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Let $X_1,...X_n$ be an iid sample distributed as $\mathcal{N}(\mu,1)$. We have the following tests $$H_0: \mu=0~~~\text{vs.}~~~H_1:\mu\neq 0$$and we use the statistic $T=\frac{1}{n}\sum_{k=1}^n X_k$ and the test function $$\delta=\Bbb{1}_{|T|\geq Q}$$ for some $Q>0$. I want to find the probability of committing a type 2 error, ie compute $\Bbb{P}_\mu(\delta=0)$

Let observe that $T\sim \mathcal{N}(\mu, 1/n)$ therefore we define a random variable $Z$ s.t. $Z\sim \mathcal{N}(\mu, 1/n)$.

$$\begin{align}\Bbb{P}_\mu(\delta=0)&=\Bbb{P}_\mu(T<Q)+ \Bbb{P}_\mu(T>-Q)\\&=\Bbb{P}(Z<Q)+\Bbb{P}(Z>-Q)\\&=\Bbb{P}\left(\sqrt{n}(Z-\mu)<\sqrt{n}(Q-\mu)\right)+\Bbb{P}\left(\sqrt{n}(Z-\mu)>\sqrt{n}(-Q-\mu)\right)\\&=\Phi\left(\sqrt{n}(Q-\mu)\right)+1-\Bbb{P}\left(\sqrt{n}(Z-\mu)\leq\sqrt{n}(-Q-\mu)\right)\\&=\Phi\left(\sqrt{n}(Q-\mu)\right)+1-\Phi\left(-\sqrt{n}(Q+\mu)\right)\\&=\Phi\left(\sqrt{n}(Q-\mu)\right)+\Phi\left(\sqrt{n}(Q+\mu)\right)\end{align}$$

where we used that $\sqrt(n)(Z-\mu)\sim \mathcal{N}(0,1)$, and where $\Phi$ is the cumulative distribution function of a standard normal random variable. I also used that $\Phi(-z)=1-\Phi(z)$. But I don't think that this is correct since in class our prof. told us that the solution is $\Phi\left(\sqrt{n}(Q-\mu)\right)-\Phi\left(\sqrt{n}(-Q-\mu)\right)$.

Can someone tell me where my error is?

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    $\begingroup$ What is your justification for $\Bbb{P}_\mu(\delta=0)=\Bbb{P}_\mu(T<Q)+ \Bbb{P}_\mu(T>-Q)$ rather than $=\Bbb{P}_\mu(-Q <T<Q)=\Bbb{P}_\mu(T<Q)- \Bbb{P}_\mu(T\le -Q)$? $\endgroup$
    – Henry
    Jun 13, 2023 at 14:57
  • $\begingroup$ @Henry hmm maybe here is my mistake. I somehow didn't thought about it, because in the exercise before we also have $\Bbb{P}(|T|\geq Q)=\Bbb{P}(T\leq -Q)+\Bbb{P}(T\geq Q)$ so I did it in the same way. Why does it then work in one but not the other case? $\endgroup$ Jun 13, 2023 at 15:02
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    $\begingroup$ So $\Bbb{P}(|T|\lt Q) $ $=1-\Bbb{P}(|T|\geq Q)$ $=1-\Bbb{P}(T\leq -Q)-\Bbb{P}(T\geq Q) $ $= \Bbb{P}(T\lt Q)-\Bbb{P}(T\leq -Q)$ remembering that that you can add probabilities of disjoint events but not of intersecting events $\endgroup$
    – Henry
    Jun 13, 2023 at 15:05
  • $\begingroup$ @Henry sorry I'm still confused why wouln't I then have $\Bbb{P}(|T|\geq Q)=\Bbb{P}(T\leq -Q)-\Bbb{P}(T\geq Q)$ when I do it as in your first comment? $\endgroup$ Jun 13, 2023 at 15:08
  • $\begingroup$ Subtraction of probability only works when one event is a subset of another event: for example $(T\leq -Q) \subset (T\lt Q)$ assuming $Q$ is positive but $(T\ge Q) \not\subset (T\le -Q)$ $\endgroup$
    – Henry
    Jun 13, 2023 at 15:12

1 Answer 1

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Your $$\Bbb{P}_\mu(\delta=0)=\Bbb{P}_\mu(T<Q)+ \Bbb{P}_\mu(T>-Q)$$

should be $$\Bbb{P}_\mu(\delta=0)=\Bbb{P}_\mu(-Q <T<Q)=\Bbb{P}_\mu(T<Q)- \Bbb{P}_\mu(T\le -Q)$$

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  • $\begingroup$ But is it then true that for example $\Bbb{P}(|T|\leq Q)=\Bbb{P}(-Q\leq T\leq Q)=\Bbb{P}(T\leq Q)-\Bbb{P}(T<-Q)$ $\endgroup$ Jun 13, 2023 at 15:42
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    $\begingroup$ Yes indeed it is $\endgroup$
    – Henry
    Jun 13, 2023 at 16:00
  • $\begingroup$ Perfect thanks! $\endgroup$ Jun 13, 2023 at 16:41
  • $\begingroup$ I have asked a question here stats.stackexchange.com/questions/618852/… would it be possible if you could take a look at it? $\endgroup$ Jun 15, 2023 at 13:11

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