Consider a random vector $X \sim N_2(\mu,\Sigma)$. Define a region $R \subset \mathbb{R}^2$ such that $P(x \in R) = 0.95$ Hint: The $0.95-$quantile of $\chi^2$ is about $5.99$ .
I think we are supposed to use that $\chi_2^2 = Z_1^2+Z_2^2$, where the are $Z_i \sim N(0,1)$. But I do not know how to go from here. Could you help me?
I suppose that $\Sigma$ is a positive definite matrix.
Consider $Y =\Sigma^{-1/2} (X-\mu), $ which follows standard bivariate normal. Now, if $Y = (Y_1, Y_2)'$, then $Y_1, Y_2$ are iid standard normal. So, we can choose a circle $C$ centred at the origin with radius $r$ given by $r^2 = \chi^2_{2, 0.95}$ (i.e.the 0.95-quantile of the $\chi^2_2$ distribution). Note that $$\Pr(Y\in C) = \Pr(Y_1^2 + Y_2^2 \le r^2) = 0.95,$$ since $Y_1^2 + Y_2^2$ follows $\chi^2_2.$
Finally, to find the confidence interval $R$ for $X,$ set $R = \Sigma^{1/2}(\mu+C)$, that is, $$R =\left\{\Sigma^{1/2}\left(\mu + {\binom{x}{y}}\right): {x^2 + y^2} \le r^2\right\}.$$