Module ii sp

MODULE II
1Vijaya Laxmi, Dept. of EEE, BIT, Mesra

RANDOM VARIABLE
• A real random variable X is a real function whose
domain is the space S (i.e. a process of assigning a real
number to every outcome of the experiment
and such that
 The set is an event for any real number x
 The probability of the events equals
to zero.
)(X  
 xX 
    XandX
    0 XPXP

Classification of random variables
• Real random variable:
A random variable ‘X’ is a function that assigns a real number,
X (ξ) to each outcome ξ in sample space of a random
experiment, whose domain is the sample space and the set Sx
is a of all values taken on by X is the range of the variable .
Thus, Sx is a subset of the set of all real numbers.
P{X=+∞} =0
P{X=-∞}=0.
• Complex random variable:
Complex random variable is a process of assigning a complex
number to every outcome Z(ξ) .
Z(ξ) = X(ξ) +jY(ξ)
where X and Y are real random variables

Classification of random variables
Discrete random variable:
It has a finite number of random values and it is defined as a
random variable who’s CDF is right continuous, staircase
function of x with jumps at countable set of points
Sx = {x0 , x1…….xn}
Continuous random variable:
It has infinite number of random values and it is defined as a
random variable who’s CDF Fx(x) is continuous every where
and which in addition is sufficiently smooth that it can be
written as an integral of some non-negative function f(x).

• Mixed type:
 It is a random variable with a CDF that has jumps on
a countable set of points {x0, x1….xn}, but that also
increases continuously over at least one interval of value
x.
 It has distribution function with discontinuities but not of
staircase form.
• Lattice Type:
If random variable is of discrete type and numbers xi is
in arithmetic progression,
Then X is Lattice type random variable.
,...2,1,  ibiaxi

SAMPLE SPACE OF RANDOM VARIABLE:
The sample space of real random variable is Sx={X: X= X (ξ) for ξЄS}
X (ξ) → is the number assign to outcome ‘ξ’
X → is the rule of correspondence between any element of the set ‘S’
and the number assign to it.
Ex: X (fi) =10i i=1, 2, ..., 6.
Find
If i=1;
X (fi) =10*1= 10.
{X ≤35} = {f1, f2, f3}.
{X ≤5}= {0}.
{20 ≤X ≤ 45}= {f2, f3, f4}.
{X=35}=0.
{X=40}= {f4}.
     40,35,4520,5,35  XXXXX

CUMMULATIVE DISTRIBUTION FUNCTION (CDF)
The cumulative distribution function (CDF) of a random
variable ‘X’ is defined as the probability of the event {X ≤x}
Fx(x)=P[X ≤ x]
For -∞ < x < +∞, the distribution function is the probability of
the event if consisting all the outcome ‘ξ’ such that {X (ξ) ≤ x }
CDF of random variables X, Y and Z is given by Fx, Fy, Fz.
For x y z between -∞ < x < +∞
Fx(x) =P (X≤x), Fy(y) =P(Y ≤ y), Fz(z)=P(Z≤z),
Fx(w) , Fy(w), Fz(w) these three are pdf of x , y ,z

Example: Tossing a coin the sample space is given by
S= {h,t}, and P(h)=p , P(t)=q.
Define X(h)=1, X(t)=0. Find F(x).
Solution:
Case 1: If x ≥ 1 i.e., certain event
X (h) =1 and X (t) =0, it contains both head and tail,
F(x)=P(X ≤ x)=P(h,t)=1
Case 2: If the x is lies between 0 and 1, 0 ≤ x< 1
In this case it has only tail x(t) =0.
F(x) =P(X ≤ x) =P(t) =q
Case 3: If x is less than ‘0’ x < 0, i.e., null event
F(x)=P(X ≤ x)=0
Fx(x) = {1 for x ≥ 1
q For 0 ≤ x< 1
0 for x < 0

Cummulative Distribution Function
Probability Density Function

PROBABILITY DENSITY FUNCTION
• The probability density function of X(PDF), if it
exist it is defined as the derivative of Fx(x), i.e.,
dx
xdF
xf X
X
)(
)( 

Properties of PDF
• f(x) ≥ 0
• The CDF of X can be obtained by integrating the PDF
• By letting x tending to infinity we obtained normalization condition
for PDF’s
  
b
a
X dxxfbXaP )(


x
XX dttfxF )()(



 dttfX )(1

Problem1: Tossing of two coins
P(HH) = P (HT) = P (TH) = P(TT) = .25
P(X=0) = P(TT) = .25
P(X=1) = P(TH) + P (HT) = .5
P(X=2) = P(HH) = .25
Solution:

DISTRIBUTION FUNCTION OF DISCRETE RANDOM
VARIABLE
• It is defined as P(X≤x) = F(x), where x is any number
from -∞<x<∞
• F(x) = { 0 for -∞<x<x1
= f(x1) for x1 < x < x2
= f(x1) + f(x2) for x2 < x <x3
=f(x1) + f(x2) + ….+ f(xn) for xn < x < ∞ }

F(x) = { 0 for x<0
.25 for 0≤x<1
.75 for 1≤x<2
1 for x<0 }

Observations
• Magnitudes of jumps at 0,1,2 are ¼, ½ and ¼, which
precisely are the coordinates. This enables to obtain
the prob. function from the distribution function.
• It is a staircase function or step function. Hence,
continuous from the right at 0,1 and 2.
• As we proceed from left to right, the distribution
function neither remain same or increases taking on
values from 0 to 1. it is a monotonically increasing
function.

Problem 2: Rolling of a die
X(i) = 10i ; for i= 1 to 6
S = {10, 20, 30, 40, 50, 60}
CDF
17
   
     
      1
6
6
,,,,,100100
6
2
,2020
6
3
,,35)35(
654321
21
321



ffffffPXPF
ffPXPF
fffPXPF
Vijaya Laxmi, Dept. of EEE, BIT, Mesra

The value of X(ζ)=a for any ζ . Find out he CDF and PDF.
Problem
Sol:
Case 1: x ≥ a then, {X≤x}={S}
F(x)=P(S)=1
Case 2: x<a then, {X≤x}=Φ
F(x)= P(Φ)=0
18
F(x) is a unit step function.
U(x-a)=F(x)=1 for x≥a
=0 for x<a

The function X (t) = t , t= [0, T]
Here t has double meaning, it gives the outcome of the
experiment and also random variable.
Find the prob. density function of X
Problem
19
Sol: Case 1: if x ≥T then X≤x
{X ≤ x} ={0≤t≤T}=S
Hence, F(x)= P{S} = 1
F(x) =1
Case 2: if 0 ≤ x < T then {X ≤ x} = {0 ≤ x < T}
Hence, F(x) =P {0 ≤ t ≤ x} = x/T.
Case 3: if {X ≤ x} =Ф then {X ≤ x} is an empty set
hence, F(x) =P (Ф) =0
F(x) = { 1 for x ≥T
x/T for 0 ≤ x < T
0 for x<0

Properties of distribution function
21
    1,0  PP
    2121
sin
xxforxFxF
xoffunctiongnondecreaaisIt

   xFxF
rightthefromcontinuousisIt

.
   
    0,0,lim
0,0,lim,






xFxF
xFxFwhere

Density function
• The density function is defined as
22
dx
xdF
xf
)(
)( 

Discrete probability distribution
• If X is a discrete RV which assumes values as
• Then
• This is the probability that these values are assumed
with and known as probability distribution function
or probability function.
23
 kxxxx ...,,,, 321
    )..(...3,2,1 ikforxfxXP kk 

• If
• In general,
24
   
  0
)(


xfotherwise
iequationtoreducesxxforxfxXP k
 
  

x
xf
xf
1.2
0.1
Property 2 shows that the sum for all possible values of x must
be equal to 1

Continuous probability distribution
• If X is a continuous RV, the probability that x takes on any
one particular value is generally zero. Therefore, we cannot
define a probability function in the same manner as for a
discrete RV.
• Probability that X lies between two different values.
• Example: selection of an individual male from a large number
of males. Probability that the height X is precisely 68 inches.
• The probability is greater than zero that X lies between 68.5
and 68.5006 inches.

Properties
26
 
 




1.2
0.1
dxxf
xf
The property 2 shows that real valued RV must certainly lie between -∞ and +∞
   
b
a
dxxfbXaP
bandabetweenliesXthatprob.
The function f(x) which satisfies the above condition is called probability
function or probability distribution for continuous RV or more often called
probability density function or density function.

Find the constant ‘c’ for which the density function given as
Problem
27


 

otherwise
xforcx
xf
0
30
)(
2
9
1
1
3
1
3
0
3
3
0
2








c
cx
dxcx
 
27
7
3
1
3
8
3
9
1
21
2
1
3
2
1
2













 
x
dxxXP
Solution:

Distribution function for Continuous RV
• For cont. RV,
28
     
 




byreplacedbecan
duuf
xXPxXPxF
x

Find the distribution function for the random variable. And find the probability P
(1<x≤2).
Problem
• Case 1:
• Case 2:
• Case 3:
29


 

otherwise
xforcx
xf
0
30
)(
2
Solution:
    021,0
0


XPxF
x
   
27
9
1
30
3
0
2
0
x
dxxdxxfxF
x
xx




     
1
39
1
.0
9
1
3
3
0
3
3
2
3
0
3
3
0












u
duduu
dxxfdxxfxF
x
x
x
 











00
30
27
31
3
xfor
xfor
x
xfor
xF

• We have,
30
     
   
27
7
27
1
27
2
12
1221
33



FF
XPXPXP

• Probability that x lies between x and x+∆x
• The derivative of distribution function is the density
function.
31
   
   
    .)(int continuousisxfwherespoallatxf
dx
xdF
and
xxfxxXxP
xif
duufxxXxP
xx
x



 


1 2 2 1( ) ( ) ( )P x X x F x F x   
1)
2)
3)
4) It is a non decreasing function of ‘x’ if a<b
F(a) <F(b)
5) P(X<x) =1-P[X>x]
P(X>x) =1-P [X≤x]
=1-F(x)
6) The function continuous from the right, i.e., h>0
7) The probability that
8) The probability at X=b,
Properties of Distribution Function:
32
  10  xFX
  1lim 

xFX
x
  0lim 

xFX
x
     

 bFhbFbF XX
h
X
0
lim
     aFbFbXaPor XX ,
     
 bFbFbXP XX

0 0
( ) ( ) (0)
( ) ( ) ( )
x x dx
x x x dx x






  
   


0 0( ) ( )F x F x 

Representation of density function in terms of impulse/delta function
Assume k is the magnitude of discontinuity of discrete function
and K=
33
The derivative of a discontinuous function F(x) at a discontinuity point x0
exists and equals kᵟ (x-x0)
       
dx
xdU
xthenxUxFIf  ,

( )
d
F x
dx
( )i
i
Pi x x 
 
1
( 10) ( 20) ............ ( 60)
6
x x x       
1 1 1
( ) ( 1) ( 2)
4 2 4
U x U x U x   
f(x)= ,f(x)=
For tossing die the unit impulse representation is
f(x)=
Unit step representation for tossing of two coins:
F(x)=
.
34
If X is discrete RV, then its density function consists of impulses.
at point xi

:
In the density function, impulses are known as point mass Pi placed at xi .
If density is positive in entire X-axis has total mass =1
35
Probability Masses
The probability that random variable x takes values in a certain region of the x-axis
equals the mass in that region.
This implies that the distribution function equals the mass in the interval (-∞,x)

PROBABILITY MASS FUNCTION
• If the density function f(x) is finite, mass in the interval [x,
x+ dx] equals to the f(x)dx .
• The impulse
in the density function are known as point mass pi
placed at xi .
• density is positive in entire X-axis has total mass =1
•Probability that RV X takes values in a certain region
of x-axis equals the mass in that region, i.e.,
Distribution function F(x) equals the mass in the
interval (-∞,x)
)( ii xxp 

Random variable of continuous type
• If distribution function F(x) is a continuous function of x
(might have corners), i.e., the number of points at which F(x)
in not differentiable is countable, then random variable x is
continuous and its density function f(x) at all points given by
dF(x)/dx exists.

Properties:
38
 
     
   
     
   
   
  xeveryforxXPthen
RVcontisXIf
x
xxXxP
xfand
xxfxxXxPsmallxIf
obabilityIntervalduufxFxF
duufxF
FFdxxf
xf
x
x
x
x
0
,..6
lim
,.5
)Pr(.4
.3
1.2
0.1
0
12
2
1

















If A and B are two events, such that P (A)>0, the
conditional probability of B given A is given by.
0)( APIf
)(
)(
)|(
AP
BAP
ABP


39
CONDITIONAL PROBABILITY

Example: Tossing a die
The event ‘A’ that the number is odd and event ‘B’
that it is less than ‘4’. Find the conditional probability
of B
Sol: P(1)=P(2)=P(3)=P(4)=P(5)=P(6)= 1/6
P (A) = P (1) +P (3) +P (5) =3/6.
P (B) = P (1) +P (2) +P (3) =3/6
P (A∩B) = P (1) +P (3) =2/6.
3/2
6/3
6/2
)(
)(
)/( 


AP
BAP
ABP

Total probability
       
             nn
n
BPBAPBPBAPBPBAPAP
BAPBAPBAPAP
|......||
.....
2211
21


41
Let B1, B2,…Bn be mutually exclusive events whose union equals
the sample space S, then any event A can be represented by
 
     n
n
BABABA
BBBASAA


....
...
21
21
The probability of A is given by

INDEPENDENT PROBABILITY:
If ‘A ‘and ‘B’ are two independent events, then
the conditional probability of ‘B’ given ‘A’ is
P (B/A) = P (A) P (B) = P (B).
P (A)

Conditional CDF
  
 
  0,)/( 

 APif
AP
AxXP
AxFX

Conditional PDF
)/()/( AxF
dx
d
Axf XX 

Example 1
• Waiting time, X of a customer in a queuing system is 0, if he
finds the system idle and exponentially distributed random
length of time, if he find the system busy.
• i.e. P[idle]=p, P[busy]=1-p
• Find CDF of X

Solution
• F(x) can be expressed as the sum of step function with
amplitude p and a continuous function of x
 
   









0)1)(1(
00
)1(||
)(
xforepp
xfor
pbusyxXPpidlexXP
xXPxF
x
X

x
epxFxf 
 
 )1()()( '

Example
• PDF of samples of amplitude of speech waveforms is
found to decay exponentially at a rate 
 vxPandcFind
xcexf
x
X



)(

Solution
2/
1/22
1
0













cgives
cdxce
dxce
x
x
48
  


v
v
vdxv
eevxP 
 12/

Example
• Lifetime of a machine has continuous CDF F(x)
• Find conditional PDF and CDF
• Event A={X>t} , i.e., machine is still working at
time t.

Solution
• Conditional CDF
 
    
 














txfor
tF
tFxF
txfor
tXP
tXxXP
tXxXPtXxF
X
XX
)(1
)()(
0
|)|(
50
tx
tF
xf
tXxfPDFlConditiona
X
X
X 

 ,
)(1
)(
)|(:

Special Random Variables

1) Gaussian (Normal) Random Variable : The PDF is given by













 


x
y
x
y
X
x
X
dyexGwhere
x
GdyexFCDF
aroundlsymmetricaandshapedbellexf
2/
2/)(
2
2/)(
2
2
22
22
2
1
)(,
2
1
)(:
,
2
1
)(








2
2
2
1  
52
Continuous type Random Variable

• Constant is normalization constant and
maintains the area under the curve f(x) to be unity.
•
2
2
2
0
2
2
0 0
2/2/)(2
2/
22
,,,
,
22222
22











  












due
rdrddxdyrSinyrCosxwhere
rdrdedxdyeQ
dxeQ
u
ryx
x

=0 and σ=1 is called standard normal random variable
Kurtosis: It is a measure of peakedness of the density function
The upper one has large Kurtosis than the lower one.

SkewnessoftCoefficien
X
,3
3
3


 positive3
negative3
54
Skewness:
It is the measure of asymmetry of density function about the mean.

The density function of an exponentially distributed RV is given by
The parameter is the rate at which events occur, i.e., the probability
of an event occurring by time x increases as the rate increases


55
2) Exponential distribution
 


 


otherwise
xfore
xf
x
X
0
0

If occurrence of events over non overlapping intervals are independent.
i.e., arrival time of telephone calls or bus arrival time at bus stop,
waiting time distribution.

Example
• If q(t) is the probability that in time interval ‘t’
no event has occurred
•
)()()(
1)(1)(1)(),()(
2121 ttqtqtq
etqtxPtXPtqtXP t

 

It has memoryless property
  
   
 
 
 
}{
)(1
)(1
1
1
][
}][{}]{[
|
.,0,
)(
tXP
e
e
e
sF
stF
sXP
stXP
sXP
stXP
sXP
sXstXP
sXstXP
eventstwoaresxstxandtsIf
t
s
st





















57
Left side is the probability of having to wait at least ‘t’ additional
seconds given that one has already been waiting ‘s’ seconds.
Right side is the probability of waiting at least ‘t’ seconds when
one first begins to wait.
Thus, the probability of waiting at least an additional t seconds is
the same regardless of how long one has already been waiting.Vijaya Laxmi, Dept. of EEE, BIT, Mesra

Example
• Waiting time of a customer spending at a
restaurant, which has mean value= 5 min.
• Find the probability that the customer will
spend more than 10 min.

Solution
• Probability that the customer will spend an
additional 10 min in the restaurant given that he has
been there for more than 10 minutes.
1353.0)10( 5/10/10
 
eeXP 
pastondependnotdoes
exPXXP

 
1353.0)10()10|10( 2

3)Gamma Distribution:
Exponential is a special type of gamma distribution.
ondistributiErlangmegeranmIf
ondistributisquareChinIf


int,
2,2/


60
 
 
otherwise
xfore
x
xf
0
0
1












  



0
1
dxexwhere x


The density function for the uniform distribution is
61
 






otherwise
bxafor
abxfX
0
1
4) Uniform Distribution

Only two possible outcomes in this random variable
The value of x is 0 or 1
P(x=1) =p,
P(x=0)=q=1-p
p→ is the probability of the successes in each experiment of
independent trail.
q →is the probability of failure in each experiment.
62
Discrete type random variable
1)Bernoulli Random Variable:

2)Binomial Random Variable :
• Probability that an event occurs exactly ‘x’
times out of ‘n’ times.
x→ number of successes
n-x→ number of failure’s
63
    nxpp
x
n
PxXP
xnx
x ...2,1,0,1 








Example
• Prob. of getting 2 heads in 6 tosses
• Solution.
64
 
64
15
2
1
2
1
2
6
2
262




















XP

3)Poisson Random Variable
• This is closely related to the Binomial Distribution where there
are number of event’s occurrence in a large number of
event’s.
• Examples:
 No of count’s of emission from the radioactive substance
 Number of demands for telephone connection.
 Number of call’s at telephone exchange at a time.
 Number of printing error’s in book.

4)Bernoulli RV
• There are only two possible outcomes-success and
failure
• Hence X=0 or 1
• Here, p is the probability of success in each experiment of
independent trials and q is the probability of failures in each
experiment of independent trials
66
 
  pqXP
pXP


10
1

Function of random variable
• X→ Random variable
• g(X)→ real valued function
• Define Y=g(X)
• The probability of ‘Y’ is depends upon probability of
‘X’ and cdf of ‘X’

Example
• A Linear function Y=aX+b, a≠0
• If CDF of X is F(x), find F(y)

Solution
• The event occurs when occurs
• If a>0 ,
• thus,
• If a<0,
}{ yY  }{ ybaXA 
}{ 




 

a
by
XA
0,)( 




 



 
 a
a
by
F
a
by
XPyF XY











 

a
by
XA
0,1)( 




 



 
 a
a
by
F
a
by
XPyF XY






 






 







 



a
by
f
a
yfor
a
a
by
f
a
yfand
a
a
by
f
a
yf
a
by
uif
dy
du
du
dF
dy
dF
XY
XY
XY
1
)(,
0,
1
)(,
0,
1
)(
,.
)(
1
)()( xf
ady
dx
xfyf XXY 

Example
• y=x2 ‘x’ is any continuous random variable
• Find cdf and pdf of ‘y’

Solution
• Event occurs when for y is
nonnegative
• Event is null when y<0
 yY     yXyoryX 2
 






0)(
00
)(
yforyFyF
yfor
yF
XX
Y

PDF
( )
( ) ( ) .
1 1
( ). ( )( )
2 2
1
( ) ( ) ( )
2
Y
Y Y
x x
Y x x
dF yd du
f y F y
dx du dy
f y f y
y y
f y f y f y
y

  
   
    
0y

Example
• Y=cos(X), X is uniformly distributed random
variable in interval (0,2π), i.e., Y is uniformly
distributed random variable over the period of
sinusoid.
• Find PDF of Y

Example







cxcx
cxcx
cxcforXg
,
,
0)(
    )(
,0
:1
cyFcyXPyYP
yIf
Case
X 

75
    )(
,0
:2
cyFcyXPyYP
yIf
Case
X 


Example






0,1
0,1
)(
x
x
Xg
,1, yhavewe
76
    )0(101:2 XFXPyPCase 
    )0(01:1 XFXPyPCase 

Expected value of random variable or
Mathematical Expectation
• It is an estimation of a random variable.
• It is a measure of central tendency.

• If all probabilities are equal,
• If X has infinite number of values, then
n
n
xxxofMeanmeanArithmatic
n
xxx
XE ,...,/
.....
)( 21
21







1
inf)()(
j
jj convergesseriesinitetheprovidedxfxXE

Expected value of discrete random
variable
• A discrete RV X having the possible values
• Expected value of a X is defined as
nxxx ,......., 21



n
j
jj
nn
xXPx
xXPxxXPxxXPxXE
1
2211
)(
)(....)()()(
79
)()(, jj xfxXPIfor 
 


n
j
jj
nn
xxfxfx
xfxxfxxfxXE
1
2211
)()(
)(.....)()()(

Expected value of continuous random
variable
• The expected value of a cont. RV is given by



 dxxxfXE )()(
80
absolutelyconvergesegraltheprovided int

Function of Random Variable (Discrete )
• If X is a RV with probability function f(x)
• Define Y=g(X), whose probability function h(y) is given by
• If
• Then,
  
  

yxgx yxgx
xfxXPyYPyh
)(| )(|
)()()()(
nmforyyyYandxxxX mn  ,...,,,...,, 2121
 







)()(
)()(
)()(....)()()()()(
)()(...)()()()()(...)()(
1
2211
22112211
xfxg
xfxg
xfxgxfxgxfxgXgE
xfxgxfxgxfxgyhyyhyyhy
n
j
jj
nn
nnmm

Function of Random Variable(Continuous)
• If X is a cont. RV having probability function f(x)
  


 dxxfxgXgE )()()(

Problem: Fair Die Experiment
• If 2 turns up, one wins Rs. 20/-, if 4 turns up, one wins Rs. 40/-
, if 6 turns up, one loses Rs. 30/- .
• Find E(X), if the RV, X is the amount of money won or lost

Solution
0 +20 0 +40 0 -30
f(x) 1/6 1/6 1/6 1/6 1/6 1/6
jx
5
)6/1)(30()6/1)(0()6/1)(40()6/1)(0()6/1)(20()6/1)(0()(

XE

Problem
• If X is a RV whose density function is given by
•
• Find E(X)
otherwise
xforxxf
0
20
2
1
)( 





Solution
3
4
2
2
1
)()(
2
0
2
2
0












dx
x
dxxxdxxxfXE

Problem: Uniform RV
• Find the expected value for a uniformly
distributed random variable in the interval
[a,b].

Problem: Exponential RV
• The time X between customer arrivals at a
service station has an exponential pdf with
parameter .
• Find the mean interarrival time.
88


89
 
 








11
lim
00lim
intsin
0
0
0























t
t
t
t
t
tt
t
e
e
te
dteteXE
partsbyegrationgu
dtetXE
e-λt and te-λt go to zero as t approaches infinity.
λ is the customer arrival rate in customers per second.
The result shows that mean interarrival time E[X] =1/λ seconds per customer.

Problem
• Find
• If
)23( 2
XXE 

otherwise
xforxxf
0
20
2
1)( 

Solution
3/10
)
2
1
)(23()23( 22

 


dxxxxXXE

Theorems on Expected value
Theorem 1: If C is a constant,
Theorem 2:
Theorem 3:
Theorem 4:
)()( XcEcXE 
)()()( YEXEYXE 
RVstindependentwoareYandXIfYEXEXYE ),()()( 
cXEcXE  )()(

Variance and Standard Deviation
• It is a parameter to measure the spread the
PDF of random variable about the mean.
• It is defined as
  .,)(
2
numbernegativenonaiswhichXEXVar
meanarounddeviation










 

  
)(
tan)(
2
rootsquarepositive
DeviationdardSXEXVarX  

Discrete Random Variable
• If X is a discrete RV taking values as
and having probability function
f(x), then variance is given by
nxxx ,.....,, 21
    





)()(
)(
2
1
222
xfx
xfxXE j
n
j
jX



• If all the probabilities are equal,
• If X takes infinite number of terms,
       nxxx n /.....
22
2
2
1
2
 
,...., 21 xx
    convergesseriesthethatprovidedxfx j
j
jX 



1
22


Continuous Random Variable
• If X is a continuous RV having density function
f(x)
     convergesegraltheprovideddxxfxXEX int)(
222



 

• The variance or standard deviation is the measure of
the dispersion or scatter (spread of PDF) of the
values of RV about the mean.

Small variance
Large variance

Problem
• If X is cont. RV with PDF
• Find the variance and standard deviation
otherwise
xforxxf
0
20
2
1
)( 





Solution
 
3
4
 XE
* If unit of X is cm, unit if variance is cm2 and unit of standard deviation
is cm
99
  
9
2
2
1
3
4
)(
3
4
,
2
0
2
2
22
























dxxx
dxxfxXEVariance 
3
2
9
2,tan DeviationdardS

Theorems on Variance
• Theorem 1:
• Proof:
• Theorem 2:
    
      XEwhereXEXE
XEXE




,
22
2222
        
    22222
22222
2
22




XEXE
XEXEXXEXE
0)()()(  cVarXVarcXVar 

• Theorem 3:
• Theorem 4: The quality of is minimum , when
• Proof:
)()( 2
XVarccXVar 
  2
aXE 
 XEa  
        
       
         
    
     0
2
2
22
22
22
22










XEXESince
aXE
aXEaXE
aaXXE
aXEaXE
101
    
a
awhenoccursaXEofvalueimumSo



 0min,
22

• Theorem 5:
If X and Y are independent RVs
• Var(X+Y)=Var(X)+Var(Y)
• Var(X-Y)=Var(X)+Var(Y)

Problem: Uniform RV
• Find the variance of Uniform RV.

Solution
 
12
1
2
1
)(
22
2
2
2
ab
dyy
ab
dx
ba
x
ab
XVar
ab
ab
b
a









 










 





 

104
 
2
ba
XE



Standardized Random Variable
• These are the random variables that mean value is ‘0’and the
variance is ‘1’
• The standardized random variable are given by X*
• E[x*] =0 where x*=x-m
Var[x*]=1

Moments
• The rth moment of a RV X about the mean is defined
as
• It is also called the rth central moment, where
r=0,1,2….
  r
r XE  
meantheaboutmomentSecond
ormomentcentralSecond

2
2
10 ,0,1


Second moment about the mean is variance

  )(xfX
r
r    Discrete Random Variable
Continuous Random
Variable
107
 


 dxxfx
r
r )(

Moment about origin
• The rth moment of X about the origin or rth
Raw moment is defined as
  .....2,1,0,'
 rwhereXE r
r

Skewness
• It is the measure of degree of asymmetry
about the mean and is given by
  
3
3
3
3
3




 


XE

Kurtosis
• It is the measure of degree of peakedness of
the density function

Moment Generating Function
• It is used to generate the moments of a
random variable and is given by
 )()( xfetM tX
X Discrete Random Variable
Continuous Random Variable
111
 tX
X eEtM )(



 dxxfetM tX
X )()(

...
!
...
!3!2
1)( '
3
'
3
2
'
2 
r
ttt
ttM
r
rX 
 
     
...
!3!2
1
...
!3!2
1
....
!3!2
1)(:Pr
3
'
3
2
'
2
3
3
2
2
3322









tt
t
XE
t
XE
t
XtE
XtXt
tXEeEtMoof tX
X

The Coefficients of this expression enables to find the moments,
hence called moment generating function.

• is the rth derivative of evaluated at t=0
'
r )(tMX
0
'
|)(  tXr
r
r tM
dt
d


Theorems on Moment generating
function
• Theorem 1: If Mx(t) is moment generating function of RV X
and ‘a’ and ‘b’ are constants, then
• Proof:












 
b
t
MetM X
b
at
b
aX
)(

































 
b
t
Me
eEeeeE
eEtM
X
b
at
b
Xt
b
at
t
b
a
t
b
X
t
b
aX
b
aX
)(

• Theorem 2: If X and Y are two independent RVs having
moment generating function Mx(t) and My(t), then
• Proof:
)()()( tMtMtM YXYX 
   
    )()(
)( )(
tMtMeEeE
eeEeEtM
YX
tYtX
tYtXYXt
YX

 


• Theorem 3: The two RVs X and Y have same probability
distribution if and only if
)()( tMtM YX  Uniqueness theorem

Problem
• A RV assumes values 1 and -1 with probabilties ½
each.
• Find (a) Moment Generating function
• (b) First 4 moments about the origin

Solution
• (a)
• (b)
   tttttX
eeeeeE 













2
1
2
1
2
1 )1()1(
...
!4!3!2
1
...
!4!3!2
1
432
432


 ttt
te
ttt
te
t
t
118
...
!4!3!2
1)(
4
'
4
3
'
3
2
'
2 
ttt
ttM X 
  ...
!4!2
1
2
1 42
  tt
ee tt
.....,1,0,1,0 '
4
'
3
'
2  

 2
2 0
0( )
x
e forx
otherwisef x



2
0
( 2 )
0
( 2 )
0
[ ] 2
2
2
( 2 )
2 2
( 2 ) 2
tx tx x
t x
t x
E e e e dx
e dx
e
t
t t



 
 



   

 
  


Find the moment generating function and first 4 moments about the
origin
Solution:
119
Problem

• If |t|<2
...
16842
1
2/1
1
2
2 432




tttt
tt
120
....
!4!3!2
1)(
4
'
4
3
'
3
2
'
2 
ttt
ttM 
...,
2
3
,
4
3
,
2
1
,
2
1 '
4
'
3
'
2  

Problem
• Find the moment generating function for a general normal
distribution.
• Solution:
121
   




 dxeeeEtM xtxtX 22
2/
2
1
)( 

Let (x-μ)/σ =v in the integral so that x=μ+σv, dx=σdv
We have
 















































2
2/2
2/
2
2
22
2
22
2
22
2
2
1
)(
2
2
1
)(
tt
w
tt
tv
tt
vvtt
e
dweetM
wtvLet
dve
e
dvetM










( ) [ ] ( )
( )
( )
iwx
x x
iwx
iwx
iw E e iw
e f x fordiscretecase
e f x dx forcontinuouscase



  
 
 


Simply replacing t =iw in the moment generating function ,
where i is the imaginary term
122
Characteristic Function

Theorems on Characteristic Function
• Theorem 1: For random variable x the characteristic
function of random variable
• Theorem 2: ‘x’ and ‘y’ has same distribution if and only
if
123
( )
( ) ( )
aiw
b
x a x
b
w
w e
b
  
Proof: ( )
( )
( ) ( ) ( )
( )
( )
x a
i w
b b
x a
b b
x a a w
i w w i w i x
b b b b
aiw
b
x
w e
e e e
w
e
b



 
 
 
( ) ( )x yw w  
Uniqueness theorem

• Theorem 3: If X and Y are independent RVs
)()()(  YXYX 

...
!
....
!2
1)( '
2
'
2 
r
ii
r
r
r
X




125
0
'
|)()1(  

 Xr
r
rr
r
d
d
iand

Problem
• If X is a RV with values -1 and 1 with ½ prob. Each. Find the Ch.
Fn.
• Solution:
126
 
 



Cos
ee
eeeE
ii
iiXi























2
1
2
1
2
1 )1()1(

Problem
• If X is a RV with PDF
• Find the Ch. Fn.
• Solution:






otherwise
axfor
axf
0
2
1
)(
127
 





a
Sina
ai
ee
i
e
a
dxe
a
dxxfeeE
iaxiax
a
a
xi
a
a
xixiXi







 
 
2
|
2
1
2
1
)(

Markov and Chebyshev Inequality
• In general, mean and variance does not provide
enough information to determine CDF/PDF. However,
they allow us to obtain bounds for probabilities of
the form  tXP 

Markov Inequality
    enonnegativXfor
a
XE
aXP 
129
 
 )(1)(
.)(
)()()()(
0 0
xFadxxfa
anosmallbyreplacedxdxxaf
dxxxfdxxxfdxxxfdxxxfXE
a
a
a
a a





   


  
   aXaPXE 
Proof:
   


a
dxxfaXPwhere,

Problem
• Mean height of children in a class is 3 feet, 6 inches.
• Find the bound on the probability that a kid in the class is
taller than 9 feet.
• Solution:
  389.0
9
5.3
9 HP

Chebyshev Inequality
• In Markov inequality, the knowledge about the variability of
random variable about the mean is not provided.
• If   2
)(,  XVarmXE
If RV has zero variance, Chebyshev inequality implies that P[X=m]=1
i.e. , RV is equal to its mean with probability 1
131
  2
2
:
a
amXPinequalityChebyshev


 
    
2
2
2
2
22
22
aa
mXE
aDP
mXDLet





    .22
termsequivalentareamXandaD 

Problem
• The mean response time and the standard deviation in a
multi-user computer system are known to be 15 seconds and
3 seconds, respectively. Estimate the probability that the
response time is more than 5 seconds from the mean.
• Solution:
132
  36.0
25
9
515
.5.3.,15


XP
gives
SecaandSecSecm
withinequalityChebyshevFrom


Transform Methods
• These are useful computational aids in the solution of
equations that involve derivatives and integral of functions.
 Characteristic Function
 Probability Generating Function
 Laplace Transform of PDF

Characteristic Function
• It is defined as
• It is the Fourier transform of PDF of X with reversal in
sign of exponent.
• The PDF of X is given by
  Xixi
X
Xi
X eofvalueExpecteddxexfeE 
  

)()(
0




 


dexf xi
XX )(
2
1
)(
Every PDF and Characteristic function form a unique Fourier
Transform pair.

Problem
• Find the Ch. Fn. of an exponentially distributed RV
with parameter
• Solution:
•

 


 
i
dxedxee xixix
X


  
 

0 0
)(

Characteristic Function for Discrete RV
• It is given by
• If discrete RV are integer valued, the Ch. Fn. is given
by
• It is the Fourier Transform of the Sequence




k
xi
kXX
k
exp 
 )()(




k
ki
XX ekp 
 )()(
)(kpX

Probability Generating Function
• The Probability Generating Function of a nonnegative
integer valued RV N is defined as
• The pmf of N is given by
 
)exp()(
,)(
0
onentinchangesignwithpmfoftransformZzkp
zNoffunctionofvalueExpectedzEzG
k
N
N
NN
N





0|)(
!
1
)(  zNr
r
N zG
dz
d
k
kp

Laplace Transform of PDF
• Useful for queueing system, where one deals with
service times, waiting times, delays etc.
• The Laplace transform of pdf is given by
 sxsx
X eEdxexfsX 


 0
*
)()(

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