[Math] Obtaining a level-$\alpha$ likelihood ratio test for $H_0: \theta = \theta_0$ vs. $H_1: \theta \neq \theta_0$ for $f_\theta (x) = \theta x^{\theta-1}$

hypothesis testingmaximum likelihoodstatistical-inferencestatistics

Suppose I have a sequence of iid random variables $X_1, \ldots, X_n$ following the pdf:

$$
f_\theta (x) = \theta x^{\theta-1}
$$

for $\theta >0$ and $0 <x<1$.

I would like to obtain a level-$\alpha$ likelihood ratio test for the null hypothesis $H_0: \theta = \theta_0$ versus the two-sided alternative $H_1: \theta \neq \theta_0$ where $\theta_0$ is a known constant.

MY ATTEMPT:

I first construct the ratio:

\begin{align}
\lambda(x) &= \frac{\sup_{\theta=\theta_0}L(\theta\mid X)}{\sup_{\theta\neq\theta_0}L(\theta\mid X)} \\[10pt]
&= \frac{\theta_0^n \left(e^{\sum \log x_i}\right)^{\theta_0-1}}{\left(\frac n {-\sum \log x_i}\right)^n \left(e^{\sum \log x_i}\right)^{\left(\frac n {-\sum \log x_i} – 1\right)}} \\[10pt] &= \left(\frac{-\theta_0 \sum \log x_i} n \right)^n e^{n+\theta_0 \sum \log x_i}
\end{align}

The denominator is calculated using the MLE of $\theta$ which is $\theta^\text{MLE} = \frac n {-\sum \log x_i}$. Now, I'd like to find the likelihood ratio test, in that I would like to choose a constant $c$ such that:

$$
\alpha = \sup_{\theta = \theta_0} P_\theta (\lambda(x) \leq c)
$$

Now, $-\sum \log x_i \sim \operatorname{Gamma}(n, \theta)$ but I CANNOT isolate the above equation due to the $\log$ form. What is the right answer here? Thanks!

Best Answer

Let $T=-\log(X_1 \cdots X_n).$ This is positive with probability $1$. The likelihood-ratio test statistic as derived in the question is an increasing function of $T^n \exp(-\theta_0 T).$ This has its largest possible value when $T = n/\theta_0 >0$ and it decreases as $T$ moves away from $n/\theta_0$ in either direction, approaching $0$ as $T\downarrow 0$ or $T\to\infty.$ Thus you reject $H_0$ if $T$ is either too big or too small. If you find $c_1$ and $c_2$ so that $\Pr(T<c_1) = \alpha/2 = \Pr(T>c_2),$ then you have a test at the level that you want.