# Confidence set for parameter vector in linear regression

This questions is in reference to equation 3.15 in the book Elements of Statistical Learning by Tibshirani and coll.

I do understand the individual beta confidence interval estimation as provided in equation 3.14, but equation 3.15 just bowls me:

$$\beta|(\hat{\beta} -\beta)^TX^TX(\hat{\beta} -\beta)\leq \sigma ^2\chi ^2$$

What is the idea being expressed here? What is a confidence set? Can we not estimate the intervals of all the betas as per equation 3.14?

• Please make the question self-contained. I do not have access to the book. – varty Nov 13 '11 at 21:13
• @varty: The book is freely available (legally) at the link provided by the OP. That said, I still agree the question should be made self-contained if for no other reason than that future versions, or even printings, of the text may have different equation numbering.) – cardinal Nov 13 '11 at 21:31
• Sorry this is the eqn I am referring to – bgbgh Nov 13 '11 at 22:47
• $$\beta|(\hat{\beta} -\beta)^TX^TX(\hat{\beta} -\beta)\leq \sigma ^2\chi ^2$$ – bgbgh Nov 13 '11 at 22:48
• @varty: I agree. My intent was not to be argumentative, but simply to (kindly) point out that you did have access to the book, in case you were interested. – cardinal Nov 14 '11 at 0:24

To make things clearer recall that

$$\hat{\beta}\sim N(\beta,\hat{\sigma}^2(X^TX)^{-1}),$$

When you isolate $$\beta_j$$ you get that

$$\hat{\beta}_j-\beta_j\sim N(0, \sigma^2 v_j)$$

where $$v_j$$ are the diagonal elements of $$(X^TX)^{-1}$$. We can write this alternatively as

$$\frac{\hat{\beta}_j-\beta_j}{\sqrt{v_j}}\sim N(0,\sigma),$$

which is the same as

$$\left(\frac{\hat{\beta}_j-\beta_j}{\sqrt{v_j}}\right)^2= (\hat{\beta}_j-\beta_j)(v_j)^{-1}(\hat{\beta}_j-\beta_j)\sim \sigma\chi_1^2.$$

Note that those $$\beta_j$$ that satisfy the condition

$$\left(\frac{\hat{\beta}_j-\beta_j}{\sqrt{v_j}}\right)^2\le \sigma^2\chi_{1,1-\alpha}^2$$

fall in the confidence interval described in the equation 3.14. Hence the confidence interval is the set in real line.

Now similarly we get $$(X^TX)^{1/2}(\hat\beta-\beta)\sim N(0, \sigma^2 I),$$

so

$$(\hat\beta-\beta)X^TX(\hat\beta-\beta)\sim \sigma^2\chi_{p+1}^2$$

where $$p$$ is the number of the regressors. Using the same analogy we can look for vector points $$\beta\in\mathbb{R}^{p+1}$$ which satisfy the condition

$$(\hat\beta-\beta)X^TX(\hat\beta-\beta)\le \sigma^2\chi_{p+1,1-\alpha}^2.$$

For $$p=1$$ this set will be the interior of the ellipsis.

The confidence set is used since it accounts for interactions between $$\beta_i$$ and $$\beta_j$$. Look at the scatter plot of two independent normal variables (which would be the case for orthogonal regressors with the same variance):

The circular shape is evident. Using the univariate confidence intervals the confidence set would be square, and this graph illustrates that it will actualy estimate the confidence incorrectly.

• Thanks a lot for the response mpiktas. Here are some follow up questions Can you please elaborate on the step in which you were able to show that (betahat-beta)XtX(betahat-beta) has a chi-squared distribution? Also how does one show that for p = 1, we get an ellipse? What ellipse are you talking about? Then is that wrong to build individual tests on the isolated coefficients? – user7413 Nov 17 '11 at 3:21
• Like also the equation in the book says that its an "approximate" confidence interval, what makes it approximate. Forgive my ignorance but I wasnt able to see how you got the scatter plot, are you just plotting you independent normally distributed random variables? – user7413 Nov 17 '11 at 3:22

To supplement, if $X\in \mathbb{R}^{N\times (p+1)}$ and $\hat\beta$ is the LS estimation for $\beta$ in the linear regression model $Y=X\beta+\epsilon$ with $\epsilon\sim\mathcal{N}(0,\sigma^2)$,

$$\frac{(\hat{\beta}-\beta)^TX^TX(\hat{\beta}-\beta)}{\hat{\sigma}^2}\sim \chi_{p+1}^2$$ holds asymptotically when $N\to+\infty$. To see this, we first have \begin{align} (\hat{\beta}-\beta)^TX^TX(\hat{\beta}-\beta)\sim & \sigma^2\chi^2_{p+1}\quad\mbox{(from $\hat\beta\sim\mathcal{N}(\beta, \sigma^2(X^TX)^{-1})$)}\\ (N-p-1)\hat{\sigma}^2\sim & \sigma^2\chi^2_{N-p-1} \end{align} which gives $$\frac{(\hat{\beta}-\beta)^TX^TX(\hat{\beta}-\beta)}{(p+1)\hat{\sigma}^2}\sim F_{p+1,N-p-1}$$ On the other hand, one can prove if $S\sim F_{m,n}$, $T=\lim_{n\to+\infty}mS\sim\chi_m^2$ by directly computing the limit of $mS$'s PDF, with the help of the relation between gamma function and beta function and Stirling's formula. With this claim, we have $$\frac{(\hat{\beta}-\beta)^TX^TX(\hat{\beta}-\beta)}{\hat{\sigma}^2}\sim \chi_{p+1}^2\quad(N\to+\infty)$$.