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.


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!

  • 1
    $\begingroup$ I am baffled by where the denominator in $\lambda(x)$ comes from: it doesn't look at all like the correct likelihood. $\endgroup$
    – whuber
    Dec 30, 2016 at 21:38
  • $\begingroup$ @whuber I made a mistake in the second line, I've changed it since. Thanks! $\endgroup$
    – user321627
    Dec 30, 2016 at 21:52
  • 1
    $\begingroup$ 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. $\endgroup$
    – whuber
    Dec 30, 2016 at 21:54
  • $\begingroup$ @whuber When you say take logarithms at the outset, are you saying that: $\lambda(x) = \dfrac{sup_{\theta=\theta_0}L(\theta|X)}{sup_{\theta\neq\theta_0}L(\theta|X)} = \dfrac{sup_{\theta=\theta_0}\log L(\theta|X)}{sup_{\theta\neq\theta_0}\log L(\theta|X)}$? $\endgroup$
    – user321627
    Dec 30, 2016 at 22:00
  • $\begingroup$ Logs don't work that way. $\log\left(\frac{a}{b}\right)=\log(a)-\log(b)$. $\endgroup$
    – whuber
    Dec 30, 2016 at 22:27

1 Answer 1


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 $- 2 \theta_0\sum_{i=1}^n \ln X_i \ge \chi^2_{\alpha/2}(2n)$.

  • $\begingroup$ 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 $-n \ln \theta_{MLE} - \frac{\theta_0}{\theta_{MLE}} \le k \\$? $\endgroup$
    – user321627
    Dec 31, 2016 at 2:53
  • $\begingroup$ Yes it won't change anything though. I'll edit. $\endgroup$
    – user365239
    Dec 31, 2016 at 3:01
  • $\begingroup$ Thanks! Just curious, is there a reason why you worked with the chi square distribution instead of the gamma? $\endgroup$
    – user321627
    Dec 31, 2016 at 3:02
  • $\begingroup$ 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 $\endgroup$
    – user365239
    Dec 31, 2016 at 3:09
  • $\begingroup$ My edit changed the function $f(z)$ slightly, but the conclusion remains the same. Should have worded that better $\endgroup$
    – user365239
    Dec 31, 2016 at 3:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.