Here is the problem I am studying:
"Consider a distribution with pdf $f(x;\theta)=\theta x^{\theta-1}$ if $0<x<1$ and $0$ otherwise. Derive the most powerful test for the hypotheses of $H_0: \theta=1$ vs $H_a: \theta=2$ based on a random sample of size $n$."
The answer was given to me as:
"Reject $H_0$ if $\prod_{i=1}^nx_i \geq c$ where $P(\prod_{i=1}^nX_i \geq c |\theta=1)=\alpha$."
But I don't understand how to arrive at that answer. In fact, I am stuck on trying to simplify the Neyman-Pearson lemma. So, I have:
$\lambda(x;\theta_0,\theta_1) = \frac{ (\theta_0 x_i^{\theta_0-1})^n }{ (\theta_1 x_i^{\theta_1-1)^n } } = \frac{ \theta_0^n \prod_{i=1}^nx_i^{\theta_0-1} }{ \theta_1^n \prod_{i=1}^nx_i^{\theta_1-1} } $
Here is where I am stuck. Can I simplify this to:
$ \frac{\theta_0^n}{\theta_1^n} \prod_{i=1}^n x_i^{\theta_0-\theta_1} $
I don't really know how the division of two products with power terms is supposed to work. Actually, to be honest, I'm not 100% sure how these products are supposed to work; if all the $x_i$'s are the same, we can add the power terms together. But that doesn't work here, right? Even if I can simplify it this far, I don't know how to get from that to the answer given above.
Can anybody give me some pointers, here?
Ryan,
Going with the notation you are using, note that you are given $H_0: \theta_0 = 1$ and $H_1: \theta_1 = 2$. Plug those in for $\theta_0$ and $\theta_1$ and then you will see that the likelihood ratio simplifies to something proportional $\left( \prod_{i=1}^n x_i \right)^{-1}$.
Thus you would reject $H_0$ when $\prod_{i=1}^n x_i$ is large, which explains the answer.