Assume that $X_{1}, \ldots, X_{n}$ is a random sample from $N(\mu, 1)$, with $\mu \in \mathbb{R}$. We want to test $H_{0}: \mu=0$ against $H_{1}: \mu=1$. For a fixed integer $m \in{1, \ldots, n}$, the following statistics are defined:
$\begin{aligned} T_{1} &= \frac{\left(X_{1}+\ldots+X_{m}\right)}{m} \\ T_{2} &= \frac{\left(X_{2}+\ldots+X_{m+1}\right)} {m} \\ \vdots &=\vdots \\ T_{n-m+1} &= \frac{\left(X_{n-m+1}+\ldots+X_{n}\right)}{m} . \end{aligned}$
Fix $\alpha \in(0,1)$.
Consider the test
Reject $H_{0}$ if $\max \{T_{i}: 1 \leq i \leq n-m+1\}>c_{m, \alpha}$
Find a choice of $c_{m, \alpha} \in \mathbb{R}$ in terms of the standard normal distribution function $\Phi$ that ensures that the size of the test is at most $\alpha$.
Show that the problem is equivalent to finding that $P_{\mu = 0}(\max \{T_{i}: 1 \leq i \leq n-m+1\}\\>c_{m, \alpha}) \leq \alpha$
$P_{\mu = 0}(\max \{T_{i}: 1 \leq i \leq n-m+1\}\\>c_{m, \alpha})$
$= P_{\mu = 0}( T_1 > c_{m, \alpha} \cup T_2 > c_{m, \alpha} \cdots T_{n-m+1}\\ > c_{m, \alpha})$
Use Boole's Inequality o get
$P_{\mu = 0}( T_1 > c_{m, \alpha} \cup T_2 > c_{m, \alpha} \cdots T_{n-m+1}\\ > c_{m, \alpha}) \leq \sum_{i = 1}^{n-m+1} P(T_i > c_{m, \alpha}) = \alpha $
Show that under $H_0$, $T_i$ ~ $N(0,\frac{1}{m})$. Hence, find $c_{m, \alpha}$
See the full solution below.
