This document provides an introduction to the Expectation Maximization (EM) algorithm. EM is used to estimate parameters in statistical models when data is incomplete or has missing values. It is a two-step process: 1) Expectation step (E-step), where the expected value of the log likelihood is computed using the current estimate of parameters; 2) Maximization step (M-step), where the parameters are re-estimated to maximize the expected log likelihood found in the E-step. EM is commonly used for problems like clustering with mixture models and hidden Markov models. Applications of EM discussed include clustering data using mixture of Gaussian distributions, and training hidden Markov models for natural language processing tasks. The derivation of the EM algorithm and

1. Introduction to Machine Learning
Expectation Maximization
Andres Mendez-Vazquez
September 16, 2017
1 / 126

2. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
2 / 126

3. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
3 / 126

4. Images/
Maximum-Likelihood
We have a density function p (x|Θ)
Assume that we have a data set of size N, X = {x1, x2, ..., xN }
This data is known as evidence.
We assume in addition that
The vectors are independent and identically distributed (i.i.d.) with
distribution p under parameter θ.
4 / 126

5. Images/
Maximum-Likelihood
We have a density function p (x|Θ)
Assume that we have a data set of size N, X = {x1, x2, ..., xN }
This data is known as evidence.
We assume in addition that
The vectors are independent and identically distributed (i.i.d.) with
distribution p under parameter θ.
4 / 126

6. Images/
What Can We Do With The Evidence?
We may use the Bayes’ Rule to estimate the parameters θ
p (Θ|X) =
P (X|Θ) P (Θ)
P (X)
(1)
Or, given a new observation ˜x
p (˜x|X) (2)
I.e. to compute the probability of the new observation being supported by
the evidence X.
Thus
The former represents parameter estimation and the latter data prediction.
5 / 126

7. Images/
What Can We Do With The Evidence?
We may use the Bayes’ Rule to estimate the parameters θ
p (Θ|X) =
P (X|Θ) P (Θ)
P (X)
(1)
Or, given a new observation ˜x
p (˜x|X) (2)
I.e. to compute the probability of the new observation being supported by
the evidence X.
Thus
The former represents parameter estimation and the latter data prediction.
5 / 126

8. Images/
What Can We Do With The Evidence?
We may use the Bayes’ Rule to estimate the parameters θ
p (Θ|X) =
P (X|Θ) P (Θ)
P (X)
(1)
Or, given a new observation ˜x
p (˜x|X) (2)
I.e. to compute the probability of the new observation being supported by
the evidence X.
Thus
The former represents parameter estimation and the latter data prediction.
5 / 126

9. Images/
Focusing First on the Estimation of the Parameters θ
We can interpret the Bayes’ Rule
p (Θ|X) =
P (X|Θ) P (Θ)
P (X)
(3)
Interpreted as
posterior =
likelihood × prior
evidence
(4)
Thus, we want
likelihood = P (X|Θ)
6 / 126

10. Images/
Focusing First on the Estimation of the Parameters θ
We can interpret the Bayes’ Rule
p (Θ|X) =
P (X|Θ) P (Θ)
P (X)
(3)
Interpreted as
posterior =
likelihood × prior
evidence
(4)
Thus, we want
likelihood = P (X|Θ)
6 / 126

11. Images/
Focusing First on the Estimation of the Parameters θ
We can interpret the Bayes’ Rule
p (Θ|X) =
P (X|Θ) P (Θ)
P (X)
(3)
Interpreted as
posterior =
likelihood × prior
evidence
(4)
Thus, we want
likelihood = P (X|Θ)
6 / 126

12. Images/
What we want...
We want to maximize the likelihood as a function of θ
7 / 126

13. Images/
Maximum-Likelihood
We have
p (x1, x2, ..., xN |Θ) =
N
i=1
p (xi|Θ) (5)
Also known as the likelihood function.
Because multiplication of quantities p (xi|Θ) ≤ 1 can be problematic
L (Θ|X) = log
N
i=1
p (xi|Θ) =
N
i=1
log p (xi|Θ) (6)
8 / 126

14. Images/
Maximum-Likelihood
We have
p (x1, x2, ..., xN |Θ) =
N
i=1
p (xi|Θ) (5)
Also known as the likelihood function.
Because multiplication of quantities p (xi|Θ) ≤ 1 can be problematic
L (Θ|X) = log
N
i=1
p (xi|Θ) =
N
i=1
log p (xi|Θ) (6)
8 / 126

15. Images/
Maximum-Likelihood
We want to find a Θ∗
Θ∗
= argmaxΘL (Θ|X) (7)
The classic method
∂L (Θ|X)
∂θi
= 0 ∀θi ∈ Θ (8)
9 / 126

16. Images/
Maximum-Likelihood
We want to find a Θ∗
Θ∗
= argmaxΘL (Θ|X) (7)
The classic method
∂L (Θ|X)
∂θi
= 0 ∀θi ∈ Θ (8)
9 / 126

17. Images/
What happened if we have incomplete data
Data could have been split
1 X = observed data or incomplete data
2 Y = unobserved data
For this type of problems
We have the famous Expectation Maximization (EM)
10 / 126

18. Images/
What happened if we have incomplete data
Data could have been split
1 X = observed data or incomplete data
2 Y = unobserved data
For this type of problems
We have the famous Expectation Maximization (EM)
10 / 126

19. Images/
What happened if we have incomplete data
Data could have been split
1 X = observed data or incomplete data
2 Y = unobserved data
For this type of problems
We have the famous Expectation Maximization (EM)
10 / 126

20. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
11 / 126

21. Images/
The Expectation Maximization
The EM algorithm
It was first developed by Dempster et al. (1977).
Its popularity comes from the fact
It can estimate an underlying distribution when data is incomplete or has
missing values.
Two main applications
1 When missing values exists.
2 When a likelihood function can be simplified by assuming extra
parameters that are missing or hidden.
12 / 126

22. Images/
The Expectation Maximization
The EM algorithm
It was first developed by Dempster et al. (1977).
Its popularity comes from the fact
It can estimate an underlying distribution when data is incomplete or has
missing values.
Two main applications
1 When missing values exists.
2 When a likelihood function can be simplified by assuming extra
parameters that are missing or hidden.
12 / 126

23. Images/
The Expectation Maximization
The EM algorithm
It was first developed by Dempster et al. (1977).
Its popularity comes from the fact
It can estimate an underlying distribution when data is incomplete or has
missing values.
Two main applications
1 When missing values exists.
2 When a likelihood function can be simplified by assuming extra
parameters that are missing or hidden.
12 / 126

24. Images/
The Expectation Maximization
The EM algorithm
It was first developed by Dempster et al. (1977).
Its popularity comes from the fact
It can estimate an underlying distribution when data is incomplete or has
missing values.
Two main applications
1 When missing values exists.
2 When a likelihood function can be simplified by assuming extra
parameters that are missing or hidden.
12 / 126

25. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
13 / 126

26. Images/
Clustering
Given a series of data sets
Given the fact that Radial Gaussian Functions are Universal Approximators
.
Samples {x1, x2, ..., xN } are the visible parameters
The Gaussian distributions generating each of the samples are the
hidden parameters
Then, we model the cluster as a mixture of Gaussian’s
14 / 126

27. Images/
Clustering
Given a series of data sets
Given the fact that Radial Gaussian Functions are Universal Approximators
.
Samples {x1, x2, ..., xN } are the visible parameters
The Gaussian distributions generating each of the samples are the
hidden parameters
Then, we model the cluster as a mixture of Gaussian’s
14 / 126

28. Images/
Natural Language Processing
Unsupervised induction of probabilistic context-free grammars
Here given a series of words o1, o2, o3, ... and normalized Context-Free
Grammar
We want to know the probabilities of each rule P (i → jk)
Thus
Here the you have two variables:
The Visible Ones: The sequence of words
The Hidden Ones: The rule that produces the possible sequence
oi → oj
15 / 126

29. Images/
Natural Language Processing
Unsupervised induction of probabilistic context-free grammars
Here given a series of words o1, o2, o3, ... and normalized Context-Free
Grammar
We want to know the probabilities of each rule P (i → jk)
Thus
Here the you have two variables:
The Visible Ones: The sequence of words
The Hidden Ones: The rule that produces the possible sequence
oi → oj
15 / 126

30. Images/
Natural Language Processing
Baum-Welch Algorithm for Hidden Markov Models
Visible Data
Hidden Data
Here
Hidden Variables: The circular nodes producing the data
Visible Variables: The square nodes representing the samples.
16 / 126

31. Images/
Natural Language Processing
Baum-Welch Algorithm for Hidden Markov Models
Visible Data
Hidden Data
Here
Hidden Variables: The circular nodes producing the data
Visible Variables: The square nodes representing the samples.
16 / 126

32. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
17 / 126

33. Images/
Incomplete Data
We assume the following
Two parts of data
1 X = observed data or incomplete data
2 Y = unobserved data
Thus
Z = (X,Y)=Complete Data (9)
Thus, we have the following probability
p (z|Θ) = p (x, y|Θ) = p (y|x, Θ) p (x|Θ) (10)
18 / 126

34. Images/
Incomplete Data
We assume the following
Two parts of data
1 X = observed data or incomplete data
2 Y = unobserved data
Thus
Z = (X,Y)=Complete Data (9)
Thus, we have the following probability
p (z|Θ) = p (x, y|Θ) = p (y|x, Θ) p (x|Θ) (10)
18 / 126

35. Images/
Incomplete Data
We assume the following
Two parts of data
1 X = observed data or incomplete data
2 Y = unobserved data
Thus
Z = (X,Y)=Complete Data (9)
Thus, we have the following probability
p (z|Θ) = p (x, y|Θ) = p (y|x, Θ) p (x|Θ) (10)
18 / 126

36. Images/
Incomplete Data
We assume the following
Two parts of data
1 X = observed data or incomplete data
2 Y = unobserved data
Thus
Z = (X,Y)=Complete Data (9)
Thus, we have the following probability
p (z|Θ) = p (x, y|Θ) = p (y|x, Θ) p (x|Θ) (10)
18 / 126

37. Images/
Incomplete Data
We assume the following
Two parts of data
1 X = observed data or incomplete data
2 Y = unobserved data
Thus
Z = (X,Y)=Complete Data (9)
Thus, we have the following probability
p (z|Θ) = p (x, y|Θ) = p (y|x, Θ) p (x|Θ) (10)
18 / 126

38. Images/
New Likelihood Function
The New Likelihood Function
L (Θ|Z) = L (Θ|X, Y) = p (X, Y|Θ) (11)
Note: The complete data likelihood.
Thus, we have
L (Θ|X, Y) = p (X, Y|Θ) = p (Y|X, Θ) p (X|Θ) (12)
Did you notice?
p (X|Θ) is the likelihood of the observed data.
p (Y|X, Θ) is the likelihood of the no-observed data under the
observed data!!!
19 / 126

39. Images/
New Likelihood Function
The New Likelihood Function
L (Θ|Z) = L (Θ|X, Y) = p (X, Y|Θ) (11)
Note: The complete data likelihood.
Thus, we have
L (Θ|X, Y) = p (X, Y|Θ) = p (Y|X, Θ) p (X|Θ) (12)
Did you notice?
p (X|Θ) is the likelihood of the observed data.
p (Y|X, Θ) is the likelihood of the no-observed data under the
observed data!!!
19 / 126

40. Images/
New Likelihood Function
The New Likelihood Function
L (Θ|Z) = L (Θ|X, Y) = p (X, Y|Θ) (11)
Note: The complete data likelihood.
Thus, we have
L (Θ|X, Y) = p (X, Y|Θ) = p (Y|X, Θ) p (X|Θ) (12)
Did you notice?
p (X|Θ) is the likelihood of the observed data.
p (Y|X, Θ) is the likelihood of the no-observed data under the
observed data!!!
19 / 126

41. Images/
Rewriting
This can be rewritten as
L (Θ|X, Y) = hX,Θ (Y) (13)
This basically signify that X, Θ are constant and the only random part is
Y.
In addition
L (Θ|X) (14)
It is known as the incomplete-data likelihood function.
20 / 126

42. Images/
Rewriting
This can be rewritten as
L (Θ|X, Y) = hX,Θ (Y) (13)
This basically signify that X, Θ are constant and the only random part is
Y.
In addition
L (Θ|X) (14)
It is known as the incomplete-data likelihood function.
20 / 126

43. Images/
Thus
We can connect both incomplete-complete data equations by doing
the following
L (Θ|X) =p (X|Θ)
=
Y
p (X, Y|Θ)
=
Y
p (Y|X, Θ) p (X|Θ)
=
N
i=1
p (xi|Θ)
Y
p (Y|X, Θ)
21 / 126

44. Images/
Thus
We can connect both incomplete-complete data equations by doing
the following
L (Θ|X) =p (X|Θ)
=
Y
p (X, Y|Θ)
=
Y
p (Y|X, Θ) p (X|Θ)
=
N
i=1
p (xi|Θ)
Y
p (Y|X, Θ)
21 / 126

45. Images/
Thus
We can connect both incomplete-complete data equations by doing
the following
L (Θ|X) =p (X|Θ)
=
Y
p (X, Y|Θ)
=
Y
p (Y|X, Θ) p (X|Θ)
=
N
i=1
p (xi|Θ)
Y
p (Y|X, Θ)
21 / 126

46. Images/
Thus
We can connect both incomplete-complete data equations by doing
the following
L (Θ|X) =p (X|Θ)
=
Y
p (X, Y|Θ)
=
Y
p (Y|X, Θ) p (X|Θ)
=
N
i=1
p (xi|Θ)
Y
p (Y|X, Θ)
21 / 126

47. Images/
Remarks
Problems
Normally, it is almost impossible to obtain a closed analytical solution for
the previous equation.
However
We can use the expected value of log p (X, Y|Θ), which allows us to find
an iterative procedure to approximate the solution.
22 / 126

48. Images/
Remarks
Problems
Normally, it is almost impossible to obtain a closed analytical solution for
the previous equation.
However
We can use the expected value of log p (X, Y|Θ), which allows us to find
an iterative procedure to approximate the solution.
22 / 126

49. Images/
The function we would like to have
The Q function
We want an estimation of the complete-data log-likelihood
log p (X, Y|Θ) (15)
Based in the info provided by X, Θn−1 where Θn−1 is a previously
estimated set of parameters at step n.
Think about the following, if we want to remove Y
ˆ
[log p (X, Y|Θ)] p (Y|X, Θn−1) dY (16)
Remark: We integrate out Y - Actually, this is the expected value of
log p (X, Y|Θ).
23 / 126

50. Images/
The function we would like to have
The Q function
We want an estimation of the complete-data log-likelihood
log p (X, Y|Θ) (15)
Based in the info provided by X, Θn−1 where Θn−1 is a previously
estimated set of parameters at step n.
Think about the following, if we want to remove Y
ˆ
[log p (X, Y|Θ)] p (Y|X, Θn−1) dY (16)
Remark: We integrate out Y - Actually, this is the expected value of
log p (X, Y|Θ).
23 / 126

51. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
24 / 126

52. Images/
Use the Expected Value
Then, we want an iterative method to guess Θ from Θn−1
Q (Θ, Θn−1) = E [log p (X, Y|Θ) |X, Θn−1] (17)
Take in account that
1 X, Θn−1 are taken as constants.
2 Θ is a normal variable that we wish to adjust.
3 Y is a random variable governed by distribution
p (Y|X, Θn−1)=marginal distribution of missing data.
25 / 126

53. Images/
Use the Expected Value
Then, we want an iterative method to guess Θ from Θn−1
Q (Θ, Θn−1) = E [log p (X, Y|Θ) |X, Θn−1] (17)
Take in account that
1 X, Θn−1 are taken as constants.
2 Θ is a normal variable that we wish to adjust.
3 Y is a random variable governed by distribution
p (Y|X, Θn−1)=marginal distribution of missing data.
25 / 126

54. Images/
Use the Expected Value
Then, we want an iterative method to guess Θ from Θn−1
Q (Θ, Θn−1) = E [log p (X, Y|Θ) |X, Θn−1] (17)
Take in account that
1 X, Θn−1 are taken as constants.
2 Θ is a normal variable that we wish to adjust.
3 Y is a random variable governed by distribution
p (Y|X, Θn−1)=marginal distribution of missing data.
25 / 126

55. Images/
Use the Expected Value
Then, we want an iterative method to guess Θ from Θn−1
Q (Θ, Θn−1) = E [log p (X, Y|Θ) |X, Θn−1] (17)
Take in account that
1 X, Θn−1 are taken as constants.
2 Θ is a normal variable that we wish to adjust.
3 Y is a random variable governed by distribution
p (Y|X, Θn−1)=marginal distribution of missing data.
25 / 126

56. Images/
Another Interpretation
Given the previous information
E [log p (X, Y|Θ) |X, Θn−1] =
´
Y∈Y log p (X, Y|Θ) p (Y|X, Θn−1) dY
Something Notable
1 In the best of cases, this marginal distribution is a simple analytical
expression of the assumed parameter Θn−1.
2 In the worst of cases, this density might be very hard to obtain.
Actually, we use
p (Y, X|Θn−1) = p (Y|X, Θn−1) p (X|Θn−1) (18)
which is not dependent on Θ.
26 / 126

57. Images/
Another Interpretation
Given the previous information
E [log p (X, Y|Θ) |X, Θn−1] =
´
Y∈Y log p (X, Y|Θ) p (Y|X, Θn−1) dY
Something Notable
1 In the best of cases, this marginal distribution is a simple analytical
expression of the assumed parameter Θn−1.
2 In the worst of cases, this density might be very hard to obtain.
Actually, we use
p (Y, X|Θn−1) = p (Y|X, Θn−1) p (X|Θn−1) (18)
which is not dependent on Θ.
26 / 126

58. Images/
Another Interpretation
Given the previous information
E [log p (X, Y|Θ) |X, Θn−1] =
´
Y∈Y log p (X, Y|Θ) p (Y|X, Θn−1) dY
Something Notable
1 In the best of cases, this marginal distribution is a simple analytical
expression of the assumed parameter Θn−1.
2 In the worst of cases, this density might be very hard to obtain.
Actually, we use
p (Y, X|Θn−1) = p (Y|X, Θn−1) p (X|Θn−1) (18)
which is not dependent on Θ.
26 / 126

59. Images/
Another Interpretation
Given the previous information
E [log p (X, Y|Θ) |X, Θn−1] =
´
Y∈Y log p (X, Y|Θ) p (Y|X, Θn−1) dY
Something Notable
1 In the best of cases, this marginal distribution is a simple analytical
expression of the assumed parameter Θn−1.
2 In the worst of cases, this density might be very hard to obtain.
Actually, we use
p (Y, X|Θn−1) = p (Y|X, Θn−1) p (X|Θn−1) (18)
which is not dependent on Θ.
26 / 126

60. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
27 / 126

61. Images/
Back to the Q function
The intuition
We have the following analogy:
Consider h (θ, Y ) a function
θ a constant
Y ∼ pY (y), a random variable with distribution pY (y).
Thus, if Y is a discrete random variable
q (θ) = EY [h (θ, Y )] =
y
h (θ, y) pY (y) (19)
28 / 126

62. Images/
Back to the Q function
The intuition
We have the following analogy:
Consider h (θ, Y ) a function
θ a constant
Y ∼ pY (y), a random variable with distribution pY (y).
Thus, if Y is a discrete random variable
q (θ) = EY [h (θ, Y )] =
y
h (θ, y) pY (y) (19)
28 / 126

63. Images/
Back to the Q function
The intuition
We have the following analogy:
Consider h (θ, Y ) a function
θ a constant
Y ∼ pY (y), a random variable with distribution pY (y).
Thus, if Y is a discrete random variable
q (θ) = EY [h (θ, Y )] =
y
h (θ, y) pY (y) (19)
28 / 126

64. Images/
Back to the Q function
The intuition
We have the following analogy:
Consider h (θ, Y ) a function
θ a constant
Y ∼ pY (y), a random variable with distribution pY (y).
Thus, if Y is a discrete random variable
q (θ) = EY [h (θ, Y )] =
y
h (θ, y) pY (y) (19)
28 / 126

65. Images/
Back to the Q function
The intuition
We have the following analogy:
Consider h (θ, Y ) a function
θ a constant
Y ∼ pY (y), a random variable with distribution pY (y).
Thus, if Y is a discrete random variable
q (θ) = EY [h (θ, Y )] =
y
h (θ, y) pY (y) (19)
28 / 126

66. Images/
Why E-step!!!
From here the name
This is basically the E-step
The second step
It tries to maximize the Q function
Θn = argmaxΘQ (Θ, Θn−1) (20)
29 / 126

67. Images/
Why E-step!!!
From here the name
This is basically the E-step
The second step
It tries to maximize the Q function
Θn = argmaxΘQ (Θ, Θn−1) (20)
29 / 126

68. Images/
Why E-step!!!
From here the name
This is basically the E-step
The second step
It tries to maximize the Q function
Θn = argmaxΘQ (Θ, Θn−1) (20)
29 / 126

69. Images/
Derivation of the EM-Algorithm
The likelihood function we are going to use
Let X be a random vector which results from a parametrized family:
L (Θ) = ln P (X|Θ) (21)
Note: ln (x) is a strictly increasing function.
We wish to compute Θ
Based on an estimate Θn (After the nth) such that L (Θ) > L (Θn)
Or the maximization of the difference
L (Θ) − L (Θn) = ln P (X|Θ) − ln P (X|Θn) (22)
30 / 126

70. Images/
Derivation of the EM-Algorithm
The likelihood function we are going to use
Let X be a random vector which results from a parametrized family:
L (Θ) = ln P (X|Θ) (21)
Note: ln (x) is a strictly increasing function.
We wish to compute Θ
Based on an estimate Θn (After the nth) such that L (Θ) > L (Θn)
Or the maximization of the difference
L (Θ) − L (Θn) = ln P (X|Θ) − ln P (X|Θn) (22)
30 / 126

71. Images/
Derivation of the EM-Algorithm
The likelihood function we are going to use
Let X be a random vector which results from a parametrized family:
L (Θ) = ln P (X|Θ) (21)
Note: ln (x) is a strictly increasing function.
We wish to compute Θ
Based on an estimate Θn (After the nth) such that L (Θ) > L (Θn)
Or the maximization of the difference
L (Θ) − L (Θn) = ln P (X|Θ) − ln P (X|Θn) (22)
30 / 126

72. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
31 / 126

73. Images/
Introducing the Hidden Features
Given that the hidden random vector Y exits with y values
P (X|Θ) =
y
P (X|y, Θ) P (y|Θ) (23)
Thus, using our first constraint L (Θ) − L (Θn)
L (Θ) − L (Θn) = ln
y
P (X|y, Θ) P (y|Θ) − ln P (X|Θn) (24)
32 / 126

74. Images/
Introducing the Hidden Features
Given that the hidden random vector Y exits with y values
P (X|Θ) =
y
P (X|y, Θ) P (y|Θ) (23)
Thus, using our first constraint L (Θ) − L (Θn)
L (Θ) − L (Θn) = ln
y
P (X|y, Θ) P (y|Θ) − ln P (X|Θn) (24)
32 / 126

75. Images/
Here, we introduce some concepts of convexity
For Convexity
Theorem (Jensen’s inequality)
Let f be a convex function defined on an interval I. If x1, x2, ..., xn ∈ I
and λ1, λ2, ..., λn ≥ 0 with n
i=1 λi = 1, then
f
n
i=1
λixi ≤
n
i=1
λif (xi) (25)
33 / 126

76. Images/
Proof:
For n = 1
We have the trivial case
For n = 2
The convexity definition.
Now the inductive hypothesis
We assume that the theorem is true for some n.
34 / 126

77. Images/
Proof:
For n = 1
We have the trivial case
For n = 2
The convexity definition.
Now the inductive hypothesis
We assume that the theorem is true for some n.
34 / 126

78. Images/
Proof:
For n = 1
We have the trivial case
For n = 2
The convexity definition.
Now the inductive hypothesis
We assume that the theorem is true for some n.
34 / 126

79. Images/
Now, we have
The following linear combination for λi
f
n+1
i=1
λixi = f λn+1xn+1 +
n
i=1
λixi
= f λn+1xn+1 +
(1 − λn+1)
(1 − λn+1)
n
i=1
λixi
≤ λn+1f (xn+1) + (1 − λn+1) f
1
(1 − λn+1)
n
i=1
λixi
35 / 126

80. Images/
Now, we have
The following linear combination for λi
f
n+1
i=1
λixi = f λn+1xn+1 +
n
i=1
λixi
= f λn+1xn+1 +
(1 − λn+1)
(1 − λn+1)
n
i=1
λixi
≤ λn+1f (xn+1) + (1 − λn+1) f
1
(1 − λn+1)
n
i=1
λixi
35 / 126

81. Images/
Now, we have
The following linear combination for λi
f
n+1
i=1
λixi = f λn+1xn+1 +
n
i=1
λixi
= f λn+1xn+1 +
(1 − λn+1)
(1 − λn+1)
n
i=1
λixi
≤ λn+1f (xn+1) + (1 − λn+1) f
1
(1 − λn+1)
n
i=1
λixi
35 / 126

82. Images/
Did you notice?
Something Notable
n+1
i=1
λi = 1
Thus
n
i=1
λi = 1 − λn+1
Finally
1
(1 − λn+1)
n
i=1
λi = 1
36 / 126

83. Images/
Did you notice?
Something Notable
n+1
i=1
λi = 1
Thus
n
i=1
λi = 1 − λn+1
Finally
1
(1 − λn+1)
n
i=1
λi = 1
36 / 126

84. Images/
Did you notice?
Something Notable
n+1
i=1
λi = 1
Thus
n
i=1
λi = 1 − λn+1
Finally
1
(1 − λn+1)
n
i=1
λi = 1
36 / 126

85. Images/
Now
We have that
f
n+1
i=1
λixi ≤ λn+1f (xn+1) + (1 − λn+1) f
1
(1 − λn+1)
n
i=1
λixi
≤ λn+1f (xn+1) + (1 − λn+1)
1
(1 − λn+1)
n
i=1
λif (xi)
≤ λn+1f (xn+1) +
n
i=1
λif (xi) Q.E.D.
37 / 126

86. Images/
Now
We have that
f
n+1
i=1
λixi ≤ λn+1f (xn+1) + (1 − λn+1) f
1
(1 − λn+1)
n
i=1
λixi
≤ λn+1f (xn+1) + (1 − λn+1)
1
(1 − λn+1)
n
i=1
λif (xi)
≤ λn+1f (xn+1) +
n
i=1
λif (xi) Q.E.D.
37 / 126

87. Images/
Now
We have that
f
n+1
i=1
λixi ≤ λn+1f (xn+1) + (1 − λn+1) f
1
(1 − λn+1)
n
i=1
λixi
≤ λn+1f (xn+1) + (1 − λn+1)
1
(1 − λn+1)
n
i=1
λif (xi)
≤ λn+1f (xn+1) +
n
i=1
λif (xi) Q.E.D.
37 / 126

88. Images/
Thus, for concave functions
It is possible to shown that
Given ln (x) a concave function:
ln
n
i=1
λixi ≥
n
i=1
λi ln (xi)
If we take in consideration
Assume that the λi = P (y|X, Θn). We know that
1 P (y|X, Θn) ≥ 0
2
y P (y|X, Θn) = 1
38 / 126

89. Images/
Thus, for concave functions
It is possible to shown that
Given ln (x) a concave function:
ln
n
i=1
λixi ≥
n
i=1
λi ln (xi)
If we take in consideration
Assume that the λi = P (y|X, Θn). We know that
1 P (y|X, Θn) ≥ 0
2
y P (y|X, Θn) = 1
38 / 126

90. Images/
We have
First
L (Θ) − L (Θn) = ln
y
P (X|y, Θ) P (y|Θ) − ln P (X|Θn)
= ln
y
P (X|y, Θ) P (y|Θ)
P (y|X, Θn)
P (y|X, Θn)
− ln P (X|Θn)
= ln
y
P (y|X, Θn)
P (X|y, Θ) P (y|Θ)
P (y|X, Θn)
− ln P (X|Θn)
≥
y
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn)
− ...
y
P (y|X, Θn) ln P (X|Θn) Why this?
39 / 126

91. Images/
We have
First
L (Θ) − L (Θn) = ln
y
P (X|y, Θ) P (y|Θ) − ln P (X|Θn)
= ln
y
P (X|y, Θ) P (y|Θ)
P (y|X, Θn)
P (y|X, Θn)
− ln P (X|Θn)
= ln
y
P (y|X, Θn)
P (X|y, Θ) P (y|Θ)
P (y|X, Θn)
− ln P (X|Θn)
≥
y
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn)
− ...
y
P (y|X, Θn) ln P (X|Θn) Why this?
39 / 126

92. Images/
We have
First
L (Θ) − L (Θn) = ln
y
P (X|y, Θ) P (y|Θ) − ln P (X|Θn)
= ln
y
P (X|y, Θ) P (y|Θ)
P (y|X, Θn)
P (y|X, Θn)
− ln P (X|Θn)
= ln
y
P (y|X, Θn)
P (X|y, Θ) P (y|Θ)
P (y|X, Θn)
− ln P (X|Θn)
≥
y
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn)
− ...
y
P (y|X, Θn) ln P (X|Θn) Why this?
39 / 126

93. Images/
We have
First
L (Θ) − L (Θn) = ln
y
P (X|y, Θ) P (y|Θ) − ln P (X|Θn)
= ln
y
P (X|y, Θ) P (y|Θ)
P (y|X, Θn)
P (y|X, Θn)
− ln P (X|Θn)
= ln
y
P (y|X, Θn)
P (X|y, Θ) P (y|Θ)
P (y|X, Θn)
− ln P (X|Θn)
≥
y
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn)
− ...
y
P (y|X, Θn) ln P (X|Θn) Why this?
39 / 126

94. Images/
Next
Because
y
P (y|X, Θn) = 1
Then
L (Θ) − L (Θn) ≥
y
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn) P (X|Θn)
=∆ (Θ|Θn)
40 / 126

95. Images/
Next
Because
y
P (y|X, Θn) = 1
Then
L (Θ) − L (Θn) ≥
y
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn) P (X|Θn)
=∆ (Θ|Θn)
40 / 126

96. Images/
Then, we have
Then, we have proved that
L (Θ) ≥ L (Θn) + ∆ (Θ|Θn) (26)
Then, we define a new function
l (Θ|Θn) = L (Θn) + ∆ (Θ|Θn) (27)
Thus l (Θ|Θn)
It is bounded from above by L (Θ) i.e l (Θ|Θn) ≤ L (Θ)
41 / 126

97. Images/
Then, we have
Then, we have proved that
L (Θ) ≥ L (Θn) + ∆ (Θ|Θn) (26)
Then, we define a new function
l (Θ|Θn) = L (Θn) + ∆ (Θ|Θn) (27)
Thus l (Θ|Θn)
It is bounded from above by L (Θ) i.e l (Θ|Θn) ≤ L (Θ)
41 / 126

98. Images/
Then, we have
Then, we have proved that
L (Θ) ≥ L (Θn) + ∆ (Θ|Θn) (26)
Then, we define a new function
l (Θ|Θn) = L (Θn) + ∆ (Θ|Θn) (27)
Thus l (Θ|Θn)
It is bounded from above by L (Θ) i.e l (Θ|Θn) ≤ L (Θ)
41 / 126

99. Images/
Now, we can do the following
We evaluate in Θn
l (Θn|Θn) =L (Θn) + ∆ (Θn|Θn)
=L (Θn) +
y
P (y|X, Θn) ln
P (X|y, Θn) P (y|Θn)
P (y|X, Θn) P (X|Θn)
=L (Θn) +
y
P (y|X, Θn) ln
P (X, y|Θn)
P (X, y|Θn)
=L (Θn)
This means that
For Θ = Θn, functions L (Θ) and l (Θ|Θn) are equal
42 / 126

100. Images/
Now, we can do the following
We evaluate in Θn
l (Θn|Θn) =L (Θn) + ∆ (Θn|Θn)
=L (Θn) +
y
P (y|X, Θn) ln
P (X|y, Θn) P (y|Θn)
P (y|X, Θn) P (X|Θn)
=L (Θn) +
y
P (y|X, Θn) ln
P (X, y|Θn)
P (X, y|Θn)
=L (Θn)
This means that
For Θ = Θn, functions L (Θ) and l (Θ|Θn) are equal
42 / 126

101. Images/
Now, we can do the following
We evaluate in Θn
l (Θn|Θn) =L (Θn) + ∆ (Θn|Θn)
=L (Θn) +
y
P (y|X, Θn) ln
P (X|y, Θn) P (y|Θn)
P (y|X, Θn) P (X|Θn)
=L (Θn) +
y
P (y|X, Θn) ln
P (X, y|Θn)
P (X, y|Θn)
=L (Θn)
This means that
For Θ = Θn, functions L (Θ) and l (Θ|Θn) are equal
42 / 126

102. Images/
Now, we can do the following
We evaluate in Θn
l (Θn|Θn) =L (Θn) + ∆ (Θn|Θn)
=L (Θn) +
y
P (y|X, Θn) ln
P (X|y, Θn) P (y|Θn)
P (y|X, Θn) P (X|Θn)
=L (Θn) +
y
P (y|X, Θn) ln
P (X, y|Θn)
P (X, y|Θn)
=L (Θn)
This means that
For Θ = Θn, functions L (Θ) and l (Θ|Θn) are equal
42 / 126

103. Images/
Now, we can do the following
We evaluate in Θn
l (Θn|Θn) =L (Θn) + ∆ (Θn|Θn)
=L (Θn) +
y
P (y|X, Θn) ln
P (X|y, Θn) P (y|Θn)
P (y|X, Θn) P (X|Θn)
=L (Θn) +
y
P (y|X, Θn) ln
P (X, y|Θn)
P (X, y|Θn)
=L (Θn)
This means that
For Θ = Θn, functions L (Θ) and l (Θ|Θn) are equal
42 / 126

104. Images/
Therefore
The function l (Θ|Θn) has the following properties
1 It is bounded from above by L (Θ) i.e l (Θ|Θn) ≤ L (Θ).
2 For Θ = Θn, functions L (Θ) and l (Θ|Θn) are equal.
3 The function l (Θ|Θn) is concave... How?
43 / 126

105. Images/
Therefore
The function l (Θ|Θn) has the following properties
1 It is bounded from above by L (Θ) i.e l (Θ|Θn) ≤ L (Θ).
2 For Θ = Θn, functions L (Θ) and l (Θ|Θn) are equal.
3 The function l (Θ|Θn) is concave... How?
43 / 126

106. Images/
Therefore
The function l (Θ|Θn) has the following properties
1 It is bounded from above by L (Θ) i.e l (Θ|Θn) ≤ L (Θ).
2 For Θ = Θn, functions L (Θ) and l (Θ|Θn) are equal.
3 The function l (Θ|Θn) is concave... How?
43 / 126

107. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
44 / 126

108. Images/
First
We have the value L (Θn)
We know that L (Θn) is constant i.e. an offset value
What about ∆ (Θ|Θn)
y
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn) P (X|Θn)
We have that the ln is a concave function
ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn) P (X|Θn)
45 / 126

109. Images/
First
We have the value L (Θn)
We know that L (Θn) is constant i.e. an offset value
What about ∆ (Θ|Θn)
y
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn) P (X|Θn)
We have that the ln is a concave function
ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn) P (X|Θn)
45 / 126

110. Images/
First
We have the value L (Θn)
We know that L (Θn) is constant i.e. an offset value
What about ∆ (Θ|Θn)
y
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn) P (X|Θn)
We have that the ln is a concave function
ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn) P (X|Θn)
45 / 126

111. Images/
Therefore
Each element is concave
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn) P (X|Θn)
Therefore, the sum of concave functions is a concave function
y
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn) P (X|Θn)
46 / 126

112. Images/
Therefore
Each element is concave
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn) P (X|Θn)
Therefore, the sum of concave functions is a concave function
y
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn) P (X|Θn)
46 / 126

113. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
47 / 126

114. Images/
Given the Concave Function
Thus, we have that
1 We can select Θn such that l (Θ|Θn) is maximized.
2 Thus, given a Θn, we can generate Θn+1.
The process can be seen in the following graph
48 / 126

115. Images/
Given the Concave Function
Thus, we have that
1 We can select Θn such that l (Θ|Θn) is maximized.
2 Thus, given a Θn, we can generate Θn+1.
The process can be seen in the following graph
48 / 126

116. Images/
Given the Concave Function
Thus, we have that
1 We can select Θn such that l (Θ|Θn) is maximized.
2 Thus, given a Θn, we can generate Θn+1.
The process can be seen in the following graph
48 / 126

117. Images/
Given
The Previous Constraints
1 l (Θ|Θn) is bounded from above by L (Θ)
l (Θ|Θn) ≤ L (Θ)
2 For Θ = Θn, functions L (Θ) and l (Θ|Θn) are equal
L (Θn) = l (Θ|Θn)
3 The function l (Θ|Θn) is concave
49 / 126

118. Images/
Given
The Previous Constraints
1 l (Θ|Θn) is bounded from above by L (Θ)
l (Θ|Θn) ≤ L (Θ)
2 For Θ = Θn, functions L (Θ) and l (Θ|Θn) are equal
L (Θn) = l (Θ|Θn)
3 The function l (Θ|Θn) is concave
49 / 126

119. Images/
Given
The Previous Constraints
1 l (Θ|Θn) is bounded from above by L (Θ)
l (Θ|Θn) ≤ L (Θ)
2 For Θ = Θn, functions L (Θ) and l (Θ|Θn) are equal
L (Θn) = l (Θ|Θn)
3 The function l (Θ|Θn) is concave
49 / 126

120. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
50 / 126

121. Images/
From
The following
Θn+1 =argmaxΘ {l (Θ|Θn)}
=argmaxΘ L (Θn) +
y
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn) P (X|Θn)
The terms with Θn are constants.
≈argmaxΘ
y
P (y|X, Θn) ln (P (X|y, Θ) P (y|Θ))
=argmaxΘ
y
P (y|X, Θn) ln
P (X, y|Θ)
P (y|Θ)
P (y, Θ)
P (Θ)
=argmaxΘ


 y
P (y|X, Θn) ln


P(X,y,Θ)
P(Θ)
P(y,Θ)
P(Θ)
P (y, Θ)
P (Θ)





51 / 126

122. Images/
From
The following
Θn+1 =argmaxΘ {l (Θ|Θn)}
=argmaxΘ L (Θn) +
y
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn) P (X|Θn)
The terms with Θn are constants.
≈argmaxΘ
y
P (y|X, Θn) ln (P (X|y, Θ) P (y|Θ))
=argmaxΘ
y
P (y|X, Θn) ln
P (X, y|Θ)
P (y|Θ)
P (y, Θ)
P (Θ)
=argmaxΘ


 y
P (y|X, Θn) ln


P(X,y,Θ)
P(Θ)
P(y,Θ)
P(Θ)
P (y, Θ)
P (Θ)





51 / 126

123. Images/
From
The following
Θn+1 =argmaxΘ {l (Θ|Θn)}
=argmaxΘ L (Θn) +
y
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn) P (X|Θn)
The terms with Θn are constants.
≈argmaxΘ
y
P (y|X, Θn) ln (P (X|y, Θ) P (y|Θ))
=argmaxΘ
y
P (y|X, Θn) ln
P (X, y|Θ)
P (y|Θ)
P (y, Θ)
P (Θ)
=argmaxΘ


 y
P (y|X, Θn) ln


P(X,y,Θ)
P(Θ)
P(y,Θ)
P(Θ)
P (y, Θ)
P (Θ)





51 / 126

124. Images/
From
The following
Θn+1 =argmaxΘ {l (Θ|Θn)}
=argmaxΘ L (Θn) +
y
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn) P (X|Θn)
The terms with Θn are constants.
≈argmaxΘ
y
P (y|X, Θn) ln (P (X|y, Θ) P (y|Θ))
=argmaxΘ
y
P (y|X, Θn) ln
P (X, y|Θ)
P (y|Θ)
P (y, Θ)
P (Θ)
=argmaxΘ


 y
P (y|X, Θn) ln


P(X,y,Θ)
P(Θ)
P(y,Θ)
P(Θ)
P (y, Θ)
P (Θ)





51 / 126

125. Images/
From
The following
Θn+1 =argmaxΘ {l (Θ|Θn)}
=argmaxΘ L (Θn) +
y
P (y|X, Θn) ln
P (X|y, Θ) P (y|Θ)
P (y|X, Θn) P (X|Θn)
The terms with Θn are constants.
≈argmaxΘ
y
P (y|X, Θn) ln (P (X|y, Θ) P (y|Θ))
=argmaxΘ
y
P (y|X, Θn) ln
P (X, y|Θ)
P (y|Θ)
P (y, Θ)
P (Θ)
=argmaxΘ


 y
P (y|X, Θn) ln


P(X,y,Θ)
P(Θ)
P(y,Θ)
P(Θ)
P (y, Θ)
P (Θ)





51 / 126

126. Images/
Thus
Then
θn+1 =argmaxΘ
y
P (y|X, Θn) ln
P (X, y, Θ)
P (y, Θ)
P (y, Θ)
P (Θ)
=argmaxΘ
y
P (y|X, Θn) ln
P (X, y, Θ)
P (Θ)
=argmaxΘ
y
P (y|X, Θn) ln (P (X, y|Θ))
=argmaxΘ Ey|X,Θn
[ln (P (X, y|Θ))]
Then argmaxΘ {l (Θ|Θn)} ≈ argmaxΘ Ey|X,Θn
[ln (P (X, y|Θ))]
52 / 126

127. Images/
Thus
Then
θn+1 =argmaxΘ
y
P (y|X, Θn) ln
P (X, y, Θ)
P (y, Θ)
P (y, Θ)
P (Θ)
=argmaxΘ
y
P (y|X, Θn) ln
P (X, y, Θ)
P (Θ)
=argmaxΘ
y
P (y|X, Θn) ln (P (X, y|Θ))
=argmaxΘ Ey|X,Θn
[ln (P (X, y|Θ))]
Then argmaxΘ {l (Θ|Θn)} ≈ argmaxΘ Ey|X,Θn
[ln (P (X, y|Θ))]
52 / 126

128. Images/
Thus
Then
θn+1 =argmaxΘ
y
P (y|X, Θn) ln
P (X, y, Θ)
P (y, Θ)
P (y, Θ)
P (Θ)
=argmaxΘ
y
P (y|X, Θn) ln
P (X, y, Θ)
P (Θ)
=argmaxΘ
y
P (y|X, Θn) ln (P (X, y|Θ))
=argmaxΘ Ey|X,Θn
[ln (P (X, y|Θ))]
Then argmaxΘ {l (Θ|Θn)} ≈ argmaxΘ Ey|X,Θn
[ln (P (X, y|Θ))]
52 / 126

129. Images/
Thus
Then
θn+1 =argmaxΘ
y
P (y|X, Θn) ln
P (X, y, Θ)
P (y, Θ)
P (y, Θ)
P (Θ)
=argmaxΘ
y
P (y|X, Θn) ln
P (X, y, Θ)
P (Θ)
=argmaxΘ
y
P (y|X, Θn) ln (P (X, y|Θ))
=argmaxΘ Ey|X,Θn
[ln (P (X, y|Θ))]
Then argmaxΘ {l (Θ|Θn)} ≈ argmaxΘ Ey|X,Θn
[ln (P (X, y|Θ))]
52 / 126

130. Images/
Thus
Then
θn+1 =argmaxΘ
y
P (y|X, Θn) ln
P (X, y, Θ)
P (y, Θ)
P (y, Θ)
P (Θ)
=argmaxΘ
y
P (y|X, Θn) ln
P (X, y, Θ)
P (Θ)
=argmaxΘ
y
P (y|X, Θn) ln (P (X, y|Θ))
=argmaxΘ Ey|X,Θn
[ln (P (X, y|Θ))]
Then argmaxΘ {l (Θ|Θn)} ≈ argmaxΘ Ey|X,Θn
[ln (P (X, y|Θ))]
52 / 126

131. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
53 / 126

132. Images/
The EM-Algorithm
Steps of EM
1 Expectation under hidden variables.
2 Maximization of the resulting formula.
E-Step
Determine the conditional expectation, Ey|X,Θn
[ln (P (X, y|Θ))].
M-Step
Maximize this expression with respect to Θ.
54 / 126

133. Images/
The EM-Algorithm
Steps of EM
1 Expectation under hidden variables.
2 Maximization of the resulting formula.
E-Step
Determine the conditional expectation, Ey|X,Θn
[ln (P (X, y|Θ))].
M-Step
Maximize this expression with respect to Θ.
54 / 126

134. Images/
The EM-Algorithm
Steps of EM
1 Expectation under hidden variables.
2 Maximization of the resulting formula.
E-Step
Determine the conditional expectation, Ey|X,Θn
[ln (P (X, y|Θ))].
M-Step
Maximize this expression with respect to Θ.
54 / 126

135. Images/
The EM-Algorithm
Steps of EM
1 Expectation under hidden variables.
2 Maximization of the resulting formula.
E-Step
Determine the conditional expectation, Ey|X,Θn
[ln (P (X, y|Θ))].
M-Step
Maximize this expression with respect to Θ.
54 / 126

136. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
55 / 126

137. Images/
Notes and Convergence of EM
Gains between L (Θ) and l (Θ|Θn)
Using the hidden variables it is possible to simplify the optimization of
L (Θ) through l (Θ|Θn).
Convergence
Remember that Θn+1is the estimate for Θ which maximizes the
difference ∆ (Θ|Θn).
56 / 126

138. Images/
Notes and Convergence of EM
Gains between L (Θ) and l (Θ|Θn)
Using the hidden variables it is possible to simplify the optimization of
L (Θ) through l (Θ|Θn).
Convergence
Remember that Θn+1is the estimate for Θ which maximizes the
difference ∆ (Θ|Θn).
56 / 126

139. Images/
Notes and Convergence of EM
Gains between L (Θ) and l (Θ|Θn)
Using the hidden variables it is possible to simplify the optimization of
L (Θ) through l (Θ|Θn).
Convergence
Remember that Θn+1is the estimate for Θ which maximizes the
difference ∆ (Θ|Θn).
56 / 126

140. Images/
Therefore
Then, we have
Given the initial estimate of Θ by Θn
∆ (Θn|Θn) = 0
Now
If we choose Θn+1 to maximize the ∆ (Θ|Θn), then
∆ (Θn+1|Θn) ≥ ∆ (Θn|Θn) = 0
We have that
The Likelihood L (Θ) is not a decreasing function with respect to Θ.
57 / 126

141. Images/
Therefore
Then, we have
Given the initial estimate of Θ by Θn
∆ (Θn|Θn) = 0
Now
If we choose Θn+1 to maximize the ∆ (Θ|Θn), then
∆ (Θn+1|Θn) ≥ ∆ (Θn|Θn) = 0
We have that
The Likelihood L (Θ) is not a decreasing function with respect to Θ.
57 / 126

142. Images/
Therefore
Then, we have
Given the initial estimate of Θ by Θn
∆ (Θn|Θn) = 0
Now
If we choose Θn+1 to maximize the ∆ (Θ|Θn), then
∆ (Θn+1|Θn) ≥ ∆ (Θn|Θn) = 0
We have that
The Likelihood L (Θ) is not a decreasing function with respect to Θ.
57 / 126

143. Images/
Notes and Convergence of EM
Properties
When the algorithm reaches a fixed point for some Θn, the value
maximizes l (Θ|Θn).
Definition
A fixed point of a function is an element on domain that is mapped to
itself by the function:
f (x) = x
Basically the EM algorithm does the following
EM [Θ∗
] = Θ∗
58 / 126

144. Images/
Notes and Convergence of EM
Properties
When the algorithm reaches a fixed point for some Θn, the value
maximizes l (Θ|Θn).
Definition
A fixed point of a function is an element on domain that is mapped to
itself by the function:
f (x) = x
Basically the EM algorithm does the following
EM [Θ∗
] = Θ∗
58 / 126

145. Images/
Notes and Convergence of EM
Properties
When the algorithm reaches a fixed point for some Θn, the value
maximizes l (Θ|Θn).
Definition
A fixed point of a function is an element on domain that is mapped to
itself by the function:
f (x) = x
Basically the EM algorithm does the following
EM [Θ∗
] = Θ∗
58 / 126

146. Images/
At this moment
We have that
The algorithm reaches a fixed point for some Θn, the value Θ∗ maximizes
l (Θ|Θn).
Then, when the algorithm
It reaches a fixed point for some Θn the value maximizes l (Θ|Θn).
Basically Θn+1 = Θn.
59 / 126

147. Images/
At this moment
We have that
The algorithm reaches a fixed point for some Θn, the value Θ∗ maximizes
l (Θ|Θn).
Then, when the algorithm
It reaches a fixed point for some Θn the value maximizes l (Θ|Θn).
Basically Θn+1 = Θn.
59 / 126

148. Images/
Therefore
We have
60 / 126

149. Images/
Then
If L and l are differentiable at Θn
Since L and l are equal at Θn
Then, Θn is a stationary point of L i.e. the derivative of L vanishes at
that point.
Local Maxima
61 / 126

150. Images/
However
You could finish with the following case, no local maxima
Saddle Point
62 / 126

151. Images/
For more on the subject
Please take a look to
Geoffrey McLachlan and Thriyambakam Krishnan, “The EM Algorithm
and Extensions,” John Wiley & Sons, New York, 1996.
63 / 126

152. Images/
Finding Maximum Likelihood Mixture Densities Parameters
via EM
Something Notable
The mixture-density parameter estimation problem is probably one of the
most widely used applications of the EM algorithm in the computational
pattern recognition community.
We have
p (x|Θ) =
M
i=1
αipi (x|θi) (28)
where
1 Θ = (α1, ..., αM , θ1, ..., θM )
2 M
i=1 αi = 1
3 Each pi is a density function parametrized by θi.
64 / 126

153. Images/
Finding Maximum Likelihood Mixture Densities Parameters
via EM
Something Notable
The mixture-density parameter estimation problem is probably one of the
most widely used applications of the EM algorithm in the computational
pattern recognition community.
We have
p (x|Θ) =
M
i=1
αipi (x|θi) (28)
where
1 Θ = (α1, ..., αM , θ1, ..., θM )
2 M
i=1 αi = 1
3 Each pi is a density function parametrized by θi.
64 / 126

154. Images/
Finding Maximum Likelihood Mixture Densities Parameters
via EM
Something Notable
The mixture-density parameter estimation problem is probably one of the
most widely used applications of the EM algorithm in the computational
pattern recognition community.
We have
p (x|Θ) =
M
i=1
αipi (x|θi) (28)
where
1 Θ = (α1, ..., αM , θ1, ..., θM )
2 M
i=1 αi = 1
3 Each pi is a density function parametrized by θi.
64 / 126

155. Images/
Finding Maximum Likelihood Mixture Densities Parameters
via EM
Something Notable
The mixture-density parameter estimation problem is probably one of the
most widely used applications of the EM algorithm in the computational
pattern recognition community.
We have
p (x|Θ) =
M
i=1
αipi (x|θi) (28)
where
1 Θ = (α1, ..., αM , θ1, ..., θM )
2 M
i=1 αi = 1
3 Each pi is a density function parametrized by θi.
64 / 126

156. Images/
A log-likelihood for this function
We have
log L (Θ|X) = log
N
i=1
p (xi|Θ) =
N
i=1
log


M
j=1
αjpj (xi|θj)

 (29)
Note: This is too difficult to optimize due to the log function.
However
We can simplify this assuming the following:
1 We assume that each unobserved data Y = {yi}N
i=1 has a the
following range yi ∈ {1, ..., M}
2 yi = k if the ith samples was generated by the kth mixture.
65 / 126

157. Images/
A log-likelihood for this function
We have
log L (Θ|X) = log
N
i=1
p (xi|Θ) =
N
i=1
log


M
j=1
αjpj (xi|θj)

 (29)
Note: This is too difficult to optimize due to the log function.
However
We can simplify this assuming the following:
1 We assume that each unobserved data Y = {yi}N
i=1 has a the
following range yi ∈ {1, ..., M}
2 yi = k if the ith samples was generated by the kth mixture.
65 / 126

158. Images/
A log-likelihood for this function
We have
log L (Θ|X) = log
N
i=1
p (xi|Θ) =
N
i=1
log


M
j=1
αjpj (xi|θj)

 (29)
Note: This is too difficult to optimize due to the log function.
However
We can simplify this assuming the following:
1 We assume that each unobserved data Y = {yi}N
i=1 has a the
following range yi ∈ {1, ..., M}
2 yi = k if the ith samples was generated by the kth mixture.
65 / 126

159. Images/
A log-likelihood for this function
We have
log L (Θ|X) = log
N
i=1
p (xi|Θ) =
N
i=1
log


M
j=1
αjpj (xi|θj)

 (29)
Note: This is too difficult to optimize due to the log function.
However
We can simplify this assuming the following:
1 We assume that each unobserved data Y = {yi}N
i=1 has a the
following range yi ∈ {1, ..., M}
2 yi = k if the ith samples was generated by the kth mixture.
65 / 126

160. Images/
A log-likelihood for this function
We have
log L (Θ|X) = log
N
i=1
p (xi|Θ) =
N
i=1
log


M
j=1
αjpj (xi|θj)

 (29)
Note: This is too difficult to optimize due to the log function.
However
We can simplify this assuming the following:
1 We assume that each unobserved data Y = {yi}N
i=1 has a the
following range yi ∈ {1, ..., M}
2 yi = k if the ith samples was generated by the kth mixture.
65 / 126

161. Images/
Now
We have
log L (Θ|X, Y) = log [P (X, Y|Θ)] (30)
Remember that X = {x1, x2, ..., xN } with Y = {y1, y2, ..., yN } and
assuming independence
log [P (X, Y|Θ)] = log [P (x1, x2, ..., xN , y1, y2, ..., yN |Θ)]
= log [P (x1, y1, ..., xi, yi, ..., xN , yN |Θ)]
= log
N
i=1
P (xi, yi|Θ)
=
N
i=1
log P (xi, yi|Θ)
66 / 126

162. Images/
Now
We have
log L (Θ|X, Y) = log [P (X, Y|Θ)] (30)
Remember that X = {x1, x2, ..., xN } with Y = {y1, y2, ..., yN } and
assuming independence
log [P (X, Y|Θ)] = log [P (x1, x2, ..., xN , y1, y2, ..., yN |Θ)]
= log [P (x1, y1, ..., xi, yi, ..., xN , yN |Θ)]
= log
N
i=1
P (xi, yi|Θ)
=
N
i=1
log P (xi, yi|Θ)
66 / 126

163. Images/
Now
We have
log L (Θ|X, Y) = log [P (X, Y|Θ)] (30)
Remember that X = {x1, x2, ..., xN } with Y = {y1, y2, ..., yN } and
assuming independence
log [P (X, Y|Θ)] = log [P (x1, x2, ..., xN , y1, y2, ..., yN |Θ)]
= log [P (x1, y1, ..., xi, yi, ..., xN , yN |Θ)]
= log
N
i=1
P (xi, yi|Θ)
=
N
i=1
log P (xi, yi|Θ)
66 / 126

164. Images/
Now
We have
log L (Θ|X, Y) = log [P (X, Y|Θ)] (30)
Remember that X = {x1, x2, ..., xN } with Y = {y1, y2, ..., yN } and
assuming independence
log [P (X, Y|Θ)] = log [P (x1, x2, ..., xN , y1, y2, ..., yN |Θ)]
= log [P (x1, y1, ..., xi, yi, ..., xN , yN |Θ)]
= log
N
i=1
P (xi, yi|Θ)
=
N
i=1
log P (xi, yi|Θ)
66 / 126

165. Images/
Now
We have
log L (Θ|X, Y) = log [P (X, Y|Θ)] (30)
Remember that X = {x1, x2, ..., xN } with Y = {y1, y2, ..., yN } and
assuming independence
log [P (X, Y|Θ)] = log [P (x1, x2, ..., xN , y1, y2, ..., yN |Θ)]
= log [P (x1, y1, ..., xi, yi, ..., xN , yN |Θ)]
= log
N
i=1
P (xi, yi|Θ)
=
N
i=1
log P (xi, yi|Θ)
66 / 126

166. Images/
Then
Thus, by the chain Rule
N
i=1
log P (xi, yi|Θ) =
N
i=1
log [P (xi|yi, θyi ) P (yi|θyi )] (31)
Question Do you need yi if you know θyi or the other way around?
Finally
N
i=1
log [P (xi|yi, θyi ) P (yi|θyi )] =
N
i=1
log [P (yi) pyi (xi|θyi )] (32)
NOPE: You do not need yi if you know θyi or the other way around.
67 / 126

167. Images/
Then
Thus, by the chain Rule
N
i=1
log P (xi, yi|Θ) =
N
i=1
log [P (xi|yi, θyi ) P (yi|θyi )] (31)
Question Do you need yi if you know θyi or the other way around?
Finally
N
i=1
log [P (xi|yi, θyi ) P (yi|θyi )] =
N
i=1
log [P (yi) pyi (xi|θyi )] (32)
NOPE: You do not need yi if you know θyi or the other way around.
67 / 126

168. Images/
Finally, we have
Making αyi
= P (yi)
log L (Θ|X, Y) =
N
i=1
log [αyi P (xi|yi, θyi )] (33)
68 / 126

169. Images/
Problem
Which Labels?
We do not know the values of Y.
We can get away by using the following idea
Assume the Y is a random variable.
69 / 126

170. Images/
Problem
Which Labels?
We do not know the values of Y.
We can get away by using the following idea
Assume the Y is a random variable.
69 / 126

171. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
70 / 126

172. Images/
Thus
You do a first guess for the parameters at the beginning of EM
Θg
= (αg
1, ..., αg
M , θg
1, ..., θg
M ) (34)
Then, it is possible to calculate given the parametric probability
pj xi|θg
j
Therefore
The mixing parameters αj can be though of as a prior probabilities of each
mixture:
αj = p (component j) (35)
71 / 126

173. Images/
Thus
You do a first guess for the parameters at the beginning of EM
Θg
= (αg
1, ..., αg
M , θg
1, ..., θg
M ) (34)
Then, it is possible to calculate given the parametric probability
pj xi|θg
j
Therefore
The mixing parameters αj can be though of as a prior probabilities of each
mixture:
αj = p (component j) (35)
71 / 126

174. Images/
Thus
You do a first guess for the parameters at the beginning of EM
Θg
= (αg
1, ..., αg
M , θg
1, ..., θg
M ) (34)
Then, it is possible to calculate given the parametric probability
pj xi|θg
j
Therefore
The mixing parameters αj can be though of as a prior probabilities of each
mixture:
αj = p (component j) (35)
71 / 126

175. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
72 / 126

176. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
73 / 126

177. Images/
We want to calculate the following probability
We want to calculate
p (yi|xi, Θg
)
Basically
We want a Bayesian formulation of this probability.
Assuming that the y = (y1, y2, ..., yN ) are samples identically
independent samples from a distribution.
74 / 126

178. Images/
We want to calculate the following probability
We want to calculate
p (yi|xi, Θg
)
Basically
We want a Bayesian formulation of this probability.
Assuming that the y = (y1, y2, ..., yN ) are samples identically
independent samples from a distribution.
74 / 126

179. Images/
We want to calculate the following probability
We want to calculate
p (yi|xi, Θg
)
Basically
We want a Bayesian formulation of this probability.
Assuming that the y = (y1, y2, ..., yN ) are samples identically
independent samples from a distribution.
74 / 126

180. Images/
Using Bayes’ Rule
Compute
p (yi|xi, Θg
) =
p (yi, xi|Θg)
p (xi|Θg)
=
p (xi|Θg) p yi|θg
yi
p (xi|Θg)
We know θg
yi
⇒ Drop it
=
αg
yi
pyi xi|θg
yi
p (xi|Θg)
=
αg
yi
pyi xi|θg
yi
M
k=1 αg
kpk xi|θg
k
75 / 126

181. Images/
Using Bayes’ Rule
Compute
p (yi|xi, Θg
) =
p (yi, xi|Θg)
p (xi|Θg)
=
p (xi|Θg) p yi|θg
yi
p (xi|Θg)
We know θg
yi
⇒ Drop it
=
αg
yi
pyi xi|θg
yi
p (xi|Θg)
=
αg
yi
pyi xi|θg
yi
M
k=1 αg
kpk xi|θg
k
75 / 126

182. Images/
Using Bayes’ Rule
Compute
p (yi|xi, Θg
) =
p (yi, xi|Θg)
p (xi|Θg)
=
p (xi|Θg) p yi|θg
yi
p (xi|Θg)
We know θg
yi
⇒ Drop it
=
αg
yi
pyi xi|θg
yi
p (xi|Θg)
=
αg
yi
pyi xi|θg
yi
M
k=1 αg
kpk xi|θg
k
75 / 126

183. Images/
Using Bayes’ Rule
Compute
p (yi|xi, Θg
) =
p (yi, xi|Θg)
p (xi|Θg)
=
p (xi|Θg) p yi|θg
yi
p (xi|Θg)
We know θg
yi
⇒ Drop it
=
αg
yi
pyi xi|θg
yi
p (xi|Θg)
=
αg
yi
pyi xi|θg
yi
M
k=1 αg
kpk xi|θg
k
75 / 126

184. Images/
As in Naive Bayes
We have the fact that there is a probability per probability at the
mixture and sample
p (yi|xi, Θg
) =
αg
yi
pyi xi|θg
yi
M
k=1 αg
kpk xi|θg
k
∀xi, yi and k ∈ {1, ..., M}
This is going to be updated at each iteration of the EM algorithm
After the initial Guess!!! Until convergence!!!
76 / 126

185. Images/
As in Naive Bayes
We have the fact that there is a probability per probability at the
mixture and sample
p (yi|xi, Θg
) =
αg
yi
pyi xi|θg
yi
M
k=1 αg
kpk xi|θg
k
∀xi, yi and k ∈ {1, ..., M}
This is going to be updated at each iteration of the EM algorithm
After the initial Guess!!! Until convergence!!!
76 / 126

186. Images/
Additionally
We assume again that the samples yis are identically and independent
samples
p (y|X, Θg
) =
N
i=1
p (yi|xi, Θg
) (36)
Where y = (y1, y2, ..., yN )
77 / 126

187. Images/
Now, using equation 17
Then
Q (Θ|Θg
) =
y∈Y
log (L (Θ|X, y)) p (y|X, Θg
)
=
y∈Y
N
i=1
log [αyi pyi (xi|θyi )]
N
j=1
p (yj|xj, Θg
)
78 / 126

188. Images/
Now, using equation 17
Then
Q (Θ|Θg
) =
y∈Y
log (L (Θ|X, y)) p (y|X, Θg
)
=
y∈Y
N
i=1
log [αyi pyi (xi|θyi )]
N
j=1
p (yj|xj, Θg
)
78 / 126

189. Images/
Here, a small stop
What is the meaning of y∈Y
It is actually a summation of all possible states of the random vector y.
Then, we can rewrite the previous summation as
y∈Y
=
M
y1=1
M
y2=1
· · ·
M
yN =1
N
Running over all the samples {x1, x2, ..., xN }.
79 / 126

190. Images/
Here, a small stop
What is the meaning of y∈Y
It is actually a summation of all possible states of the random vector y.
Then, we can rewrite the previous summation as
y∈Y
=
M
y1=1
M
y2=1
· · ·
M
yN =1
N
Running over all the samples {x1, x2, ..., xN }.
79 / 126

191. Images/
Then
We have
Q (Θ|Θg
) =
M
y1=1
M
y2=1
· · ·
M
yN =1
N
i=1

log [αyi pyi (xi|θyi )]
N
j=1
p (yj|xj, Θg
)


80 / 126

192. Images/
We introduce the following
We have the following function
δl,yi
=
1 I = yi
0 I = yi
Therefore, we can do the following
αi =
M
j=1
δi,jαj
Then
log [αyi pyi (xi|θyi )]
N
j=1
p (yj|xj, Θg
) =
M
l=1
δl,yi log [αlpl (xi|θl)]
N
j=1
p (yj|xj, Θg
)
81 / 126

193. Images/
We introduce the following
We have the following function
δl,yi
=
1 I = yi
0 I = yi
Therefore, we can do the following
αi =
M
j=1
δi,jαj
Then
log [αyi pyi (xi|θyi )]
N
j=1
p (yj|xj, Θg
) =
M
l=1
δl,yi log [αlpl (xi|θl)]
N
j=1
p (yj|xj, Θg
)
81 / 126

194. Images/
We introduce the following
We have the following function
δl,yi
=
1 I = yi
0 I = yi
Therefore, we can do the following
αi =
M
j=1
δi,jαj
Then
log [αyi pyi (xi|θyi )]
N
j=1
p (yj|xj, Θg
) =
M
l=1
δl,yi log [αlpl (xi|θl)]
N
j=1
p (yj|xj, Θg
)
81 / 126

195. Images/
Thus
We have that for
M
y1=1 · · · M
yN =1
N
i=1 log [αyi
pyi
(xi|θyi
)] N
j=1 p (yj|xj, Θg
) = ∗
∗ =
M
y1=1
M
y2=1
· · ·
M
yN =1
N
i=1
M
l=1
δl,yi
log [αlpl (xi|θl)]
N
j=1
p (yj|xj, Θg
)
=
N
i=1
M
l=1
log [αlpl (xi|θl)]
M
y1=1
M
y2=1
· · ·
M
yN =1

δl,yi
N
j=1
p (yj|xj, Θg
)


Because
M
y1=1
M
y2=1 · · · M
yN =1 applies only to δl,yi
N
j=1 p (yj|xj, Θg)
82 / 126

196. Images/
Thus
We have that for
M
y1=1 · · · M
yN =1
N
i=1 log [αyi
pyi
(xi|θyi
)] N
j=1 p (yj|xj, Θg
) = ∗
∗ =
M
y1=1
M
y2=1
· · ·
M
yN =1
N
i=1
M
l=1
δl,yi
log [αlpl (xi|θl)]
N
j=1
p (yj|xj, Θg
)
=
N
i=1
M
l=1
log [αlpl (xi|θl)]
M
y1=1
M
y2=1
· · ·
M
yN =1

δl,yi
N
j=1
p (yj|xj, Θg
)


Because
M
y1=1
M
y2=1 · · · M
yN =1 applies only to δl,yi
N
j=1 p (yj|xj, Θg)
82 / 126

197. Images/
Thus
We have that for
M
y1=1 · · · M
yN =1
N
i=1 log [αyi
pyi
(xi|θyi
)] N
j=1 p (yj|xj, Θg
) = ∗
∗ =
M
y1=1
M
y2=1
· · ·
M
yN =1
N
i=1
M
l=1
δl,yi
log [αlpl (xi|θl)]
N
j=1
p (yj|xj, Θg
)
=
N
i=1
M
l=1
log [αlpl (xi|θl)]
M
y1=1
M
y2=1
· · ·
M
yN =1

δl,yi
N
j=1
p (yj|xj, Θg
)


Because
M
y1=1
M
y2=1 · · · M
yN =1 applies only to δl,yi
N
j=1 p (yj|xj, Θg)
82 / 126

198. Images/
Then, we have that
First notice the following
M
y1=1
M
y2=1
· · ·
M
yN =1
δl,yi
N
j=1
p (yj|xj, Θg
) =
=


M
y1=1
· · ·
M
yi−1=1
M
yi+1=1
· · ·
M
yN =1
M
yi=1
δl,yi
p (yi|xi, Θg
)
N
j=1,j=i,
p (yj|xj, Θg
)


Then, we have
M
yi=1
δl,yi
p (yi|xi, Θg
) = p (l|xi, Θg
)
83 / 126

199. Images/
Then, we have that
First notice the following
M
y1=1
M
y2=1
· · ·
M
yN =1
δl,yi
N
j=1
p (yj|xj, Θg
) =
=


M
y1=1
· · ·
M
yi−1=1
M
yi+1=1
· · ·
M
yN =1
M
yi=1
δl,yi
p (yi|xi, Θg
)
N
j=1,j=i,
p (yj|xj, Θg
)


Then, we have
M
yi=1
δl,yi
p (yi|xi, Θg
) = p (l|xi, Θg
)
83 / 126

200. Images/
Then, we have that
First notice the following
M
y1=1
M
y2=1
· · ·
M
yN =1
δl,yi
N
j=1
p (yj|xj, Θg
) =
=


M
y1=1
· · ·
M
yi−1=1
M
yi+1=1
· · ·
M
yN =1
M
yi=1
δl,yi
p (yi|xi, Θg
)
N
j=1,j=i,
p (yj|xj, Θg
)


Then, we have
M
yi=1
δl,yi
p (yi|xi, Θg
) = p (l|xi, Θg
)
83 / 126

201. Images/
In this way
Plugging back the previous equation
M
y1=1
M
y2=1
· · ·
M
yN =1
δl,yi
N
j=1
p (yj|xj, Θg
) =
=


M
y1=1
· · ·
M
yi−1=1
M
yi+1=1
· · ·
M
yN =1
p (l|xi, Θg
)
N
j=1,j=i
p (yj|xj, Θg
)


=


M
y1=1
· · ·
M
yi−1=1
M
yi+1=1
· · ·
M
yN =1
N
j=1,j=i
p (yj|xj, Θg
)

 p (l|xi, Θg
)
84 / 126

202. Images/
In this way
Plugging back the previous equation
M
y1=1
M
y2=1
· · ·
M
yN =1
δl,yi
N
j=1
p (yj|xj, Θg
) =
=


M
y1=1
· · ·
M
yi−1=1
M
yi+1=1
· · ·
M
yN =1
p (l|xi, Θg
)
N
j=1,j=i
p (yj|xj, Θg
)


=


M
y1=1
· · ·
M
yi−1=1
M
yi+1=1
· · ·
M
yN =1
N
j=1,j=i
p (yj|xj, Θg
)

 p (l|xi, Θg
)
84 / 126

203. Images/
In this way
Plugging back the previous equation
M
y1=1
M
y2=1
· · ·
M
yN =1
δl,yi
N
j=1
p (yj|xj, Θg
) =
=


M
y1=1
· · ·
M
yi−1=1
M
yi+1=1
· · ·
M
yN =1
p (l|xi, Θg
)
N
j=1,j=i
p (yj|xj, Θg
)


=


M
y1=1
· · ·
M
yi−1=1
M
yi+1=1
· · ·
M
yN =1
N
j=1,j=i
p (yj|xj, Θg
)

 p (l|xi, Θg
)
84 / 126

204. Images/
Now, what about...?
The left part of the equation
M
y1=1
· · ·
M
yi−1=1
M
yi+1=1
· · ·
M
yN =1
N
j=1,j=i
p (yj|xj, Θg
) =
=


M
y1=1
p (y1|x1, Θg
)

 · · ·


M
yi−1=1
p (yi−1|xi−1, Θg
)

 × ...


M
yi+1=1
p (yi+1|xi+1, Θg
)

 · · ·


M
yN =1
p (yN |xN , Θg
)


=
N
j=1,j=i


M
yj=1
p (yj|xj, Θg
)


85 / 126

205. Images/
Now, what about...?
The left part of the equation
M
y1=1
· · ·
M
yi−1=1
M
yi+1=1
· · ·
M
yN =1
N
j=1,j=i
p (yj|xj, Θg
) =
=


M
y1=1
p (y1|x1, Θg
)

 · · ·


M
yi−1=1
p (yi−1|xi−1, Θg
)

 × ...


M
yi+1=1
p (yi+1|xi+1, Θg
)

 · · ·


M
yN =1
p (yN |xN , Θg
)


=
N
j=1,j=i


M
yj=1
p (yj|xj, Θg
)


85 / 126

206. Images/
Now, what about...?
The left part of the equation
M
y1=1
· · ·
M
yi−1=1
M
yi+1=1
· · ·
M
yN =1
N
j=1,j=i
p (yj|xj, Θg
) =
=


M
y1=1
p (y1|x1, Θg
)

 · · ·


M
yi−1=1
p (yi−1|xi−1, Θg
)

 × ...


M
yi+1=1
p (yi+1|xi+1, Θg
)

 · · ·


M
yN =1
p (yN |xN , Θg
)


=
N
j=1,j=i


M
yj=1
p (yj|xj, Θg
)


85 / 126

207. Images/
Then, we have that
Plugging back to the original equation



M
y1=1
· · ·
M
yi−1=1
M
yi+1=1
· · ·
M
yN =1
N
j=1,j=i
p (yj|xj, Θg
)



p (l|xi, Θg
) =
=



N
j=1,j=i


M
yj=1
p (yj|xj, Θg
)





p (l|xi, Θg
)
86 / 126

208. Images/
Then, we have that
Plugging back to the original equation



M
y1=1
· · ·
M
yi−1=1
M
yi+1=1
· · ·
M
yN =1
N
j=1,j=i
p (yj|xj, Θg
)



p (l|xi, Θg
) =
=



N
j=1,j=i


M
yj=1
p (yj|xj, Θg
)





p (l|xi, Θg
)
86 / 126

209. Images/
We can use properties of probability
We know that
M
yi=1
p (yi|xi, Θg
) = 1 (37)
Then



N
j=1,j=i


M
yj=1
p (yj|xj, Θg
)





p (l|xi, Θg
) =
=



N
j=1,j=i
1



p (l|xi, Θg
)
=p (l|xi, Θg
)
=
αg
l pyi xi|θg
l
M
k=1 αg
kpk xi|θg
k
87 / 126

210. Images/
We can use properties of probability
We know that
M
yi=1
p (yi|xi, Θg
) = 1 (37)
Then



N
j=1,j=i


M
yj=1
p (yj|xj, Θg
)





p (l|xi, Θg
) =
=



N
j=1,j=i
1



p (l|xi, Θg
)
=p (l|xi, Θg
)
=
αg
l pyi xi|θg
l
M
k=1 αg
kpk xi|θg
k
87 / 126

211. Images/
We can use properties of probability
We know that
M
yi=1
p (yi|xi, Θg
) = 1 (37)
Then



N
j=1,j=i


M
yj=1
p (yj|xj, Θg
)





p (l|xi, Θg
) =
=



N
j=1,j=i
1



p (l|xi, Θg
)
=p (l|xi, Θg
)
=
αg
l pyi xi|θg
l
M
k=1 αg
kpk xi|θg
k
87 / 126

212. Images/
We can use properties of probability
We know that
M
yi=1
p (yi|xi, Θg
) = 1 (37)
Then



N
j=1,j=i


M
yj=1
p (yj|xj, Θg
)





p (l|xi, Θg
) =
=



N
j=1,j=i
1



p (l|xi, Θg
)
=p (l|xi, Θg
)
=
αg
l pyi xi|θg
l
M
k=1 αg
kpk xi|θg
k
87 / 126

213. Images/
Thus
We can write Q in the following way
Q (Θ, Θg
) =
N
i=1
M
l=1
log [αlpl (xi|θl)] p (l|xi, Θg
)
=
N
i=1
M
l=1
log (αl) p (l|xi, Θg
) + ...
N
i=1
M
l=1
log (pl (xi|θl)) p (l|xi, Θg
) (38)
88 / 126

214. Images/
Thus
We can write Q in the following way
Q (Θ, Θg
) =
N
i=1
M
l=1
log [αlpl (xi|θl)] p (l|xi, Θg
)
=
N
i=1
M
l=1
log (αl) p (l|xi, Θg
) + ...
N
i=1
M
l=1
log (pl (xi|θl)) p (l|xi, Θg
) (38)
88 / 126

215. Images/
Thus
We can write Q in the following way
Q (Θ, Θg
) =
N
i=1
M
l=1
log [αlpl (xi|θl)] p (l|xi, Θg
)
=
N
i=1
M
l=1
log (αl) p (l|xi, Θg
) + ...
N
i=1
M
l=1
log (pl (xi|θl)) p (l|xi, Θg
) (38)
88 / 126

216. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
89 / 126

217. Images/
A Method
That could be used as a general framework
To solve problems set as EM problem.
First, we will look at the Lagrange Multipliers setup
Then, we will look at a specific case using the mixture of Gaussian’s
Note
Not all the mixture of distributions will get you an analytical solution.
90 / 126

218. Images/
A Method
That could be used as a general framework
To solve problems set as EM problem.
First, we will look at the Lagrange Multipliers setup
Then, we will look at a specific case using the mixture of Gaussian’s
Note
Not all the mixture of distributions will get you an analytical solution.
90 / 126

219. Images/
A Method
That could be used as a general framework
To solve problems set as EM problem.
First, we will look at the Lagrange Multipliers setup
Then, we will look at a specific case using the mixture of Gaussian’s
Note
Not all the mixture of distributions will get you an analytical solution.
90 / 126

220. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
91 / 126

221. Images/
Lagrange Multipliers
The method of Lagrange multipliers
Gives a set of necessary conditions to identify optimal points of equality
constrained optimization problems.
This is done by converting a constrained problem
To an equivalent unconstrained problem with the help of certain
unspecified parameters known as Lagrange multipliers.
92 / 126

222. Images/
Lagrange Multipliers
The method of Lagrange multipliers
Gives a set of necessary conditions to identify optimal points of equality
constrained optimization problems.
This is done by converting a constrained problem
To an equivalent unconstrained problem with the help of certain
unspecified parameters known as Lagrange multipliers.
92 / 126

223. Images/
Lagrange Multipliers
The classical problem formulation
min f (x1, x2, ..., xn)
s.t h1 (x1, x2, ..., xn) = 0
It can be converted into
min L (x1, x2, ..., xn, λ) = min {f (x1, x2, ..., xn) − λh1 (x1, x2, ..., xn)}
(39)
where
L(x, λ) is the Lagrangian function.
λ is an unspecified positive or negative constant called the Lagrange
Multiplier.
93 / 126

224. Images/
Lagrange Multipliers
The classical problem formulation
min f (x1, x2, ..., xn)
s.t h1 (x1, x2, ..., xn) = 0
It can be converted into
min L (x1, x2, ..., xn, λ) = min {f (x1, x2, ..., xn) − λh1 (x1, x2, ..., xn)}
(39)
where
L(x, λ) is the Lagrangian function.
λ is an unspecified positive or negative constant called the Lagrange
Multiplier.
93 / 126

225. Images/
Lagrange Multipliers
The classical problem formulation
min f (x1, x2, ..., xn)
s.t h1 (x1, x2, ..., xn) = 0
It can be converted into
min L (x1, x2, ..., xn, λ) = min {f (x1, x2, ..., xn) − λh1 (x1, x2, ..., xn)}
(39)
where
L(x, λ) is the Lagrangian function.
λ is an unspecified positive or negative constant called the Lagrange
Multiplier.
93 / 126

226. Images/
Lagrange Multipliers are applications of Py
Think about this
Remember First Law of Newton!!!
Yes!!!
94 / 126

227. Images/
Lagrange Multipliers are applications of Py
Think about this
Remember First Law of Newton!!!
Yes!!!
A system in equilibrium does not move
Static Body
94 / 126

228. Images/
Lagrange Multipliers
Definition
Gives a set of necessary conditions to identify optimal points of equality
constrained optimization problem
95 / 126

229. Images/
Lagrange was a Physicists
He was thinking in the following formula
A system in equilibrium has the following equation:
F1 + F2 + ... + FK = 0 (40)
But functions do not have forces?
Are you sure?
Think about the following
The Gradient of a surface.
96 / 126

230. Images/
Lagrange was a Physicists
He was thinking in the following formula
A system in equilibrium has the following equation:
F1 + F2 + ... + FK = 0 (40)
But functions do not have forces?
Are you sure?
Think about the following
The Gradient of a surface.
96 / 126

231. Images/
Lagrange was a Physicists
He was thinking in the following formula
A system in equilibrium has the following equation:
F1 + F2 + ... + FK = 0 (40)
But functions do not have forces?
Are you sure?
Think about the following
The Gradient of a surface.
96 / 126

232. Images/
Gradient to a Surface
After all a gradient is a measure of the maximal change
For example the gradient of a function of three variables:
f (x) = i
∂f (x)
∂x
+ j
∂f (x)
∂y
+ k
∂f (x)
∂z
(41)
where i, j and k are unitary vectors in the directions x, y and z.
97 / 126

233. Images/
Example
We have f (x, y) = x exp {−x2
− y2
}
98 / 126

234. Images/
Example
With Gradient at the the contours when projecting in the 2D plane
99 / 126

235. Images/
Now, Think about this
Yes, we can use the gradient
However, we need to do some scaling of the forces by using parameters λ
Thus, we have
F0 + λ1F1 + ... + λKFK = 0 (42)
where F0 is the gradient of the principal cost function and Fi for
i = 1, 2, .., K.
100 / 126

236. Images/
Now, Think about this
Yes, we can use the gradient
However, we need to do some scaling of the forces by using parameters λ
Thus, we have
F0 + λ1F1 + ... + λKFK = 0 (42)
where F0 is the gradient of the principal cost function and Fi for
i = 1, 2, .., K.
100 / 126

237. Images/
Thus
If we have the following optimization:
min f (x)
s.tg1 (x) = 0
g2 (x) = 0
101 / 126

238. Images/
Geometric interpretation in the case of minimization
What is wrong? Gradients are going in the other direction, we can fix
by simple multiplying by -1
Here the cost function is f (x, y) = x exp −x2 − y2 we want to minimize
f (−→x )
g1 (−→x )
g2 (−→x )
−∇f (−→x ) + λ1∇g1 (−→x ) + λ2∇g2 (−→x ) = 0
Nevertheless: it is equivalent to f −→x − λ1 g1
−→x − λ2 g2
−→x = 0
101 / 126

239. Images/
Geometric interpretation in the case of minimization
What is wrong? Gradients are going in the other direction, we can fix
by simple multiplying by -1
Here the cost function is f (x, y) = x exp −x2 − y2 we want to minimize
f (−→x )
g1 (−→x )
g2 (−→x )
−∇f (−→x ) + λ1∇g1 (−→x ) + λ2∇g2 (−→x ) = 0
Nevertheless: it is equivalent to f −→x − λ1 g1
−→x − λ2 g2
−→x = 0
101 / 126

240. Images/
Method
Steps
1 Original problem is rewritten as:
1 minimize L (x, λ) = f (x) − λh1 (x)
2 Take derivatives of L (x, λ) with respect to xi and set them equal to
zero.
3 Express all xi in terms of Lagrangian multiplier λ.
4 Plug x in terms of λ in constraint h1 (x) = 0 and solve λ.
5 Calculate x by using the just found value for λ.
From the step 2
If there are n variables (i.e., x1, · · · , xn) then you will get n equations with
n + 1 unknowns (i.e., n variables xi and one Lagrangian multiplier λ).
102 / 126

241. Images/
Method
Steps
1 Original problem is rewritten as:
1 minimize L (x, λ) = f (x) − λh1 (x)
2 Take derivatives of L (x, λ) with respect to xi and set them equal to
zero.
3 Express all xi in terms of Lagrangian multiplier λ.
4 Plug x in terms of λ in constraint h1 (x) = 0 and solve λ.
5 Calculate x by using the just found value for λ.
From the step 2
If there are n variables (i.e., x1, · · · , xn) then you will get n equations with
n + 1 unknowns (i.e., n variables xi and one Lagrangian multiplier λ).
102 / 126

242. Images/
Method
Steps
1 Original problem is rewritten as:
1 minimize L (x, λ) = f (x) − λh1 (x)
2 Take derivatives of L (x, λ) with respect to xi and set them equal to
zero.
3 Express all xi in terms of Lagrangian multiplier λ.
4 Plug x in terms of λ in constraint h1 (x) = 0 and solve λ.
5 Calculate x by using the just found value for λ.
From the step 2
If there are n variables (i.e., x1, · · · , xn) then you will get n equations with
n + 1 unknowns (i.e., n variables xi and one Lagrangian multiplier λ).
102 / 126

243. Images/
Method
Steps
1 Original problem is rewritten as:
1 minimize L (x, λ) = f (x) − λh1 (x)
2 Take derivatives of L (x, λ) with respect to xi and set them equal to
zero.
3 Express all xi in terms of Lagrangian multiplier λ.
4 Plug x in terms of λ in constraint h1 (x) = 0 and solve λ.
5 Calculate x by using the just found value for λ.
From the step 2
If there are n variables (i.e., x1, · · · , xn) then you will get n equations with
n + 1 unknowns (i.e., n variables xi and one Lagrangian multiplier λ).
102 / 126

244. Images/
Method
Steps
1 Original problem is rewritten as:
1 minimize L (x, λ) = f (x) − λh1 (x)
2 Take derivatives of L (x, λ) with respect to xi and set them equal to
zero.
3 Express all xi in terms of Lagrangian multiplier λ.
4 Plug x in terms of λ in constraint h1 (x) = 0 and solve λ.
5 Calculate x by using the just found value for λ.
From the step 2
If there are n variables (i.e., x1, · · · , xn) then you will get n equations with
n + 1 unknowns (i.e., n variables xi and one Lagrangian multiplier λ).
102 / 126

245. Images/
Method
Steps
1 Original problem is rewritten as:
1 minimize L (x, λ) = f (x) − λh1 (x)
2 Take derivatives of L (x, λ) with respect to xi and set them equal to
zero.
3 Express all xi in terms of Lagrangian multiplier λ.
4 Plug x in terms of λ in constraint h1 (x) = 0 and solve λ.
5 Calculate x by using the just found value for λ.
From the step 2
If there are n variables (i.e., x1, · · · , xn) then you will get n equations with
n + 1 unknowns (i.e., n variables xi and one Lagrangian multiplier λ).
102 / 126

246. Images/
Example
We can apply that to the following problem
min f (x, y) = x2
− 8x + y2
− 12y + 48
s.t x + y = 8
103 / 126

247. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
104 / 126

248. Images/
Lagrange Multipliers for Q
We can us the following constraint for that
l
αl = 1 (43)
We have the following cost function
Q (Θ, Θg
) + λ
l
αl − 1 (44)
Deriving by αl
∂
∂αl
Q (Θ, Θg
) + λ
l
αl − 1 = 0 (45)
105 / 126

249. Images/
Lagrange Multipliers for Q
We can us the following constraint for that
l
αl = 1 (43)
We have the following cost function
Q (Θ, Θg
) + λ
l
αl − 1 (44)
Deriving by αl
∂
∂αl
Q (Θ, Θg
) + λ
l
αl − 1 = 0 (45)
105 / 126

250. Images/
Lagrange Multipliers for Q
We can us the following constraint for that
l
αl = 1 (43)
We have the following cost function
Q (Θ, Θg
) + λ
l
αl − 1 (44)
Deriving by αl
∂
∂αl
Q (Θ, Θg
) + λ
l
αl − 1 = 0 (45)
105 / 126

251. Images/
Thus
The Q function
Q (Θ, Θg
) =
N
i=1
M
l=1
log (αl) p (l|xi, Θg
) + ...
N
i=1
M
l=1
log (pl (xi|θl)) p (l|xi, Θg
)
106 / 126

252. Images/
Deriving
We have
∂
∂αl
Q (Θ, Θg
) + λ
l
αl − 1 =
N
i=1
1
αl
p (l|xi, Θg
) + λ
107 / 126

253. Images/
Finally
We have making the previous equation equal to 0
N
i=1
1
αl
p (l|xi, Θg
) + λ = 0 (46)
Thus
N
i=1
p (l|xi, Θg
) = −λαl (47)
Summing over l, we get
λ = −N (48)
108 / 126

254. Images/
Finally
We have making the previous equation equal to 0
N
i=1
1
αl
p (l|xi, Θg
) + λ = 0 (46)
Thus
N
i=1
p (l|xi, Θg
) = −λαl (47)
Summing over l, we get
λ = −N (48)
108 / 126

255. Images/
Finally
We have making the previous equation equal to 0
N
i=1
1
αl
p (l|xi, Θg
) + λ = 0 (46)
Thus
N
i=1
p (l|xi, Θg
) = −λαl (47)
Summing over l, we get
λ = −N (48)
108 / 126

256. Images/
Lagrange Multipliers
Thus
αl =
1
N
N
i=1
p (l|xi, Θg
) (49)
About θl
It is possible to get an analytical expressions for θl as functions of
everything else.
This is for you to try!!!
For more, please look at
“Geometric Idea of Lagrange Multipliers” by John Wyatt.
109 / 126

257. Images/
Lagrange Multipliers
Thus
αl =
1
N
N
i=1
p (l|xi, Θg
) (49)
About θl
It is possible to get an analytical expressions for θl as functions of
everything else.
This is for you to try!!!
For more, please look at
“Geometric Idea of Lagrange Multipliers” by John Wyatt.
109 / 126

258. Images/
Lagrange Multipliers
Thus
αl =
1
N
N
i=1
p (l|xi, Θg
) (49)
About θl
It is possible to get an analytical expressions for θl as functions of
everything else.
This is for you to try!!!
For more, please look at
“Geometric Idea of Lagrange Multipliers” by John Wyatt.
109 / 126

259. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
110 / 126

260. Images/
Remember?
Gaussian Distribution
pl (x|µl, Σl) =
1
(2π)
d/2
|Σl|
1/2
exp −
1
2
(x − µl)T
Σ−1
l (x − µl) (50)
111 / 126

261. Images/
How to use this for Gaussian Distributions
For this, we need to refresh some linear algebra
1 tr (A + B) = tr (A) + tr (B)
2 tr (AB) = tr (BA)
3
i xT
i Axi = tr (AB) where B = i xixT
i .
4 A−1 = 1
|A|
Now, we need the derivative of a matrix function f (A)
Thus, ∂f(A)
∂A is going to be the matrix with i, jth entry ∂f(A)
∂ai,j
where ai,j
is the i, jth entry of A.
112 / 126

262. Images/
How to use this for Gaussian Distributions
For this, we need to refresh some linear algebra
1 tr (A + B) = tr (A) + tr (B)
2 tr (AB) = tr (BA)
3
i xT
i Axi = tr (AB) where B = i xixT
i .
4 A−1 = 1
|A|
Now, we need the derivative of a matrix function f (A)
Thus, ∂f(A)
∂A is going to be the matrix with i, jth entry ∂f(A)
∂ai,j
where ai,j
is the i, jth entry of A.
112 / 126

263. Images/
How to use this for Gaussian Distributions
For this, we need to refresh some linear algebra
1 tr (A + B) = tr (A) + tr (B)
2 tr (AB) = tr (BA)
3
i xT
i Axi = tr (AB) where B = i xixT
i .
4 A−1 = 1
|A|
Now, we need the derivative of a matrix function f (A)
Thus, ∂f(A)
∂A is going to be the matrix with i, jth entry ∂f(A)
∂ai,j
where ai,j
is the i, jth entry of A.
112 / 126

264. Images/
How to use this for Gaussian Distributions
For this, we need to refresh some linear algebra
1 tr (A + B) = tr (A) + tr (B)
2 tr (AB) = tr (BA)
3
i xT
i Axi = tr (AB) where B = i xixT
i .
4 A−1 = 1
|A|
Now, we need the derivative of a matrix function f (A)
Thus, ∂f(A)
∂A is going to be the matrix with i, jth entry ∂f(A)
∂ai,j
where ai,j
is the i, jth entry of A.
112 / 126

265. Images/
How to use this for Gaussian Distributions
For this, we need to refresh some linear algebra
1 tr (A + B) = tr (A) + tr (B)
2 tr (AB) = tr (BA)
3
i xT
i Axi = tr (AB) where B = i xixT
i .
4 A−1 = 1
|A|
Now, we need the derivative of a matrix function f (A)
Thus, ∂f(A)
∂A is going to be the matrix with i, jth entry ∂f(A)
∂ai,j
where ai,j
is the i, jth entry of A.
112 / 126

266. Images/
In addition
If A is symmetric
∂ |A|
∂A
=
Ai,j if i = j
2Ai,j if i = j
(51)
Where Ai,j is the i, jth cofactor of A.
Note: The determinant obtained by deleting the row and column of
a given element of a matrix or determinant. The cofactor is
preceded by a + or – sign depending whether the element is
in a + or – position.
Thus
∂ log |A|
∂A
=



Ai,j
|A| if i = j
2Ai,j if i = j
= 2A−1
− diag A−1
(52)
113 / 126

267. Images/
In addition
If A is symmetric
∂ |A|
∂A
=
Ai,j if i = j
2Ai,j if i = j
(51)
Where Ai,j is the i, jth cofactor of A.
Note: The determinant obtained by deleting the row and column of
a given element of a matrix or determinant. The cofactor is
preceded by a + or – sign depending whether the element is
in a + or – position.
Thus
∂ log |A|
∂A
=



Ai,j
|A| if i = j
2Ai,j if i = j
= 2A−1
− diag A−1
(52)
113 / 126

268. Images/
In addition
If A is symmetric
∂ |A|
∂A
=
Ai,j if i = j
2Ai,j if i = j
(51)
Where Ai,j is the i, jth cofactor of A.
Note: The determinant obtained by deleting the row and column of
a given element of a matrix or determinant. The cofactor is
preceded by a + or – sign depending whether the element is
in a + or – position.
Thus
∂ log |A|
∂A
=



Ai,j
|A| if i = j
2Ai,j if i = j
= 2A−1
− diag A−1
(52)
113 / 126

269. Images/
In addition
If A is symmetric
∂ |A|
∂A
=
Ai,j if i = j
2Ai,j if i = j
(51)
Where Ai,j is the i, jth cofactor of A.
Note: The determinant obtained by deleting the row and column of
a given element of a matrix or determinant. The cofactor is
preceded by a + or – sign depending whether the element is
in a + or – position.
Thus
∂ log |A|
∂A
=



Ai,j
|A| if i = j
2Ai,j if i = j
= 2A−1
− diag A−1
(52)
113 / 126

270. Images/
Finally
The last equation we need
∂tr (AB)
∂A
= B + BT
− diag (B) (53)
In addition
∂xT Ax
∂x
(54)
114 / 126

271. Images/
Finally
The last equation we need
∂tr (AB)
∂A
= B + BT
− diag (B) (53)
In addition
∂xT Ax
∂x
(54)
114 / 126

272. Images/
Thus, using last part of equation 38
We get, after ignoring constant terms
Remember they disappear after derivatives
N
i=1
M
l=1
log (pl (xi|µl, Σl)) p (l|xi, Θg
)
=
N
i=1
M
l=1
−
1
2
log (|Σl|) −
1
2
(xi − µl)T
Σ−1
l (xi − µl) p (l|xi, Θg
) (55)
115 / 126

273. Images/
Thus, using last part of equation 38
We get, after ignoring constant terms
Remember they disappear after derivatives
N
i=1
M
l=1
log (pl (xi|µl, Σl)) p (l|xi, Θg
)
=
N
i=1
M
l=1
−
1
2
log (|Σl|) −
1
2
(xi − µl)T
Σ−1
l (xi − µl) p (l|xi, Θg
) (55)
115 / 126

274. Images/
Finally
Thus, when taking the derivative with respect to µl
N
i=1
Σ−1
l (xi − µl) p (l|xi, Θg
) = 0 (56)
Then
µl =
N
i=1 xip (l|xi, Θg)
N
i=1 p (l|xi, Θg)
(57)
116 / 126

275. Images/
Finally
Thus, when taking the derivative with respect to µl
N
i=1
Σ−1
l (xi − µl) p (l|xi, Θg
) = 0 (56)
Then
µl =
N
i=1 xip (l|xi, Θg)
N
i=1 p (l|xi, Θg)
(57)
116 / 126

276. Images/
Now, if we derive with respect to Σl
First, we rewrite equation 55
N
i=1
M
l=1
−
1
2
log (|Σl|) −
1
2
(xi − µl)T
Σ−1
l
(xi − µl) p (l|xi, Θg
)
=
M
l=1
−
1
2
log (|Σl|)
N
i=1
p (l|xi, Θg
) −
1
2
N
i=1
p (l|xi, Θg
) tr Σ−1
l
(xi − µl) (xi − µl)T
=
M
l=1
−
1
2
log (|Σl|)
N
i=1
p (l|xi, Θg
) −
1
2
N
i=1
p (l|xi, Θg
) tr Σ−1
l
Nl,i
Where Nl,i = (xi − µl) (xi − µl)T
.
117 / 126

277. Images/
Now, if we derive with respect to Σl
First, we rewrite equation 55
N
i=1
M
l=1
−
1
2
log (|Σl|) −
1
2
(xi − µl)T
Σ−1
l
(xi − µl) p (l|xi, Θg
)
=
M
l=1
−
1
2
log (|Σl|)
N
i=1
p (l|xi, Θg
) −
1
2
N
i=1
p (l|xi, Θg
) tr Σ−1
l
(xi − µl) (xi − µl)T
=
M
l=1
−
1
2
log (|Σl|)
N
i=1
p (l|xi, Θg
) −
1
2
N
i=1
p (l|xi, Θg
) tr Σ−1
l
Nl,i
Where Nl,i = (xi − µl) (xi − µl)T
.
117 / 126

278. Images/
Now, if we derive with respect to Σl
First, we rewrite equation 55
N
i=1
M
l=1
−
1
2
log (|Σl|) −
1
2
(xi − µl)T
Σ−1
l
(xi − µl) p (l|xi, Θg
)
=
M
l=1
−
1
2
log (|Σl|)
N
i=1
p (l|xi, Θg
) −
1
2
N
i=1
p (l|xi, Θg
) tr Σ−1
l
(xi − µl) (xi − µl)T
=
M
l=1
−
1
2
log (|Σl|)
N
i=1
p (l|xi, Θg
) −
1
2
N
i=1
p (l|xi, Θg
) tr Σ−1
l
Nl,i
Where Nl,i = (xi − µl) (xi − µl)T
.
117 / 126

279. Images/
Now, if we derive with respect to Σl
First, we rewrite equation 55
N
i=1
M
l=1
−
1
2
log (|Σl|) −
1
2
(xi − µl)T
Σ−1
l
(xi − µl) p (l|xi, Θg
)
=
M
l=1
−
1
2
log (|Σl|)
N
i=1
p (l|xi, Θg
) −
1
2
N
i=1
p (l|xi, Θg
) tr Σ−1
l
(xi − µl) (xi − µl)T
=
M
l=1
−
1
2
log (|Σl|)
N
i=1
p (l|xi, Θg
) −
1
2
N
i=1
p (l|xi, Θg
) tr Σ−1
l
Nl,i
Where Nl,i = (xi − µl) (xi − µl)T
.
117 / 126

280. Images/
Deriving with respect to Σ−1
l
We have that
∂
∂Σ−1
l
M
l=1
−
1
2
log (|Σl|)
N
i=1
p (l|xi, Θg
) −
1
2
N
i=1
p (l|xi, Θg
) tr Σ−1
l
Nl,i
=
1
2
N
i=1
p (l|xi, Θg
) (2Σl − diag (Σl)) −
1
2
N
i=1
p (l|xi, Θg
) 2Nl.i − diag Nl,i
=
1
2
N
i=1
p (l|xi, Θg
) 2Ml,i − diag Ml,i
=2S − diag (S)
Where Ml,i = Σl − Nl,i and S = 1
2
N
i=1
p (l|xi, Θg) Ml,i
118 / 126

281. Images/
Deriving with respect to Σ−1
l
We have that
∂
∂Σ−1
l
M
l=1
−
1
2
log (|Σl|)
N
i=1
p (l|xi, Θg
) −
1
2
N
i=1
p (l|xi, Θg
) tr Σ−1
l
Nl,i
=
1
2
N
i=1
p (l|xi, Θg
) (2Σl − diag (Σl)) −
1
2
N
i=1
p (l|xi, Θg
) 2Nl.i − diag Nl,i
=
1
2
N
i=1
p (l|xi, Θg
) 2Ml,i − diag Ml,i
=2S − diag (S)
Where Ml,i = Σl − Nl,i and S = 1
2
N
i=1
p (l|xi, Θg) Ml,i
118 / 126

282. Images/
Deriving with respect to Σ−1
l
We have that
∂
∂Σ−1
l
M
l=1
−
1
2
log (|Σl|)
N
i=1
p (l|xi, Θg
) −
1
2
N
i=1
p (l|xi, Θg
) tr Σ−1
l
Nl,i
=
1
2
N
i=1
p (l|xi, Θg
) (2Σl − diag (Σl)) −
1
2
N
i=1
p (l|xi, Θg
) 2Nl.i − diag Nl,i
=
1
2
N
i=1
p (l|xi, Θg
) 2Ml,i − diag Ml,i
=2S − diag (S)
Where Ml,i = Σl − Nl,i and S = 1
2
N
i=1
p (l|xi, Θg) Ml,i
118 / 126

283. Images/
Deriving with respect to Σ−1
l
We have that
∂
∂Σ−1
l
M
l=1
−
1
2
log (|Σl|)
N
i=1
p (l|xi, Θg
) −
1
2
N
i=1
p (l|xi, Θg
) tr Σ−1
l
Nl,i
=
1
2
N
i=1
p (l|xi, Θg
) (2Σl − diag (Σl)) −
1
2
N
i=1
p (l|xi, Θg
) 2Nl.i − diag Nl,i
=
1
2
N
i=1
p (l|xi, Θg
) 2Ml,i − diag Ml,i
=2S − diag (S)
Where Ml,i = Σl − Nl,i and S = 1
2
N
i=1
p (l|xi, Θg) Ml,i
118 / 126

284. Images/
Thus, we have
Thus
If 2S − diag (S) = 0 =⇒ S = 0
Implying
1
2
N
i=1
p (l|xi, Θg
) [Σl − Nl,i] = 0 (58)
Or
Σl =
N
i=1 p (l|xi, Θg) Nl,i
N
i=1 p (l|xi, Θg)
=
N
i=1 p (l|xi, Θg) (xi − µl) (xi − µl)T
N
i=1 p (l|xi, Θg)
(59)
119 / 126

285. Images/
Thus, we have
Thus
If 2S − diag (S) = 0 =⇒ S = 0
Implying
1
2
N
i=1
p (l|xi, Θg
) [Σl − Nl,i] = 0 (58)
Or
Σl =
N
i=1 p (l|xi, Θg) Nl,i
N
i=1 p (l|xi, Θg)
=
N
i=1 p (l|xi, Θg) (xi − µl) (xi − µl)T
N
i=1 p (l|xi, Θg)
(59)
119 / 126

286. Images/
Thus, we have
Thus
If 2S − diag (S) = 0 =⇒ S = 0
Implying
1
2
N
i=1
p (l|xi, Θg
) [Σl − Nl,i] = 0 (58)
Or
Σl =
N
i=1 p (l|xi, Θg) Nl,i
N
i=1 p (l|xi, Θg)
=
N
i=1 p (l|xi, Θg) (xi − µl) (xi − µl)T
N
i=1 p (l|xi, Θg)
(59)
119 / 126

287. Images/
Thus, we have the iterative updates
They are
αNew
l =
1
N
N
i=1
p (l|xi, Θg
)
µNew
l =
N
i=1 xip (l|xi, Θg)
N
i=1 p (l|xi, Θg)
ΣNew
l =
N
i=1 p (l|xi, Θg) (xi − µl) (xi − µl)T
N
i=1 p (l|xi, Θg)
120 / 126

288. Images/
Thus, we have the iterative updates
They are
αNew
l =
1
N
N
i=1
p (l|xi, Θg
)
µNew
l =
N
i=1 xip (l|xi, Θg)
N
i=1 p (l|xi, Θg)
ΣNew
l =
N
i=1 p (l|xi, Θg) (xi − µl) (xi − µl)T
N
i=1 p (l|xi, Θg)
120 / 126

289. Images/
Thus, we have the iterative updates
They are
αNew
l =
1
N
N
i=1
p (l|xi, Θg
)
µNew
l =
N
i=1 xip (l|xi, Θg)
N
i=1 p (l|xi, Θg)
ΣNew
l =
N
i=1 p (l|xi, Θg) (xi − µl) (xi − µl)T
N
i=1 p (l|xi, Θg)
120 / 126

290. Images/
Outline
1 Introduction
Maximum-Likelihood
Expectation Maximization
Examples of Applications of EM
2 Incomplete Data
Introduction
Using the Expected Value
Analogy
3 Derivation of the EM-Algorithm
Hidden Features
Proving Concavity
Using the Concave Functions for Approximation
From The Concave Function to the EM
The Final Algorithm
Notes and Convergence of EM
4 Finding Maximum Likelihood Mixture Densities
The Beginning of The Process
Bayes’ Rule for the components
Mixing Parameters
Maximizing Q using Lagrange Multipliers
Lagrange Multipliers
In Our Case
Example on Mixture of Gaussian Distributions
The EM Algorithm
121 / 126

291. Images/
EM Algorithm for Gaussian Mixtures
Step 1
Initialize:
The means µl
Covariances Σl
Mixing coefficients αl
122 / 126

292. Images/
Evaluate
Step 2 - E-Step
Evaluate the the probabilities of component l given xi using the
current parameter values:
p (l|xi, Θg
) =
αg
l pyi xi|θg
l
M
k=1 αg
kpk xi|θg
k
123 / 126

293. Images/
Now
Step 3 - M-Step
Re-estimate the parameters using the current iteration values:
αNew
l =
1
N
N
i=1
p (l|xi, Θg
)
µNew
l =
N
i=1 xip (l|xi, Θg)
N
i=1 p (l|xi, Θg)
ΣNew
l =
N
i=1 p (l|xi, Θg) (xi − µl) (xi − µl)T
N
i=1 p (l|xi, Θg)
124 / 126

294. Images/
Now
Step 3 - M-Step
Re-estimate the parameters using the current iteration values:
αNew
l =
1
N
N
i=1
p (l|xi, Θg
)
µNew
l =
N
i=1 xip (l|xi, Θg)
N
i=1 p (l|xi, Θg)
ΣNew
l =
N
i=1 p (l|xi, Θg) (xi − µl) (xi − µl)T
N
i=1 p (l|xi, Θg)
124 / 126

295. Images/
Now
Step 3 - M-Step
Re-estimate the parameters using the current iteration values:
αNew
l =
1
N
N
i=1
p (l|xi, Θg
)
µNew
l =
N
i=1 xip (l|xi, Θg)
N
i=1 p (l|xi, Θg)
ΣNew
l =
N
i=1 p (l|xi, Θg) (xi − µl) (xi − µl)T
N
i=1 p (l|xi, Θg)
124 / 126

296. Images/
Evaluate
Step 4
Evaluate the log likelihood:
log p (X|µ, Σ, α) =
N
i=1
log
M
l=1
αNew
l pl xi|µNew
l , Σl
New
Step 6
Check for convergence of either the parameters or the log likelihood.
If the convergence criterion is not satisfied return to step 2.
125 / 126

297. Images/
Evaluate
Step 4
Evaluate the log likelihood:
log p (X|µ, Σ, α) =
N
i=1
log
M
l=1
αNew
l pl xi|µNew
l , Σl
New
Step 6
Check for convergence of either the parameters or the log likelihood.
If the convergence criterion is not satisfied return to step 2.
125 / 126

298. Images/
References I
S. Borman, “The expectation maximization algorithm-a short
tutorial,” Submitted for publication, pp. 1–9, 2004.
J. Bilmes, “A gentle tutorial of the EM algorithm and its application
to parameter estimation for Gaussian mixture and hidden Markov
models,” International Computer Science Institute, vol. 4, 1998.
F. Dellaert, “The expectation maximization algorithm,” tech. rep.,
Georgia Institute of Technology, 2002.
G. McLachlan and T. Krishnan, The EM algorithm and extensions,
vol. 382.
John Wiley & Sons, 2007.
126 / 126