# Finding the uniformly most powerful test for hypothesis

Let $$\mathbf{X}=(X_1,...,X_n)^T$$ is a simple sample where $$X$$ belongs to exponential distribution family $$\mathcal{P}=\{ f(x;\mu,\sigma \}, -\infty<\mu<\infty, 0<\sigma<\infty.$$ Density is $$f(x;\mu,\sigma)=\sigma e^{-\sigma(x-\mu)}, \mu

I need to find the uniformly most powerful test for hypothesis $$H:\sigma=\sigma_0$$ when $$\mu$$ is known.

So I think alternative hypothesis is $$\overline{H}: \sigma>\sigma_0.$$

I also know that $$\frac{f_1(X)}{f_0(X)}>c.$$

And $$\varphi(x)= \begin{cases} 1, \text{ when } f_1(X)>cf_0(X) \\ \gamma, \text{ when } f_1(X)=cf_0(X) \\ 0, \text{ when } f_1(X)

In my case I got $$\frac{f_1(X)}{f_0(X)}=\frac{\sigma_1 e^{-\sigma_1(x-\mu)}}{\sigma_0 e^{-\sigma_0(x-\mu)}}$$

$$\frac{\sigma_1 e^{-\sigma_1(x-\mu)}}{\sigma_0 e^{-\sigma_0(x-\mu)}}>c$$

$$\log \frac{\sigma_1 e^{-\sigma_1(x-\mu)}}{\sigma_0 e^{-\sigma_0(x-\mu)}}> \log c$$

$$\frac{\sigma_1}{\sigma_0}>\sqrt{c}.$$

Am I right? But what to do next? From theory I know that $$\overline{X}>c'$$. But how to apply it here?

• If this is some sort of homework, please add the 'self-study' tag. – StubbornAtom Mar 20 '19 at 11:38
• You need to replace the unknown parameters with their MLEs under each of the hypotheses – Glen_b -Reinstate Monica Mar 20 '19 at 12:03

You should be working with the density of the sample $$X_1,X_2,\ldots,X_n$$, given by

$$f_{\sigma}(x_1,x_2,\ldots,x_n)=\sigma^n \exp\left[-\sigma\sum_{i=1}^n(x_i-\mu)\right]\mathbf1_{x_{(1)}>\mu}\quad,\sigma>0,\mu\in\mathbb R$$

You are testing the hypothesis $$H_0:\sigma=\sigma_0$$ against the alternative $$H_1:\sigma=\sigma_1(>\sigma_0)$$.

The likelihood ratio is therefore

\begin{align} \frac{f_{H_1}(x_1,x_2,\ldots,x_n)}{f_{H_0}(x_1,x_2,\ldots,x_n)}&=\left(\frac{\sigma_1}{\sigma_0}\right)^n e^{-n(\bar x-\mu)(\sigma_1-\sigma_0)} \\&=ke^{-n\bar x(\sigma_1-\sigma_0)}\qquad,\small\text{ for some positive constant }k \end{align}

Since you are to reject $$H_0$$ for large values of this ratio, the critical region is of the form

\begin{align} ke^{-n\bar x(\sigma_1-\sigma_0)}>\text{ constant }\implies \bar x<\text{ constant } \end{align}