# How can I find the value $Q$ for which the given region is a confidence region for $\mu$ with confidence level $1-\alpha$?

Let $$X_{11},...,X_{1n}\sim \mathcal{N}(\mu_1,\sigma^2)$$ and $$X_{21},...,X_{2n}\sim \mathcal{n}(\mu_2,\sigma^2)$$ be iid distributed and such that $$\{X_{i1}\}$$ are independent from $$\{X_{i2}\}$$ and $$\sigma^2$$ is known. Now we consider the region $$R=\{(\mu_1,\mu_2): \frac{n}{\sigma^2}\left((\bar X_1-\mu_1)^2+(\bar X_2-\mu_2)^2\right)\leq Q\}$$ I want to find $$Q$$ for which $$R$$ is a confidence region with level $$1-\alpha$$. ($$\bar X_1=\frac{1}{n}\sum_{k=1}^n X_{1k}$$)

I know that I need to show that $$\Bbb{P}_{(\mu_1,\mu_2)} ((\mu_1,\mu_2)\in R)=1-\alpha$$. therefore I first tried to compute the probability: \begin{align}\Bbb{P}_{(\mu_1,\mu_2)} ((\mu_1,\mu_2)\in R)&=\Bbb{P}_{(\mu_1,\mu_2)}\left(\frac{n}{\sigma^2}\left((\bar X_1-\mu_1)^2+(\bar X_2-\mu_2)^2\right)\leq Q\right)\\&=\Bbb{P}_{(\mu_1,\mu_2)}\left(\frac{\bar X_1-\mu_1}{\sigma/\sqrt{n}}+\frac{\bar X_2-\mu_2}{\sigma/\sqrt{n}}\leq \sqrt{Q}\right)\\&=\Bbb{P}_{(\mu_1,\mu_2)}\left(Z\leq \sqrt{Q}\right)\\&=\Bbb{P}_{(\mu_1,\mu_2)}\left(\frac{Z}{\sqrt{2}}\leq \sqrt{\frac{Q}{2}}\right)\\&=\Phi\left(\sqrt{\frac{Q}{2}}\right)\end{align} where I used that $$\frac{\bar X_i-\mu_1}{\sigma/\sqrt{n}}\sim\mathcal{N}(0,1)$$ and the sum of two $$\mathcal{N}(0,1)$$ random variables has distribution $$\mathcal{N}(0,2)$$ so I took $$Z\sim \mathcal{N}(0,2)$$ and additionally I used in the last step that $$\frac{Z}{\sqrt 2}\sim \mathcal{N}(0,1)$$. So I only need to solve $$1-\alpha=\Phi\left(\sqrt{\frac{Q}{2}}\right)$$ which is equivalent to say that $$\sqrt{\frac{Q}{2}}=z_{1-\alpha}$$ where $$z_{1-\alpha}$$ is the $$1-\alpha$$ quantile of a normal distribution. But then I would have said that $$Q=2(z_{1-\alpha})^2$$.

Now somehow our prof. told us that $$Q=\chi^2_{2,1-\alpha}$$ where $$\chi^2_{2,1-\alpha}$$ is the $$1-\alpha$$ quantile of a $$\chi^2_{2}$$ distribution. I don't see how to get from my solution to his solution. Am I doing something wrong?

Note, that for any standard normal variable $$Z \sim N(0,1)$$ it holds that $$Z^2 \sim \chi_1^2$$. I suppose this is what you need here.
• @whuber can I say that since $Z^2\sim \chi^2_1$ then also $z_{1-\alpha}^2=\chi^2_{1,1-\alpha}$ and since I got $2z_{1-\alpha}^2=2\chi^2_{1,1-\alpha}=\chi^2_{1,1-\alpha}+\chi^2_{1,1-\alpha}=\chi^2_{2,1-\alpha}$ since sums of $\chi^2$ distributions is again a $\chi^2$ distribution with degree of freedom to be the sum of the degrees of freedom? Commented Jun 15, 2023 at 14:03
• @user1294729 the comment from whuber technically solves your question. The sum of $\chi_1^2$ variables is again $\chi_n^2$ distributed. Equipped with that you should take a look at your computations again. Commented Jun 15, 2023 at 14:16