How to construct a compact confidence region of $(\mu,\sigma)$ look for a normal distribution? 
I am working on an exercise, where given $n$ iid observations $X_1,X_2,\ldots X_n\sim\mathcal{N}(\mu,\sigma^2)$, construct a compact $95\%$-confidence region $C \subseteq \mathbb{R} \times \mathbb{R}^+$ for the parameter $(\mu,\sigma)$.

I did the following: I constructed two $97.5\%$-confidence intervals for $\mu$, via the $t$-distribution and $\bar{X}$, and for $\sigma^2$, via the $\chi^2$-distribution and $S^2$, and used that Fisher's Theorem says that $\bar{X}$ and $S^2$ are independent. Than I "combined" both intervals into a rectangle (which is of course compact) and noted that the probability that $(\mu,\sigma)$ is inside this rectangle is $0.975 \cdot 0.975 > 0.95$. Is this correct?
 A: Your confidence region will take on the shape of a trapezoid if you proceed in this manner.
First note if $X,Y$ are independent random variables such that $X\sim \chi^2_{n-1}$ and $Y\sim \mathcal{N}(0,1)$ then $$\mathbb{P}\left(\chi^2_{1-\alpha/2}<X<\chi^2_{\alpha/2},-z_{\alpha/2}<Y<z_{\alpha/2}\right)=(1-\alpha)^2$$ Since $\frac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}$ and $\sqrt{n}\left(\frac{\overline{X}-\mu}{\sigma}\right)\sim\mathcal{N}(0,1)$ are independent, $$\mathbb{P}\left(\chi^2_{1-\alpha/2}<\frac{(n-1)S^2}{\sigma^2}<\chi^2_{\alpha/2},-z_{\alpha/2}<\sqrt{n}\left(\frac{\overline{X}-\mu}{\sigma}\right)<z_{\alpha/2}\right)=(1-\alpha)^2$$ Manipulating these inequalities yields $$\mathbb{P}\left(S\sqrt{\frac{n-1}{\chi^2_{\alpha/2}}}<\sigma<S\sqrt{\frac{n-1}{\chi^2_{1-\alpha/2}}},\sigma>\frac{\sqrt{n}}{z_{\alpha/2}}\big|\mu-\overline{X}\big|\right)=(1-\alpha)^2$$ This means a $100(1-\alpha)^2$% confidence region is $$C=\Bigg\{(\mu,\sigma)\in \mathbb{R}\times [0,\infty):\sigma>\frac{\sqrt{n}}{z_{\alpha/2}}\big|\mu-\overline{x}\big|,S\sqrt{\frac{n-1}{\chi^2_{\alpha/2}}}<\sigma<S\sqrt{\frac{n-1}{\chi^2_{1-\alpha/2}}}\Bigg\}$$ Taking $\alpha=1-\sqrt{0.95}$ yields your $95$% confidence region $C$.
