lecture_06

Report
ECE 8443 – Pattern Recognition
LECTURE 06: MAXIMUM LIKELIHOOD AND
BAYESIAN ESTIMATION
• Objectives:
Bias in ML Estimates
Bayesian Estimation
Example
• Resources:
D.H.S.: Chapter 3 (Part 2)
Wiki: Maximum Likelihood
M.Y.: Maximum Likelihood Tutorial
J.O.S.: Bayesian Parameter Estimation
J.H.: Euro Coin
URL:
Audio:
Gaussian Case: Unknown Mean (Review)
• Consider the case where only the mean,  = , is unknown:
   ln  px k    0
n
k 1
ln( p (xk ))  ln[
1
(2 ) d / 2 
exp[
1/ 2
1
(x k  ) t  1 (x k  )]
2
1
1
  ln[( 2 ) d  ]  (x k  ) t  1 (x k  )
2
2
which implies:   ln( p (xk ))   1 (x k  )
because:
  1
1

d
t 1
[

ln[(
2

)

]

(
x


)

(
x


)]


k
k
  2
2

 1
 1
 [ ln[( 2 ) d  ]  [ (x k  ) t  1 (x k  )]
 2
 2
  1 (x k  )
ECE 8443: Lecture 06, Slide 1
Gaussian Case: Unknown Mean (Review)
• Substituting into the expression for the total likelihood:
 l    ln  px k     1 (x k  )  0
n
n
k 1
k 1
• Rearranging terms:
n
1
  (x k  ˆ)  0
k 1
n
 (x k  ˆ)  0
k 1
n
n
 x k   ˆ  0
k 1
n
k 1
 x k  n ˆ  0
k 1
n
1
ˆ   x k
n k 1
• Significance???
ECE 8443: Lecture 06, Slide 2
Gaussian Case: Unknown Mean and Variance (Review)
• Let  = [,2]. The log likelihood of a SINGLE point is:
1
1
ln( p ( xk ))   ln[( 2 ) 2 ] 
( xk  1 )t  2 1(xk  1 )
2
2 2
1


(
x


)
k
1
 

2
θl  θ ln( p( xk θ))  
2
 1  ( xk  1) 
 2 2
2 22 

• The full likelihood leads to:
n
1
 ˆ ( xk  ˆ1 )  0
k 1 2
n
n
1
( xk  ˆ1 ) 2
2
ˆ
 0   ( xk  1 )   ˆ2
 ˆ 
2
2ˆ2
k 1 2 2
k 1
k 1
n
ECE 8443: Lecture 06, Slide 3
Gaussian Case: Unknown Mean and Variance (Review)
1 n
ˆ
• This leads to these equations: 1  ˆ   xk
n k 1
• In the multivariate case:
1 n
2
ˆ
 2  ˆ  ( xk  ˆ ) 2
n k 1
1 n
ˆ   x k
n k 1
ˆ 2 
1 n
t
 x k  ˆ  x k  ˆ 
n k 1
• The true covariance is the expected value of the matrix  x k  ˆ  x k  ˆ  ,
which is a familiar result.
t
ECE 8443: Lecture 06, Slide 4
Convergence of the Mean (Review)
• Does the maximum likelihood estimate of the variance converge to the true
value of the variance? Let’s start with a few simple results we will need later.
• Expected value of the ML estimate of the mean:
1 n
E[ ˆ ]  E[  xi ]
n i 1

n
1
 E[ xi ]
n i 1
1 n
   
n i 1
ECE 8443: Lecture 06, Slide 5
var[ ˆ ]  E[ ˆ 2 ]  ( E[ ˆ ]) 2
 E[ ˆ 2 ]   2

 1 n  1 n
 E[  xi   x j ]   2
 n i 1  n j 1 
2

1  n n

 2    E[ xi x j ]    2
n  i 1 j 1

Variance of the ML Estimate of the Mean (Review)
• The expected value of xixj will be 2 for j  k since the two random variables are
independent.
• The expected value of xi2 will be 2 + 2.
• Hence, in the summation above, we have n2-n terms with expected value 2
and n terms with expected value 2 + 2.
• Thus,
var[ ˆ ] 
1
n
2
n
2


 n   n  
2
2
2
 
2

2
n
which implies:
E[ ˆ ]  var[ ˆ ]  ( E[ ˆ ]) 
2
2
2
n
 2
• We see that the variance of the estimate goes to zero as n goes to infinity, and
our estimate converges to the true estimate (error goes to zero).
ECE 8443: Lecture 06, Slide 6
Variance Relationships
• We will need one more result:
 2  E[( x   ) 2  E[ x 2 ]  2 E[ x]  E[  2 ]
 E[ x 2 ]  2  2  E[  2 ]
 E[ x 2 ]   2
n
  xi2
i 1
1 n 2
 (  xi )
n i 1
Note that this implies:
n
2
2
2
 xi    
i 1
• Now we can combine these results. Recall our expression for the ML
estimate of the variance:
1 n
n i 1
ˆ 2  E[   xi  ˆ 2 ]
ECE 8443: Lecture 06, Slide 7
Covariance Expansion
• Expand the covariance and simplify:
n
1 n
2 1
ˆ  E[   xi  ˆ   E[  ( xi2  2 xi ˆ  ˆ 2 )]
n i 1
n i 1
2
1 n
  ( E[ xi2 ]  2 E[ xi ˆ ]  E[ ˆ 2 ])
n i 1
1 n
  (( 2   2 )  2 E[ xi ˆ ]  (  2   2 n))
n i 1
• One more intermediate term to derive:
n
1 n
1 n
E[ xi ˆ ]  E[ xi  x j ]   E[ xi x j ]  (  E[ xi x j ]  E[ xi xi ])
n j 1
n j 1
j 1
i j


2
1
1
2
2
2
2
2
2 
 (( n  1)      )  (( n   )   
n
n
n
ECE 8443: Lecture 06, Slide 8
Biased Variance Estimate
• Substitute our previously derived expression for the second term:
1 n
ˆ   (( 2   2 )  2 E[ xi ˆ ]  (  2   2 n))
n i 1
2
1 n
  (( 2   2 )  2(  2   2 n)  (  2   2 n))
n i 1
1 n
  ( 2   2  2 2   2  2  2 n   2 n)
n i 1
1 n
  ( 2   2 n)
n i 1
1 n 2 (n  1)
1 n 2
1 n
2
2
  (   n)    (1  1 / n)   
n
n i 1
n i 1
n i 1
(n  1) 2


n
ECE 8443: Lecture 06, Slide 9
Expectation Simplification
• Therefore, the ML estimate is biased:
n 1 2
1 n
ˆ  E[  xi  ˆ 2 ] 
 2
n
ni 1
2
However, the ML estimate converges (and is MSE).
• An unbiased estimator is:
C
1 n
t
 x i  ˆ x i  ˆ 
n  1 i 1
• These are related by:
ˆ  (n  1) C

n
which is asymptotically unbiased. See Burl, AJWills and AWM for excellent
examples and explanations of the details of this derivation.
ECE 8443: Lecture 06, Slide 10
Introduction to Bayesian Parameter Estimation
• In Chapter 2, we learned how to design an optimal classifier if we knew the
prior probabilities, P(i), and class-conditional densities, p(x|i).
• Bayes: treat the parameters as random variables having some known prior
distribution. Observations of samples converts this to a posterior.
• Bayesian learning: sharpen the a posteriori density causing it to peak near
the true value.
• Supervised vs. unsupervised: do we know the class assignments of the
training data.
• Bayesian estimation and ML estimation produce very similar results in many
cases.
• Reduces statistical inference (prior knowledge or beliefs about the world) to
probabilities.
ECE 8443: Lecture 06, Slide 11
Class-Conditional Densities
• Posterior probabilities, P(i|x), are central to Bayesian classification.
• Bayes formula allows us to compute P(i|x) from the priors, P(i), and the
likelihood, p(x|i).
• But what If the priors and class-conditional densities are unknown?
• The answer is that we can compute the posterior, P(i|x), using all of the
information at our disposal (e.g., training data).
• For a training set, D, Bayes formula becomes:
P(i | x, D) 
likelihood  prior

evidence
p(x i , D) P(i D)
c
 p(x  j , D) P( j D)
j 1
• We assume priors are known: P(i|D) = P(i).
• Also, assume functional independence:
Di have no influence on
This gives:
P(i | x, D) 
p(x  j , D) if i  j
p(x i , Di ) P(i )
c
 p(x  j , D j ) P( j )
j 1
ECE 8443: Lecture 06, Slide 12
The Parameter Distribution
• Assume the parametric form of the evidence, p(x), is known: p(x|).
• Any information we have about  prior to collecting samples is contained in a
known prior density p().
• Observation of samples converts this to a posterior, p(|D), which we hope is
peaked around the true value of .
• Our goal is to estimate a parameter vector:
p(x D)   p(x, D)d
• We can write the joint distribution as a product:
p(x D)   p(x  , D ) p( D)d
  p(x  ) p( D)d
because the samples are drawn independently.
• This equation links the class-conditional density p(x D)
to the posterior, p( D) . But numerical solutions are typically required!
ECE 8443: Lecture 06, Slide 13
Univariate Gaussian Case
• Case: only mean unknown
p(x  )  N (, 2 )
• Known prior density:
p( )  N (0 , 02 )
• Using Bayes formula:
p(  D) p( D)  p( D  ) p(  )
• Rationale: Once a value of  is
known, the density for x is
completely known.  is a
normalization factor that
p(  D) 

p( D  ) p(  )
p( D)
p( D  ) p( )
 p ( D  ) p (  ) d
  [ p ( D  ) p (  )]
n
depends on the data, D.
ECE 8443: Lecture 06, Slide 14
   p(x k  ) p( )
k 1
Univariate Gaussian Case
• Applying our Gaussian assumptions:
2
 n  1
 1 

x


1
 k
    1
0

p  | D     
exp 
exp 
  
 k 1  2 
 2   0
 2       2  0
 



2
 

 
 1      2 n  x    2 
0
    k
   exp 
 
  
 2   0 
k 1 


 1    2  2  0   02  n  x 2k
x k   2  
    2  2 2  2  
   exp  
2

0

  
 2  
 k 1  
n
 1   02
x 2k
   exp  2   2
 2   0 k 1 
 1    2 2  0

 exp   2 

 02
 2    0

 1    2 2  0
   exp   2 

 02
 2    0
ECE 8443: Lecture 06, Slide 15
  n
x k   2  
    (2 2  2 )   



 
  k 1
  n
x k   2  
    (2 2  2 )   




 
  k 1
Univariate Gaussian Case (Cont.)
• Now we need to work this into a simpler form:
 1    2 2  0
p | D     exp   2 

 02
 2    0
 1  n  2
   exp     2
 2    k 1 
  n
x k   2  
    (2 2  2 )   




 
  k 1
 2
  2
 0
  n 
x  

      2 k 2    2 20
  k 1 
  
0
 
 1 2 2

   exp  n 2  2  2 2
0

 2  
 1  n
1
   exp(  2  2
0
 2  
 1  n
1
   exp(  2  2
0
 2  
1 n
where ˆ n   x k
n k 1
ECE 8443: Lecture 06, Slide 16
n
 x k  2
k 1




 0 

2 
 0 
 2
 1  n
0

   2
x

n
2
 2  k 


k

1


0


 2


   2 1 (nˆ n )  0
 2

 02


 
   
  
 
   
  
Univariate Gaussian Case (Cont.)
• p(|D) is an exponential of a quadratic function, which makes it a normal
distribution. Because this is true for any n, it is referred to as a reproducing
density.
• p() is referred to as a conjugate prior.
• Write p(|D) ~ N(n,n
2):
p(  D) 
1   n 2
exp[  (
) ]
2  n
2 n
1
• Expand the quadratic term:
1   n 2
1
1  2  2 n   n2
p( D) 
exp[ (
) ]
exp[ (
)]
2
2 n
2
n
2  n
2  n
1
• Equate coefficients of our two functions:
 1   2  2 n   n2  
exp  
  
2

n
2  n

 2
1
 1  n
 1

1  2




 exp   2  2   2 2 (nˆ n )  02
 2  
0
0 



ECE 8443: Lecture 06, Slide 17
  
 
 
  
Univariate Gaussian Case (Cont.)
• Rearrange terms so that the dependencies on  are clear:
 1   n2
exp   2
 2 
2  n
 n

 1 1

 
  exp   2  2  2 n2    

 2 
 n  

 n

1
 1  n
1
  exp   2  2
 2  
0


 2


   2 1 (nˆ n )  0
 2

 02


  
 
 
  
• Associate terms related to 2 and :
 n2   2 :
1

n

1
 n2  2  02
n
0
n
ˆ
n   :



n
 n2  2
 02
• There is actually a third equation involving terms not related to :
 1   n2  
exp   2       or 
 2  
2  n
 n 

1
 1   n2  
1
exp   2    
 2  
2  n
2  0
 n 

1
 1

 2 



n
but we can ignore this since it is not a function of  and is a complicated
equation to solve.
ECE 8443: Lecture 06, Slide 18
Univariate Gaussian Case (Cont.)
• Two equations and two unknowns. Solve for n and n2. First, solve for n2 :
 2 02
 02 2
 


n
1
n 02   2 n 02   2
 2
2

0
1
2
n
• Next, solve for n:
 n  ˆ n (
n n2
2
  n2
)   0  2
0
n   02 2
 ˆ n ( 2 )
  n 02   2
 n 02
 ˆ n 
2
2
 n 0  





 1
  0  2



 0
  02 2

 n 2   2
 0

 2
  0 

 n 2   2

 0




• Summarizing:
 2



n  ( 2
) ˆ 
2 n 
2
2 0
n 0  
 n 0   
n 02
 n2
 02 2

n 02   2
ECE 8443: Lecture 06, Slide 19




Bayesian Learning
• n represents our best guess after n samples.
• n2 represents our uncertainty about this guess.
• n2 approaches 2/n for large n – each additional observation decreases our
uncertainty.
• The posterior, p(|D), becomes more sharply peaked as n grows large. This is
known as Bayesian learning.
ECE 8443: Lecture 06, Slide 20
“The Euro Coin”
• Getting ahead a bit, let’s see how we can put these ideas to work on a simple
example due to David MacKay, and explained by Jon Hamaker.
ECE 8443: Lecture 06, Slide 21
Summary
• Review of maximum likelihood parameter estimation in the Gaussian case,
with an emphasis on convergence and bias of the estimates.
• Introduction of Bayesian parameter estimation.
• The role of the class-conditional distribution in a Bayesian estimate.
• Estimation of the posterior and probability density function assuming the only
unknown parameter is the mean, and the conditional density of the “features”
given the mean, p(x|), can be modeled as a Gaussian distribution.
ECE 8443: Lecture 06, Slide 22

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