I'm having trouble to find a UMP test after finding a MP test.

Consider one observation $X$ from CDF $F_\theta(x) = x^\theta$ where $x \in [0, 1]$ and $\theta > 0$.

I found the MP test for testing $H_0: \theta = 1$ against $H_1: \theta = 2$ with significance level $\alpha=0.05$ using the Neyman Pearson lemma: $$\lambda_{\theta_0, \theta_1}(x) = \frac{f_{\theta_0}}{f_{\theta_1}}= \cdots = \frac{1}{2x}$$

Reject $H_0$ if $\lambda_{\theta_0, \theta_1}(x)\leq\frac{1}{c}$, hence if $X\geq \tilde{c}$, where $\tilde{c} = 0.95$.

Now I'm asked to find the UMP test for testing $H_0: \theta \in [\frac{1}{2}, 1]$ against $H_1: \theta=2$ for significance level $\alpha = 0.05$. How to proceed?

  • $\begingroup$ The MP correct and the UMP is 1-(beta) $\endgroup$ Commented Jun 27, 2015 at 16:17

1 Answer 1


Before trying to find a UMP test, one needs to first check if there exists one. To do this one needs to find the likelihood ratio function


This function must be monotone non-decreasing in $x$ for every $\theta_1\geq \theta_0$. In the given question $\theta_1=2$, and the density function is $$f_{2}(x)=2x.$$ Similarly for $\theta_0\in[1/2,1]$, $$f_{\theta_0}(x)=\theta_0x^{\theta_0-1}$$ Hence, the likelihood ratio function is


Since this function is increasing in $x$ for all $\theta_0\in[1/2,1]$, there exists a UMP test of level $\alpha$.

By definition of UMP test, the significance level $\alpha$ is the expected value of the decision rule (which is the likelihood ratio test with a certain threshold $\lambda$), for which the false alarm probability lies below $\alpha$, for every $\theta_0$


Now, we have a nice simplification (Why?) $${\{x:l_{\theta_0}(x)>\lambda\}}\equiv {\{x:x>\lambda^{'}\}}$$



It is known that $\lambda^{'}\in[0,1]$ and $\theta_0\in[1/2,1]$. Now what value of $\theta_0$ maximizes $1-{\lambda^{'}}^{\theta_0}$ or similarly minimizes ${\lambda^{'}}^{\theta_0}$?

The UMP test is then $$\phi(x)=\begin{cases}1,\quad x>\lambda^{'}\\0,\quad x\leq \lambda^{'}\end{cases}$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .