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# Neyman Welcomes You | ISI MStat 2018 PSB Problem 8

This is a problem from ISI MStat Examination,2018.
It involves construction of a most powerful test of size alpha using Neyman Pearson Lemma. The aim is to find its critical region in terms of quantiles of a standard distribution.

This is a problem from ISI MStat Examination,2018. It involves the construction of the most powerful test of size alpha using Neyman Pearson Lemma. The aim is to find its critical region in terms of quantiles of a standard distribution.

## Problem

Let $X_1 ,X_2,.. X_n$ be an i.i.d sample from $f(x;\theta) , \theta \in {0,1 }$, with

$f(x;0) = \begin{cases} 1 & \text{if} \ 0<x<1 \\ 0 & \text{otherwise} \\ \end{cases}$

and $f(x,1)= \begin{cases} \frac{1}{2 \sqrt{x}} & \text{if} \ 0<x<1 \\ 0 & \text{otherwise} \\ \end{cases}$
Based on the above sample,obtain the most powerful test for testing $H_0:\theta=0$ against $H_1: \theta=1$, at level $\alpha$, with $0 < \alpha <1$.Find the critical region in terms of quantiles of a standard distribution.

## Prerequisites

1. The Fundamental Neyman Pearson Lemma

2. Useful Transformations of Random Variables

3. Properties of standard probability distributions (e.g. Normal,Chi-squared etc)

All these topics are included in the regular coursework of undergraduate statistics students. If not, one may refer standard texts like Casella Berger.

## Solution

As, $X_1,X_2,..X_n$ is a random sample, they are independent by definition.
So, their joint pdf when $\theta=0$ is given by $f(\textbf{x},0)= 1 . \prod_{i=1}^{n} 1_{0<x_i<1}$, where $1_{0<x_i<1}$ denotes the indicator function over the interval $[0,1]$.

Similarly, the joint pdf when $\theta=1$ is given by:
$f(\textbf{x},1)=\frac{1}{2^n \prod_{i=1}^{n}\sqrt{x_i}} . \prod_{i=1}^{n}1_{0 <x_i<1}$

According to the Fundamental Neyman Pearson Lemma}, the most powerful size $\alpha$ test for testing $H_{0}$ vs $H_{1}$ is given by the test function $\phi$ as follows:

$\phi=\begin{cases} 1 & \text{if} \ \frac{f(\textbf{x},1)}{f(\textbf{x},0)} > k \\ 0 & \text{otherwise} \\ \end{cases}$

where k is such that $E_{H_0}(\phi)=\alpha$.

So, our test criterion is $\frac{f(\textbf{x},1)}{f(\textbf{x},0)} > k$
Plugging in the pdfs, we get the criterion as $\prod_{i=1}^{n} X_i < \frac{1}{2^{2n }k^2} = \lambda$(say)

Our aim now is to find the value of $\lambda$ from the given size $\alpha$ criterion,
Thus,

$P_{H_0}(\prod_{i=1}^{n}X_i < \lambda)=\alpha$

$\iff P_{H_{0}}(\sum_{i=1}^{n} \ln{X_i} < \ln{\lambda}) =\alpha$

$\iff P_{H_{0}}(-2.\sum_{i=1}^{n} \ln{X_i} >-2. \ln{\lambda}) =\alpha$

Now, we state a result: If $X_i \sim U(0,1)$ ,then $-2 \ln{X_i} \sim \chi^2_{2}$ distribution (Prove it yourself!)

As $X_i$’s are independent, due to reproductive property of chi-squared distribution, $-2.\sum_{i=1}^{n} \ln{X_i} \sim \chi^2_{2n}$
Hence , we simply need that value of $\lambda$ such that the quantity $P_{H_0}(\chi^2=-2.\sum_{i=1}^{n} \ln{X_i} > -2 \ln{\lambda})=\alpha$
The obvious choice is $-2 \ln{\lambda} = \chi^2_{2n , \alpha}$ , where $\chi^2_{2n , \alpha}$ is the upper $\alpha$ point of $\chi^2_{2n}$ distribution.

So, we have $-2 \ln{\lambda} = \chi^2_{\alpha,2n}$ implies $\lambda =e^{-\frac{1}{2}\chi^2_{\alpha,2n}}$
So, our critical region for this test is $\prod_{i=1}^{n} X_i < e^{-\frac{1}{2} \chi^2_{\alpha,2n}}$

## Food For Thought

In this problem , look at the supports of the two distributions under the null and alternative hypotheses.
See that both the supports are the same and hence the quantity $\frac{f_1}{f_0}$ is defined everywhere.
But suppose for a problem the two supports are not the same and they are not disjoint then try constructing a most powerful test using the Neyman Pearson Lemma.
For Example:
Let the family of distributions be ${\theta:X \sim U(0,\theta)}$
Find the most powerful test for testing $H_0 : \theta=1$ against $H_1: \theta=2$
Note that the supports under null and alternative hypotheses are not the same in this case.
Give it a try!

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