Assume $X_1$ .. $X_n$ ~ iid beta($\theta$,1)

I'm supposed to construct a Maximum likelihood ratio test for:

$H_0$: $\theta$ = $\theta_0$

$H_1$: $\theta$ != $\theta_0$

But I'm stuck at the last step of simplifying the rejection region.

My steps are as follows:

1) Obtain the MLE (skipping derivation)

$$\theta_{mle} = \frac{-n}{\sum_{i=1}^n log(x_i)}$$

2) Likelihood ratio:

$$ \frac{\prod_{i=1}^n \frac{\Gamma(\theta_0+1)}{\Gamma(\theta_0)\Gamma(1)} x_i^{\theta_0-1} (1-x_i)^{1-1}}{\prod_{i=1}^n\frac{\Gamma(\theta_{mle}+1)}{\Gamma(\theta_{mle})\Gamma(1)}x_i^{\theta_{mle}-1} (1-x_i)^{1-1}}$$

Simplify the gamma terms and Cancel the last term:

$$ \frac{\theta_0\prod_{i=1}^n x_i^{\theta_0-1}}{\theta_{mle}\prod_{i=1}^nx_i^{\theta_{mle}-1}}$$

Combine the product terms:

$$ \frac{\theta_0}{\theta_{mle}}\prod_{i=1}^n{x_i^{\theta_0 - \theta_{mle}}} $$

Rejection Region:

$$ P (\frac{\theta_0}{\theta_{mle}}\prod_{i=1}^n{x_i^{\theta_0 - \theta_{mle}}} < K) $$

And now I'm stuck. I know i must get to something like

$$ P (\bar{x} < c) $$

But I'm unsure how to proceed. Any help will really be appreciated!!

  • $\begingroup$ 1. Under "2)" ... how did you get from the first line to the second? ... 2. Why do you think you must necessarily get a closed form on the LHS of the inequality? (It's not always true) $\endgroup$
    – Glen_b
    Jun 5 '17 at 9:39
  • $\begingroup$ @Glen_b Gamma(thetha + 1) = thetha!, so it cancels out to leave thetha_0 and thetha_mle Err what do you mean for the second part? $\endgroup$
    – Wboy
    Jun 5 '17 at 9:44
  • $\begingroup$ 1. Yes, I know $\Gamma(\theta+1) = \theta!$ and how it cancels, leaving a $\theta_0$ on the numerator in front of the term with $x_i$ in it and $\theta_{mle}$ on the denominator. How do you get from that $\prod \theta_0 x_i^{\theta_0-1}$ (and a similar term on the denominator) to your second line? 2. You said "I know I must get something like ..." ... how do you "know" this? Why must it come out to be in a simple form like that? (I am not saying it doesn't - nor that it does - in this case) $\endgroup$
    – Glen_b
    Jun 5 '17 at 9:48
  • $\begingroup$ @Glen_b Oh i see, I assumed since they have the same base Xi, I can just subtract the powers instead ($\theta_0 - 1$) - $(\theta_{mle} - 1)$ $\endgroup$
    – Wboy
    Jun 5 '17 at 9:57
  • $\begingroup$ @Glen_b For all the previous likelihoods I've done, It always gets to X bar or x, I was taught to reduce everything until that form $\endgroup$
    – Wboy
    Jun 5 '17 at 9:58