2013-1 Machine Learning Lecture 02 - Andrew Moore: Entropy

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Entropy and
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Information Gain
Andrew W. Moore
Professor
School of Computer Science
Carnegie Mellon University
www.cs.cmu.edu/~awm
awm@cs.cmu.edu
412-268-7599

Copyright © 2001, 2003, Andrew W. Moore

Bits
You are watching a set of independent random samples of X
You see that X has four possible values
P(X=A) = 1/4 P(X=B) = 1/4 P(X=C) = 1/4 P(X=D) = 1/4

So you might see: BAACBADCDADDDA…
You transmit data over a binary serial link. You can encode
each reading with two bits (e.g. A = 00, B = 01, C = 10, D =
11)

0100001001001110110011111100…
Copyright © 2001, 2003, Andrew W. Moore Information Gain: Slide 2

Fewer Bits
Someone tells you that the probabilities are not equal

P(X=A) = 1/2 P(X=B) = 1/4 P(X=C) = 1/8 P(X=D) = 1/8

It’s possible…

…to invent a coding for your transmission that only uses
1.75 bits on average per symbol. How?


Fewer Bits
Someone tells you that the probabilities are not equal

P(X=A) = 1/2 P(X=B) = 1/4 P(X=C) = 1/8 P(X=D) = 1/8
It’s possible…
…to invent a coding for your transmission that only uses
1.75 bits on average per symbol. How?
A 0
B 10
C 110
D 111

(This is just one of several ways)

Fewer Bits
Suppose there are three equally likely values…

P(X=A) = 1/3 P(X=B) = 1/3 P(X=C) = 1/3

Here’s a naïve coding, costing 2 bits per symbol

A 00
B 01
C 10

Can you think of a coding that would need only 1.6 bits
per symbol on average?

In theory, it can in fact be done with 1.58496 bits per
symbol.

General Case
Suppose X can have one of m values… V1, V2, … Vm

P(X=V1) = p1 P(X=V2) = p2 …. P(X=Vm) = pm
What’s the smallest possible number of bits, on average, per
symbol, needed to transmit a stream of symbols drawn from
X’s distribution? It’s
H ( X )   p1 log 2 p1  p2 log 2 p2    pm log 2 pm
m
  p j log 2 p j
j 1

H(X) = The entropy of X (Shannon, 1948)
• “High Entropy” means X is from a uniform (boring) distribution
• “Low Entropy” means X is from varied (peaks and valleys) distribution

General Case

P(X=V1) = p1 P(X=V2) = p2 …. P(X=Vm) = pm
A histogram of the
What’s the smallest possible number of frequency average, per
bits, on distribution of
symbol, needed to transmit a stream values of X would have
A histogram of the of symbols drawn from
frequency distribution of many lows and one or
values log would be flat p   highs
H(X )   p of X p  p log two
p log p
1 2 1 2 2 2 m 2 m
m
  p j log 2 p j
j 1

H(X) = The entropy of X

General Case

P(X=V1) = p1 P(X=V2) = p2 …. P(X=Vm) = pm
A histogram of the
What’s the smallest possible number of frequency average, per
bits, on distribution of
symbol, needed to transmit a stream values of X would have
A histogram of the of symbols drawn from
frequency distribution of many lows and one or
values log would be flat p   highs
H(X )   p of X p  p log two
p log p
1 2 1 2 2 2 m 2 m
m
  p ..and sop j values
j log 2 the ..and so the values
j 1 sampled from it would sampled from it would be
be all over the place more predictable
H(X) = The entropy of X

Entropy in a nut-shell

Low Entropy High Entropy


Entropy in a nut-shell

Low Entropy High Entropy
..the values (locations of
..the values (locations soup) unpredictable...
of soup) sampled almost uniformly sampled
entirely from within the throughout our dining room
soup bowl

Entropy of a PDF

Entropy of X  H [ X ]    p( x) log p( x)dx
x  

Natural log (ln or loge)

The larger the entropy of a distribution…
…the harder it is to predict
…the harder it is to compress it
…the less spiky the distribution


1
The “box”  w if
p( x)  
| x |
w
2
distribution  0 if

| x |
w
2

1/w

-w/2 0 w/2
 w/ 2 w/ 2
1 1 1 1
H [ X ]    p( x) log p( x)dx    log dx   log wdx  log w
x   x  w / 2
w w w w x / 2

1
Unit variance  w if
p( x)  
| x |
w
2
box distribution  0 if

| x |
w
2

E[ X ]  0
1
w2
2 3 Var[ X ] 
12

 3 0 3
if w  2 3 then Var[ X ]  1 and H [ X ]  1.242

The Hat  w | x |

p ( x)   w2
if |x|  w
distribution  0
 if |x|  w

E[ X ]  0
1 2
w
w Var[ X ] 
6

w 0 w


Unit variance hat  w | x |

p ( x)   w2
if |x|  w
distribution  0
 if |x|  w

E[ X ]  0
1 2
w
6 Var[ X ] 
6

 6 0 6
if w  6 then Var[ X ]  1 and H [ X ]  1.396

Dirac Delta
The “2 spikes”  ( x  1)   ( x  1)
p ( x) 
distribution 2

1 1 E[ X ]  0

 ( x  1)  ( x  1)
2 2
2 Var[ X ]  1

-1 0 1

H[ X ]    p( x) log p( x)dx  
x  


Entropies of unit-variance
distributions
Distribution Entropy

Box 1.242

Hat 1.396

2 spikes -infinity

??? 1.4189 Largest possible
entropy of any unit-
variance distribution


Unit variance p( x) 
1  x2 
exp   
 2
Gaussian 2  

E[ X ]  0

Var[ X ]  1


H[ X ]    p( x) log p( x)dx  1.4189
x  


Specific Conditional Entropy H(Y|X=v)
Suppose I’m trying to predict output Y and I have input X
X = College Major Let’s assume this reflects the true
probabilities
Y = Likes “Gladiator”
X Y E.G. From this data we estimate
Math Yes • P(LikeG = Yes) = 0.5
History No • P(Major = Math & LikeG = No) = 0.25
CS Yes • P(Major = Math) = 0.5
Math No
• P(LikeG = Yes | Major = History) = 0
Math No
Note:
CS Yes
History No • H(X) = 1.5
Math Yes •H(Y) = 1

X = College Major Definition of Specific Conditional
Y = Likes “Gladiator” Entropy:
H(Y |X=v) = The entropy of Y
X Y among only those records in which
Math Yes
X has value v
History No
CS Yes
Math No
Math No
CS Yes
History No
Math Yes

X = College Major Definition of Specific Conditional
H(Y |X=v) = The entropy of Y
X Y among only those records in which
Math Yes
X has value v
History No Example:
CS Yes
• H(Y|X=Math) = 1
Math No
• H(Y|X=History) = 0
Math No
CS Yes • H(Y|X=CS) = 0
History No
Math Yes

Conditional Entropy H(Y|X)
X = College Major Definition of Conditional
H(Y |X) = The average specific
X Y
conditional entropy of Y
Math Yes
History No = if you choose a record at random what
CS Yes will be the conditional entropy of Y,
Math No conditioned on that row’s value of X
Math No = Expected number of bits to transmit Y if
CS Yes both sides will know the value of X
History No
Math Yes = Σj Prob(X=vj) H(Y | X = vj)

Conditional Entropy
X = College Major Definition of Conditional Entropy:
H(Y|X) = The average conditional
entropy of Y
= ΣjProb(X=vj) H(Y | X = vj)
X Y
Math Yes Example:
History No vj Prob(X=vj) H(Y | X = vj)
CS Yes
Math No
Math 0.5 1
Math No History 0.25 0
CS Yes CS 0.25 0
History No
Math Yes H(Y|X) = 0.5 * 1 + 0.25 * 0 + 0.25 * 0 = 0.5

Information Gain
X = College Major Definition of Information Gain:
IG(Y|X) = I must transmit Y.
How many bits on average
would it save me if both ends of
X Y
the line knew X?
Math Yes IG(Y|X) = H(Y) - H(Y | X)
History No
CS Yes Example:
Math No • H(Y) = 1
Math No
• H(Y|X) = 0.5
CS Yes
History No • Thus IG(Y|X) = 1 – 0.5 = 0.5
Math Yes

Relative Entropy:Distance Kullback-
Leibler

p( x)
D( p, q)   p( x) log 2 ( )
x q ( x)


Mutual Information
A quantity that measures the mutual dependence of
the two random variables.
p(x , y )
I (X ,Y )    p(x , y )log2( )
p(x )q (y )

p(x , y )
I (X ,Y )    p(x , y )log2( )dxdy
Y X p(x )q (y )

p(x , y |c )
I (X ,Y |C )    p(x , y |c )log2( p(x |c )q (y |c )
)


Mutual Information
I(X,Y)=H(Y)-H(Y/X)

p(y / x )
I (X ,Y )    p(x , y )log2(
x y q (y )
)

I (X ,Y )    p(x , y )log2(p(y ))    p(x , y )log2(p(y / x ))
x y x y

I ( X , Y )   q( y) log 2 (q( y))   p( x) p( y / x) log 2 ( p( y / x))
y x y


Mutual information
• I(X,Y)=H(Y)-H(Y/X)
• I(X,Y)=H(X)-H(X/Y)
• I(X,Y)=H(X)+H(Y)-H(X,Y)
• I(X,Y)=I(Y,X)
• I(X,X)=H(X)


Information Gain Example


Another example


Relative Information Gain
X = College Major Definition of Relative Information
Y = Likes “Gladiator” Gain:
RIG(Y|X) = I must transmit Y, what
fraction of the bits on average would
X Y
it save me if both ends of the line
knew X?
Math Yes
History No RIG(Y|X) = [H(Y) - H(Y | X) ]/ H(Y)
CS Yes
Math No Example:
Math No • H(Y|X) = 0.5
CS Yes
• H(Y) = 1
History No
Math Yes • Thus IG(Y|X) = (1 – 0.5)/1 = 0.5

What is Information Gain used for?
Suppose you are trying to predict whether someone
is going live past 80 years. From historical data you
might find…
•IG(LongLife | HairColor) = 0.01
•IG(LongLife | Smoker) = 0.2
•IG(LongLife | Gender) = 0.25
•IG(LongLife | LastDigitOfSSN) = 0.00001
IG tells you how interesting a 2-d contingency table is
going to be.

Cross Entropy
Sea X una variable aleatoria con distribucion conocida p(x) y distribucion
estimada q(x), la “cross entropy” mide la diferencia entre las dos
distribuciones y se define por
HC ( x)  E[ log( q( x)]  H ( x)  KL( p, q)
donde H(X) es la entropia de X con respecto a la distribucion p y KL es
la distancia Kullback-Leibler ente p y q.
Si p y q son discretas se reduce a :

H C ( X )   p( x) log 2 (q( x))
x
y para p y q continuas se tiene


Bivariate Gaussians
X
Write r.v. X   
Y  Then define X ~ N (μ, Σ) to mean
 

p ( x) 
1
1

exp  1 (x  μ)T Σ 1 (x  μ)
2

2 || Σ || 2

Where the Gaussian’s parameters are…

 x   2 x  xy 
μ 
  Σ 
 y  2 
 y
 xy

Where we insist that S is symmetric non-negative definite


Bivariate Gaussians
X
Write r.v. X   
Y  Then define X ~ N (μ, Σ) to mean
 

p ( x) 
1
1

exp  1 (x  μ)T Σ 1 (x  μ)
2

2 || Σ || 2

Where the Gaussian’s parameters are…

 x   2 x  xy 
μ 
  Σ 
 y  2 
 y
 xy


It turns out that E[X] =  and Cov[X] = S. (Note that this is a
resulting property of Gaussians, not a definition)


Evaluating p ( x) 
1

exp  1 (x  μ)T Σ 1 (x  μ) 
p(x): Step 1
1 2
2 || Σ || 2

1. Begin with vector x

x




1

exp  1 (x  μ)T Σ 1 (x  μ) 
p(x): Step 2
1 2
2 || Σ || 2

2. Define  = x - 
x





1

exp  1 (x  μ)T Σ 1 (x  μ) 
p(x): Step 3
1 2
2 || Σ || 2

Contours defined by
1. Begin with vector x sqrt(TS-1) = constant
2. Define  = x - 
x
3. Count the number of contours
crossed of the ellipsoids 
formed S-1 
D = this count = sqrt(TS-1)
= Mahalonobis Distance
between x and 


1

exp  1 (x  μ)T Σ 1 (x  μ) 
p(x): Step 4
1 2
2 || Σ || 2

2. Define  = x - 

exp(-D 2/2)
crossed of the ellipsoids
formed S-1
between x and 
4. Define w = exp(-D 2/2)
D2
x close to  in squared Mahalonobis
space gets a large weight. Far away gets
a tiny weight

1

exp  1 (x  μ)T Σ 1 (x  μ) 
p(x): Step 5
1 2
2 || Σ || 2

2. Define  = x - 

exp(-D 2/2)
crossed of the ellipsoids
formed S-1
between x and 
4. Define w = exp(-D 2/2)
1
5. Multiply w by 1 to ensure p(x)dx  1 D2
2 || Σ || 2


Normal Bivariada NB(0,0,1,1,0)

persp(x,y,a,theta=30,phi=10,zlab="f(x,y)",box=FALSE,col=4)


Normal Bivariada NB(0,0,1,1,0)

3 0.20

2
0.15
1

0 0.10

-1
0.05
-2

-3 0.00
-3 -2 -1 0 1 2 3

filled.contour(x,y,a,nlevels=4,col=2:5)

Multivariate Gaussians
 X1 
 
 X2 
Write r.v. X   Then define X ~ N (μ, Σ) to mean
 
 
X 
 m

p ( x)  m
1
1

exp  1 (x  μ)T Σ 1 (x  μ)
2

(2 ) 2
|| Σ || 2

 1    21  12   1m 
Where the Gaussian’s    
parameters have…  2    12  2 2   2m 
μ  Σ

      
 
   2 
 m  1m  2 m   m
Again, E[X] =  and Cov[X] = S. (Note that this is a resulting property of Gaussians, not a definition)

General Gaussians
 1    21  12   1m 
   
 2    12  2 2   2m 
μ  Σ

      
 
   2 
 m  1m  2 m   m

x2

x1

Axis-Aligned Gaussians
  21 0 0  0 0 
 
 1   0  22 0  0 0 
   0
 2   0  23  0 0  
μ  Σ
        
   
    2 m 1
 m  0 0 0 0 
 0   2m 
 0 0 0 
X i  X i for i  j
x2

x1

Spherical Gaussians
 2 0 0  0 0 
 
 1   0 2 0  0 0 
   0
 2  0 2  0 0 
μ  Σ 
        
   
   2
 m  0 0 0 0 
 0  0 2
 0 0 
X i  X i for i  j
x2

x1

Subsets of variables
 X1 
 
 X1  U  
  X 
 X2   U  m (u ) 
Write X   as X    where
V
     X m ( u ) 1 
   
X  V   
 m
 X 
 m 

This will be our standard notation for breaking an m-
dimensional distribution into subsets of variables


Gaussian Marginals U
 
Margin-
U
V
are Gaussian
alize
 

 X1 
   X1   X m (u ) 1 
 X2   U    
Write X  
  as X   V  where U    , V    
 
    X   X 
X   m(u )   m 
 m
 U   μ u   Σuu Σuv  
IF   ~ N  ,  T
V μ   Σ 
    v   uv Σ vv  


THEN U is also distributed as a Gaussian

U ~ Nμu , Σuu 

 
Margin-
U
V
are Gaussian
alize
 

 X1 
   X1   X m (u ) 1 
 X2   U    
Write X  
  as X   V  where U    , V    
 
    X   X 
X   m(u )   m 
 m
 U   μ u   Σuu Σuv  
IF   ~ N  ,  T
V μ   Σ 
    v   uv Σ vv  
 This fact is not
immediately obvious
THEN U is also distributed as a Gaussian
Obvious, once we know
U ~ Nμu , Σuu  it’s a Gaussian (why?)


 
Margin-
U
V
are Gaussian
alize
 

 X1 
   X1   X m (u ) 1 
 X2   U    
Write X  
  as X   V  where U    , V    
 
    X   X 
X   m(u )   m 
 m How would you prove
this?
 U   μ u   Σuu Σuv  
IF   ~ N  ,  T
V μ   Σ 
    v   uv Σ vv  

p (u)
THEN U is also distributed as a Gaussian   p(u, v)dv
v
U ~ Nμu , Σuu   (snore...)

Matrix A
Linear Transforms X Multiply AX
remain Gaussian
Assume X is an m-dimensional Gaussian r.v.

X ~ Nμ, Σ
Define Y to be a p-dimensional r. v. thusly (note p  m):

Y  AX

…where A is a p x m matrix. Then…


Y ~ N Aμ, AΣ AT 


Adding samples of 2
independent Gaussians X
+ XY
is Gaussian Y

if X ~ Nμ x , Σ x  and Y ~ Nμ y , Σ y  and X  Y

then X  Y ~ Nμ x  μ y , Σ x  Σ y 

Why doesn’t this hold if X and Y are dependent?
Which of the below statements is true?
If X and Y are dependent, then X+Y is Gaussian but possibly
with some other covariance
If X and Y are dependent, then X+Y might be non-Gaussian


 U   μ u   Σuu Σuv    w   2977   849 2  967  
IF   ~ N  ,  T
V μ   Σ  IF   ~ N 
 y  76 ,   967 3.682  


    v   uv Σ vv  
   
 


THEN U | V ~ Nμu|v , Σu|v  where THEN w | y ~ Nμ w| y , Σ w| y  where
976( y  76)
1
μu|v  μu  Σ Σ (V  μ v )
T
μ w| y  2977 
uv vv
3.682
 967 2
uv Σ w| y  8492   8082
3.682

P(w|m=82)

P(w|m=76)

P(w)


 U   μ u   Σuu Σuv    w   2977   849 2  967  
IF   ~ N  ,  T
V μ   Σ  IF   ~ N 
 y  76 ,   967 3.682  


    v   uv Σ vv  
  Note:  
  
when given value of  


THEN U | V ~ Nμu|v , Σu|v  where THEN v isy~, Nμ w| y , Σ w| y  where
w | v the conditional
mean of u is u
976( y  76)
1
μu|v  μu  ΣT Σvv (V  μ v ) μ w| y  2977 
uv
3.682
 967 2
Σu|v  Σuu  ΣT Σvv1Σuv Σ w| y  8492   8082
uv
Note: marginal 2
3.68 mean is
a linear function of v

P(w|m=82)
Note: conditional
variance can only be
equal to or smaller than P(w|m=76)
marginal variance
Note: conditional
variance is independent
of the given value of v

P(w)


Gaussians and the U|V Chain U
 
V
chain rule
Rule
V  

Let A be a constant matrix

IF U | V ~ NAV , Σu|v  and V ~ Nμv , Σvv 
 U
THEN   ~ Nμ, Σ , with
V
 
 Aμ v   AΣ vv AT  Σu|v AΣ vv 
μ
 μ   Σ
 ( AΣ )T

 v   vv Σ vv 



Available Gaussian tools
U Margin-  U μ   Σ
IF   ~ N  u ,  uu
Σuv  
 THEN U ~ Nμu , Σuu 
 
V alize
U V
 
  μ   ΣT
  v   uv Σ vv  

 
Matrix A
IF X ~ Nμ, Σ AND Y  AX THEN Y ~ N Aμ, AΣ AT  
X Multiply AX
if X ~ Nμ x , Σ x  and Y ~ Nμ y , Σ y  and X  Y
X then X  Y ~ Nμ x  μ y , Σ x  Σ y 
Y
+ XY
U | V ~ Nμu|v , Σu|v 
 U μ   Σ Σuv   THEN
IF   ~ N  u ,  uu
V   μ   ΣT 

Σ vv  
U Condition-     v   uv
 
V alize
U | V where 1
  uv 
uv

U|V Chain U
Rule
 
V  U  AΣ vv AT  Σu|v AΣ vv 
V   THEN   ~ Nμ, Σ , with Σ  
V  ( AΣ )T

   vv Σ vv 



Assume…
• You are an intellectual snob
• You have a child


Intellectual snobs with children
• …are obsessed with IQ
• In the world as a whole, IQs are drawn from
a Gaussian N(100,152)


IQ tests
• If you take an IQ test you’ll get a score that,
on average (over many tests) will be your
IQ
• But because of noise on any one test the
score will often be a few points lower or
higher than your true IQ.

SCORE | IQ ~ N(IQ,102)


Assume…
• You drag your kid off to get tested
• She gets a score of 130
• “Yippee” you screech and start deciding how
to casually refer to her membership of the
top 2% of IQs in your Christmas newsletter.

P(X<130|=100,2=152) =
P(X<2| =0,2=1) =
erf(2) = 0.977


Assume…
• You drag your kid off to get tested
You are thinking:
• She gets a score of 130
Well sure the test isn’t accurate, so
• “Yippee” you screech andan IQ of 120 or she how
she might have start deciding
might have an 1Q of 140, but the
to casually refermost her IQ given the evidenceof the
to likely membership
top 2% of IQs in“score=130” is, of course, newsletter.
your Christmas 130.

P(X<130|=100,2=152) =
P(X<2| =0,2=1) =
erf(2) = 0.977

Can we trust
this reasoning?

What we really want:
• IQ~N(100,152)
• S|IQ ~ N(IQ, 102)
• S=130

• Question: What is
IQ | (S=130)?

Called the Posterior
Distribution of IQ

Which tool or tools?
• IQ~N(100,152) U Margin-
 
V alize
U
• S|IQ ~ N(IQ, 102)  
Matrix A
• S=130
X Multiply AX

• Question: What is X
+ XY
IQ | (S=130)? Y
U Condition-
 
V alize
U|V
 

U|V Chain U
Rule
 
V
V  

Plan
• IQ~N(100,152)
• S|IQ ~ N(IQ, 102)
• S=130

• Question: What is
IQ | (S=130)?

S | IQ Chain  S   IQ  Condition-
Rule
 
 IQ  Swap  
 S  alize
IQ | S
IQ    


Working…  U μ   Σ
IF   ~ N  u ,  uu
V   μ   ΣT
Σuv   THEN

Σ vv  
    v   uv 
IQ~N(100,152) 1
S|IQ ~ N(IQ, 102) uv

S=130

Question: What is IQ | (S=130)?  U  AΣ vv AT  Σu|v AΣ vv 
THEN   ~ Nμ, Σ , with Σ  
V 
   ( AΣ )T Σ vv 
 vv 


2013-1 Machine Learning Lecture 02 - Andrew Moore: Entropy

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2013-1 Machine Learning Lecture 02 - Andrew Moore: Entropy