How can I show that $\prod^n_{i=1}X_i$ has a monotone likelihood ratio? 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:
$\phi(\mathbf{x})=1,T(\mathbf{x})>\frac{\theta}{2}\chi^2_{2n;\alpha}$
 A: 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?
