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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:

$\phi(\mathbf{x})=1,T(\mathbf{x})>\frac{\theta}{2}\chi^2_{2n;\alpha}$

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    $\begingroup$ 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

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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?

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  • $\begingroup$ 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
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    $\begingroup$ @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

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