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# ISI MStat PSB 2009 Problem 8 | How big is the Mean?

This is a very simple and regular sample problem from ISI MStat PSB 2009 Problem 8. It It is based on testing the nature of the mean of Exponential distribution. Give it a Try it !

## Problem- ISI MStat PSB 2009 Problem 8

Let $$X_1,.....,X_n$$ be i.i.d. observation from the density,

$$f(x)=\frac{1}{\mu}exp(-\frac{x}{\mu}) , x>0$$

where $$\mu >0$$ is an unknown parameter.

Consider the problem of testing the hypothesis $$H_o : \mu \le \mu_o$$ against $$H_1 : \mu > \mu_o$$.

(a) Show that the test with critical region $$[\bar{X} \ge \mu_o {\chi_{2n,1-\alpha}}^2/2n]$$, where $${\chi^2}_{2n,1-\alpha}$$ is the $$(1-\alpha)$$th quantile of the $${\chi^2}_{2n}$$ distribution, has size $$\alpha$$.

(b) Give an expression of the power in terms of the c.d.f. of the $${\chi^2}_{2n}$$ distribution.

### Prerequisites

Likelihood Ratio Test

Exponential Distribution

Chi-squared Distribution

## Solution :

This problem is quite regular and simple, from the given form of the hypotheses , it is almost clear that using Neyman-Pearson can land you in trouble. So, lets go for something more general , that is Likelihood Ratio Testing.

Hence, the Likelihood function of the $$\mu$$ for the given sample is ,

$$L(\mu | \vec{X})=(\frac{1}{\mu})^n exp(-\frac{\sum_{i=1}^n X_i}{\mu}) , \mu>0$$, also observe that sample mean $$\vec{X}$$ is the MLE of $$\mu$$.

So, the Likelihood Ratio statistic is,

$$\lambda(\vec{x})=\frac{\sup_{\mu \le \mu_o}L(\mu |\vec{x})}{\sup_\mu L(\mu |\vec{x})} \\ =\begin{cases} 1 & \mu_o \ge \bar{X} \\ \frac{L(\mu_o|\vec{x})}{L(\bar{X}|\vec{x})} & \mu_o < \bar{X} \end{cases}$$

So, our test function is ,

$$\phi(\vec{x})=\begin{cases} 1 & \lambda(\vec{x})<k \\ 0 & otherwise \end{cases}$$.

We, reject $$H_o$$ at size $$\alpha$$, when $$\phi(\vec{x})=1$$, for some $$k$$, $$E_{H_o}(\phi) \le \alpha$$,

Hence, $$\lambda(\vec{x}) < k \\ \Rightarrow L(\mu_o|\vec{x})<kL(\bar{X}|\vec{x}) \\ \ln k_1 -\frac{1}{\mu_o}\sum_{i=1}^n X_i < \ln k -n \ln \bar{X} -\frac{1}{n} \\ n \ln \bar{X}-\frac{n\bar{X}}{\mu_o} < K*$$.

for some constant, $$K*$$.

Let $$g(\bar{x})=n\ln \bar{x} -\frac{n\bar{x}}{\mu_o}$$, and observe that $$g$$ is,

decreasing function of $$\bar{x}$$ for $$\bar{x} \ge \mu_o$$,

Hence, there exists a $$c$$ such that $$\bar{x} \ge c$$,we have $$g(\bar) < K*$$. See the figure.

So, the critical region of the test is of form $$\bar{X} \ge c$$, for some $$c$$ such that,

$$P_{H_o}(\bar{X} \ge c)=\alpha$$, for some $$0 \le \alpha \le 1$$, where $$\alpha$$ is the size of the test.

Now, our task is to find $$c$$, and for that observe, if $$X \sim Exponential(\theta)$$, then $$\frac{2X}{\theta} \sim {\chi^2}_2$$,

Hence, in this problem, since the $$X_i$$'s follows $$Exponential(\mu)$$, hence, $$\frac{2n\bar{X}}{\mu} \sim {\chi^2}_{2n}$$, we have,

$$P_{H_o}(\bar{X} \ge c)=\alpha \\ P_{H_o}(\frac{2n\bar{X}}{\mu_o} \ge \frac{2nc}{\mu_o})=\alpha \\ P_{H_o}({\chi^2}{2n} \ge \frac{2nc}{\mu_o})=\alpha$$,

which gives $$c=\frac{\mu_o {\chi^2}_{2n;1-\alpha}}{2n}$$,

Hence, the rejection region is indeed, $$[\bar{X} \ge \frac{\mu_o {\chi^2}_{2n;1-\alpha}}{2n}$$.

Hence Proved !

(b) Now, we know that the power of the test is,

$$\beta= E_{\mu}(\phi) \\ = P_{\mu}(\lambda(\bar{x})>k)=P(\bar{X} \ge \frac{\mu_o {\chi_{2n;1-\alpha}}^2}{2n}) \\ \beta = P_{\mu}({\chi^2}_{2n} \ge \frac{mu_o}{\mu}{\chi^2}_{2n;1-\alpha})$$.

Hence, the power of the test is of form of a cdf of chi-squared distribution.

## Food For Thought

Can you use any other testing procedure to conduct this test ?

Think about it !!

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