We have a random sample $X_1,\cdots,X_n \sim \mathrm{Beta}(\theta,1), \theta > 0$ is unknown. My ultimate goal is to find a UMP size $\alpha$ test for $H_0: \theta \le \theta_0$ v. $H_1: \theta > \theta_0$, where $\theta_0$ is specified. I recognize that I will need to use Karlin-Rubin Theorem to find the UMP size $\alpha$ test; the use of this theorem requires a sufficient statistic $T$ for $\theta$ and $\{g(t|\theta):\theta \in \Theta\}$ needs to have an monotone likelihood ratio.

my work:

I have found through the Factorization Theorem that $T=\prod^n_{i=1}X_i$ is sufficient for $\theta$. I am pretty confident that I can complete this problem after I show that $g(t|\theta)$ has the MLR property; and I will add the completed problem after this MLR property is shown.

How do you show that $g(t|\theta)$ has the MLR property? I know that a few papers exist that discuss the distribution of a product of independent Beta distributions, but I think that is above what we are expected to know to solve this problem.

updated work:

Based off of jld's answer, I am getting that the family for $T_n=\sum^n_{i=1}(-\log(X_i))$ has increasing MLR property.

$\frac{2}{\theta}\sum^N_{i=1}Y_i \sim \chi^2_{2n}$, where $Y_i=-\log(X_i)\sim \mathrm{Gamma}(n,\theta)$.

By applying the Karlin-Rubin Theorem, we get the UMP size $\alpha$ test to be:


  • 1
    $\begingroup$ You are to show that the ratio $f_{\theta_2}(x_1,\ldots,x_n)/f_{\theta_1}(x_1,\ldots,x_n)$ is monotone (either non-decreasing or non-increasing) as a function of $\prod x_i$ for every $\theta_2>\theta_1$ where $f_{\theta}$ is the joint density. Surely, this doesn't require knowing the distribution of the product. $\endgroup$ Apr 2, 2020 at 19:10

1 Answer 1


Typically sums are easier than products, so in this case I'd recommend using $$ T_n := -\sum_{i=1}^n \log X_i. $$ I'll start by working out the pdf of one of these RVs. Let $X \sim \text{Beta}(\theta,1)$ so $$ f_X(x) = \theta x^{\theta-1}\mathbf 1_{0 \leq x \leq 1}. $$ This transformation is a bijection on the support so we can use the Jacobian theorem to conclude $$ f_Y(y) = f_X(e^{-y})e^{-y} \\ = \theta e^{-(\theta-1)y - y}\mathbf 1_{y \geq 0} \\ = \theta e^{-\theta y}\mathbf 1_{y \geq 0} $$ which is $\text{Exp}(\theta)$ (depending on the parameterization).

For $T_n$ we can use MGFs: $$ M_Y(t) = \text{E}(e^{tY}) = \theta \int_0^\infty e^{-(\theta - t)y}\,\text dy \\ = \frac{\theta}{\theta - t}, \;\;t < \theta $$ (convolutions and induction could have also worked here if you don't want to use MGFs).

This means that $$ M_{T_n}(t) = \left(\frac{\theta}{\theta-t}\right)^n $$

which can be recognized as the MGF of a $\Gamma(n,\theta)$ random variable, therefore $$ f_{T_n}(t; \theta) = \frac{\theta^n}{\Gamma(n)} t^{n-1}e^{-\theta t}\mathbf 1_{t\geq 0}. $$

Note that $$ \frac{f_T(t;\theta_0)}{f_T(t; \theta_1)} = \left(\frac{\theta_0}{\theta_1}\right)^n e^{-(\theta_0-\theta_1)t}\mathbf 1_{t\geq 0}. $$

Does this help?

  • $\begingroup$ Thank you for your comment. I am a bit interested in knowing how you can make $T_n$ negative. Is it because all we need is a monotone transformation? $\endgroup$
    – Ron Snow
    Apr 2, 2020 at 21:35
  • $\begingroup$ I added an updated work section in my answer that uses your gracious work. Thank you! $\endgroup$
    – Ron Snow
    Apr 2, 2020 at 22:08
  • 1
    $\begingroup$ @Edison the function $x \mapsto -\log x$ is a bijection on $(0,1)$ so I'm just considering $-\log\left(\prod_i X_i\right)$ which is my $T_n$ instead of just the product (and the negative is because otherwise we'd have the negative of a gamma distribution since $\log x < 0$ for $x\in(0,1)$. It's typical safe to transform statistics like this with bijections $\endgroup$
    – jld
    Apr 2, 2020 at 22:20
  • $\begingroup$ Awesome- thank you so much! $\endgroup$
    – Ron Snow
    Apr 2, 2020 at 22:37
  • $\begingroup$ @Edison absolutely, no prob! $\endgroup$
    – jld
    Apr 2, 2020 at 23:00

Your Answer

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

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