Neyman Pearson Theorem

Suppose $X_1, \ldots, X_n$ is a random sample with joint density function $f({\bf X};\, \theta)$. For simple $H_0$ and simple $H_1$, the joint density function can be written as $f_0({\bf x}; \, \theta)$, $f_1({\bf x}; \,\theta)$, respectively. Alternatively, we could use the likelihood notation, $L(\theta_0; \, {\bf x})$, $L(\theta_1;\, {\bf x})$.

Theorem 9..1

In testing $H_0:\theta=\theta_0$ against $H_1:\theta=\theta_1$, the critical region

$$C_K=\{{\bf x}: \,f_0({\bf x})/f_1({\bf x}) < K \}$$

is most powerful (where K $\geq$ 0).

[Or, in terms of likelihood, for a given $\alpha$, the test that maximizes the power at $\theta_1$ has rejection region determined by

$$L(\theta_0;x_1, \ldots, x_n)/L(\theta_1;x_1,\ldots,x_n) \ < \ K.$$

Such a test will be most powerful for testing $H_0$ against $H_1$.]

Proof. See Hogg and Craig. (Not required).

Example 9..5
Suppose $X$ represents a single observation from the probability density function given by

$$f(x;\theta)=\theta x^{\theta-1}, \ 0 \ < \ x \ < \theta.$$

Find the most powerful (MP) test with significance level $\alpha=.05$ to test $H_0:\theta=1$ versus $H_1: \theta=2$.

Solution. Since both $H_0$ and $H_1$ are simple, the previous Theorem can be applied to derive the test. Here

$$\frac{L(\theta_0)}{L(\theta_a)}=\frac{f(x;\theta_0)}{f(x;\theta_a)} =\frac{1 \times x^{1-1}}{2 \times x^{2-1}}=\frac{1}{2x}.$$

The form of the rejection region for the MP test is

$$\frac{1}{2x}<k.$$

Equivalently, $x > 1/2k$ or, since $1/2k$ is a constant ($k'$ say), the critical region is $x \ > \ k'$.

The value of $k'$ is determined by

$$\begin{eqnarray*} .05&=&P(X \mbox{ is in the critical region when }\theta=1)\\ ... ... k' \mbox{ when }\theta=1)\\ &=& \int_{k'}^{1}1.dy \\ &=& 1-k' \end{eqnarray*}$$

So $k' \ = \ .95$. So the rejection region is C $=\{y:y > .95 \}$. Among all tests for $H_0$ versus $H_1$ based on a sample of size 1 and $\alpha=.05$, this test has smallest Type II error probability.
[ Note that the form of the test statistic and rejection region depends on both $H_0$ and $H_1$. If $H_1$ is changed to $\theta=3$, the MP test is based on $Y^2$ and we reject $H_0$ in favour of $H_1$ if $Y^2 > k'$ for some $k'$].