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 !!