Neyman–Pearson lemma

Section of Most powerful tests, part of Statistical inference

Neyman–Pearson lemma

Type Section
Taught in Most powerful tests
Part Statistical inference
Questions using it 1

Among all tests of H_0:\theta=\theta_0 against H_1:\theta=\theta_1 with type I error at most \alpha, the likelihood-ratio test has the greatest power.

Let R be the likelihood-ratio region \{x:L(\theta_1;x)>k\,L(\theta_0;x)\}, with k chosen so that \int_R L(\theta_0;x)\,dx=\alpha. Let R^\star be any other region with the same size, \int_{R^\star} L(\theta_0)=\alpha. Write \pi(R)=\int_R L(\theta_1) for the power. Split each region into the part it shares and the part it does not:

\pi(R)-\pi(R^\star)=\int_{R\cap (R^\star)^c}L(\theta_1)-\int_{(R)^c\cap R^\star}L(\theta_1).

On R\cap(R^\star)^c the likelihood ratio exceeds k, and on (R)^c\cap R^\star it does not. Replacing L(\theta_1) by k\,L(\theta_0) therefore lowers the first integral and raises the second, and the difference is at least

k\int_{R\cap(R^\star)^c}L(\theta_0)-k\int_{(R)^c\cap R^\star}L(\theta_0)=k\int_R L(\theta_0)-k\int_{R^\star}L(\theta_0)=k\alpha-k\alpha=0.

Hence \pi(R)\ge\pi(R^\star). A discrete distribution may not hit \alpha exactly. The same argument applies to a randomised test that rejects with a fractional probability on the boundary point where the likelihood ratio equals k; without randomisation the most powerful test is the largest likelihood-ratio region whose size does not exceed \alpha. The binomial serve test on the hypothesis-testing page is that discrete case: at level 0.05 the region \{13,14,15\} has size 0.027, and the next larger region already has size 0.091.

The lemma compares two completely specified distributions. It does not by itself choose a test that is good for every alternative at once. That is the next definition.

Used in

See also

Uniformly most powerful test · Uniformly most powerful unbiased test · Neyman–Pearson approach

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