Neyman-Pearson Lemma

Study guide: Exam MAS-I

The Neyman-Pearson Lemma says how to build the most powerful test of a simple null hypothesis $H_0: f = f_0$ against a simple alternative $H_1: f = f_1$ : reject $H_0$ where the likelihood ratio $L(\mathbf{x}) = f_0(\mathbf{x})/f_1(\mathbf{x})$ is small. Among all tests whose Type I error probability is at most $\alpha$ , this best critical region has the greatest power.

$C = \left\{\mathbf{x} : \frac{f_0(\mathbf{x})}{f_1(\mathbf{x})} \le k\right\}$

$P(\mathbf{X} \in C \mid H_0) = \alpha$

$f_0$ and $f_1$ are the joint densities (or PMFs) of the sample under each hypothesis; for a random sample they are products, $f_j(\mathbf{x}) = \prod_i f_j(x_i)$ . The constant $k > 0$ is whatever makes the size equal $\alpha$ .
The guarantee: if any other critical region $A$ has $P_0(A) \le P_0(C)$ , then $P_1(A) \le P_1(C)$ .
In practice the inequality $L \le k$ simplifies to a condition on a sufficient statistic, such as $\sum X_i \ge c$ , and $c$ is set from that statistic's distribution under $H_0$ . With discrete data only certain sizes $\alpha$ can be attained exactly.
Uniformly most powerful (UMP) tests. If the best critical region for $\theta_0$ against $\theta_1$ is the same for every $\theta_1$ in a composite alternative, it is most powerful against each of them, and so is UMP. The one-sided tests for an exponential mean, a binomial $p$ , or a normal mean or variance are UMP; the two-sided tests are not, because the best region for $\theta_1 > \theta_0$ is the opposite tail from the best region for $\theta_1 < \theta_0$ .
The Likelihood Ratio Test generalizes the ratio to composite hypotheses by maximizing each likelihood over its parameter set.

Most Powerful Test for an Exponential Claim Size A single claim $X$ is exponential. Test $H_0$ : mean $100$ against $H_1$ : mean $200$ with $\alpha = 0.05$ . Find the best critical region and its power.
Answer The likelihood ratio is
 $\begin{align*} \frac{f_0(x)}{f_1(x)} &= \frac{\tfrac{1}{100}e^{-x/100}}{\tfrac{1}{200}e^{-x/200}} \\ &= 2\,e^{-x/200} \end{align*}$ 
which decreases in $x$ , so $L \le k$ is the same as $X \ge c$ . Set the size:
 $\begin{align*} P(X \ge c \mid H_0) &= e^{-c/100} \\ &= 0.05 \\ c &= 100 \ln 20 \\ &= 299.57 \end{align*}$ 
The power is $P(X \ge 299.57 \mid H_1) = e^{-299.57/200} = 0.224$ . No other test with a 5% Type I error rate detects the larger mean more often.

A Uniformly Most Powerful Test for Mean Severity Claim sizes are normal with $\sigma = 250$ . From $n = 25$ claims, test $H_0: \mu = 1{,}000$ against $H_1: \mu = 1{,}100$ at $\alpha = 0.05$ . Find the best critical region and its power, and explain why the same test is UMP for $H_1: \mu > 1{,}000$ .
Answer For normal data the ratio $f_0/f_1$ is a decreasing function of $\bar{x}$ when $\mu_1 > \mu_0$ , so the best region is $\bar{X} \ge c$ . Under $H_0$ , $\bar{X}$ is normal with standard deviation $250/\sqrt{25} = 50$ :
 $\begin{align*} c &= 1{,}000 + 1.645(50) \\ &= 1{,}082.25 \end{align*}$ 
Power at $\mu = 1{,}100$ :
 $\begin{align*} P(\bar{X} \ge 1{,}082.25 \mid \mu = 1{,}100) &= P\left(Z \ge \frac{1{,}082.25 - 1{,}100}{50}\right) \\ &= P(Z \ge -0.355) \\ &= 0.639 \end{align*}$ 
The cutoff $1{,}082.25$ was fixed by $H_0$ alone and never used the value $1{,}100$ , so the same region is most powerful against every $\mu_1 > 1{,}000$ — it is the UMP test of $H_0: \mu = 1{,}000$ against $H_1: \mu > 1{,}000$ .