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

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) &= \dfrac{sup_{\theta=\theta_0}L(\theta|X)}{sup_{\theta\neq\theta_0}L(\theta|X)} \\ &= \dfrac{\theta_0^n \left(e^{\sum log x_i}\right)^{\theta_0-1}}{\left(\frac{n}{-\sum logx_i}\right)^n \left(e^{\sum logx_i}\right)^{\left(\frac{n}{-\sum log x_i} - 1\right)}} \\ &= \left(\dfrac{-\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 $\theta^{MLE} = \frac{n}{-\sum log x_i}$. Now, I'd like to find the likelihood ratio test, in that I would like to choose a constant $c$ such that:

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

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

• I am baffled by where the denominator in $\lambda(x)$ comes from: it doesn't look at all like the correct likelihood.
– whuber
Dec 30 '16 at 21:38
• @whuber I made a mistake in the second line, I've changed it since. Thanks! Dec 30 '16 at 21:52
• I think you might be making things too hard for yourself. Take logarithms at the outset and keep expressing the MLE as $\hat\theta$ rather than as an explicit function of the data. That will make the essential simplicity of the problem more evident.
– whuber
Dec 30 '16 at 21:54
• @whuber When you say take logarithms at the outset, are you saying that: $\lambda(x) = \dfrac{sup_{\theta=\theta_0}L(\theta|X)}{sup_{\theta\neq\theta_0}L(\theta|X)} = \dfrac{sup_{\theta=\theta_0}\log L(\theta|X)}{sup_{\theta\neq\theta_0}\log L(\theta|X)}$? Dec 30 '16 at 22:00
• Logs don't work that way. $\log\left(\frac{a}{b}\right)=\log(a)-\log(b)$.
– whuber
Dec 30 '16 at 22:27

Using whuber's method, we'll reject if $\ell(\theta_0; \vec x) - \ell(\theta_{MLE}; \vec x) \le k$ for some constant $k$, where $\ell(\theta; \vec x) = \ln L(\theta ; \vec x) = \ln \left(\theta^n \prod_{i=1}^n x_i^\theta \right) = n \ln \theta + \theta \sum_{i=1}^n \ln x_i$.

We see that $\ell(\theta_0; \vec x) - \ell(\theta_{MLE}; \vec x) = n (\ln \theta_0 - \ln \theta_{MLE}) + (\theta_0 - \theta_{MLE}) \sum_{i=1}^n \ln x_i$.

Using some generic constant $k$ at each step, we reject if \begin{align*} n (\ln \theta_0 - \ln \theta_{MLE}) + (\theta_0 - \theta_{MLE}) \sum_{i=1}^n \ln x_i \le k \\ -n \ln \theta_{MLE} + (\theta_0 - \theta_{MLE}) \left( - \frac{n}{\theta_{MLE}} \right) \le k \\ -n \ln \theta_{MLE} - \frac{n\theta_0}{\theta_{MLE}} \le k \\ \ln \theta_{MLE} + \frac{\theta_0}{\theta_{MLE}} \ge k \end{align*}

Now, consider the function $f(z) = \ln(z) + \frac{\theta_0}{z}$. This has a minimum at $z = \theta_0$ and is concave up and $\lim \limits_{z \to 0^+} f(z) = \lim \limits_{z \to \infty} f(z) = \infty$. So $f(z) \ge k$ means $z$ is either sufficiently small or sufficiently large.

That is, $- \frac{n}{\sum_{i=1}^n \ln(x_i)}$ is either sufficiently small or sufficiently large; i.e. $- \sum_{i=1}^n \ln x_i \le c_1$ or $- \sum_{i=1}^n \ln x_i \ge c_2$. We want to choose $c_1$ and $c_2$ so that the test is size $\alpha$ under $H_0$.

Note that under $H_0$, $- \sum_{i=1}^n \ln X_i \sim \Gamma(n,\theta_0) \implies -2\theta_0 \sum_{i=1}^n \ln X_i \sim \Gamma(n,\frac{1}{2} ) = \chi^2(2n).$

So one LRT of size $\alpha$ is to reject $H_0$ in favor of $H_1$ if $- 2 \theta_0 \sum_{i=1}^n \ln X_i \le \chi^2_{1-\alpha/2}(2n)$ or if $- 2 \theta_0\sum_{i=1}^n \ln X_i \ge \chi^2_{\alpha/2}(2n)$.

• For the above, are we missing an $n$ from step $-n \ln \theta_{MLE} + (\theta_0 - \theta_{MLE}) \left( - \frac{n}{\theta_{MLE}} \right) \le k$ to $-n \ln \theta_{MLE} - \frac{\theta_0}{\theta_{MLE}} \le k \\$? Dec 31 '16 at 2:53
• Yes it won't change anything though. I'll edit. Dec 31 '16 at 3:01
• Thanks! Just curious, is there a reason why you worked with the chi square distribution instead of the gamma? Dec 31 '16 at 3:02
• Mainly because working with and integrating the gamma pdf can be a pain (two parameters instead of one, and integrating to some value $< \infty$ and then solving for your constant might pose a problem) .Overall the chi-square distribution is less cumbersome. Also, chi-square tables are well known & tabulated IRL Dec 31 '16 at 3:09
• My edit changed the function $f(z)$ slightly, but the conclusion remains the same. Should have worded that better Dec 31 '16 at 3:12