1
$\begingroup$

Cross posting from math.stackexchange...

Suppose that $X_1, ...X_n$ is a random sample of size $n$ drawn from an exponential distribution with unknown parameter $\theta$. Suppose that it is desired to test the hypotheses $H_0:\theta=\frac{1}{2}$ and $H_1:\theta>\frac{1}{2}$. Find the UMP test that specifies the rejection region and also find the constant $K$ such that the probability of Type-I Error is $\alpha=0.05$

Using the maximum likelihood ratio test, I come up with the following:

$\lambda=\frac{max L_0}{max L_\Omega}$ where $max L_0$ is the maximum likelihood under the null $(\theta=\frac{1}{2}$) and $max L_\Omega$ is the maximum likelihood under the whole set ($\theta\geq\frac{1}{2}$)

$\lambda=(\frac{\bar{x}}{2})^ne^{n-\frac{n\bar{x}}2}$

And I want to find the probability that $\lambda\leq k$ given $\theta=\frac{1}{2}$

Doing some algebraic manipulation, I get $\lambda=\frac{\bar{x}}{4}e^{-\frac{\bar{x}}{2}}$, which, as it turns out, is a Gamma distribution with shape parameter=2 and scale parameter=2 ($\alpha=2,\beta=2$)

Here is where I'm beginning to get stuck. If Gamma($\alpha=2, \beta=\frac{\nu}{2}$) follows Chi Square distribution with $\nu$ d.f, then it follows that $\lambda$ follows chi square with 4 d.f., right?

With an exponential distribution, if $\theta=\frac{1}{2}$ then that is a chi square distribution with 2 d.f.,right?

So, ultimately, I need to find a value K such that $P(\lambda\leq K|\theta=\frac{1}{2})=0.05$. I think that both $\lambda$ and $\theta$ follow chi square, but from there I'm lost.

$\endgroup$
  • $\begingroup$ You should not crosspost. Try what you think is the most appropriate site first. $\endgroup$ – Michael Chernick Apr 12 '17 at 13:27

Your Answer

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

Browse other questions tagged or ask your own question.