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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 Simmons
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1 Answers1

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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.

gogurt
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