machine learning: a statistics and optimization perspective · machine learning machine learning...

Machine Learning: A Statistics andOptimization Perspective

Nan Ye

Mathematical Sciences SchoolQueensland University of Technology

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What is Machine Learning?

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Machine Learning

• Machine learning turns data into insight, predictions and/ordecisions.

• Numerous applications in diverse areas, including natural languageprocessing, computer vision, recommender systems, medicaldiagnosis.

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A Much Sought-after Technology

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Enabled Applications

Make reminders by talking to your phone.

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Tell the car where you want to go to and the car takes you there.

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Check your emails, see some spams in your inbox, mark them as spams,and similar spams will not show up.

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Video recommendations.

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Play Go against the computer.

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Tutorial Objective

Essentials for crafting basic machine learning systems.

Formulate applications as machine learning problems

Classification, regression, density estimation, clustering...

Understand and apply basic learning algorithms

Least squares regression, logistic regression, support vectormachines, K-means,...

Theoretical understanding

Position and compare the problems and algorithms in a unifyingstatistical framework

Have fun...

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Outline

• A statistics and optimization perspective

• Statistical learning theory

• Regression

• Model selection

• Classification

• Clustering

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Hands-on

• An exercise on using WEKA.

WEKA Java http://www.cs.waikato.ac.nz/ml/weka/H2O Java http://www.h2o.ai/

scikit-learn python http://scikit-learn.org/CRAN R https://cran.r-project.org/web/views/MachineLearning.html

• Some technical details are left as exercises. These are tagged with(verify).

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A Statistics and OptimizationPerspective

Illustrations

• Learning a binomial distribution

• Learning a Gaussian distribution

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Learning a Binomial Distribution

I pick a coin with the probability of heads being θ. I flip it 100 times foryou and you see a dataset D of 70 heads and 30 tails, can you learn θ?

Maximum likelihood estimationThe likelihood of θ is

P(D | θ) = θ70(1− θ)30.

Learning θ is an optimization problem.

θml = arg maxθ

P(D | θ)

= arg maxθ

lnP(D | θ)

= arg maxθ

(70 ln θ + 30 ln(1− θ)).

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Learning a Binomial Distribution

I pick a coin with the probability of heads being θ. I flip it 100 times foryou and you see a dataset D of 70 heads and 30 tails, can you learn θ?

Maximum likelihood estimationThe likelihood of θ is

P(D | θ) = θ70(1− θ)30.

Learning θ is an optimization problem.

θml = arg maxθ

P(D | θ)

= arg maxθ

lnP(D | θ)

= arg maxθ

(70 ln θ + 30 ln(1− θ)).

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θml = arg maxθ

(70 ln θ + 30 ln(1− θ)).

Set derivative of log-likelihood to 0,

θ− 30

1− θ= 0,

we have

θml = 70/(70 + 30).

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Learning a Gaussian distribution

I pick a Gaussian N(µ, σ2) andgive you a bunch of data D ={x1, . . . , xn} independently drawnfrom it. Can you learn µ and σ.

f (X )

P(x | µ, σ) =1

2πe−

(x−µ)2

2σ2 .

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Learning a Gaussian distribution

I pick a Gaussian N(µ, σ2) andgive you a bunch of data D ={x1, . . . , xn} independently drawnfrom it. Can you learn µ and σ.

f (X )

P(x | µ, σ) =1

2πe−

(x−µ)2

2σ2 .

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Maximum likelihood estimation

lnP(D | µ, σ)

(−∑i

(xi − µ)2

= −n ln(σ√

2π)−∑i

(xi − µ)2

Set derivative w.r.t. µ to 0,

−∑i

xi − µσ2

= 0 ⇒ µml =1

Set derivative w.r.t. σ to 0,

(xi − µ)2

σ3= 0 ⇒ σ2ml =

n∑i=1

(xi − µml)2.

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What You Need to Know...

Learning is...

• Collect some data, e.g. coin flips.

• Choose a hypothesis class, e.g. binomial distribution.

• Choose a loss function, e.g. negative log-likelihood.

• Choose an optimization procedure, e.g. set derivative to 0.

• Have fun...

Statistics and optimization provide powerful tools for formulating andsolving machine learning problems.

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Statistical Learning Theory• The framework

• Applications in classification, regression, and density estimation

• Does empirical risk minimization work?

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There is nothing more practical than a good theory.Kurt Lewin

...at least in the problems of statistical inference.Vladimir Vapnik

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Learning...

• H. Simon: Any process by which a system improves itsperformance.

• M. Minsky: Learning is making useful changes in our minds.

• R. Michalsky: Learning is constructing or modifying representationsof what is being experienced.

• L. Valiant: Learning is the process of knowledge acquisition in theabsence of explicit programming.

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A Probabilistic Framework

Training examples z1, . . . , zn are i.i.d. drawn from a fixed butunknown distribution P(Z ) on Z.

e.g. outcomes of coin flips.

Hypothesis space H

e.g. head probability θ ∈ [0, 1].

Loss function

L(z , h) measures the penalty for hypothesis h on example z .

e.g. log-loss L(z , θ) = − lnP(z | θ) =

{− ln(θ), z = H,

− ln(1− θ), z = T .

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Expected risk

• The expected risk of h is E(L(Z , h)).

• We want to find the hypothesis with minimum expected risk,

arg minh∈H

E(L(Z , h)).

Empirical risk minimization (ERM)

Minimize empirical risk Rn(h)def= 1

∑i L(zi , h) over h ∈ H.

e.g. choose θ to minimize −70 ln θ − 30 ln(1− θ).

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This provides a unified formulation for many machine learning problems,which differ in

• the data domain Z,

• the choice of the hypothesis space H, and

• the choice of loss function L.

Most algorithms that we see later can be seen as special cases of ERM.

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Classificationpredict a discrete class

Digit recognition: image to {0, 1, . . . , 9}.

Spam filter: email to {spam, not spam}.

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Given D = {(x1, y1), . . . , (xn, yn)} ⊆ X × Y, find a classifier f thatmaps an input x ∈ X to a class y ∈ Y.

We usually use the 0/1 loss

L((x , y), h) = I(h(x) = y) =

{1, h(x) 6= y ,

0, h(x) = y ..

ERM chooses the classifier with minimum classification error

minh∈H

I(h(xi ) = yi ).

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Regressionpredict a numerical value

Stock market prediction: predict stock price using recent trading data

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Given D = {(x1, y1), . . . , (xn, yn)} ⊆ X × R, find a function f thatmaps an input x ∈ X to a value y ∈ R.

We usually use the quadratic loss

L((x , y), h) = (y − h(x))2.

ERM is often called the method of least squares in this case

minh∈H

(yi − h(xi ))2.

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Density Estimation

E.g. learning a binomial distribution, or a Gaussian distribution.

f (X )

We often use the log-loss

L(x , h) = − ln p(x | h).

ERM is MLE in this case.

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Does ERM Work?

Estimation error

• How does the empirically best hypothesis hn = arg minh∈H Rn(h)compare with the best in the hypothesis space? Specifically, howlarge is the estimation error R(hn)− infh∈H R(h)?

• Consistency: Does R(hn) converge to infh∈H R(h) as n→∞?

If |H| is finite, ERM is likely to pick the function with minimal expectedrisk when n is large, because then Rn(h) is close to R(h) for all h ∈ H.

If |H| is infinite, we can still show that ERM is likely to choose anear-optimal hypothesis if H has finite complexity (such asVC-dimension).

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Does ERM Work?

Estimation error

• How does the empirically best hypothesis hn = arg minh∈H Rn(h)compare with the best in the hypothesis space? Specifically, howlarge is the estimation error R(hn)− infh∈H R(h)?

• Consistency: Does R(hn) converge to infh∈H R(h) as n→∞?

If |H| is finite, ERM is likely to pick the function with minimal expectedrisk when n is large, because then Rn(h) is close to R(h) for all h ∈ H.

If |H| is infinite, we can still show that ERM is likely to choose anear-optimal hypothesis if H has finite complexity (such asVC-dimension).

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Approximation error

How good is the best hypothesis in H? That is, how large is theapproximation error infh∈H R(h)− infh R(h)?

Trade-off between estimation error and approximation error:

• Larger hypothesis space implies smaller approximation error, butlarger estimation error.

• Smaller hypothesis space implies larger approximation error, butsmaller estimation error.

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Approximation error

How good is the best hypothesis in H? That is, how large is theapproximation error infh∈H R(h)− infh R(h)?

Trade-off between estimation error and approximation error:

• Larger hypothesis space implies smaller approximation error, butlarger estimation error.

• Smaller hypothesis space implies larger approximation error, butsmaller estimation error.

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Optimization error

Is the optimization algorithm computing the empirically besthypothesis exact?

While ERM can be efficiently implemented in many cases, there are alsocomputationally intractable cases, and efficient approximations aresought. The performance gap between the sub-optimal hypothesis andthe empirically best hypothesis is the optimization error.

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Optimization error

Is the optimization algorithm computing the empirically besthypothesis exact?

While ERM can be efficiently implemented in many cases, there are alsocomputationally intractable cases, and efficient approximations aresought. The performance gap between the sub-optimal hypothesis andthe empirically best hypothesis is the optimization error.

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Recognise machine learning problems as special cases of the generalstatistical learning problem.

Understand that the performance of ERM depends on theapproximation error, estimation error and optimization error.

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Regression• Ordinary least squares

• Ridge regression

• Basis function method

• Regression function

• Nearest neighbor regression

• Kernel regression

• Classification as regression

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Ordinary Least Squares

Find a best fitting hyperplane for (x1, y1), . . . , (xn, yn) ∈ Rd × R.

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OLS finds a hyperplane minimizing the sum of squared errors

βn = arg minβ∈Rd

n∑i=1

(xTi β − yi )2.

A special case of function learning using ERM

• The input set is X = Rd , and the output set is Y = R.

• The hypothesis space are hyperplanes H = {xTβ : β ∈ Rd}.• Quadratic loss is used, as typically in regression.

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Empirically best hyperplane. The solution to OLS is

βn = (XTX)−1XTy,

where X is the n × d matrix with xi as the i-th row, andy = (y1, . . . , yn)T .

The formula holds when XTX is non-singular. When XTX is singular, there are

infinitely many possible values for βn. They can be obtained by solving the linear

systems (XTX)β = XTy.

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Proof. The empirical risk is (ignoring a factor of 1n )

Rn(β) =n∑

(xTi β − yi )2 = ||Xβ − y||22.

Set the gradient of Rn to 0

∇Rn = 2XT (Xβ − y) = 0, (verify)

we have

βn = (XTX)−1XTy.

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Optimal hyperplane. The hyperplane β∗ = E(XXT )−1 E(XY )minimizes the expected quadratic loss among all hyperplanes.

Proof. The expected quadratic loss of a hyperplane β is

R(β) = E((βTX − Y )2)

= E(βTXXTβ − 2βTXY + Y 2)

= βT E(XXT )β − 2βT E(XY ) + E (Y 2).

Set the gradient of R to 0, we have

∇R(β) = 2E(XXT )β − 2E(XY ) = 0 ⇒ β∗ = E(XXT )−1 E(XY ).

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Optimal hyperplane. The hyperplane β∗ = E(XXT )−1 E(XY )minimizes the expected quadratic loss among all hyperplanes.

Proof. The expected quadratic loss of a hyperplane β is

R(β) = E((βTX − Y )2)

= E(βTXXTβ − 2βTXY + Y 2)

= βT E(XXT )β − 2βT E(XY ) + E (Y 2).

Set the gradient of R to 0, we have

∇R(β) = 2E(XXT )β − 2E(XY ) = 0 ⇒ β∗ = E(XXT )−1 E(XY ).

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Consistency. We can show that least squares linear regression is

consistent, that is, R(βn)P→ R(β∗), by using the law of large numbers.

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Least Squares as MLE

• Consider the class of conditional distributions {pβ(Y |X ) : β ∈ Rd},where

pβ(Y | X = x) = N(Y ; xTβ, σ)def=

1√2πσ

e−(Y−xTβ)2/2σ2,

with σ being a constant.

• The (conditional) likelihood of β is

Ln(β) = pβ(y1 | x1) . . . pβ(yn | xn).

• Maximizing the likelihood Ln(β) gives the same βn as given by themethod of least squares. (verify)

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Ridge Regression

When collinearity is present, the matrix XTX may be singular or closeto singular, making the solution unreliable.

Ridge regression

We add a regularizer λ||β||22 to OLS objective, where λ > 0 is afixed constant.

( n∑i=1

(xTi β − yi )2 +

quadratic/`2 regularizer︷︸︸︷λ||β||22

Empirically optimal hyperplane

βn = (λI + XTX)−1XTy. (verify)

The matrix λI + XTX is non-singular (verify), and thus there isalways a unique solution.

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Regression with a Bias

So far we have only considered hyperplanes of the form y = xTβ, whichpasses through the origin (green line).

Considering hyperplanes with a bias term, that is, hyperplanes of theform y = xTβ + b is more useful (red line).

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OLS with a bias solves

(bn, βn) = arg minb∈R,β∈Rd

n∑i=1

(xTi β +

bias︷︸︸︷b − yi )

Solution. Reduce it to regression without a bias term by replacing

xTβ + b with (1 xT )

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OLS with a bias solves

n∑i=1

(xTi β +

bias︷︸︸︷b − yi )

Solution. Reduce it to regression without a bias term by replacing

xTβ + b with (1 xT )

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Ridge regression with a bias solves

( n∑i=1

(xTi β + b − yi )2 + λ||β||22

Solution. Reduce it to ridge regression without a bias term as follows.Let xi = xi − x, and yi = yi − y , where x =

∑ni=1 xi/n, and

y =∑n

i=1 yi/n, then

( n∑i=1

(xTi β − yi )2 + λ||β||22

bn = y − xTβn. (verify)

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Ridge regression with a bias solves

( n∑i=1

(xTi β + b − yi )2 + λ||β||22

Solution. Reduce it to ridge regression without a bias term as follows.Let xi = xi − x, and yi = yi − y , where x =

∑ni=1 xi/n, and

y =∑n

i=1 yi/n, then

( n∑i=1

(xTi β − yi )2 + λ||β||22

bn = y − xTβn. (verify)

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Basis Function Method

We can use linear regression to learn complex regression functions

• Choose some basis functions g1, . . . , gk : Rd → R.

• Transform each input x to (g1(x), . . . , gk(x)).

• Perform linear regression on the transformed data.

Examples

• Linear regression: use basis functions g1, . . . , gd with gi (x) = xi ,and g0(x) = 1.

• Quadratic functions: use basis functions of the above form,together with basis functions of the form gij(x) = xixj for all1 ≤ i ≤ j ≤ d .

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Basis Function Method

We can use linear regression to learn complex regression functions

• Choose some basis functions g1, . . . , gk : Rd → R.

• Transform each input x to (g1(x), . . . , gk(x)).

• Perform linear regression on the transformed data.

Examples

• Linear regression: use basis functions g1, . . . , gd with gi (x) = xi ,and g0(x) = 1.

• Quadratic functions: use basis functions of the above form,together with basis functions of the form gij(x) = xixj for all1 ≤ i ≤ j ≤ d .

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Regression Function

The minimizer of the expected quadratic loss is the regression function

h∗(x) = E(Y | x).

Proof. The expected quadratic loss of a function h is

E((h(X )− Y )2) = EX

(h(X )2 − 2h(X )E(Y | X ) + E(Y 2 | X )

Hence we can set the value of h(x) independently for each x bychoosing it to minimize the expression under expectation. This leads toh∗(x) = E(Y | x).

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Regression Function

The minimizer of the expected quadratic loss is the regression function

h∗(x) = E(Y | x).

Proof. The expected quadratic loss of a function h is

E((h(X )− Y )2) = EX

(h(X )2 − 2h(X )E(Y | X ) + E(Y 2 | X )

Hence we can set the value of h(x) independently for each x bychoosing it to minimize the expression under expectation. This leads toh∗(x) = E(Y | x).

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k nearest neighbor (kNN)

kNN approximates the regression function using

hn(x) = avg(yi | xi ∈ Nk(x)),

which is the average of the values of the set Nk(x) of the k nearestneighbors of x in the training data.

• Under mild conditions, as k →∞ and n/k →∞, hn(x)→ h∗(x),for any distribution P(X ,Y ).

• (Curse of dimensionality) The number of samples required foraccurate approximation is exponential in the dimension.

• kNN is a non-parametric method, while linear regression is aparametric method.

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k nearest neighbor (kNN)

kNN approximates the regression function using

hn(x) = avg(yi | xi ∈ Nk(x)),

which is the average of the values of the set Nk(x) of the k nearestneighbors of x in the training data.

• kNN is a non-parametric method, while linear regression is aparametric method.

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Kernel Regression

hn(x) =n∑

K (x, xi )yi/ n∑

K (x, xi ),

where K (x, x′) is a function measuring the similarity between x and x′,and is often called a kernel function.

Example kernel functions

• Gaussian kernel Kλ(x, x′) = 1λ exp(− ||x

′−x||222λ ).

• kNN kernel Kk(x, x′) = I(||x′ − x|| ≤ maxx′′∈Nk (x) ||x′′ − x||). Note

that this kernel is data-dependent and non-symmetric.

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Kernel Regression

hn(x) =n∑

K (x, xi )yi/ n∑

K (x, xi ),

where K (x, x′) is a function measuring the similarity between x and x′,and is often called a kernel function.

Example kernel functions

• Gaussian kernel Kλ(x, x′) = 1λ exp(− ||x

′−x||222λ ).

• kNN kernel Kk(x, x′) = I(||x′ − x|| ≤ maxx′′∈Nk (x) ||x′′ − x||). Note

that this kernel is data-dependent and non-symmetric.

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Binary Classification as Regression

• Label one class as −1 and and the other as +1.

• Fit a function f (x) using least squares regression.

• Given a test example x, predict −1 if f (x) < 0 and +1 otherwise.

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• Regression function.

• Parametric methods: Ordinary least squares, ridge regression, basisfunction method.

• Non-parametric methods: kNN, kernel regression.

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Model Selection

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Bias-Variance Tradeoff

The predicted value Y ′ at a fixed point x can be considered a randomfunction of the training set. Let Y be the true value at x. The expectedprediction error E((Y ′ − Y )2) is a property of the model class.

Bias-variance decomposition

expected prediction error︷︸︸︷E((Y ′ − Y )2

variance︷︸︸︷E((Y ′ − E(Y ′))2

bias (squared)︷︸︸︷(E(Y ′)− E(Y )

irreducible noise︷︸︸︷E((Y − E(Y ))2

Proof. Expand the RHS and simplify.

Bias-variance tradeoff

In general, as model complexity increases (i.e., the hypothesisbecomes more complex), variance tends to increase, and bias tendsto decrease.

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Bias-variance Tradeoff in kNN

Assumption

Suppose Y | X ∼ N(f (X ), σ) for some function f and some fixedσ. In addition, suppose x1, . . . , xn are fixed.

Bias and variance

At x, Y ′ = 1k

∑xi∈Nk (x)

yi is predicted and the true value is Y .

bias = E(Y ′)− E(Y ) =1

∑xi∈Nk (x)

f (xi )− f (x),

variance = E((Y ′ − E(Y ′))2) = σ2/k .

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Bias-variance Tradeoff in kNN

Assumption

Suppose Y | X ∼ N(f (X ), σ) for some function f and some fixedσ. In addition, suppose x1, . . . , xn are fixed.

Bias and variance

At x, Y ′ = 1k

∑xi∈Nk (x)

yi is predicted and the true value is Y .

bias = E(Y ′)− E(Y ) =1

∑xi∈Nk (x)

f (xi )− f (x),

variance = E((Y ′ − E(Y ′))2) = σ2/k .

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bias =1

∑xi∈Nk (x)

f (xi )− f (x),

variance = σ2/k.

1k as a complexity measure

This is because with smaller 1k , hn(x) = avg(yi | xi ∈ Nk(x)) is

closer to a constant, and thus model complexity is smaller.

Bias-variance trade-off

As 1k increases (or as model complexity increases), bias is likely to

decrease, and variance increases.

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Model Selection

Assume model complexity is controlled by some parameter θ. How topick the best value among candidates θ0, . . . , θm?

Using a development set

• Split available data into a training set T and a development set D.

• For each θi , train a model on T , and test it on D.

• Choose the parameter with best performance.

A lot of data is needed, while the amount may be limited.

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Model Selection

Assume model complexity is controlled by some parameter θ. How topick the best value among candidates θ0, . . . , θm?

Using a development set

• Split available data into a training set T and a development set D.

• For each θi , train a model on T , and test it on D.

• Choose the parameter with best performance.

A lot of data is needed, while the amount may be limited.

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K -fold cross validation

• Split the training data into K folds.

• For each θi , train K models, with each trained on K − 1 folds andtested on the remaining fold.

• Choose the parameter with best average performance.

Computationally more expensive than using a development set.

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The expected prediction error is a function of the model complexity.

The bias-variance decomposition implies that minimizing the expectedprediction error requires careful tuning of the model complexity.

Using a development set and cross-validation are two basic methods formodel selection.

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• A statistics and optimization perspective

Learning a binomial and a Gaussian

• Statistical learning theory

Data, hypotheses, loss function, expected risk, empirical risk...

• Regression

Regression function, parametric regression, non-parametricregression...

• Model selection

Bias-variance tradeoff, development set, cross-validation...

• Classification

• Clustering

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Classification• Bayes optimal classifier

• Nearest neighbor classifier

• Naive Bayes classifier

• Logistic regression

• The perceptron

• Support vector machines

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Recall

Classification

Output a function f : X → Y where Y is a finite set.

0/1 loss

L((x , y), h) = I(y 6= h(x)) =

{1, y 6= h(x),

0, y = h(x).

Expected/True risk

R(h) = E(L((X ,Y ), h).

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Bayes Optimal Classifier

The expected 0/1 loss is minimized by the Bayes optimal classifier

h∗(x) = arg maxy∈Y

P(y | x).

Proof. The expected 0/1 loss of a classifier h(x) is

E(L((X ,Y ), h)) = EX EY |X (I(Y 6= h(X )))

= EX P(Y 6= h(X ) | X ).

Hence we can set the value of h(x) independently for each x bychoosing it to minimize the expression under expectation. This leads toh∗(x) = arg maxy∈Y P(y | x).

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Bayes Optimal Classifier

The expected 0/1 loss is minimized by the Bayes optimal classifier

P(y | x).

Proof. The expected 0/1 loss of a classifier h(x) is

E(L((X ,Y ), h)) = EX EY |X (I(Y 6= h(X )))

= EX P(Y 6= h(X ) | X ).

Hence we can set the value of h(x) independently for each x bychoosing it to minimize the expression under expectation. This leads toh∗(x) = arg maxy∈Y P(y | x).

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The Bayes optimal classifier is

P(y | x).

However, P(Y | x) is unknown...

Idea. Estimate P(y | x) from data.

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Nearest Neighbor Classifier

Approximate P(y | x) using the label distribution in {yi | xi ∈ Nk(x)},where Nk(x) consists of the k nearest examples of x (with respect tosome distance measure), and predict the majority label

hn(x) = majority{yi | xi ∈ Nk(x)}.

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Nearest Neighbor Classifier

Approximate P(y | x) using the label distribution in {yi | xi ∈ Nk(x)},where Nk(x) consists of the k nearest examples of x (with respect tosome distance measure), and predict the majority label

hn(x) = majority{yi | xi ∈ Nk(x)}.

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1-NN classifier 15-NN classifier Bayes optimal classifier

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Naive Bayes Classifier (NB)

• X = X1 × . . .×Xd , where each Xi is a finite set.

• A model p(X ,Y ) satisfies the independence assumption

p(x1, . . . , xd | y) = p(x1 | y) . . . p(xd | y).

Classification

An example x = (x1, . . . , xd) is classified as

y = arg maxy ′∈Y

p(y ′ | x).

This is equivalent to

p(y ′, x) = arg maxy ′∈Y

p(y ′)p(x1 | y)...p(xd | y),

by the independence assumption.

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Naive Bayes Classifier (NB)

• X = X1 × . . .×Xd , where each Xi is a finite set.

• A model p(X ,Y ) satisfies the independence assumption

p(x1, . . . , xd | y) = p(x1 | y) . . . p(xd | y).

Classification

An example x = (x1, . . . , xd) is classified as

p(y ′ | x).

This is equivalent to

p(y ′, x) = arg maxy ′∈Y

p(y ′)p(x1 | y)...p(xd | y),

by the independence assumption.

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Learning (MLE)

The maximum likelihood Naive Bayes model is p(X ,Y ) given by

p(y) = ny/n,

p(xi | y) = ny ,xi/ny ,

where ny is the number of times class y appears in the training set,and ny ,xi is the number of times attribute i takes value xi when theclass label is y . (verify)

Issues

• Independence assumption unlikely to be satisified.

• The counts ny may be 0, making the estimates undefined.

• The counts may be very small, leading to unstable estimates.

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Learning (MLE)

The maximum likelihood Naive Bayes model is p(X ,Y ) given by

p(y) = ny/n,

p(xi | y) = ny ,xi/ny ,

where ny is the number of times class y appears in the training set,and ny ,xi is the number of times attribute i takes value xi when theclass label is y . (verify)

Issues

• Independence assumption unlikely to be satisified.

• The counts ny may be 0, making the estimates undefined.

• The counts may be very small, leading to unstable estimates.

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Laplace correction

p(y) = (ny + c0)/∑y∈Y

(ny ′ + c0),

p(xi | y) = (ny ,xi + c1)/∑x ′i ∈Xi

(ny ,x ′i + c1),

where c0 > 0 and c1 > 0 are user-chosen constants. Laplace correctionmakes NB more stable, but still relies on strong independenceassumption.

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Logistic Regression (LR)

• X = Rd .

• Logistic regression estimates conditional distributions of the form

p(y | x, θ) = exp(xT θy )/∑

y ′∈Yexp(xT θy ′),

where θy = (θy1, . . . , θyd) ∈ Rd , and θ is the concatenation of θy ’s.

Classification

An example x is classified as

p(y ′ | x, θ).

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Logistic Regression (LR)

• X = Rd .

• Logistic regression estimates conditional distributions of the form

p(y | x, θ) = exp(xT θy )/∑

y ′∈Yexp(xT θy ′),

where θy = (θy1, . . . , θyd) ∈ Rd , and θ is the concatenation of θy ’s.

Classification

An example x is classified as

p(y ′ | x, θ).

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Learning

• Training is often done by maximizing regularized log-likelihood

L(θ) = logn∏

p(yi | xi , θ)− λ||θ||22.

That is, the parameter estimate is

θn = arg maxθ

L(θ).

• L(θ) is a concave function, and can be optimized using standardnumerical methods (such as L-BFGS).

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Comparing kNN, NB and LR

All approximates the Bayes-optimal classifier

• Learn an approximation P(X ,Y ) to P(X ,Y ) (or learn anapproximation P(Y | X ) to P(Y | X )).

• Choose the function hn(x) = arg maxy P(y | x).

kNN and logistic regression estimate P(Y | X ), while naive Bayesestimates P(X ,Y ).

kNN is a non-parametric method, while naive Bayes and logisticregression are both parametric methods.

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Application in Digit Recognition

Assume each digit image is a binary image, represented as a vector withthe pixel values as its elements.

Applying the algorithms

• kNN: use the Euclidean distance between the examples.

• NB can be applied because each example is a discrete vector.

• LR can be applied because each example is a continuous vector.

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The Perceptron

x0 = 1x1

xTw y = sgn(xTw)

A perceptron maps an input x ∈ Rd+1 to

h(x) = sgn(xTw) =

1, xTw > 0,

0, xTw = 0,

−1, xTw < 0.

Here x = (1, x1, . . . , xd) includes a dummy variable 1.75 / 109

−−

A perceptron corresponds to a linear decision boundary (i.e., theboundary between the regions for examples of the same class).

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It is NP-hard to minimize the empirical 0/1 loss of a perceptron, that is,given a training set {(x1, y1), . . . , (xn, yn)}, it is NP-hard to solve

nI(sgn(xTi w) 6= yi )

Idea: Can we use (xTi w − yi )2 as a surrogate loss for the 0/1 loss?

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Least Squares May Fail

Recall: Binary classification as regression

Least squares fitting may not find a separating hyperplane (i.e., ahyperplane which puts the positive and negative examples ondifferent sides of it) even there is one.

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Least Squares May Fail

Recall: Binary classification as regression

Least squares fitting may not find a separating hyperplane (i.e., ahyperplane which puts the positive and negative examples ondifferent sides of it) even there is one.

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The decision boundary learned using least squares fitting (red line)wrongly classifies the negative example, while there exists separatinghyperplanes (like the blue line).

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Perceptron Algorithm

Require: (x1, y1), . . . , (xn, yn) ∈ Rd+1 × {−1,+1}, η ∈ (0, 1].Ensure: Weight vector w.

Randomly or smartly initialize w.while there is any misclassified example do

Pick a misclassified example (xi , yi ).w← w + ηyixi .

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Why the update rule w← w + ηyixi?

• w classifies (xi , yi ) correctly if and only if yixTi w > 0.

• If w classifies (xi , yi ) wrongly, then the update rule moves yixTi w

towards positive, because

yixTi (w + ηyixi ) = yix

Ti w + η||xi ||2 > yix

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Perceptron convergence theorem

If the training data is linearly separable (i.e., there exists some w∗

such that yixTi w∗ > 0 for all i), then the perceptron algorithm

terminates with all training examples correctly classified.

Proof. Suppose w∗ separates the data. We can scale w∗ such that |xTi w∗| ≥ ||xi ||22.

If w classifies (xi , yi ) wrongly, then it will be updated to w′ = w + ηyixi .

We show that ||w∗ − w′||22 ≤ ||w∗ − w||22 − ηR2, where R = mini ||xi ||2. This impliesthat only finitely many updates is possible.

The inequality can be shown as follows.

||w∗ − w′||22 = ||w∗ − w − ηyixi ||22= ||w∗ − w||22 − 2ηyix

Ti (w∗ − w) + η2y 2

i ||xi ||22≤ ||w∗ − w||22 − 2η(|xT

i w∗|+ |xTi w|) + η||xi ||22

≤ ||w∗ − w||22 − 2η|xTi w∗|+ η||xi ||22

≤ ||w∗ − w||22 − η|xTi w∗|

≤ ||w∗ − w||22 − ηR2.

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Perceptron convergence theorem

If the training data is linearly separable (i.e., there exists some w∗

such that yixTi w∗ > 0 for all i), then the perceptron algorithm

terminates with all training examples correctly classified.

Proof. Suppose w∗ separates the data. We can scale w∗ such that |xTi w∗| ≥ ||xi ||22.

If w classifies (xi , yi ) wrongly, then it will be updated to w′ = w + ηyixi .

We show that ||w∗ − w′||22 ≤ ||w∗ − w||22 − ηR2, where R = mini ||xi ||2. This impliesthat only finitely many updates is possible.

The inequality can be shown as follows.

||w∗ − w′||22 = ||w∗ − w − ηyixi ||22= ||w∗ − w||22 − 2ηyix

Ti (w∗ − w) + η2y 2

i ||xi ||22≤ ||w∗ − w||22 − 2η(|xT

i w∗|+ |xTi w|) + η||xi ||22

≤ ||w∗ − w||22 − 2η|xTi w∗|+ η||xi ||22

≤ ||w∗ − w||22 − η|xTi w∗|

≤ ||w∗ − w||22 − ηR2.

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Issues

• When the data is separable, the hyperplane found by theperceptron algorithm depends on the initial weight and is thusarbitrary.

• Convergence can be very slow, especially when the gap betweenthe positive and negative examples is small.

• When the data is not separable, the algorithm does not stop, butthis can be difficult to detect.

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Support Vector Machines (SVMs)Separable data

wTx + w0 = 0

yi (wT xi+w0)||w||2−

(xi , yi )

Geometric intuitionFind a separating hyperplane with maximalmargin (i.e., the minimum distance from thepoints to it).

Algebraic formulationmaxM,w,w0

subject to yi (wT xi+w0)||w||2

≥ M, i = 1, . . . , n.

Equivalent formulation (add M||w||2 = 1). minw,w0

12||w||22

subject to yi (wT xi + w0) ≥ 1, i = 1, . . . , n.

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wTx + w0 = 0

(xi , yi )

≥ M, i = 1, . . . , n.

12||w||22

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wTx + w0 = 0

(xi , yi )

≥ M, i = 1, . . . , n.

12||w||22

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Soft-margin SVMsNon-separable data

Algebraic formulation

minw,w0,ξ1,...,ξn

2||w||22 + C

subject to yi (wTxi + w0) ≥ 1− ξi , i = 1, . . . , n,

ξi ≥ 0, i = 1, . . . , n.

• C > 0 is a user chosen constant.

• Introducing ξi allows (xi , yi ) to be misclassified with a penalty ofCξi in the original objective function 1

2 ||w||22.

An SVM always have a unique solution that can be found efficiently.

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SVM as minimizing regularized hinge loss

Soft-margin SVMs can be equivalently written as

minw,w0

2C||w||22 +

max(0, 1− yi (wTxi + w0)),

where max(0, 1− y(wTx + w0)) is the hinge loss

Lhinge((x, y), h) = max(0, 1− yh(x))

of the classifier h(x) = wTx + w0, and upper bounds the 0/1 loss

L0/1((x, y), h) = I(y 6= sgn(h(x)))

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• Bayes optimal classifier.

• Probabilistic classifiers: NN, NB and LR.

• Perceptrons and SVMs.

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Clustering

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Clustering

Clustering is unsupervised function learning which learns a functionfrom X to [K ] = {1, . . . ,K}. The objective generally depends on somesimilarity/distance measure between the items, so that similar items aregrouped together.

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K -means Clustering

Given observations x1, . . . , xn, find a K -clustering f (a surjective from[n] to [K ]) to minimize the cost

n∑i=1

||xi − cf (i)||2,

where ck is the centroid of cluster k , that is, the average of xi ’s withf (i) = k .

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K -means algorithm

Randomly initialize c1:K , and set each f (i) = 0.repeat

(Assignment) Set each f (i) to be the index of the cj closest to xi .(Update) Set each ci as the centroid of cluster i given by f .

until f does not change

Initialization

• Forgy method: randomly choose K observations as centroids, andinitialize f as in the assignment step.

• Random Partition: assign a random cluster to each example.

Random partition is preferable.

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K -means algorithm

Randomly initialize c1:K , and set each f (i) = 0.repeat

(Assignment) Set each f (i) to be the index of the cj closest to xi .(Update) Set each ci as the centroid of cluster i given by f .

until f does not change

Initialization

• Forgy method: randomly choose K observations as centroids, andinitialize f as in the assignment step.

• Random Partition: assign a random cluster to each example.

Random partition is preferable.

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Illustration

Initialize centroids

Generated using http://www.naftaliharris.com/blog/visualizing-k-means-clustering/

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Illustration

Iter 1: Assignment

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Illustration

Iter 1: Update

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Illustration

Iter 2: Assignment

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Illustration

Iter 2: Update (converged)

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K -means as Coordinate Descent

K -means is a coordinate descent algorithm for the cost function

C (f , c1:K ) =n∑

||xi − cf (i)||2.

Specifically, we have (verify)

(Assignment) f ← arg minf ′ C (f ′, c1:K ), where f ′ is a K -clustering.

(Update) c1:K ← arg minc′1:KC (f , c′1:K ),

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Convergence

• The cost decreases at each iteration before termination.

• The cost converges to a local minimum by the monotoneconvergence theorem.

• Convergence may be very slow, taking exponential time in somecases, but such cases do not seem to arise in practice.

Dealing with poor local minimum

• Restart multiple times and pick the minimum cost clustering found.

K -means gives hard clusterings. Can we give probabilistic assignmentsto clusters?

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Convergence

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Convergence

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Soft Clustering with Gaussian

Mixtures

Assumption

• Each cluster is represented as a Gaussian

N(x;µk ,Σk) =1√|2πΣk |

exp(−1

2(x− µk)TΣ−1k (x− µk)

)• Cluster k has weight wk ≥ 0, with

∑Kk=1 wk = 1.

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Equivalently, we assume the probability of observing x and it is incluster z = k is

p(x, z = k | θ) = wkN(x;µk ,Σk),

where θ = {w1:K , µ1:K ,Σ1:K}.

The distribution p(x | θ) =∑

k wkN(x;µk ,Σk) is called a Gaussianmixture.

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Equivalently, we assume the probability of observing x and it is incluster z = k is

p(x, z = k | θ) = wkN(x;µk ,Σk),

where θ = {w1:K , µ1:K ,Σ1:K}.

The distribution p(x | θ) =∑

k wkN(x;µk ,Σk) is called a Gaussianmixture.

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Computing a soft clustering

p(Z = k | x, θ) =p(x,Z = k | θ)∑K

k ′=1 p(x,Z = k ′ | θ).

Learning a Gaussian mixture

Given the observations D = {x1, . . . , xn}, we choose θ bymaximizing the log-likelihood L(D | θ) =

∑ni=1 ln p(xi | θ)

L(D | θ).

This is solved using the EM algorithm.

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The EM Algorithm

Let zi be the random variable representing the cluster from which xi isdrawn from. EM starts with some initial parameter θ(0), and repeats thefollowing steps:

Expectation step: Q(θ | θ(t)) = E(∑

ln p(xi , zi | θ) | D, θ(t)),

Maximization step: θ(t+1) = arg maxθ

Q(θ | θ(t)).

In words...

• E-step: Expectation of the log-likelihood for the complete data,w.r.t. the conditional distribution p(z1:n | D, θ(t)).

• M-step: Maximization of the expectation

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The EM Algorithm

Let zi be the random variable representing the cluster from which xi isdrawn from. EM starts with some initial parameter θ(0), and repeats thefollowing steps:

Expectation step: Q(θ | θ(t)) = E(∑

ln p(xi , zi | θ) | D, θ(t)),

Maximization step: θ(t+1) = arg maxθ

Q(θ | θ(t)).

Data completion interpretation

• E-step: Create complete data (xi , zi ) with weight p(zi | xi , θ(t)),for each xi and each zi ∈ [K ].

• M-step: Perform maximum likelihood estimation on the completedata set.

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EM algorithm is iterative likelihood maximization

EM algorithm iteratively improves the likelihood function,

L(D | θ(t+1)) ≥ L(D | θ(t)).

Proof. With some algebraic manipulation, we have

Q(θ(t+1) | θ(t))− Q(θ(t) | θ(t))

= L(D | θ(t+1))− L(D | θ(t))−∑i

KL(p(Zi | xi , θ(t)) | p(Zi | xi , θ(t+1))),

where KL(q||q′) is the KL-divergence∑

x q(x) ln q(x)q′(x) .

The result follows by noting that the LHS is non-negative by the choiceof θ(t+1) in the M-step, and the nonnegativeness of the KL-divergence.

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EM algorithm is iterative likelihood maximization

EM algorithm iteratively improves the likelihood function,

L(D | θ(t+1)) ≥ L(D | θ(t)).

Proof. With some algebraic manipulation, we have

Q(θ(t+1) | θ(t))− Q(θ(t) | θ(t))

= L(D | θ(t+1))− L(D | θ(t))−∑i

KL(p(Zi | xi , θ(t)) | p(Zi | xi , θ(t+1))),

where KL(q||q′) is the KL-divergence∑

x q(x) ln q(x)q′(x) .

The result follows by noting that the LHS is non-negative by the choiceof θ(t+1) in the M-step, and the nonnegativeness of the KL-divergence.

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Q(θ(t+1) | θ(t))− Q(θ(t) | θ(t))

= E(∑

ln p(xi , zi | θ(t+1)) | D, θ(t))− E

ln p(xi , zi | θ(t)) | D, θ(t))

= E(∑

(ln p(xi | θ(t+1)) + ln p(zi | xi , θ(t+1))

− ln p(xi | θ(t))− ln p(zi | xi , θ(t) | D, θ(t)))

ln p(xi | θ(t+1))−∑i

ln p(xi | θ(t))− E(

lnp(zi | xi , θ(t))p(zi | xi , θ(t)

) | D, θ(t))

= L(D | θ(t+1))− L(D | θ(t))−∑i

KL(p(Zi | xi , θ(t)) | p(Zi | xi , θ(t+1))).

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Update Equations for Gaussian

Mixtures

Scalar covariance matrices

Assume each covariance matrix Σk is a scalar matrix σ2k Id . Let

w(t,i)k = p(zi = k | xi , θ(t)). Then given θ(t), θ(t+1) can be

computed using

w(t+1)k =

w(t,i)k /n,

µ(t+1)k =

w(t,i)k xi/

w(t,i)k ,

(σ(t+1)k )2 =

w(t,i)k

(xij − µ(t+1)kj )2/(d

w(t,i)k ).

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Data completion interpretation

• Split example xi into K complete examples (xi , 1), . . . , (xi ,K ),

where example (xi , k) has weight w(t,i)k .

• Apply maximum likelihood estimation to the complete data

• w(t+1)k is total weight of examples in cluster k .

• µ(t+1)k is the mean x value of the (weighted) examples in cluster k.

• σ(t+1)k is the standard deviation of all the attributes of the

(weighted) examples in cluster k .

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Diagonal covariance matrices

Assume each covariance matrix Σk is a diagonal matrix with

diagonal entries σ2k1, . . . , σ2kd . Let w

(t,i)k = p(zi = k | xi , θ(t)).

Then given θ(t), θ(t+1) can be computed using

w(t+1)k =

w(t,i)k /n,

µ(t+1)k =

w(t,i)k xi/

w(t,i)k ,

(σ(t+1)kj )2 =

w(t,i)k (xij − µ

(t+1)kj )2/

w(t,i)k .

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Arbitrary covariance matrices

Let w(t,i)k = p(zi = k | xi , θ(t)). Then given θ(t), θ(t+1) can be

computed using

w(t+1)k =

w(t,i)k /n,

µ(t+1)k =

w(t,i)k xi/

w(t,i)k ,

Σ(t+1)k =

w(t,i)k (xi − µ

(t+1)k )(xi − µ

(t+1)k )T/

w(t,i)k .

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• The clustering problem.

• Hard clustering with K -means algorithm.

• Soft clustering with Gaussian mixture.

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Density Estimation

• Maximum likelihood estimation.

• Naive Bayes.

• Logistic regression.

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This Tutorial...

Essentials for crafting basic machine learning systems.

Formulate applications as machine learning problems

Classification, regression, density estimation, clustering

Understand and apply basic learning algorithms

Least squares regression, logistic regression, support vectormachines, K-means,...

Theoretical understanding

Position and compare the problems and algorithms in the unifyingframework of statistical learning theory.

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Beyond This Course

• Representation: dimensionality reduction, feature selection,...

• Algorithms: decision tree, artificial neural network, Gaussianprocesses,...

• Meta-learning algorithms: boosting, stacking, bagging,...

• Learning theory: generalization performance of learning algorithms

• Many other exciting stuff...

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