A.4 The Neyman–Pearson lemma

Suppose that P and Q are two probability measures on (Ω,F), and denote by

the Lebesgue decomposition of P with respect to Q as in Theorem A.17. For fixed c ≥ 0, we let

where we make use of the convention that dP/dQ = ∞on N.

Proposition A.33 (Neyman–Pearson lemma). If A ∈ F is such that Q[ A ]≤ Q[ A0 ], then P[ A ]≤ P[ A0 ].

Proof. Let Then F ≥ 0 on N, and (dP/dQ − c)F ≥ 0. Hence

This proves the proposition.

Remark A.34. In statistical test theory, A0 is interpreted as the likelihood quotient test of the null hypothesis Q against the alternative hypothesis P: If the outcome ω of a statistical experiment is in A0, then the null hypothesis is rejected. There are two possible kinds of error which can occur in such a test. A type 1 error occurs if the null hypotheses is rejected despite the fact that Q is the “true” probability. Similarly, a type 2 error occurs if the null hypothesis is not rejected, although Q is not the “true” probability. The probability of a type 1 error is given by Q[ A0 ]. This quantity is usually called the size or the significance level of the statistical test A0. A type 2 error occurs with probability P[ (A0)c ]. The complementary probability P[ A0 ] = 1 − P[ (A0)c ] is called the power of the test A0. In this setting, the set A of Proposition A.33 can be regarded as another statistical test to which our likelihood quotient test is compared. The proposition can thus be restated as follows: A likelihood quotient test has maximal power on its significance level.

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Indicator functions of sets take only the values 0 and 1. We now generalize Proposition A.33 by considering F-measurable functions ψ : Ω → [0, 1]; let R denote the set of all such functions.

Theorem A.35. Let and define the density φ := dP/dQ as above.

(a) Take c ≥ 0, and suppose that ψ0 ∈ R satisfies R-a.s.

Then, for any ψ ∈ R,

(b) For any α0 ∈ (0, 1) there is some ψ0 ∈ R of the form (A.30) such that ∫ ψ0 dQ = α0. More precisely, if c is an (1 − α0)-quantile of φ under Q, we can define ψ0 by

where κ is defined as

(c) Suppose conversely that ψ0 ∈ R satisfies (A.31) and ∫ ψ0 dQ > 0. Then ψ0 R-a.s. is of the form (A.30) for some c ≥ 0.

Proof. (a): Take F := ψ0 −ψ and repeat the argument in the proof of Proposition A.33.

(b): Let F denote the distribution function of φ under Q. Then Q[ φ > c ] = 1 − F (c) ≤ α0 and

Hence 0 ≤ κ ≤ 1 and ψ0 belongs to R. The fact that ∫ ψ0 dQ = α0 is obvious.

(c): Suppose that ψ0 ∈ R satisfies (A.31) and α0 := ∫ ψ0 dQ > 0. The case α0 = 1 is trivial. For 0 < α0 < 1, we let := + κwhere c and κ are as in part {φ>c} {φ= c}, (b). Then α0 = ∫ ψ0 dQ = ∫ dQ. One also has that ∫ ψ0 dP = ∫ dP, as can be seen by applying (A.31) to both ψ0 and with reversed roles. Hence, for f := − ψ0 and N = {φ = ∞},

But the fact that satisfies (A.30) implies that both f ≥ 0 P-a.s. on N, and (φ − c)f ≥ 0 Q-a.s. Hence f vanishes R-a.s. on {φ ≠ c}, and so ψ0 must also be R-a.s. of the form (A.30).

Remark A.36. In the context of Remark A.34, an element ψ of R is interpreted as a randomized statistical test: If ω is the outcome of a statistical experiment and p := ψ(ω), then the null hypothesis is rejected with probability p, i.e., after performing an independent random coin toss with success probability p. Significance level and power of a randomized test are defined as above, and a test of the form (A.30) is called a generalized likelihood quotient test. Thus, the general Neyman–Pearson lemma in the form of Theorem A.35 can be stated as follows: A randomized test has maximal power on its significance level, if and only if it is a generalized likelihood quotient test.

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