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).

IIT JAM MS 2020 Section A Problem 1 Solution

Problem

If $\{x_{n}\}_{n \geq 1}$ is a sequence of real numbers such that $\lim _{n \rightarrow \infty} \frac{x_{n}}{n}=0.001$, then

(A) $\{x_{n}\}_{n \geq 1}$ is a bounded sequence
(B)$\{x_{n}\}_{n \geq 1}$ is an unbounded sequence
(C) $\{x_{n}\}_{n \geq 1}$ is a convergent sequence
(D) $\{x_{n}\}_{n \geq 1}$ is a monotonically decreasing sequence

Hints

Hint 1

If $\{x_{n}\}_{n \geq 1}$ was bounded, show that $\lim _{n \rightarrow \infty} \frac{x_{n}}{n}=0$, by sandwich theorem.

Hint 2

If $\{x_{n}\}_{n \geq 1}$ was convergent, show that $\lim _{n \rightarrow \infty} \frac{x_{n}}{n}=0$, by algebra of limits.

Hint 3

If $\{x_{n}\}_{n \geq 1}$ was motonotically decreasing and bounded below, then it would have been convergent by Monotone Convergence Theorem.

Let's consider if it is not below below, i.e. $\lim _{n \rightarrow \infty} {x_{n}} = -\infty $

• Take $ {x_{n}} = -n $
• Take $ {x_{n}} = -n^2 $
• Take $ {x_{n}} = - \sqrt{n} $

Find the limit in each of this case.

Hence, it will be unbounded. See the full solution and proof idea below.

Full Solution

Food For Thoughts

• What if, $\lim _{n \rightarrow \infty} \frac{x_{n}}{n}= 0$?
• Find examples.