Uniformly Most Powerful (UMP) Test

Suppose we sample from a population with a distribution that is completely specified except for the value of a single parameter $\theta$. If we wish to test $H_0:\theta=\theta_0$ (simple) versus $H_1:\theta > \theta_0$ (composite) there is no general theorem like Theorem 3.1 that can be applied. But it can be applied to find a MP test for $H_0:\theta=\theta_0$ versus $H_A:\theta= \theta_a$ for any single value $\theta_a \in H_1$. In many situations the form of the rejection region for the MP test does not depend on the particular choice of $\theta_a$. When a test obtained by Theorem 1.3 actually maximizes the power for every value of $\theta > \theta_0$, it is said to be uniformly most powerful, (UMP) for $H_0:\theta=\theta_0$ against $H_1:\theta > \theta_0$.

We may state the definition as follows:

Definition 9..7
The critical region C is a uniformly most powerful critical (UMPCR) of size $\alpha$ for testing the simple hypothesis $H_0$ against a composite alternative $H_1$ if the set C is a best critical region of size $\alpha$ for testing $H_0$ against each simple hypothesis in $H_1$. A test defined by this critical region C is called a uniformly most powerful test, with significance level $\alpha$, for testing the simple $H_0$ against the composite $H_1$.

Uniformly most powerful tests don't always exist, but when they do, the Neyman Pearson Theorem provides a technique for finding them.

Example 9..6
Let $X_1, \ldots, X_n$ be a random sample from a $N(0, \theta)$ distribution where the variance $\theta$ is unknown. Find a UMP test for $H_0:\theta=\theta_0$ ($> 0$) against $H_1:\theta > \theta_0$.

Solution. Now $H_0 \cup H_1=\{\theta: \theta \geq \theta_0\}$. The likelihood of the sample is

$$L(\theta)=\left(\frac{1}{2 \pi \theta}\right)^{n/2} e^{-\frac{1}{2}\sum x_i^2/\theta}.$$

Let $\theta_a$ be a number greater than $\theta_0$ and let $K > 0$. Let C be the set of points where

$$L(\theta_0;x_1,\ldots,x_n)/L(\theta_a;x_1,\ldots,x_n) \leq K.$$

That is, the set of points where

$$\left(\frac{\theta_a}{\theta_0}\right)^{n/2}e^{-\frac{1}{2}\sum x_i^2( \theta_a-\theta_0)/\theta_0\theta_a} \leq K.$$

Or equivalently,

$$\sum x_i^2 \geq \frac{2\theta_0\theta_a}{(\theta_a-\theta_0)}... ...\log(\frac{\theta_a}{\theta_0})-\log K \right]=c\mbox{ , say}.$$

The set $C=\{(x_1,\ldots,x_n): \sum x_i^2 \geq c\}$ is then a BCR for testing $H_0:\theta=\theta_0$ against $H_1: \theta=\theta_a$. It remains to determine $c$ so that this critical region has the desired $\alpha$. If $H_0$ is true, $\sum X_i^2/\theta_0$ is distributed as $\chi^2_n$. Since

$$\alpha=P(\sum X_i^2/\theta_0 \geq \frac{c}{\theta_0}\big\vert H_0),$$

$c/\theta_0$ may be found from chisquare tables.

So $C$ defined above is a BCR of size $\alpha$ for testing $H_0:\theta=\theta_0$ against $H_1: \theta=\theta_a$. We note that, for each number $\theta_a > \theta_0$, the above argument holds. So $C=\{(x_1,\ldots,x_n): \sum x_i^2 \geq c\}$ is a UMP critical region of size $\alpha$ for testing $H_0:\theta=\theta_0$ against $H_1:\theta > \theta_0$. To be specific, suppose now that $\theta_0=3$, $n=15$, $\alpha=.05$, show that $c=75$.