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Date
Nov 24, 2024
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Uploaded by CoachRiverTiger30
Assignment 6: Linear Model Selection
SDS293 - Machine Learning
Due: 1 November 2017 by 11:59pm
Conceptual Exercises
6.8.2 (p. 259 ISLR)
For parts each of the following, indicate whether each method is
more or less flexible
than least
squares. Describe how each method’s trade-off between bias and variance impacts its prediction
accuracy. Justify your answers.
(a) The lasso
Solution:
Puts a budget constraint on least squares. It is therefore less flexible. The lasso
will have improved prediction accuracy when its increase in bias is less than its decrease in
variance.
(b) Ridge regression
Solution:
For the same reason as above, this method is also less flexible. Ridge regression
will have improved prediction accuracy when its increase in bias is less than its decrease in
variance.
(c) Non-linear methods (PCR and PLS)
Solution:
Non-linear methods are more flexible and will give improved prediction accuracy
when their increase in variance are less than their decrease in bias.
6.8.5 (p. 261)
Ridge regression tends to give similar coefficient values to correlated variables, whereas the lasso
may give quite different coefficient values to correlated variables. We will now explore this property
in a very simple setting.
Suppose that
n
= 2
, p
= 2
, x
11
=
x
12
, x
21
=
x
22
.
Furthermore, suppose that
y
1
+
y
2
= 0 and
x
11
+
x
21
= 0 and
x
12
+
x
22
= 0, so that the estimate for the intercept in a least squares, ridge
regression, or lasso model is zero:
ˆ
β
0
= 0.
1
(a) Write out the ridge regression optimization problem in this setting.
Solution:
In general, Ridge regression optimization looks like:
min
1
...n
X
i
(
y
i
-
ˆ
β
0
-
1
...p
X
j
ˆ
β
j
x
j
)
2
+
λ
1
...p
X
i
ˆ
β
2
i
In this case,
ˆ
β
0
= 0
and
n
=
p
= 2
. So, the optimization simplifies to:
min
h
(
y
1
-
ˆ
β
1
x
11
-
ˆ
β
2
x
12
)
2
+ (
y
2
-
ˆ
β
1
x
21
-
ˆ
β
2
x
22
)
2
+
λ
(
ˆ
β
2
1
+
ˆ
β
2
2
)
i
(b) Argue that in this setting, the ridge coefficient estimates satisfy
ˆ
β
1
=
ˆ
β
2
.
Solution:
We know the following:
x
11
=
x
12
, so we’ll call that
x
1
, and
x
21
=
x
22
, so we’ll
call that
x
2
. Plugging this into the above, we get:
min
h
(
y
1
-
ˆ
β
1
x
1
-
ˆ
β
2
x
1
)
2
+ (
y
2
-
ˆ
β
1
x
2
-
ˆ
β
2
x
2
)
2
+
λ
(
ˆ
β
2
1
+
ˆ
β
2
2
)
i
Taking the partial derivatives of the above with respect to
ˆ
β
1
and
ˆ
β
2
and setting them equal
to 0 will give us the point at which the function is minimized. Doing this, we find:
ˆ
β
1
(
x
2
1
+
x
2
2
+
λ
) +
ˆ
β
2
(
x
2
1
+
x
2
2
)
-
y
1
x
1
-
y
2
x
2
= 0
and
ˆ
β
1
(
x
2
1
+
x
2
2
) +
ˆ
β
2
(
x
2
1
+
x
2
2
+
λ
)
-
y
1
x
1
-
y
2
x
2
= 0
Since the right-hand side of both equations is identical, we can set the two left-hand sides
equal to one another:
ˆ
β
1
(
x
2
1
+
x
2
2
+
λ
) +
ˆ
β
2
(
x
2
1
+
x
2
2
)
-
y
1
x
1
-
y
2
x
2
=
ˆ
β
1
(
x
2
1
+
x
2
2
) +
ˆ
β
2
(
x
2
1
+
x
2
2
+
λ
)
-
y
1
x
1
-
y
2
x
2
and then cancel out common terms:
ˆ
β
1
(
x
2
1
+
x
2
2
+
λ
) +
ˆ
β
2
(
x
2
1
+
x
2
2
)
-
y
1
x
1
-
y
2
x
2
=
ˆ
β
1
(
x
2
1
+
x
2
2
) +
ˆ
β
2
(
x
2
1
+
x
2
2
+
λ
)
-
y
1
x
1
-
y
2
x
2
ˆ
β
1
(
x
2
1
+
x
2
2
) +
ˆ
β
1
λ
+
ˆ
β
2
(
x
2
1
+
x
2
2
) =
ˆ
β
1
(
x
2
1
+
x
2
2
) +
ˆ
β
2
(
x
2
1
+
x
2
2
) +
ˆ
β
2
λ
ˆ
β
1
λ
+
ˆ
β
2
(
x
2
1
+
x
2
2
) =
ˆ
β
2
(
x
2
1
+
x
2
2
) +
ˆ
β
2
λ
ˆ
β
1
λ
=
ˆ
β
2
λ
Thus,
ˆ
β
1
=
ˆ
β
2
.
(c) Write out the lasso optimization problem in this setting.
Solution:
min
h
(
y
1
-
ˆ
β
1
x
11
-
ˆ
β
2
x
12
)
2
+ (
y
2
-
ˆ
β
1
x
21
-
ˆ
β
2
x
22
)
2
+
λ
(
|
ˆ
β
1
|
+
|
ˆ
β
2
|
)
i
2
(d) Argue that in this setting, the lasso coefficients
ˆ
β
1
and
ˆ
β
2
are not unique – in other words,
there are many possible solutions to the optimization problem in (c). Describe these solutions.
Solution:
One way to demonstrate that these solutions are not unique is to make a geometric
argument. To make things easier, we’ll use the alternate form of Lasso constraints that we
saw in class, namely:
|
ˆ
β
1
|
+
|
ˆ
β
2
|
< s
.
If we were to plot these constraints, they take the
familiar shape of a diamond centered at the origin
(0
,
0)
.
Next we’ll consider the squared optimization constraint, namely:
(
y
1
-
ˆ
β
1
x
11
-
ˆ
β
2
x
12
)
2
+ (
y
2
ˆ
β
1
x
21
-
ˆ
β
2
x
22
)
2
Using the facts we were given regarding the equivalence of many of the variables, we can
simplify down to the following optimization:
min
h
2(
y
1
-
(
ˆ
β
1
+
ˆ
β
2
)
x
11
i
This optimization problem has a minimum at
ˆ
β
1
+
ˆ
β
2
=
y
1
x
11
, which defines a line parallel
to one edge of the Lasso-diamond
ˆ
β
1
+
ˆ
β
2
=
s
.
As
ˆ
β
1
and
ˆ
β
2
vary along the line
ˆ
β
1
+
ˆ
β
2
=
y
1
x
11
, these contours touch the Lasso-diamond
edge
ˆ
β
1
+
ˆ
β
2
=
s
at different points. As a result, the entire edge
ˆ
β
1
+
ˆ
β
2
=
s
is a potential
solution to the Lasso optimization problem!
A similar argument holds for the opposite Lasso-diamond edge, defined by:
ˆ
β
1
+
ˆ
β
2
=
-
s
.
Thus, the Lasso coefficients are not unique.
The general form of solution can be given by
two line segments:
ˆ
β
1
+
ˆ
β
2
=
s
;
ˆ
β
1
≥
0;
ˆ
β
2
≥
0 and
ˆ
β
1
+
ˆ
β
2
=
-
s;
ˆ
β
1
≤
0;
ˆ
β
2
≤
0
3
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Applied Exercises
6.8.9 (p. 263 ISLR)
In this exercise, we will predict the number of applications received using the other variables in the
College
data set. For consistency, please use
set.seed(11)
before beginning.
(a) Split the data set into a training set and a test set.
(b) Fit a linear model using least squares on the training set, and report the test error obtained.
(c) Fit a ridge regression model on the training set, with
λ
chosen by cross-validation. Report
the test error obtained.
(d) Fit a lasso model on the training set, with
λ
chosen by cross-validation. Report the test error
obtained, along with the number of non-zero coefficient estimates.
(e) Fit a PCR model on the training set, with
M
chosen by cross-validation.
Report the test
error obtained, along with the value of
M
selected by cross-validation.
(f) Fit a PLS model on the training set, with
M
chosen by cross-validation.
Report the test
error obtained, along with the value of
M
selected by cross-validation.
(g) Comment on the results you obtained. How accurately can we predict the number of college
applications received? Is there much difference among the test errors resulting from these five
approaches?
4
A6 Applied Solutions
6.8.9 (a)
library
(ISLR)
library
(dplyr)
Check to make sure we don’t have any null values
sum
(
is.na
(College))
## [1] 0
Split the data set into a training set and a test set.
set.seed
(
1
)
train = College %>%
sample_frac
(
0.5
)
test = College %>%
setdiff
(train)
6.8.9 (b)
Fit a linear model using least squares on the training set, and report the test error obtained.
lm_fit =
lm
(Apps~.,
data =
train)
lm_pred =
predict
(lm_fit, test)
mean
((test[,
"Apps"
] - lm_pred)^
2
)
## [1] 1108531
6.8.9 (c)
Fit a ridge regression model on the training set, with
λ
chosen by cross-validation. Report the test error
obtained, along with the number of non-zero coefficient estimates.
library
(glmnet)
# Build model matrices for
# test and training data
train_mat =
model.matrix
(Apps~.,
data =
train)
test_mat =
model.matrix
(Apps~.,
data =
test)
# Find best lambda using cross-validation,
# alpha = 0 --> use ridge regression
grid =
10
^
seq
(
4
, -
2
,
length=
100
)
mod_ridge =
cv.glmnet
(train_mat, train[,
"Apps"
],
alpha =
0
,
lambda =
grid,
thresh =
1e-12
)
1
lambda_best_ridge = mod_ridge$lambda.min
# Predict on test data, report error
ridge_pred =
predict
(mod_ridge,
newx =
test_mat,
s =
lambda_best_ridge)
mean
((test[,
"Apps"
] - ridge_pred)^
2
)
## [1] 1108512
6.8.9 (d)
Fit a lasso model on the training set, with
λ
chosen by cross-validation. Report the test error obtained, along
with the number of non-zero coefficient estimates.
# Find best lambda using cross-validation,
# alpha = 1 --> use lasso
mod_lasso =
cv.glmnet
(train_mat, train[,
"Apps"
],
alpha =
1
,
lambda =
grid,
thresh =
1e-12
)
lambda_best_lasso = mod_lasso$lambda.min
# Predict on test data, report error
lasso_pred =
predict
(mod_lasso,
newx =
test_mat,
s =
lambda_best_lasso)
mean
((test[,
"Apps"
] - lasso_pred)^
2
)
## [1] 1028718
predict
(mod_lasso,
newx =
test_mat,
s =
lambda_best_lasso,
type=
"coefficients"
)
## 19 x 1 sparse Matrix of class "dgCMatrix"
##
1
## (Intercept) -4.248125e+02
## (Intercept)
.
## PrivateYes
-4.955003e+02
## Accept
1.540306e+00
## Enroll
-3.900157e-01
## Top10perc
4.779689e+01
## Top25perc
-7.926581e+00
## F.Undergrad -9.846932e-03
## P.Undergrad
.
## Outstate
-5.231286e-02
## Room.Board
1.880308e-01
## Books
1.265938e-03
## Personal
.
## PhD
-4.137294e+00
## Terminal
-3.184316e+00
## S.F.Ratio
.
## perc.alumni -2.181304e+00
## Expend
3.193679e-02
## Grad.Rate
2.877667e+00
6.8.9 (e)
Results for OLS, Lasso, Ridge are comparable. Lasso reduces the
P.Undergrad
,
Personal
and
S.F.Ratio
variables to zero and shrinks coefficients of other variables. Below are the test
R
2
values for all models.
2
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test_avg =
mean
(test[,
"Apps"
])
lm_test_r2 =
1
-
mean
((test[,
"Apps"
] - lm_pred)^
2
) /
mean
((test[,
"Apps"
] - test_avg)^
2
)
ridge_test_r2 =
1
-
mean
((test[,
"Apps"
] - ridge_pred)^
2
) /
mean
((test[,
"Apps"
] - test_avg)^
2
)
lasso_test_r2 =
1
-
mean
((test[,
"Apps"
] - lasso_pred)^
2
) /
mean
((test[,
"Apps"
] - test_avg)^
2
)
barplot
(
c
(lm_test_r2,
ridge_test_r2,
lasso_test_r2),
ylim=
c
(
0
,
1
),
names.arg =
c
(
"OLS"
,
"Ridge"
,
"Lasso"
),
main =
"Test R-squared"
)
abline
(
h =
0.9
,
col =
"red"
)
OLS
Ridge
Lasso
Test R-squared
0.0
0.2
0.4
0.6
0.8
1.0
Since the test
R
2
values for all three models are above .90, they all predict the number of college applications
with high accuracy.
3
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19
Car 9
3,308
19
Car 10
2,997
26
Car 11
2,751
27
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