04 Uncertainty inference(continuous)

Bayesian Networks
Unit 4 Uncertainty Inference
- Continuous
Wang, Yuan-Kai, 王元凱
ykwang@mails.fju.edu.tw
http://www.ykwang.tw

Department of Electrical Engineering, Fu Jen Univ.
輔仁大學電機工程系

2006~2011
Reference this document as:
Wang, Yuan-Kai, “Uncertainty Inference - Continuous,"
Lecture Notes of Wang, Yuan-Kai, Fu Jen University, Taiwan, 2011.
Fu Jen University Department of Electrical Engineering Wang, Yuan-Kai Copyright

王元凱 Unit - Uncertainty Inference (Continuous) p. 2

Goal of this Unit
 Review basic concepts of
statistics in terms of
 Image processing
 Pattern recognition



Related Units
 Previous unit(s)
 Probability Review
 Next units
 Uncertainty Inference (Discrete)
 Uncertainty Inference (Continuous)



Self-Study
 Artificial Intelligence: a modern
approach
 Russell & Norvig, 2nd, Prentice Hall,
2003. pp.462~474,
 Chapter 13, Sec. 13.1~13.3
 統計學的世界
 墨爾著，鄭惟厚譯, 天下文化，2002
 深入淺出統計學
 D. Grifiths, 楊仁和譯，2009, O’ Reilly



Contents
1. Gaussian …………………………….. 6
2. Gaussian Mixtures .......................... 36
3. Linear Gaussian .............................. 80
4. Sampling .......................................... 92
5. Markov Chain .................................. 102
6. Stochastic Process ........................ 106
7. Reference …………………………… 114



1. Gaussian Distribution
 1.1 Univariate Gaussian
 1.2 Bivariate Gaussian
 1.3 Multivariate Gaussian



Why Should We Care
 Gaussians are as natural as
Orange Juice and Sunshine
 We need them to understand
mixture models
Bayes Optimal Classifiers
Bayes Network


1.1 Univariate Gaussian
 Univaraite Gaussian is a
Gaussian with only one variable
1  x2 
p ( x)  exp  
 2 E[ X ]  0 Var[ X ]  1
2  

Unit-variance Gaussian


General Univariate Gaussian
1  (x   ) 2
 E[ X ]  μ
p ( x)  exp 
 

2   2 2
 Var[ X ]   2
=15

=100
• It is also called Normal distribution
• Bell-shape curve


Normal Distribution
=15

=100
•X~ N()
• “X is distributed as a Gaussian with
parameters  and 2”
• In this figure, X ~ N(100,152)


A Live Demo
  and  are two parameters of the
Gaussian
  : Position parameter
  : Shape parameter

Demo



Cumulative Distribution
Function
1 x x
F ( x)   p ( x)dx   e  ( x   ) 2 / 2 2
dx

2  
Density Function for the Standardized Normal Variate Cumulative Distribution Function for a Standardized Normal
Variate
0.45
1
0.4
0.9
0.35
0.8
0.3 0.7

Probabilty
Density

0.25 0.6
0.5
0.2
0.4
0.15
0.3
0.1 0.2
0.05 0.1
0 0
-5 -4 -3 -2 -1 0 1 2 3 4 5 -5 -4 -3 -2 -1 0 1 2 3 4 5

Standard Deviations Standard Deviations



The Error Function
• Assume X ~ N(0,1)
• Define ERF(x) = P(X<x)
= Cumulative Distribution of X
x
ERF ( x)   p( z )dz
z  

1
x
 z2 

2 z
 exp  2 dz







Using The Error Function
Assume X ~ N(0,1)
P(X<x| , 2) = ERF ( x   )
2



The Central Limit Theorem
 If(X1, X2, … Xn) are i.i.d. continuous
random variables
1 n
 Then define z  f ( x1 , x2 ,...xn )   xi
n i 1
 As n  , p(z)  Gaussian with
mean E[Xi] and variance Var[Xi]



Example –
Zero Mean Gaussian & Noise
 Zero mean Gaussian: N(0,)
 Usually used as noise model in
images
 An image f(x,y) with noise N(0,)
means ?
 f(x,y) = g(x,y) + N(0,)



1.2 Bivariate Gaussian



The Formula
 X1 
For random vector X  
X  
 2
If X  N(, )

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

 1    21  12 
μ 
  Σ
 2 

 2  21  2 


Gaussian Parameters
p( X ) 
1
1

exp  ( X  μ) Σ ( X  μ)
1
2
T 1

2 || Σ || 2

 &  are Gaussian’s parameters
 1    21  12 
μ 
  Σ


 2  21  22 

  : Position parameter
  : Shape parameter


Graphical Illustration
p(X) X2
Principal axis

2
2

X1
1
1

X1
  : Shape parameter: 1, 2


General Gaussian
 1    21  12 
μ 
  Σ


 2  21  22 


X2 X2

X1
X1



Axis-Aligned Gaussian
 X1 and X2 are independent or
uncorrelated
 1    21 0 
μ 
  Σ 
 0  22 
 2  
X2 X2

X1 X1
σ1 > σ2 σ1 < σ2


Spherical Gaussian
 1   2 0 
μ 
  Σ
 0

2
 2   

X2

X1



Degenerated Gaussians
 1 
μ 
  || Σ || 0
 2

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

X2

X1


Example – Clustering (1/4)
 Given a set of data points in a 2D
space
 Find the Gaussian distribution of
those points



 A 2D space example:
 Face verification of a person
 We use 2 features to verify the person
 Size
 Length
 We get 1000 face
images of the person
 Each image has 2
features: a data point
in the 2D space
 Find the mean and range of 2 features



x and y are dependent



x and y are almost x and y are
independent dependent


1.3 Multivariate Gaussian
 X1 
 
 X2 
For random vector X     ( X 1 , X 2 ,, X m )
T


 
If X  N(, ) 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 
   
 2    21  2 2   2m 
μ  Σ
      
 
   2 
 m  m1  m 2   m


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

 &  are Gaussian’s parameters
 1    21  12   1m 
   
 2    21  2 2   2m 
μ  Σ
      
  
 
 m  m1  m 2   2m 

  : Shape parameter


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 

X2 X2

X1 X1


Spherical Gaussians
 1   2

0 0  0 0 

   0 2 
 2 
0  0 0
μ   
 0 0 2  0 0 
 Σ
         
   
 m  0 0 0  2 0 
 0  0  2
 0 0 

x2

x1


Degenerate Gaussians
 1 
 
 2  || Σ || 0
μ 

 
 
 m
x2

x1


Example –
3-Variate Gaussian (1/2)
p ( x)  3
1
1

exp  (x  μ)T Σ 1 (x  μ)
1
2

(2 ) 2
|| Σ || 2

 1   21  12  13 
   
μ   2 Σ   21  2 2  23 
 3   31  32  23 
   



Example –
3-Variate Gaussian (2/2)
 Assume a simple case
 ij=0 if i≠j  21 0 0 
 
Σ 0  22 0 
 0 0  23 
 

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

?



2. Gaussian Mixture Model
• What is Gaussian Mixture
•  2 Gaussians are mixed to be a pdf
• Why Gaussian Mixture
• Single Gaussian is not enough
 Usually the distribution of your data
is assumed as one Gaussian
 Also called unimodal Gaussian
 However, sometimes the
distribution of data is not a
unimodal Gaussian


Why Is Unimodal Gaussian
not Enough (1/3)
 A univariate example
 Histogram of an image



not Enough (2/3)
 Bivariate example

One Gaussian PDF



not Enough (3/3)
 To solve it

Mixture of Three Gaussians



Gaussian Mixture Model
(GMM)
2.1 Combine Multiple
Gaussians
2.2 Formula of GMM
2.3 Parameter Estimation
of GMM



2.1 Combine Multiple
Gaussians
• Unimodal Gaussian (Single Gaussian)
1  1 
exp    x      x    
1
T
p( x ) 
 2   2 
n 2

1 2

• Multi-modal Gaussians
(Multiple Gaussians)
1  1 
exp    x  1  1 1  x  1  

T
p( x ) 
 2  1  2 
n2 12

1  1 
exp    x  2   21  x  2  

T

 2   2  2 
n2 12

...


Combine 2 Gaussians (1/4)
 Suppose
 Two Gaussians in 1-dimension
p( x)  p( x | 1 ,1)  p( x | 2 , 2 )
1   x  i 2 
p  x | i ,  i   exp    , i  1, 2
2  i  2 i 
2
 

 p(x) = p(x | C1) + p(x | C2)
 p(x|Ci)dx = 1


1-D Example (2/4)
1=4, 1=0.3 1=0.6
2=6.4, 2=0.5 2=0.4
1 ( x  4) 2
p( x )  exp( )
2  0.3 2  0.3 2

1 ( x  6.4) 2
 exp( )
2  0.5 2  0.5 2

Given x=5
1 (5  4) 2
p( x  5)  exp( )
2  0.3 2  0.3 2

1 (5  6.4) 2
 exp( )
2  0.5 2  0.52



2 Gaussians Gaussian Mixture
0.5 0.5

0.45 0.45

0.4 0.4
 N(0,1)  N(3,1)  N(0,1)  N(3,1)
0.35 0.35

0.3 0.3
p(x)

p(x)
0.25 0.25

0.2 0.2

0.15 0.15

0.1 0.1

0.05 0.05

0 0
-4 -3 -2 -1 0 1 2 3 4 5 6 7 -4 -3 -2 -1 0 1 2 3 4 5 6 7
x x

N(0,1)=p(x|0,1) p(x)=p(x|0,1)+p(x|3,1)
N(3,1)=p(x|3,1)


0.5 0.5

0.45 0.45

0.4 0.4
 N(0,1)  N(0,1)
0.35 0.35

0.3 0.3

p(x)
p(x)

0.25 0.25

0.2  N(3,4) 0.2  N(3,4)

0.15 0.15

0.1 0.1

0.05 0.05

0 0
-4 -3 -2 -1 0 1 2 3 4 5 6 7 -4 -3 -2 -1 0 1 2 3 4 5 6 7
x x

N(0,1)=p(x|0,1) p(x)=p(x|0,1)+p(x|3,4)
N(3,4)=p(x|3,4)


Combine 2 Gaussians
with Weights (2/3)
0.5 0.5

0.45 0.45

0.4 0.4
 N(0,1)  N(3,1)  N(0,1)  N(3,1)
0.35 0.35

0.3 0.3
p(x)

p(x)
0.25 0.25

0.2 0.2

0.15 0.15

0.1 0.1

0.05 0.05

0 0
-4 -3 -2 -1 0 1 2 3 4 5 6 7 -4 -3 -2 -1 0 1 2 3 4 5 6 7
x x

N(0,1)=p(x|0,1) p(x) = ½ * p(x|0,1)
N(3,1)=p(x|3,1) + ½ * p(x|3,1)


Combine 2 Gaussians
with Weights (3/3)
0.5 0.5

0.45 0.45

0.4 0.4
 N(0,1)  N(0,1)
0.35 0.35

0.3 0.3
p(x)

p(x)
0.25 0.25

0.2  N(3,4) 0.2  N(3,4)

0.15 0.15

0.1 0.1

0.05 0.05

0 0
-4 -3 -2 -1 0 1 2 3 4 5 6 7 -4 -3 -2 -1 0 1 2 3 4 5 6 7
x x

N(0,1)=p(x|0,1) p(x) = ½ * p(x|0,1)
N(3,4)=p(x|3,4)
+ ½ * p(x|3,4)


Combine 2 Gaussians with
Different Mean Distances (1/2)

 Suppose
 Two Gaussians in 1D
p( x)  p( x | 1 ,1)  p( x |  2 , 2 )
1
2
1
2

 Let   = 1
 Let =   



Different Mean Distances (2/2)
=1 =2

=3 =4



Different Weights (1/2)
 Suppose
 Two Gaussians in 1D
p( x)  0.75 p( x | 1 ,1 )  0.25 p( x |  2 , 2 )

 Let   = 1
 Let =   



Different Weights (2/2)
=1 =2

=3 =4



2D Gaussian Combination
(1/2)
4 0 4 0
p( x | C1 ), 1  (0, 0), 1    , p( x | C 2 ), 2  (0,3), 1   0 4 
0 4  



2D Gaussian Combination
(2/2)
p(x) = p(x|C1) + p(x|C2)



More Gaussians
 As no. of Gaussians, M,
increases, it can represent
any possible density
 By adjusting M, and , ,  of
each Gaussians
M
p ( x )    i p ( x| C i )
i 1
M
   i p ( x|  i ,  i )
i 1


2

p(x) 1.5 5 Gaussians
Component Models

1

0.5

0
-5 0 5 10

0.5

0.4 Mixture Model
0.3
p(x)

0.2

0.1

0
-5 0 5 10
Fu Jen University x
Department of Electrical Engineering Wang, Yuan-Kai Copyright


2.2 Formula of GMM
 A Gaussian mixture model
(GMM) is a linear combination
of M Gaussians
M
p(x)  i 1
 i p(x |Ci)

• P(x) is the probability of a point x
•x=(Cb, Cr) or (R,G,B) or ...
• i is mixing parameter (weight)
• p(x|Ci) is a Gaussian function



Comparison of Formula
1  1 
Gaussian: p( x )  exp    x      x    
1
T

 2   2 
n2

12

M

GMM : p ( x )    i p ( x| C i )
i 1
M
i  1 
 exp    x  i   i  x  i  
1
T

 2   2 
n2
i
12
i 1

 In GMM, p(x|Ci) means the
probability of x in the i Gaussian
component


Two Constraints of GMM
M
• i  i  1, and 0   i  1
i 1

• p(x|Ci)
• It is normalized,
• i.e., p(x|Ci)dx = 1



The Problem (1/5)
 Now we know any density can be
obtained by Gaussian mixture
 Given the mixture function, we can
plot its density
 But in reality, what we need to do
in computer is
 We get a lot of data point ={xj Rn,
j=1,…N} with unknown density
 Can we find the mixture function of
these data points?

0.5

0.4 Histograms of ={xj Rn, j=1,…N}
Mixture Model
0.3
p(x)

0.2

0.1

0
-5 0 5 10
x

2

1.5 5 Gaussians
Component Models
p(x)

1

0.5

0
-5
Fu Jen University Department of Electrical Engineering 5
0 Wang, Yuan-Kai Copyright
10


The Problem (3/5)
 To find the mixture function
means to estimate the parameters
of the mixture function
 Mixing parameters 
 Gaussian component densities
 Mean vector i
 Covariance matrix i
 Number of components M


The Problem (4/5)
No. of Parameters
A Gaussian
1D Gaussian 2D Gaussian 3D Gaussian
 1 2 3
 () 1 3 6
Total 2 5 9
GMM with M Gaussians
1D GMM 2D GMM 3D GMM
 M M M
 1M 2M 3M
 () 1M 3M 6M
Total 3M 6M 10M


The Problem (5/5)
 That is, given {xj Rn, j=1,…N}
M
p ( x1 )  
i 1
i p ( x1 |  i ,  i )
M
p ( x2 )  
i 1
i p ( x2 |  i ,  i )
Solve i, i, i
...

M Also called
p( xN )    i p( xN | i ,  i ) parameter estimation
i 1
 Usually we use i to denote (i, i)
M
p( x)    i p( x |  i )
i 1


2.3 Parameter Estimation
 Given
 Fixed M
 Data ={xj Rn, j=1,…N}
 We may calculate the histogram of

 We want to find the parameters
 = (1, ..., M, 1, ..., M, 1, ...,M)
that best fit the histogram of data
 Examples
 1-D example: xj R1
 Two 2-D examples: xj R2


1-D Example
={1.5, -0.2, 1.4, 1.8, ... } Histogram
•Nx=-2.5=10
•...
•Nx=1.5=40
•...
Parameter
Estimation
=1.5, =1.3
1 (x1.5)2
p(x)  exp( )
21.3 21.32



M=7 M=7
(Izenman & (Basford et al.)
Sommer)

M=3 M=4
(equal variances)



2-D Example
={(3.45,4.02), ... }
ANEMIA PATIENTS AND CONTROLS
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4.3
Red Blood Cell Hemoglobin Concentration

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Fu Jen University Department ofRed Blood CellEngineering
Electrical Volume Wang, Yuan-Kai Copyright

EM ITERATION 1
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4.3 Initialization

4.2

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Fu Jen University Red Blood Cell Volume

EM ITERATION 3
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EM ITERATION 25
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LOG-LIKELIHOOD AS A FUNCTION OF EM ITERATIONS
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Fu Jen University EM Iteration

ANEMIA DATA WITH LABELS
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Control Group
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3.9 Anemia Group

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Parameter Estimation of GMM
 Two methods
 Maximum Likelihood Estimation
(MLE)
 Expectation Maximization (EM)
 Discussed in Lecture Note 24



3. Linear Gaussian
 P(X) can belong to a distribution
 Ex, Gaussian, uniform, Exponential,
Gaussian mixture)
 P(Y|X) : conditional probability
 Can the conditional probability
belong to a distribution?
 Linear Gaussian describes
 The distribution of conditional
probability as Gaussian
 The dependence between random
variables


From Gaussian to Linear
Gaussian (1/2)
 If two variables x and y has linear
relationship, we say
 y = ax + b, y becomes a
 a and b are parameters random variable
 If y belongs to Gaussian distribution
 y ~ N(, ) = N(y;, ) N(y)
 But y = ax+b
  ax+b ~ N(, )
 = N(ax+b, )
 But we can write
-  +
N(y; ax+b, ) y


Linear Gaussian=N(y;ax+b, )
 The meaning of linear Gaussian
N(y; ax+b, )
• When y=ax+b,
N(y) N(y) is the maximum
probability
• However, yax+b
occurs
•with lower probability,
-3 -  + +3 y •decayed as in
Gaussian distribution
=

ax+b



Example 1 (1/2)
 Illumination change of an image
f(x,y) g(x,y)
g(x,y) =
af(x,y)+b

g =af+b

•Two kinds of illumination change(same a,b)
•Real light change : g1
•Changed by image processing software: g2
•g1 = g2 : No
•Because of noise, g1 is a random variable


Example 1 (2/2)
 Illumination change of an image
f(x,y) g1(x,y)

g1 =af+b

•But g1 is a random variable
•It means g1 undergoes a noise
•g1 ~ af + b + N(0,) = N(af+b, )
•Or P(g1|f) = N(af+b, )


Extension: Linear Transform
 X and Y are two vectors
 Y = (y1, y2, …, ym)T
 X = (x1, x2, …, xm)T
 X and Y are linearly dependent
 Y = AX + B : Linear transform
 If Y becomes a random vector
 P(Y|X)=N(Y; AX+B, ∑)



Example 2 (1/5)
 Illumination change of color image
F(x,y) G(x,y)

G =AF+B

•F(x,y)=(rF(x,y), gF(x,y), bF(x,y)T
•G(x,y)=(rG(x,y), gG(x,y), bG(x,y)T
 rG   a11 a12 a13   rF   b1 
G =AF+B   g   a
 G   21 a22 a23   g F   b2 
   
 bG   a31
   a32 a33   bF  b3 
   


Example 2 (2/5)
 A simple case of G=AF+B
 aij=0 if i≠j  rG  2 0 0  rF  1
 g    0 3 0   g   0 
 G   F   
 bG  0 0 1  bF  0
      
 RG=a11RF+b1,
GG=a22GF+b2,
BG=a33BF+b3
 rG   a11 a12 a13   rF   b1 
 g   a a22 a23   g F   b2 
 G   21    
 bG  a31
   a32 a33   bF  b3 
   


Example 2 (3/5)
 If G has noises
F(x,y) G1(x,y)

G1 =AF+B

•G1(x,y) is a random vector
•It means G1 undergoes a noise  11  12  13 
•G1 ~ AF + B + N(0,∑)    21  22  23 
= N(AF+B, ∑)  
 31
  32  33 

•Or P(G1|F) = N(AF+B, ∑)


Example 2 (4/5)
 G1 ~ N(AF+B, ∑)
 N is a multivariate Gaussian (3-D)
exp 2 ( X  μ) Σ ( X  μ ) 
1 1
p( X )  1
1 T

2 || Σ || 2

 a11rF  a12 g F  a13bF  b1   11  12  13 
a r  a g  a b  b     
  AF  B   21 F 22 F 23 F 2   21  22  23 
 a31rF  a32 g F  a33bF  b3 
   31  32  33 
 



Example 2 (5/5)
 Assume a simple case 4 0 0
 aij=0 if i≠j  2 0 0
0 3 0   0 8 0
 ij=0 if i≠j A   
0 0 1 
  0 0
 2

 Then
  a11rF  a12 g F  a13bF  b1   a11rF  b1 
  AF  B  a21rF  a22 g F  a23bF  b2   a22 g F  b2 
   
 a31rF  a32 g F  a33bF  b3   a33bF  b3 
   
 P(rG|rF) = N(a11rF+b1, 1)
 P(gG|gF) = N(a21gF+b2, 2)
 P(bG|bF) = N(a31bF+b3, 3)



4. Sampling
 Generate N samples S from P(X)
 S=(s1,s2, …, sN)
 X can be a random variable or a
random vector
 If X=x, si=xi
 If X=(x1,x2,…,xn), si=(x1i,x2i,…,xni)
 Why generate N samples?
 Estimate probabilities by frequencies
# samples with X  x
P( X  x) 
N


Example (1/2)
 A simple example: coin toss
 Tossing the coin, get head or tail
 It is a Boolean random variable
 coin = head or tail
 Random variable, but not
random vector
 If it is unbiased coin, head and tail
have equal probability
 A prior probability distribution
P(Coin) = <0.5, 0.5>
 Uniform distribution
 But we do not know it is unbiased


Example (2/2)
 Sampling in this example
= flipping the coin many times N
 e.g., N=1000 times
 Ideally, 500 heads, 500 tails
 P(head) = 500/1000=0.5
P(tail) = 500/1000=0.5
 Practically, 5001 heads, 499 tails
 P(head) = 501/1000=0.501
P(tail) = 499/1000=0.499
 By the sampling, we can
estimate the probability
distribution


Sampling (Math)
 For a Boolean random variable X
 P(X) is prior distribution
= <P(x), P(x)>
 Using a sampling algorithm to
generate N samples
 Say N(x) is the number of samples
that x is true, N(x) of x is false
N (x)
 Pˆ ( x ), N (  x )  P (  x )
ˆ
N N
N ( x) N (x)
lim  P( x), lim  P(x)
N  N N  N


Sampling Algorithm
 It is the algorithm to
 Generate samples from a known
probability distribution
 Estimate the approximate
probability Pˆ
 How does a sampling algorithm
generate a sample?
 C/C++: rand()
 Return 0 ~ RAND_MAX (32767)
 Java: random()


A Sampling Algorithm
of the Coin Toss
 Flip the coin 1000 times
 int coin_face;
for (i=0; i<1000; i++)
{ if (rand() > RAND_MAX/2)
coin_face = 1;
else coin_face = 0;
}
What kind of distribution?


Sampling Algorithms
for Many R.V.s (1/2)
 3 Boolean random variables X, Y, Z
 (X=1, Y=0, Z=0) is called a sample
 int X, Y, Z;
for (i=0; i<1000; i++)
{ if (rand() > RAND_MAX/2) X = 1;
else X = 0;
if (rand() > RAND_MAX/2) Y = 1;
else Y = 0;
if (rand() > RAND_MAX/2) Z = 1;
else Z = 0;
}
X, Y, Z are all
uniform distribution


Sampling Algorithms
for Many R.V.s (2/2)
 Y, Z are not uniform distribution
 P(Y)=<0.67, 0.33>,
P(Z)=<0.25,0.75>
 int X, Y, Z;
for (i=0; i<1000; i++)
{ if (rand() > RAND_MAX/2) X = 1;
else X = 0;
if (rand() > RAND_MAX/3) Y = 1;
else Y = 0;
if (rand() > RAND_MAX/4) Z = 1;
else Z = 0;
}


Various Sampling Algorithms
 For more complex P(X), we need
more complex sampling algo.
 Stochastic simulation
 Direct Sampling
 Rejection sampling
 Reject samples disagreeing with
evidence
 Likelihood weighting
 Use evidence to weight samples
 Markov chain Monte Carlo (MCMC)
 Also called Gibbs sampling


Example
 Approximate reasoning for
Bayesian networks
 TBU



5. Markov Chain
 Markov Assumption
 Each state at time t only
depends on the state (Andrei Andreyevich Markov)

at time t-1
 Ex. The weather today only
depends on the weather of
yesterday

X1 X2 X3
t=1 t=2 t=3



Deterministic v.s.
Non-Deterministic
 Deterministic patterns :
 Traffic light
 FSMs
 …

 Non-Deterministic patterns :
 Weather
 Speech
 Tracking
 …



Example –
Weather Prediction (1/2)
 Only 3 possible weather states :
 Sunny, Cloudy, Rainy

 Transition Matrix :
 A=Pr( today | yesterday)
Weather Today
Sunny Cloudy Rainy
Weather Sunny 0.5 0.25 0.25
Yesterday Cloudy 0.375 0.125 0.375
Rainy 0.125 0.625 0.375


Example –
Weather Prediction (2/2)
 Suppose we know the weather of
previous days
 t=1: rainy R S S C
 t=2: sunny X1 X2 X3 X4
 t=3: sunny t=1 t=2 t=3 t=4
 t=4: cloudy
 Predict the weather of day t=5 X5
 ? t=5



6. Stochastic Process
 Also called Random Process
 It is a collection of random
variables
 For each t in the index set T, X(t) is a
random variable
 Usually t refers to time, and X(t) is
the state of the process at time t
 X(t) can be discrete or continuous



Graphical View
of Stochastic Process



Statistics of Stochastic
Process
 Mean of X(t)
 Variance, standard deviation of X(t)
 Frequency distribution of X(t) : P(X)
 Conditional probability of X(t) :
P(X(t) | X(t-1))



Markov Chain
 A Markov chain is a stochastic
process where
 P(X(t+1) | X(t), X(t-1), …, X(0))
= P(X(t+1) | X(t)), or
 P(Xn+1 | Xn, Xn-1, …, X0)
= P(Xn+1 | Xn)
 Next state depends only on current
state
 Future and past are conditionally
independent given current



Higher-order Markov Chain
 Second-order Markov chain
 P(X(t+1) | X(t), X(t-1), …, X(0))
= P(X(t+1) | X(t), X(t-1))

X1 X2 X X4
3
t=1 t=2 t=3 t=4

 Third-order, n-order



Stationary Process
 The probability distribution of X is
independent of t
X1 X2 X X4
3
t=1 t=2 t=3 t=4



Doubly Stochastic Process
 Hidden variable

X1 X2 X3

Y1 Y2 Y3



Example - Video
 Facial expression recognition
 TBU



7. Reference
 Moments
 Hu, M. K., [1962]. “Visual Pattern Recognition by
Moment Invariants.” IRE Trans. Info. Theory,
vol. IT-8, pp. 179-187.
 Gonzalez, R.C. and R. E. Woods. [2001]. Digital
Image Processing, 2nd, Prentice Hall, pp.659-
660, 672-675.
 L. Rocha, et. al., “Image Moments-Based
Structuring and Tracking of Objects,”
Proceedings of the XV Brazilian Symposium on
Computer Graphics and Image Processing,
2002.



Reference
 Gaussian mixture
 C. M. Bishop. Neural Networks for Pattern
Recognition. Oxford University Press, 1995.
 McLachlan and Peel, Finite Mixture Models,
John Wiley & Sons,
 Rennie, “A Short Tutorial on Using EM with
Mixture Models,” MIT Tech. Report, 2004.
http://www.ai.mit.edu/
people/jrennie/writing/mixtureEM.pdf.
 Tomasi, “Estimating Gaussian Mixture Density
with EM: a tutorial,” Duke University,
 Y. Weiss. Motion segmentation using EM – a
short tutorial. 1997.
http://www.cs.huji.ac.il/˜yweiss/emTutorial.pdf



Reference
 Linear Gaussian
 Russell & Norvig, Artificial Intelligence: a
modern approach, 2nd, Prentice Hall, 2003.
Sec. 14.3, pp.501-503
 Sampling
 Russell&Norvig, Artificial Intelligence: a modern
approach, 2nd, Prentice Hall, 2003.
Sec. 14.5, pp.511-518
 Stochastic process
 Probability, Random variables and Random
Signal Principles, 4e, Peebles, McGraw Hill,
2001

04 Uncertainty inference(continuous)

04 Uncertainty inference(continuous)

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