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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.

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).

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