CSC2515

Homework 2

1. [3pts] Robust Regression. One problem with linear regression using squared error loss

is that it can be sensitive to outliers. Another loss function we could use is the Huber loss,

parameterized by a hyperparameter δ:

Lδ(y, t) = Hδ(y − t)

Hδ(a) = (

1

2

a

2

if |a| ≤ δ

δ(|a| − 1

2

δ) if |a| > δ

(a) [1pt] Sketch the Huber loss Lδ(y, t) and squared error loss LSE(y, t) = 1

2

(y − t)

2

for

t = 0, either by hand or using a plotting library. Based on your sketch, why would you

expect the Huber loss to be more robust to outliers?

(b) [1pt] Just as with linear regression, assume a linear model:

y = w>x + b.

Give formulas for the partial derivatives ∂Lδ/∂w and ∂Lδ/∂b. (We recommend you find

a formula for the derivative H0

δ

(a), and then give your answers in terms of H0

δ

(y − t).)

(c) [1pt] Write Python code to perform (full batch mode) gradient descent on this model.

Assume the training dataset is given as a design matrix X and target vector y. Initialize

w and b to all zeros. Your code should be vectorized, i.e. you should not have a for

loop over training examples or input dimensions. You may find the function np.where

helpful.

Submit your code as q1.py.

1

https://markus.teach.cs.toronto.edu/csc2515-2019-09

1

CSC2515 Fall 2019 Homework 2

2. [5pts] Locally Weighted Regression.

(a) [2pts] Given {(x

(1), y(1)), ..,(x

(N)

, y(N)

)} and positive weights a

(1), …, a(N)

show that

the solution to the weighted least squares problem

w∗ = arg min

1

2

X

N

i=1

a

(i)

(y

(i) − wT x

(i)

)

2 +

λ

2

||w||2

(1)

is given by the formula

w∗ =

XT AX + λI

−1 XT Ay (2)

where X is the design matrix (defined in class) and A is a diagonal matrix where Aii =

a

(i)

It may help you to review Section 3.1 of the csc321 notes2

.

(b) [2pts] Locally reweighted least squares combines ideas from k-NN and linear regression.

For each new test example x we compute distance-based weights for each training example a

(i) =

exp(−||x−x

(i)

||2/2τ

2

P

)

j

exp(−||x−x(j)

||2/2τ

2)

, computes w∗ = arg min 1

2

PN

i=1 a

(i)

(y

(i) − wT x

(i)

)

2 +

λ

2

||w||2 and predicts ˆy = x

T w∗

. Complete the implementation of locally reweighted least

squares by providing the missing parts for q2.py.

Important things to notice while implementing: First, do not invert any matrix, use

a linear solver (numpy.linalg.solve is one example). Second, notice that P

exp(Ai)

j

exp(Aj ) =

P

exp(Ai−B)

j

exp(Aj−B)

but if we use B = maxj Aj it is much more numerically stable as P

exp(Ai)

j

exp(Aj )

overflows/underflows easily. This is handled automatically in the scipy package with the

scipy.misc.logsumexp function3

.

(c) [1pt] Based on our understanding of overfitting and underfitting, how would you expect

the training error and the validation error to vary as a function of τ? (I.e., what do you

expect the curves to look like?)

Now run the experiment. Randomly hold out 30% of the dataset as a validation set.

Compute the average loss for different values of τ in the range [10,1000] on both the

training set and the validation set. Plot the training and validation losses as a function

of τ (using a log scale for τ ). Was your guess correct?

3. [2pts] AdaBoost. The goal of this question is to show that the AdaBoost algorithm changes

the weights in order to force the weak learner to focus on difficult data points. Here we consider

the case that the target labels are from the set {−1, +1} and the weak learner also returns a

classifier whose outputs belongs to {−1, +1} (instead of {0, 1}). Consider the t-th iteration

of AdaBoost, where the weak learner is

ht ← argmin

h∈H

X

N

i=1

wiI{h(x

(i)

) 6= t

(i)

},

the w-weighted classification error is

errt =

PN

i=1 wiI{ht(x

(i)

) 6= t

(i)}

PN

i=1 wi

,

2

http://www.cs.toronto.edu/~rgrosse/courses/csc321_2018/readings/L02%20Linear%20Regression.pdf

3

https://docs.scipy.org/doc/scipy-0.14.0/reference/generated/scipy.misc.logsumexp.html

CSC2515 Fall 2019 Homework 2

and the classifier coefficient is αt =

1

2

log 1−errt

errt

. (Here, log denotes the natural logarithm.)

AdaBoost changes the weights of each sample depending on whether the weak learner ht

classifies it correctly or incorrectly. The updated weights for sample i is denoted by w

0

i

and is

w

0

i ← wi exp

−αtt

(i)ht(x

(i)

)

.

Show that the error w.r.t. (w

0

1

, . . . , w0

N ) is exactly 1

2

. That is, show that

err0

t =

PN

i=1 w

0

i

I{ht(x

(i)

) 6= t

(i)}

PN

i=1 w0

i

=

1

2

.

Note that here we use the weak learner of iteration t and evaluate it according to the new

weights, which will be used to learn the t + 1-st weak learner. What is the interpretation of

this result?

Tips:

• Start from err0

t and divide the summation to two sets of E = {i : ht(x

(i)

) 6= t

(i)} and its

complement Ec = {i : ht(x

(i)

) = t

(i)}.

• Note that

P

i∈E wi

PN

i=1 wi

= errt

.

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