Let $X$ and $Y$ be independent $N(\mu, 1)$ r.v's such that $$ H_0: \mu=0, ~~~H_1:\mu \neq 0$$ where the tests are as follows$$f_1(x,y) = \begin{cases} 1,&|x+y|>a\\ 0, &|x+y|\le a \end{cases}$$ and $$f_2(x,y) = \begin{cases} 1,&|x+2y|>b\\ 0, &|x+2y|\le b \end{cases}$$ 1. Find $a$ and $b$ so that the size is $\alpha=5\%$ for both tests.
- Since the size of the test is the probability of incorrectly rejecting the null, then we need to find $$P(Reject H_0|\mu=0)=0.05$$ $$ \iff P(\frac{(X-\mu_0)\sqrt{n}}{\sigma}~~|~\mu=0)=0.05$$ $$ \iff P(X\sqrt{n})=0.05$$ I am unsure how to implement the function into this so that I can find $a,b$?
Note that $f_{1}$ and $f_{2}$ are test functions. A test function $f$, rejects $H_0$ if and only if $f=1$. That is, you want to find $a$ and $b$ such that that \begin{align*} \begin{cases} P(f_1=1|\mu=0) &= \alpha \\ P(f_2=1|\mu=0) &= \alpha \end{cases} \end{align*} Note that \begin{align*} P(f_1=1|\mu=0) &= P(|X+Y|>a|\mu=0) \\ &= P\left(\bigg|\frac{X+Y}{\sqrt{2}}\bigg| > \frac{a}{\sqrt{2}}\right) \end{align*} Since, under $\mu=0$, $X$ and $Y$ are independent standard normals, $Z=\frac{X+Y}{\sqrt{2}}$ is a standard normal. Therefore, you wish $P\left(|Z| > \frac{a}{\sqrt{2}}\right)=\alpha$. This happens if $\frac{a}{\sqrt{2}}=\Phi^{-1}(1-2^{-1}\alpha)$, where $\Phi$ is the cdf of the standard normal. Conclude that $a=\sqrt{2}\Phi^{-1}(1-2^{-1}\alpha)$.
Can you use the same reasoning to find $b$?