Transformation of Variables & General Bivariate Normal Distribution r Mean Covariance X a bivariate normal m X = E[ X ] = 0 0  1 0  (indep comp) N(0,1) mX =   K XX =   K XX = E[ X ⋅ X T ] = I 0  0 1  Linear Xform to Y Y = AX + b b  mY = b =  1  KYY = AAT b2  Computation mY mY = E[Y ] = E[ A ⋅ X + b] = A ⋅ {+ b = b E[ X ] r =0 Computation KYY [ ]   K YY = E (Y − mY )(Y − mY )T = E ( Y − b)(Y − b)T  = E[ AX ( AX )T ] = E[ A( XX T ) AT ] = A E[ XX T ] AT = AAT { 123 4 4  = AX +b  =I det K YY = det A ⋅ det AT = (det A) 2 Determinant KYY ⇒ det A = det KYY  ∂y     y A is Jacobian: det  i  = det{ Aij } ⇒ J   = det( A) = det KYY   ∂x j    x (A ) (A ) = (AA ) −1 T −1 T −1 −1 = K YY 1 ( )  T −  A−1 ( y − b ) ⋅ A−1 ( y − b )  1 2    New Prob Density f X ( x) e fY ( y ) =  y1 y2  = 2π J x  det KYY  1 x2   1 − 2 [( y − m y )T KYY −1 ( y − m y ) ] 1 (No Longer Independent General Bivariate e f Y ( y ) = 2π Normal Distribution Components or zero det KYY means & unit variances) 2/24/2012 132We introduced the Bivariate Gaussian distribution for the case of two independent N(0,1) Gaussians(with the same variance =1) and arrived at a zero mean vector mX and a diagonal covariance matrix KXX=diag(1,1) corresponding to a pair of uncorrelated Gaussian RVs and displayed in the first line of thetable. The second line of the table shows the results of making a linear transformation of variablesY=AX+b from the X1 X2 coordinates to the new Y1 Y2 coordinates; note that the vector b =[b1,b2]Trepresents the displaced origin of the Y1 Y2 coordinates relative to X =[0,0]T. We see that the new meanvector is no longer zero but rather mY = b and the new covariance KYY =AAT no longer has unitvariances along the diagonal, but, in general, now has non-zero off-diagonal elements as well. The factthat this linear transformation yields non-zero off-diagonal elements in the covariance matrix means thatthe new RVs Y1 Y2 are no longer uncorrelated.The computations supporting these table entries are straightforward. The new mean is obtained by takingthe expectation E[Y]= E[AX+ b] and using the fact that the original mean E[X] is zero to give mY =E[Y]= b . Substituting this value b for mY in the covariance expression KYY = E[(Y-b)(Y-b)T] yieldsKYY = E[(AX)(AX)T] = A E[XXT] AT =A AT since E[XXT] =KXX = I (i.e., the identity matrix diag(1,1)).In order to find the new Bivariate density fY1,Y2(y1,y2) we need to divide fX1,X2(x1,x2) by the Jacobiandeterminant J(Y,X) and replace X by A-1(Y-b). This Jacobian is found by differentiating thetransformation Y=AX+b to find J=det[∂Y / ∂X ] = det(A) ; note that this is easily verified by writing outthe two equations explicitly and differentiating y1 and y2 with respect to x1 and x2 to obtain the partials∂yi / ∂xj = aij and then taking the determinant to find the Jacobian. Taking the det(KYY) =det(AAT) andusing the fact that the determinant det(A) = det(AT), we find that detA = det (KYY)½ . Finally substitutingthis and X = A-1(Y-b) yields the general Bivariate Normal Distribution fY(y) given in the grey boxedequation at the bottom of the slide. Be careful to note that the inverse KYY-1 occurs in the exponentialquadratic form and that the matrix KYY occurs in the denominator det (KYY)½ ; also observe the“shorthand” vector notation for the bivariate density fY(y) in place of the more explicit fY1,Y2(y1,y2). 132
  • 43. Bivariate Gaussian Distribution & Level Surfaces −1 < ρ < +1 Ellipse in y1 – y2 space; fY1Y2 ( y1 , y2 ) ≠ fY1 ( y1 ) ⋅ fY2 ( y2 ) y1 & y2 are dependent  σ2 ρσ1σ 2  K = 1 2  ρ =0 Diagonal Terms only; ρσ1σ 2 σ2  Either Ellipse or Circle fY1Y2 ( y1 , y2 ) = fY1 ( y1 ) ⋅ fY2 ( y2 ) Principal Axes along y1 & y2 det (K ) = σ1 σ 2 (1 − ρ 2 ) ≥ 0 2 2 independent ρ = ±1 Degenerate Case: Ellipse st. line: Along one of the “Principal Axes”; y2 = ±ρ · y1 = ± y1 1 1 − yT KYY −1 y fY1Y2 ( y1 , y2 ) = e 2 y1 & y2 are “extremely dependent” correlated or anti-correlated 2π det KYY Positive ρ > 0 Negative ρ < 0 NO ρ = 0 Correlation Correlation Correlation y2 Ellipses Ellipse Along Principal Axes y2 y2 ρ >0 ρ=0 + 45o ρ<0 σ1 > σ 2 y1 y1 y1 fY1Y2 ( y1 , y2 ) - 45o Gaussian Degenerate Ellipses Circle y Probability y2 y2 2 Surface ρ=0 y2 ρ = +1 σ1 = σ 2 ρ = −1 + 45o arbitrary 2 d Ellipses y1 Ellipse Areas y1 y1 y1 collapse to a line orientation - 45o 2/24/2012 135The bivariate density fY(y) = fY1,Y2(y1,y2) is completely determined by its mean vector mY and itscovariance matrix KYY as given by the equations on the upper right. Consider the the Bivariate Gaussiandensity which is plotted as a 2d surface relative to its mean vector components mY1 and mY2 taken as theorigin. The level surfaces represented by cuts parallel to the y1-y2 plane are the ellipses given by thequadratic form equation of the last slide. The structure of these ellipses are shown in the tableauconsisting of 3 columns for positive, negative, and zero correlation coefficient ρ and by 2 rowscorresponding to general (top row) and degenerate cases.The general cases in the top row have unequal sigmas σ1> σ2 and as we go across the row we have anellipse with positive correlation (ρ > 0), one with negative correlation (ρ < 0) and an ellipse along itsprincipal axes with no correlation (ρ =0). The (red) arrows show the directions of the principal axes ofthe ellipse in each case; the zero correlation case on the extreme right has the principal axes coincidingwith y1 and y2 , while the negative correlation case has its principal axes rotated at -45o to the y1-axis andthe positive correlation case has its principal axes rotated at +45o to the y1-axis.The bottom row illustrates the two degenerate cases ρ =+1 and ρ =-1 in which the ellipse “collapses’ to astraight line corresponding to complete correlation or anti-correlation (opposite variations of Y1 and Y2)respectively, and the degenerate uncorrelated case ρ =0 in which the principal axis ellipse above itdegenerates into a circle because the two sigmas are equal (σ1=σ2 ). 135
  • 44. Ellipses of Concentration 1D Gaussian Distribution 2D Gaussian Distributions described described by two scalars: by vector & Matrix: mean vector mean µX & Var(X) intuitive Tabulate Area mX & Covariance KXX Normalized & Centered RV x 1 −t / 2 Φ( y) = ∫ e dt 2 Vector mX and KXX are not very intuitive! Standardized Distribution 2π t = −∞ 1 Tabulation of CDF 1 − xT K XX −1 x f X ( x1 , x2 ) = e 2 2π det K XX fX(x) Y= X − µX fY(y) σX Gaussian σX σX Probability Surface x y y x2 µX 0 “Level Curves” Prob Density Standardized Density 2 d Ellipses x1 “Level curves” of Zero  x2 x  2 −1 1 2ρx1 x2 Mean 2D Gaussian Surface xT K XX x =  12− + 2 2  = c 2 = const. with Covariance KXX ( 1 − ρ2 )  σ X1  σ X1 σ X 2 σ X 2   2/24/2012 138The 1-dimensional Gaussian distribution is completely described by two scalars the mean µX and thevariance σX2. The tabulation of a single integral for the cumulative distribution function FY(y) shown inthe left box is sufficient to characterize all Gaussians X: N(µX , σX2 ) if we first transform to astandardized Gaussian RV Y via Y = X- µX) / σX. The Gaussian integral representing the probabilitydistribution for the standardized Pr[Y≤y] = FY(y) is used so often it is denoted as the “Normal Integral”Φ(x).We would like to extend this concept of a single tabulated integral to describe all 2-dimensional Gaussiandistributions; however, as we have seen, the Bivariate Gaussian distribution requires more than just themeans and variances of two Gaussians as we must also characterize their “co-variation” by specifyingtheir correlation coefficient ρ. Thus we must specify the two elements of the mean vector µX and all threeelements of the (symmetric) covariance matrix KXX in order to completely characterize a BivariateGaussian fX1X2(x1,x2) given in the right box of the slide.We have seen that the level “surfaces” (actually curves) of the Gaussian PDF are ellipses centered aboutthe mean vector coordinates µX1 and µX2 and described by quadratic form xTK-1XX x in the exponent ofthe PDF. The explicit equation for the level curves with zero mean is obtained by setting this term equalto an arbitrary positive constant c2 as given by the equation in the slide. These ellipses are called ellipsesof concentration because the area contained within them measures the concentration of probability for thespecific “cut through” the PDF surface. In the next few slides we will show how this leads to a singletabulated function for the Bivariate Gaussian that is analogous to Φ(x) for the Normal Distribution. 138
  • 45. Gaussian & Bivariate (2d) Gaussian Distributions Compared Probability for x to be within an ellipse “scaled by c”: α = 68.3% Prob region 2 Prob( xT K xx −1 x < c 2 ) = FC (c ) = 1 − e − c /2 =α “slice” −1 Note: Inverse Covariance K xx x2 2 d Ellipse determines Ellipse 68.3% x1 Scale Factor c in terms of % concentration: Equivalent 1d sigma table c = − 2 ln(1 − α ) 1d sigma α (%) c fX(x) 1-σ 68.3 1.52 1-σ ≈ c=1.52 Prob Density σX σX 2-σ 95.4 2.48 µX x 3-σ 99.7 3.41 68.3% 2/24/2012 141On the last slide we found that the 2d probabilities are described in terms ellipses of concentrationspecified by the axis scale parameter c which is related to the percentage of events contained within theellipse by the expression shown in the slide. This CDF is in fact a Rayleigh distribution with “radialdistance r” replaced by the ellipse scale parameter “c”.Setting this probability within the ellipse (parameterized by the value “c”) equal to α allows us to solvefor the value of c in the boxed equation. Using this equation, we compute the table which displays thevalues of the ellipse scaling parameter “c” corresponding to the standard values of 1-σ (68.3%) , 2-σ(95.4%), and 3-σ (99.7%) associated with a 1-dimensional Gaussian distribution.These ellipses are used to specify equivalent “standard deviations” for the Bivariate Gaussian andextending this tabulation for all probabilities allows us to define a standard Bivariate Normal functionΨ(c) similar to the Φ(x) for the Normal Gaussian.The two figures illustrate this equivalence by showing the c=1.52 cut through the Bivariate Gaussiansurface yielding an equivalent “1-σ”ellipse containing α = 68.3% of the probability and then notionallycomparing the ellipse with the “1-σ” area under the standard Gaussian curve. 141
  • 46. Closure Under Bayesian Updates - Summary Summary: r X  ρ rr  X  0  1 Started with a pair of N(0,1) RVs X & Y with correlation ρ X =  µX ≡ E  =   K XY =  1 Y   Y  0  ρ  1) The joint distribution is a correlated Gaussian in X and Y  x 2 − 2 ρ xy + y 2    −  2(1− ρ 2 ) f XY ( x, y ) = 1 e 2 2π 1− ρ 2 e − y /2 2) Marginal fY(y) is found to be N(0,1): fY ( y ) = 2π ( x − ρ y )2    −  3) Bayes’ Update fX|Y(x|y) is Gaussian 2(1− ρ 2 ) N ( ρ y,1 − ρ 2 ) f X |Y ( x | y ) = 1 e 2π (1− ρ 2 ) 4) Pick off “conditional” mean & variance from fX|Y(x|y) µ X |Y ≡ E[ X | Y ] = ρy ; Var ( X | Y ) = 1 − ρ 2 Conditional Mean represents an “estimate of X given meas.Y” with Var(X|Y) obtained from Bayes’ Updated Gaussian Generalize: r X  µ   σ 2 ρσ X σY  Start with General Gaussian Vector X =  µ= X ; K XY =  X 2  Y   µY  ρσ X σY σY  with non-zero mean &Variance σX µ X |Y ≡ E[ X | Y ] = µ X + ρ ( y − µY ) Conditional Mean and Variance σY Represents the Bayes’ Update Equation 2 Var ( X | Y ) = σ X (1 − ρ 2 ) ; σ X |Y = σ X 1 − ρ 2 Note 1 “Gaussian Arena” we do not need to work Note 2: Y is irrelevant for ρ=0 with distributions directly since both X & Y indep => Conditionals do not 1) Linear Xfms & 2)Bayes’ Update Equation yield depend upon value of y: Gaussian Vector Results (surrogates for the joint µX|Y = µX & σXY2 =Var(X|Y) = σX2 and conditional distributions respectively) 2/24/2012 151Closure Under Bayesian Updates started with a pair of correlated N(0,1) Gaussian RVs with correlationcoefficient ρ. and resulted in a Gaussian conditional distribution fX|Y(x|y) with conditional mean is µX|Y= E[X|Y] = ρy and conditional variance is Var(X|Y) = σX|Y2 = 1-ρ2.If instead, we start with a pair of correlated Gaussian RVs having different means and variances given bythe mean vector µX and covariance matrix KXY shown in the middle panel of the slide yields the generalresult for a Gaussian with a conditional mean E[X|Y] = µX|Y = µX + ρσX(y- µY)/σY , and conditional variance Var(X|Y) = σX|Y2given in the boxed equation.The lower panel interprets these results in terms of a two dimensional “Gaussian Arena” in which theinput and output are related by the underlying joint Gaussian distribution which remains Gaussian for allpossible linear coordinate transformations and even maintains its Gaussian character when one of thevariables is conditioned on the other. Thus the Gaussian vector remains Gaussian under both lineartransformations and Bayes’ updates. Also note that if the correlation is zero (ρ =0) then the input andoutput variables are independent as is evident in the boxed equations which reduce to statements that theconditional mean is equal to the mean µX|Y = µX and the conditional variance is equal to the varianceσX|Y2 = σX2 .We note in passing that because the quadratic form in the joint Gaussian is symmetric in the X and Yvariables, we could just as well have computed the output Y conditioned on the input X to find analogousresults with X Y corresponding to the forward Bayesian relation.A visual interpretation of this result will be given in the next slide and further insight into the role of thecommunication channel and its inverse will be given in the slides after that. 151
  • 47. General Case: Visualization of Conditional Mean given a priori yields a posteriori Bayesian Update ρσ X Conditions X on Y µX ; σ X 2 µ X |Y = µ X + ( y − µY ) ; σ 2 = (1 − ρ 2 )σ 2 σY X |Y X fX|Y(x) Distribution is Gaussian with conditional mean µX|Y “y0-slice” conditional variance σX|Y2 σ X |Y σ X |Y Choose arb. y0 ; it is tangent to an ellipse whose max is ymax= y0 = +c x y µ X |Y x’ Recall Covariance Ellipse Construction Extremum y = y0 y’ “slice” x 2 − 2 ρ x y + y 2 = (1 − ρ 2 ) ⋅ c 2 x − µX y − µY % %% % x= % ; y= % σX σY found the corresponding x- value to be x x0 − µ X y0 − µY “origin at” x0 = µ X |Y = y0 x0 ≡ x( y0 ) = ρ y0 ⇒ % % % % = ρ⋅ σX σY ( µ X , µY ) y − µY x0 = mean x0 = µ X + ρ ⋅ σ X ⋅ 0 = µ X |Y = y “conditioned σY 0 on the y0-slice” Special Cases: σ µ X |Y = µ X + ρ X ( y − µY ) Degenerate Ellipse ρ = + 1 σY y If ρ = 0 µ X |Y = µ X Indep. (Y is irrelevant) Distribution is a Single E[ XY ] y = y0 Unique point with zero ρ= σ X ⋅ σY If ρ = +1 µ X |Y = µ X + σ X ( y − µY ) direct correlation σ variance! Y ( µ X , µY ) x σX µ X |Y = y If ρ = −1 µ X |Y = µ X − σY ( y − µY ) inverse correlation 0 2/24/2012 152The results for the conditional mean and variance can be understood graphically as follows. Starting withthe Bivariate Gaussian Density we draw the elliptical contours corresponding to the horizontal cutsthrough the density surface centered at the mean coordinates µX and µY indicated by the black dot at thecenter. If we choose a fixed value of y=y0 the line parallel to the x-axis is tangent to one of the ellipsesand hence y0 represents the maximum y-value for that ellipse as shown by the red dot. This line alsoresults from a vertical plane y=y0 cutting through the distribution and the Gaussian cut through thedistribution is shown above the contours.The x-coordinate corresponding to this maximum is obtained by dropping a perpendicular onto the x-axisat a value x0 = µX|Y=y0 as shown in the figure. Recalling the calculation used for the covariance ellipseconstruction, the x0-value corresponding to this maximum at y=y0 is given in standardized coordinates x0=ρy0 which is converted to the coordinates of the figure by letting x0 -> (x0-µX)/σX and y0 -> (y0-µY)/σYto yield (x0 –µX)/ σX = ρ (y0-µY)/σY or x0 = µX +ρ σX (y0-µY)/σY which is exactly the statement that x0is the conditional mean µX|Y=y0 .The three special cases ρ=0,+1,-1 shown in the bottom panel are:(i) ρ=0 no correlation corresponds a coordinate system along the principal axis of the ellipse for which aconstant y=y0 cut will always yield a conditional mean µX|Y=y0 = µX(ii) ρ=+1 complete positive correlation corresponds the case where the ellipse collapses to a straightline; the conditional distribution is a single point with zero variance on the line with slope (σY/σX) asshown and yields a conditional mean µX|Y=y0 = µX +σX (y0-µY)/σY(iii) ρ=-1 complete negative correlation corresponds the case where the ellipse collapses to a straightline; the conditional distribution is a single point with zero variance on the line with slope (-σY/σX) (notshown) and yields a conditional mean µX|Y=y0 = µX -σX (y0-µY)/σY 152
  • 48. Rationale for “Inverse Channel” & Generating Correlated RVs Rationale: “X=ρY+V” Given Y: N(0,1) RV (i) If Noise is not added: X=ρY: Generate X: N(0,1) correlated to Y with coeff. ρ Var(X) =Var(ρY) =ρ2 Var(Y)= ρ2 ≠ 1 (ii) If uncorrel noise is added X=ρY+”V” with Inverse Channel Method: X=ρY+V appropriate Var(V)= (1- ρ2 ) to cancel correl contrib. to Var(X) then Var(X) = Var(ρY+V) = ρ2 Var(Y) + Var(V)+2Cov(Y,V) Y=N(0,1) ρ X=N(0,1) = ρ2 . 1 + (1- ρ2 ) + 0 = 1 input output Special Cases: “X=ρY+V” ; -1 ≤ ρ ≤ +1 V=N(0,1-ρ2 ) ρ = 0: No correlation between X & Y. noise 0.Y + N(0,1-02 ) = N(0,1 ) X (i) Generate samples of RV “Y” using standard X is simply the uncorrel noise sample N(0,1). method (e.g., sum 12 uniform Variates on [-0.5, 0.5]) to yield N(0,1). ρ = ±1: Full correlation/anti-correlation (Degenerate Ellipse or St.Line) (ii) Generate zero mean Gaussian noise “V” with variance 1- ρ2 to yield N(0, 1- ρ2 ). ±1 . Y + N(0,1-(±1 )2 ) = ±Y X (iii) Multiply each RV sample “Y” by desired X is simply ±Y – value correlation coefficient ρ -1 < ρ < 1: General correlation (iv) Add noise sample “V” to obtain output “X” which is N(0,1) and has the desired correlation ρ . Y + N(0,1- ρ 2 ) X coefficient correl(X,Y)= ρ X results from multiplying Y by the correlation ρ and adding noise with variance (1- ρ 2 ) 2/24/2012 155The last couple slides considered the inverse channel and its relation to a Bayesian update which startswith an a priori value of the mean µX and variance σX2 and then updates their values as a result of anactual “measurement Y”. The conditional mean and variance formulas that we found comported withboth the Bayesian Update equation for conditional probability densities and also to those obtained byconstructing an inverse channel which creates an input X from an output Y. In this slide and the next weconsider this important “coincidence” in some detail.The box on the left uses the inverse channel model as a computer program flow diagram to actuallygenerate a RV X~N(0,1) from a linear combination of Y ~N(0,1) and noise V~N(0,1-ρ2) . Note that theinput and output RVs are both N(0,1) Gaussians with unit variance yet the noise must have a variancethat is less than unity for this to work.The rationale is simple enough, for consider what might be your first impulse to generate a pair ofcorrelated RVs by setting Y = ρ X (upper right box); taking the expectations E[Y] and E[Y2] we find µY= ρ µX = ρ *0 = 0 and σY2 = ρ2 σX2 = ρ2 ≠ 1 which this does not agree with the assumption that both Xand Y are N(0,1). Agreement is possible only if we add zero-mean noise with variance (1-ρ2) becausewhen added to ρ2 it yields the desired unit variance for the RV Y.The special cases of no correlation (ρ = 0 ) and full positive and negative correlation (ρ = ±1 ) areexplicitly shown to be in agreement this model. For no correlation the model gives X as just N(0,1)random noise which is takes on values completely independent of the y–values. On the other hand forfull positive (or negative) correlation the model gives X as N(0,1) which takes on values that are exactlythe same as those for Y (or –Y). In the general case -1 < ρ <+1 the model gives X as N(0,1) RV whichtracks Y more closely for correlations near +1 and tracks the noise more closely for correlations nearer tozero thus giving the expected intermediate behavior. 155
  • 49. Multilinear Gaussian Distribution 1 n-dimensional Gaussian 1 ( x −µ X )T K XX −1 ( x −µ X ) f X ( x) = e− 2 Vector X= [ X1, X2,... Xn]T ( 2 π) n/2 det K XX  K11 K12 K13 L K1n  K K 22 K 23 L K 2n   21   K 31 K 3n  Matrix components (K XX )rc = E [(X r − µ X r )(X c − µ Xc )] ; r , c = 1,2, L n  K 32 K 33 L   M M M K rc M   K n1  K n2 K n3 L K nn   1 T r T t K XX t + µ X T t Moment Generating Fcn φ X (t ) = E[e X ⋅t ] = e 2 r ; t = [t1 , t 2 , L t n ]T Still Gaussian After Linear Transformation: Y = AX + b µ Y = Aµ X + b KYY = AK XX AT (See Next Slide =>) 1 1 ( y −µY )T K YY −1 ( y −µY ) fY ( y ) = e− 2 Gaussian 1st and 2nd Moment Vector µX & (2π) n / 2 det KYY Covariance KXX Uniquely Defines Multivariate Gaussian Details r r r r r r µY = E[Y] = E[AX + b ] = Aµ X + b ( ) ( ( )) r r r r r r r r Y − µY = AX + b − Aµ X + b = A(X − µ X ) [( )( r r r r T )] [ r r r r T ( r r )] [ r r r r r r ] K YY = E Y − µY Y − µY = E A(X − µ X ) A(X − µ X ) = E A(X − µ X )(X − µ X )T AT = AE (X − µ X )(X − µ X )T AT = AK XX AT 144424443 [ ] = K XX 2/24/2012 157The extension to Multilinear Gaussian distributions or Vectors is straight forward; taking the product of“n” independent N(µX, σX2) Gaussians symbolized by the vector X=[X1,X2,...Xn]T yields an n-dimensional Gaussian characterized by an n-dimensional mean vector µX and n x n covariance matrixKXX whose diagonals equal the variances of the individual RVs and whose off diagonal elements are allzero.Even if we start with independent RVs, a linear transformation of the form Y= AX + b producescorrelations and the off-diagonal terms of the new covariance matrix are no longer zero. Thetransformation leaves the Gaussian structure the same, but the mean and covariance become µY = AµX +b and KYY = AKXX AT respectively.The Gaussian always has the form fX(x)=(2π)- n/2 (detKXX) )-1/2 exp(- ½ q) with the scalar quadratic q =[x-µX]T KXX-1[x-µX]. The row-column components of the covariance matrix are determined by theexpected values of the “row-col” pair products of centered deviations.The moment generating functiongeneralizes to φX(t) = E[exp(X tT )] = exp( ½ tT KXX t +µXT t) with t= [t1,t2, ...,tn]T.Note that we have reverted to the old notation in which the components of the Gaussian vectors arelabeled by indexed quantities Xi and the new components under a coordinate transformation are Yi. Thisis temporary, however, because we shall want to consider communication channels with a number ofinputs and a number of outputs and partition the n-dimensional Gaussian vector into these two distincttype of components in order to define the conditional distribution as µX|Y in a useful manner. 157
  • 50. Partitioned Multivariate Gaussian & Xfm to Block Diagonal Partition: [X(1) | X(2) ]T {Comm Channel with multiple inputs: “X”= X(1) & outputs “Y”= X(2) } 2 x 1 Partitioned Vectors 2 x 2 Partitioned Matrix  K11 L K1k K1,k +1 L K1n   x1   µ1   M x  µ   kxk M M k x (n-k) M    2   2   K (1)(1) K (1)( 2 )   K k ,1 K K kk K k ,k +1 L K kn  = K ( 2 )( 2 )   M   M   x(1)    µ (1)      K ( 2 )(1)      xk  K =  K=   µk   K k +1,1 L K k +1,k K k +1,k L K k +1,n      x( 2 )   L    µ ( 2)   L    µ k +1  M M (n-k) x k M (n-k) x (n-k) M  x  k +1       M   M   K n1  L K nk K n ,k +1 L K n ,n        xn   µn   y(1)   A11 M A12   x(1)   I k ,k Bk ,( n − k )   I k Bk ,( n − k )  Perform Linear Xfm      where, A =  = in “partitioned form”  K  =  L M L  K   A M A  x   y(2)   21  0 ( n − k ), k I ( n − k ), ( n − k )   0 ( n − k ), k   I (n−k )     22   (2)  Now drop parentheses notation for partitioned components !! T I B   K11 K12   I k B   Ik B   K11 K12   I k 0  Find “B”matrix so that new AK XX AT =  k ⋅ ⋅  =0 ⋅ I n −k   K 21 ⋅ K 22   BT I n −k   0 I n −k   K 21 K 22   0 I n −k        KYY is block diagonal  K11 + BK 21 K12 + BK 22   I k 0  =  ⋅  BT I   K 21 K 22   n−k  K 21 + K 22 B T = 0 (1)  K11 + BK 21 K12 + BK 22   + K BT + BK BT  =  12 22  K12 + BK 22 = 0 (2)      K 21 + K 22 B T K 22  2/24/2012 159Consider a multi-dimensional communication channel partitioned into two sets as follows:“X”: k-inputs X(1) = [X1, X2, ..., Xk]T and “Y”: (n-k)-outputs X(2) = [Xk+1, Xk+2, ..., Xn]T . Themean vector and covariance matrix are also partitioned in the same manner to yield 2 x 1 partitionedvector X(I) and 2 x 2 partitioned covariance matrix K(I)(J). Note that the partition dimensions of K(I)(J) arespecifically as follows:Row#1 [K11 : K12] = [ k x k : (n-k) x k ]Row#2 [K21 : K22] = [ k x (n-k) : (n-k) x (n-k)] .Now lets perform a linear transformation to a new coordinate system according to the equation Y=AX+bwhere it is now understood that the Y(I) and X(I) and b(I) are all partitioned in the same manner as 2 x 1column vectors and the matrix A(I)(J) is partitioned into a 2x2 matrix which corresponds to the partitioningof the original covariance martix K(I)(J) as shown in detail on the slide. The transformed covariancematrix KYY is defined by the following product of n x n matrices A KXX AT ; in partitioned form weinstead have a product of three 2 x 2 matrices. The sub-matrices in the partition of A(I)(J) are chosen asfollows: A(1)(J) =[ Ik, k : Bk , (n-k)] and A(2)(J) =[ 0n-k, k : I(n-k), (n-k)] (labeled by their dimensions). The problemis to find the 2x2 matrix B such that the new covariance matrix KYY is block diagonal; taking the productof the three partitioned matrices A KXX AT results in two a 2x2 matrix shown at the bottom of the slide.Forcing the two “off-diagonal” partitions (circled) to be zero yields two conditions on the matrix B andits transpose BT as follows: (1) K21 + K22 BT =0 ; (2) K12 +BK22 =0Note that the partitioned components are of the original matrix KXX so for example K21 is the 2,1partition component or (KXX)21 . On the next slide we formally solve for B and B and write down theexplicit form of the block diagonal matrix KYY with just 2 components, namely, (KYY )11 and (KYY )22 .This will allow us to factor the multivariate Gaussian and prove a very elegant generalization of Bayes’Update for the conditional mean and conditional covariance known as the Gauss-Markov Theorem. 159
  • 51. Gauss-Markov Theorem Updating Gaussian Vectors under Bayes’Rule Given X and Y are jointly Gaussian Random input and output vectors with dim k and n-k respectively Combine to form n-dim vector with partitioned mean and covariance as follows :  K XX {  K XY r  X (k )  r  µ X (k )   { µ ≡ k ×( n − k )  { ≡ Y X  {  K ≡  k ×k n×1  ( n−k )  n×1 µY ( n − k )  { K K YY  n×n ( n − kYX k { {   )× ( n − k )×( n − k )  Gauss-Markov Theorem states that the conditional PDF of ”X given Y” is also Gaussian with conditional mean & covariance given by −1 µ X |Y = µ X + K XY KYY ( y − µY ) −1 { { { 1 3 123 2 4 4 K X |Y = K XX − { { K XY K YY K YX { k ×1 k ×1 k ×( n − k ) ( n − k )×( n − k ) ( n − k )×1 13 2 k ×k k ×( n − k ) ( n − k )×( n − k ) ( n − k )×k T Note: Although Covariance K Symmetry of K requires   K  = K is symmetric, the blocks { ≠ { K XY KYX the following relationship { XY {YX themselves are not , i.e., for the off diagonal blocks  k ×(24 ( n − k )×k 4  k ×( n − k ) ( n − k )× k n−k ) 1 3 ( n − k )× k 2/24/2012 163The result of the last section for the n-dimensional Multivariate Gaussian are now cast in a form moresuitable for a communication channel. We introduce the new notation in which the 1st partition of theGaussian Vector X consists of the k inputs Xk = [X1, ...,Xk]T and the 2nd partition consists of n-k outputsYk = [Yk ...Yk]T . The mean vector µX and covariance matrix KXX are partitioned in a natural manner asshown on the slide.In this notation, the Gauss-Markov Theorem states that the conditional PDF of “vector X given vector Y”is also a Gaussian with conditional mean and covariance given by the two boxed equations. This isidentical to the results of previous slide, however in a new notation.Note that a possible source of confusion is to equate the partitions Xk and Yk (whose dimensions k +(n-k)add to “partition” n) with the transformation of coordinates Y=AX used to transform between to n-dimensional coordinate systems from X to the canonical coordinates Y.Also note that even though the full nxn covariance matrix is symmetric Kr c = Kc r with respect to itsindices (i.e., K = KT), this is no longer true for the partitioned components K(R)(C) ≠ K(C)(R) as evidencedby the fact that KXY ≠ KYX as they usually do not even have the same dimensions. The symmetry of thefull matrix requires blocks with transposed partition indices be transposes of one another, i.e., KXYT =KYX which is possible now because these two matrices now have the same dimensions.The Gauss Markov Theorem is the basis for using the conditional mean estimator µX|Y to update the apriori mean value µX = E[X] of a k-dimensional state vector X by using an (n-k) dimensionalmeasurement vector Y. The state and measurement vectors must be part of the same multivariateGaussian distribution or equivalently the must be components of a partitioned Gaussian vector whosemeans, variances, and correlations are given by the partitioned n-dimensional mean vector andcovariance matrix shown at the top of the slide. They indeed form a Gaussian “Arena”. 163
  • 52. Gauss-Markov Estimator New RVs: Note: The “Estimator” and the “Error” depend ) Estimator RV upon the specific values of X=“x” and Y=“y” µ X |Y → X = µ X + K XY KYY −1 (Y − µY ) and hence generate samples of two new random ˆ ˆ variables X & e whose statistics can be . e = X − X = X − [ µ X + K XY KYY −1 (Y − µY )] Error RV inferred from those of X and Y. Following remarkable properties can be shown for these RVs ˆ Error e and Conditional Mean Estimator X satisfy the following: ˆ 1) E[eX ] = 0 & E[eY ] = 0 ˆ e ⊥ X & e ⊥ Y i.e., e is uncorrelated with the “orthogonal” ˆ estimator X and the data Y 2) K XY = K XY ˆ ˆ Estimator X and RV X have same correlation with measurements Y 3) Distributions for ˆ X and e satisfy “Pythagorean Right Triangle Relationship”as shown ˆ −1 X = N (µ X , K XY K YY KYX ) = N (µ X , Q) 14 244 4 3 Random ≡Q ˆ X = X +e X : N ( µ X , K XX ) Variable −1 e = N (0, K XX − K XY K YY KYX ) = N (0, P) 144 2444 4 3 ≡P Gaussian Means & Variances Add e : N (0, P ) Error N (µ X , K XX ) = N (µ X , Q) + N (0, P ) ˆ X : N (µ X , Q) Gauss-Markov Recall for Scalar X & Y: Y=ρ X+V N (0,1) = N (0, ρ ) + N (0,1 − ρ ) 2 Estimator 2/24/2012 164The conditional mean is evaluated for a specific “realization” of a Gaussian RV X=“x” and Y=“y” andhence looking at many realizations allows us to consider the conditional mean µX|Y as a random variableitself. Thus we replace the specific realizations µX|Y and “y” in the update equation by RVs denotedrespectively as X-hat and Y as shown in the first equation. Now the difference between the true state Xand the conditional mean estimate of that state X-hat is a RV that represents the Estimation Error e =X-(X-hat) as shown in the second equation.These two equation can be shown to have the following remarkable properties : 1) the error isuncorrelated with either the estimator X-hat or the data Y, 2) the X-hat estimator and the true state Xcorrelate with the measurements in the same way, and 3) the distributions for the RVs X-hat and esatisfy a “Pythagorean Right Triangle Relationship between their Gaussian designations.Looking at the figure the true state X ~ N(µX , KXX) on the hypotenuse, the estimator X-hat ~ N(µX , Q)where Q= KXYKYY-1KYX in the plane, and the error e ~ N(0 , P) where P= KXX - KXY KYY-1KYXperpendicular to the plane. The vector relation is X = X-hat + e which forms the right triangle and themeans and variances add so that µX =µX +0 and KXX = P + Q = (KXX - KXY KYY-1KYX )+(KXYKYY-1KYX).For the normal distributions this may be written in the suggestive form N(µX , KXX) = N(µX , Q) + N(0, P) .Also recall this relationship showed up for the scalar case of a single input X and single output Y in theform Y=ρX+V (where V = e (noise) and solving for the error e =Y-ρX) N(0,1) = N(0,ρ) + N(0,1-ρ2) 164
  • 53. To learn more please attend this ATI course Please post your comments and questions to our blog: http://www.aticourses.com/blog/ Sign-up for ATIs monthly Course Schedule Updates :http://www.aticourses.com/email_signup_page.html

×