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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, Here, $K*, \mu_o$ are fixed quantities.

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 !!

## Subscribe to Cheenta at Youtube

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, Here, $K*, \mu_o$ are fixed quantities.

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 !!

## Subscribe to Cheenta at Youtube

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