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Neyman Welcomes You | ISI MStat 2018 PSB Problem 8

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!

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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