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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?

Thanks for your help.

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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.

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  • $\begingroup$ Thanks for your answer. Yes I know this, but in the first attempt I would not have seen it. I would have done it as above and I would like to know where my error is or how do I can conclude. $\endgroup$ Commented Jun 15, 2023 at 13:56
  • $\begingroup$ +1. This exactly responds to the question. $\endgroup$
    – whuber
    Commented Jun 15, 2023 at 14:02
  • $\begingroup$ @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? $\endgroup$ Commented Jun 15, 2023 at 14:03
  • $\begingroup$ @user1294729 I think that the squared quantile does not match with the chisq quantiles. As far as I see it, you took the root of sum of squares which should not result in a simple sum. $\endgroup$ Commented Jun 15, 2023 at 14:08
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    $\begingroup$ @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. $\endgroup$ Commented Jun 15, 2023 at 14:16

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