Here is a GPT-5 prompt I used to create some content:
How does ridge regression deal with correlated variables? Is it always better than Marrow’s C_p for prediction? Can you suggest a code with 20 variables to confirm your answer?
The following is the answer produced by GPT-5
Here’s a concise answer:
Ridge with correlated variables: Ridge regression shrinks coefficients toward zero in a way that distributes the effect across correlated predictors. Instead of picking one variable and ignoring the others, it tends to keep correlated variables together and stabilize their estimates. This reduces variance and avoids the instability of ordinary least squares.
Ridge vs. Mallows’ \(C_p\): Ridge is not always better. Ridge generally performs better when the signal is spread across many correlated predictors (dense signals). Mallows’ \(C_p\), used for subset selection, can perform better when only a few predictors truly matter (sparse signals).
In practice, which is better depends on the structure of the true underlying model. Would you like me to provide a small R code example (with 20 variables) that demonstrates both cases — dense vs. sparse signals — so you can see the difference in prediction performance?
I would like to first consider a simulation study to understand the performance. And while doing this, let’s also compare the bias and variance of the estimations. Consider the following data generator, with \(p = 20\), but \(y\) depends on only the first two variables:
\[ y = x_1 + x_2 + \epsilon \]
where \(\epsilon \sim N(0,1)\) and \(x_1\) and \(x_2\) are generated from independent standard normal distributions, while the rest of the \(p-2\) variables are generated in the following way:
\[ x_j \sim \eta_j x_1 + \phi_j x_2 + \xi_j, \quad j = 3, \ldots, p \]
where \(\eta_j, \phi_j \sim
\text{Uniform}(-1, 1)\) and \(\xi_j
\sim N(0, 0.3^2)\) are all independently generated. Write your
code to generate this data with \(n =
100\), and fit both the ridge regression (use leave-one-out CV
and lambda.min
to pick the best tuning parameter) and the
best subset selection (using Mallows’ \(C_p\)). Generate another \(1000\) testing data, and evaluate the
prediction performance of both methods.
Now, let’s consider a simulation. For this question, we will repeat the above procedure 200 times, record their prediction errors and also evaluate the bias and variance of each of the estimations of the coefficients. To do this, we need to record the estimates of the coefficients for all variables for both methods, and store them. For the Marlow’s \(C_p\) method, if a variable is not selected, then we record its coefficient as 0. After all simulations, you should have a matrix of \(200 \times 20\) for each method. The intercept is not needed. Based on this matrix, you can calculate the bias and variance of the estimations of these coefficients for each variable by
Answer the following question:
barplot()
to visualize and compare the
parameter estimates for both methods.Based on your results (b)
We will use the golub
dataset from the
multtest
package. This dataset contains 3051 genes from 38
tumor mRNA samples from the leukemia microarray study Golub et
al. (1999). This package is not included in R
, but on
bioconductor
. Install the latest version of this package
from bioconductor
, and read the documentation of this
dataset to understand the data structure of golub
and
golub.cl
.
We will not use this data for classification (the original problem). Instead, we will do a toy regression example to show how genes are highly correlated and could be used to predict each. Carry out the following tasks:
golub.cl
using mt.teststat()
. Use
t.equalvar
(two sample \(t\) test with equal variance) as the test
statistic.Based on your results, do you observe a “U” shaped curve? Can you comment on how is this related to the bias-variance trade-off?
Recall the previous homework, we have a quadratic function for minimization. We know that analytic solution exist. However, in this example, let’s use coordinate descent to solve the problem. To demonstrate this, let’s consider the following simulated dataset, with design matrix \(x\) (without intercept) and response vector \(y\):
set.seed(432)
n <- 150
x <- matrix(rnorm(n*2), n, 2)
y <- 0.3 * x[, 1] + 0.5 * x[, 2] + rnorm(n)
We will consider a model without the intercept term. In this case, our objective function (of \(\beta_1\) and \(\beta_2\) for linear regression is to minimize the sum of squared residuals:
\[ f(\beta_1, \beta_2) = \frac{1}{n} \sum_{i=1}^n (y_i - \beta_1 x_{i1} - \beta_2 x_{i2})^2 \]
where \(x_{ij}\) represents the \(j\)th variable of the \(i\)th observation.
Write down the objective function in the form of
\[ f(x,y) = a \beta_1^2 + b \beta_2^2 + c \beta_1 \beta_2 + d \beta_1 + e \beta_2 + f \]
by specifying what are coefficients a, b, c, d, e, and f, using the simulated data. Calculate them in R, using vector operations rather than for-loops.
A coordinate descent algorithm essentially does two steps:
Write down the updating rules for \(\beta_1\) and \(\beta_2\) using the coordinate descent
algorithm. Use those previously defined coefficients in your formula and
write them in Latex. Implement them in a for-loop algorithm in R that
iterates at most 100 times. Use the initial values \(\beta_1 = 0\) and \(\beta_2 = 0\). Decide your stopping
criterion based on the change in \(\beta_1\) and \(\beta_2\). Validate your solution using the
lm()
function.