0
$\begingroup$

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 $$f(x;\mu,\sigma)=\sigma e^{-\sigma(x-\mu)}, \mu<x<\infty.$$

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

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?

$\endgroup$
  • $\begingroup$ If this is some sort of homework, please add the 'self-study' tag. $\endgroup$ – StubbornAtom Mar 20 '19 at 11:38
  • $\begingroup$ You need to replace the unknown parameters with their MLEs under each of the hypotheses $\endgroup$ – Glen_b -Reinstate Monica Mar 20 '19 at 12:03
1
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.