iaml: linear regression · 2010. 10. 18. · i generalized linear regression i radial basis...

IAML: Linear Regression

Charles Sutton and Victor LavrenkoSchool of Informatics

Semester 1

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Overview

I The linear modelI Fitting the linear model to dataI Probabilistic interpretation of the error functionI Examples of regression problemsI Dealing with multiple outputsI Generalized linear regressionI Radial basis function (RBF) modelsI Curse of dimensionality

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The Regression Problem

I Classification and regression problems:I Classification: target of prediction is discreteI Regression: target of prediction is continuous

II Training data: Set D of pairs (xi , yi) for i = 1, . . . ,n, wherexi ∈ RD and yi ∈ R

I Today: Linear regression, i.e., relationship between x andy is linear, for example, . . .

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One dimensional exampleOne dimensional example

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What would a linear model with a two-dimensional input space(x1, x2) look like geometrically?

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Two dimensional example

Elements of Statistical Learning (2nd Ed.) c©Hastie, Tibshirani & Friedman 2009 Chap 3

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FIGURE 3.1. Linear least squares fitting withX ∈ IR2. We seek the linear function of X that mini-mizes the sum of squared residuals from Y .

Figure: Hastie, Tibshirani, and Friedman

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The Linear Model

I Linear model

f (x; w) = w0 + w1x1 + . . .+ wDxD

= wTφ(x)

where φ(x) = (1, x1, . . . , xD)T = (1,xT )T

I The maths of fitting linear models to data is easy. We usethe notation φ(x) to make generalisation easy later.

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Linear model (part 2)

In matrix notation:

I Design matrix is n × (D + 1)

Φ =

1 x11 x12 . . . x1D1 x21 x22 . . . x2D...

......

......

1 xn1 xn2 . . . xnD

I xij is the j th component of the training input xi

I Let y = (y1, . . . , yn)T

I Then y = Φw is ...?

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Linear model (part 2)

In matrix notation:

I Design matrix is n × (D + 1)

Φ =

1 x11 x12 . . . x1D1 x21 x22 . . . x2D...

......

......

1 xn1 xn2 . . . xnD

I xij is the j th component of the training input xi

I Let y = (y1, . . . , yn)T

I Then y = Φw is the model’s predicted values on traininginputs.

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Solving for Model Parameters

This looks like what we’ve seen in linear algebra

y = Φw

We know y and Φ but not w.

So why not take w = Φ−1y? (You can’t, but why?)

Three reasons:I Φ is not square. It is n × (D + 1).I The system is overconstrained (n equations for D + 1

parameters), in other wordsI The data has noise

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Solving for Model Parameters

This looks like what we’ve seen in linear algebra

y = Φw

We know y and Φ but not w.

So why not take w = Φ−1y? (You can’t, but why?)

Three reasons:I Φ is not square. It is n × (D + 1).I The system is overconstrained (n equations for D + 1

parameters), in other wordsI The data has noise

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Loss function

Want a loss function O(w) thatI We minimize wrt w.I At minimum, y looks like y.I (Recall: y depends on w)

y = Φw

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Fitting a linear model to data

I Squared error makes the maths easy

O(w) =n∑

i=1

(yi − f (xi ; w))2

=n∑

i=1

(yi −wT xi)2

= (y− Φw)T (y− Φw)

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Fitting procedure

I The error surface is a parabolic bowl

Fitting a linear model to data

� Given a dataset D of pairs (xi , yi) for i = 1, . . . , n� Squared error makes the maths easy

E(w) =n�

i=1

(yi − f (xi ; w))2

� The error surface is a parabolic bowl

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Figure: Tom Mitchell 5 / 17I Minimize O(w) = |y− Φw|2, analytical solution

w = (ΦT Φ)−1ΦT y

I (ΦT Φ)−1ΦT is the pseudo-inverse of Φ

Figure: Tom Mitchell

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Another Geometric Viewpoint

(Warning: slightly mind blowing)

Elements of Statistical Learning (2nd Ed.) c©Hastie, Tibshirani & Friedman 2009 Chap 3

x1

x2

y

y

FIGURE 3.2. The N-dimensional geometry of leastsquares regression with two predictors. The outcomevector y is orthogonally projected onto the hyperplanespanned by the input vectors x1 and x2. The projectiony represents the vector of the least squares predictions

I This figure depicts RN

I x1 and x2 are columns of the design matrixI The dashed line is the residual vector.

Figure: Hastie, Tibshirani, and Friedman

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Probabilistic interpretation of O(w)

I Assume that y = wT x + ε, where ε ∼ N(0,1)

I (This is an exact linear relationship plus Gaussian noise.)I This implies that y |xi ∼ N(f (xi ; w), σ2), i.e.

− log p(yi |xi) = log√

2π + logσ +(yi − f (xi ; w))2

2σ2

I So minimising O(w) equivalent to maximising likelihood!I Can view wT x as E [y |x].I Squared residuals allow estimation of σ2

σ2 =1n

n∑

i=1

(yi − f (xi ; w))2

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Fitting this into the general structure for learning algorithms:

I Define the task: regressionI Decide on the model structure: linear regression modelI Decide on the score function: squared error (likelihood)I Decide on optimization/search method to optimize the

score function: calculus (analytic solution)

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Examples of regression problems

I Robot inverse dynamics: predicting what torques areneeded to drive a robot arm along a given trajectory

I Electricity load forecasting, generate hourly forecasts twodays in advance (see W & F, §1.3)

I Predicting staffing requirements at help desks based onhistorical data and product and sales information,

I Predicting the time to failure of equipment based onutilization and environmental conditions

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Sensitivity to Outliers

I Linear regression is sensitive to outliersI Example: Suppose y = 0.5x + ε, where ε ∼ N(0,

√0.25),

and then add a point (2.5,3):

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Diagnositics

Graphical diagnostics can be useful for checking:I Is the relationship obviously nonlinear? Look for structure

in residuals?I Are there obvious outliers?

The goal isn’t to find all problems. You can’t. The goal is to findobvious, embarrassing problems.

Examples: Plot residuals by fitted values. Stats packages willdo this for you.

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Dealing with multiple outputs

I Suppose there are q different targets for each input xI We introduce a different wi for each target dimension, and

do regression separately for each oneI This is called multiple regression

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Basis expansion

I We can easily transform the original attributes xnon-linearly into φ(x) and do linear regression on them

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polynomial Gaussians sigmoids

Figure credit: Chris Bishop, PRML

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I Design matrix is n ×m

Φ =

φ1(x1) φ2(x1) . . . φm(x1)φ1(x2) φ2(x2) . . . φm(x2)

......

......

φ1(xn) φ2(xn) . . . φm(xn)

I Let y = (y1, . . . , yn)T

I Minimize E(w) = |y− Φw|2. As before we have ananalytical solution

w = (ΦT Φ)−1ΦT y

I (ΦT Φ)−1ΦT is the pseudo-inverse of Φ

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Example: polynomial regression

φ(x) = (1, x , x2, . . . , xM)T

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t

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Figure credit: Chris Bishop, PRML

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Radial basis function (RBF) models

I Set φi(x) = exp(−12 |x− ci |2/α2).

I Need to position these “basis functions” at some priorchosen centres ci and with a given width α. There aremany ways to do this but choosing a subset of thedatapoints as centres is one method that is quite effective

I Finding the weights is the same as ever: thepseudo-inverse solution.

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How big should M be?

I How many radial basis bumps do we need?I Suppose we only needed 3 for a 1D regression problem.

Then we might need 3D for a D dimensional problem.I Danger of curse of dimensionality (Bellman) – number of

basis functions may need to grow exponentially withdimensionality of x if all of the input space is explored

I If M ≥ n then we can fit the training data exactly; danger ofoverfitting

I Being able to adapt the basis functions in light of the datacan help→ neural networks (multilayer perceptrons)

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Summary

I There are many linear-in-the-parameters models; e.g.linear, polynomial, RBF

I Maximum likelihood solution is computationally efficient(pseudo-inverse)

I Danger of overfitting

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