Adjusted $R^2$ is a weaker penalty than AIC

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Abstract

This is a note of thought that I bumped into randomly.

It has been a long history in statistics on model selection that penalizes the exploding number of parameters. AIC, i.e., Alkaline information criterion, is perhaps the most famous one due to its simplicity and generality. While, when $R^2$ is introduced in the class, immediately adjusted $R^2$ is introduced. The later one does not follow the “variance explanation” per se since there is no guarantee about the ratio being kept in $[0,1]$. But, it is supposed to penalize large number of parameters by showing a lower $R^2$. In this post, I will show that adjusted $R^2$ has a weaker penalty than AIC/BIC criterion.

Introduction

Before discussion, let’s make the definition clear. $n \in \mathbb{N}$ is the total number of samples. $p$ is the number of predictors (excluding 1).

Adjusted $R^2$

The difference between common $R^2$ and adjusted $R^2$ is that, adjusted $R^2$ considers the variance explanation by taking independency into account. Therefore, the more parameter you have, the residuals of all data, would be less independent simply due to more and more constraints are imposed by the OLS formulation.

The expression for adjusted $R^2$ is

Minimal Description Length (MDL)

AIC originates in information theory. In Bayes belief net, there is a criterion called Minimal Description Length (MDL). One would like to choose the Bayes belief net models with the shortest MDL.

In general, for a given training set $D = { {x}_1,\ldots,{x}_m }$, the scoring function on Bayes net $B = \langle G, \Theta \rangle$ on the training set $D$ is

where is the bits required to describe each parameter while is the number of parameters in the Bayes net.

which is the log-likelihood of all data and $\sigma^2$ is the uncertainty in the likelihood model. Note that $\sigma^2 = SSE/m$ as a MLE estimation for the residual variance. Remember that this implies i.i.d, i.e., independently identical distribution.

MDL induces several concepts which are shown below without proof. Note that the sources are from here.

  • AIC

which assumes each parameter costs $1$ bit for description.

  • BIC

which assumes each parameter costs $\log m /2$ bits for descriptions.

Adjusted $R^2$ penalize weaker than AIC/BIC

Note that for model selection, we hope to select the one maximize criterion.

Start with $\overline{R}^2$, so it is equivalent to minimize the $\dfrac{SSE}{SST} \dfrac{n-1}{n-p-1}$. Since $SST$ is fixed by the given data and $n$ is also fixed. It is simply the one that minimize the

while it does not hurt to take the $\log$ and add/minus constants so

Second, for AIC/BIC, the equivalent quantity to minimize is

where $C = 1, \frac{\log n}{2}$ for AIC and BIC respectively.

Note that $n$ is a constant, therefore it is equivalent to minimize

Clearly, the ratio between the $\log \frac{1}{1-(p+1)/n}$ and $2 C (p+1)/n$ determines the relative penalization between adjusted $R^2$ and AIC/BIC.

First, let’s investigate this ratio as follows

To see this, note $f(0^+) = 1$ and one simply take $f’(x)$ as

Therefore, take $x = \frac{p+1}{n}$, we have it as a monotonic increasing function with respect to $(p+1)/n$ and when $p \ll n$, we have the ratio between adjusted $R^2$ and AIC/BIC as the following:

where $C = 1, \frac{\log n}{2} $.