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?
 A: 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.
A: 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
$$
\frac{(\hat{\beta}-\beta)^TX^TX(\hat{\beta}-\beta)}{\hat{\sigma}^2}\sim \chi_{p+1}^2\quad(N\to+\infty)
$$.
A: Because (under the assumption made in the text that the errors are Gaussian with mean zero and with variance $\sigma^2$) the vector $\hat\beta-\beta$ is multinormal with mean zero and covariance matrix $(X^T X)^{-1}\sigma^2$, the squared Mahalanobis Distance of $\hat\beta$ is (by definition)
$$\frac{1}{\sigma^2}(\hat\beta-\beta)^T(X^T X)(\hat\beta-\beta)$$
and as per https://en.wikipedia.org/wiki/Mahalanobis_distance#Normal_distributions, that random variable is $\chi^2$ with $p+1$ degrees of freedom where $p+1$ is the number of dimensions in $\beta$. So, $(\hat\beta-\beta)^T(X^T X)(\hat\beta-\beta)$ is distributed like $\sigma^2 \chi_{p+1}^2$. Thus, under the assumptions in the text, there is a concise answer to the original question.
