ISI MStat 2020 PSB Problem 8 Solution

Problem

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 $\alpha$ .

Give a free ISI MStat Diagonistic Test

Hint 1

Show that the problem is equivalent to finding that $P_{\mu = 0}(\max \{T_{i}: 1 \leq i \leq n-m+1\}\\>c_{m, \alpha}) \leq \alpha$

Hint 2

$P_{\mu = 0}(\max \{T_{i}: 1 \leq i \leq n-m+1\}\\>c_{m, \alpha})$

$= P_{\mu = 0}( T_1 > c_{m, \alpha} \cup T_2 > c_{m, \alpha} \cdots T_{n-m+1}\\ > c_{m, \alpha})$

Hint 3

Use Boole's Inequality o get

$P_{\mu = 0}( T_1 > c_{m, \alpha} \cup T_2 > c_{m, \alpha} \cdots T_{n-m+1}\\ > c_{m, \alpha}) \leq \sum_{i = 1}^{n-m+1} P(T_i > c_{m, \alpha}) = \alpha$

Hint 4

Show that under $H_0$ , $T_i$ ~ $N(0,\frac{1}{m})$ . Hence, find $c_{m, \alpha}$

See the full solution below.

Full Solution

Food For Thoughts

What if, $\lim _{n \rightarrow \infty} c_{m, \alpha}$ ?
What if strict $c_{m, \alpha}$ was required to find? Can you solve this using Jacobian. (You can give an approximate solution).