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A continuous likelihood-ratio test

Example

On [0,1] with restricted Lebesgue measure, let P0 have density p0(x)=1 and let P1 have density p1(x)=2x. At level α=1/4, the likelihood-ratio test rejects for x>3/4, has size 1/4, and has power 7/16.

Verification

Given: The interval data space ([0,1],A), where A is the trace of the Lebesgue sigma-algebra, restricted Lebesgue measure μ, the densities p0(x)=1 and p1(x)=2x, and α=1/4.

[F0] Statistical experiments: an experiment consists of a data measurable space, a parameter set, and one probability law for each parameter.

[A1] The Axiom of Choice: AC is assumed for the general Neyman–Pearson theorem; the example's density representatives p0=1 and p1=2x are explicit.

[A2] The Axiom of Countable Choice (ACω): CC is used for the restricted Lebesgue-measure construction and the Riemann-to-Lebesgue equality below.

[F2] Continuous functions on Euclidean spaces are Borel measurable and A measurable function between measurable spaces: the continuous extension x↦2x on R is Borel measurable, so its restriction to [0,1] is measurable for the trace sigma-algebra A.

[F3] The measure with density f relative to μ and Integrating against a density agrees with integrating the product: A↦∫A2x dμ is a measure and integrates nonnegative g as ∫g(x)2x dμ(x).

[F4] A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion: the continuous function 2x is Riemann integrable on every compact subinterval.

[F7] For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε: for each ε>0 a positive natural N can be chosen with 1/N<ε.

[F9] Likelihood-ratio tests with boundary randomization, The likelihood-ratio rule is a measurable test of exact size, and Test size and power: the candidate rejects above its likelihood-ratio cutoff and has the displayed null size.

[F10] Neyman–Pearson lemma with singular alternative mass: no test of size at most 1/4 has greater power than the candidate.

[F11] Finite, sigma-finite, and semifinite measures: a finite measure is sigma-finite by the one-set cover consisting of the whole space.

[F12] A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral: under CC, a bounded Riemann-integrable function is Lebesgue measurable and its Lebesgue and Riemann integrals agree.

[F13] Probability measures and probability spaces: a measure with total mass 1 is a probability measure.

Proof technique: direct.

1.1A2F1F2F3F13given

By [A2] and [F1], Lebesgue measure is a measure and [0,1] is measurable with measure 1. Each event in the trace sigma-algebra A is therefore Lebesgue measurable, and restricting Lebesgue measure to these events preserves countable additivity. Hence μ is a probability measure by [F13]. Define Q1(A)=∫A2x dμ(x). The density is nonnegative and measurable by [F2], so [F3] makes Q1 a measure.

1.2F6algebrainduction

Let SN=∑k<Nk. Its recursion is S0=0 and SN+1=SN+N, so induction [F6] gives 2SN=N(N−1) for every N≥0.

2.1F4F5F6step 1.2algebra

Fix 0≤a<b≤1, put d=b−a, and divide [a,b] into N≥1 equal intervals of length d/N. Since 2x is increasing, the lower and upper sums are respectively LN=2ad+d2(1−1/N) and UN=2ad+d2(1+1/N), by substituting SN=N(N−1)/2 into the finite sums in [F5].

3.1A2F5F7F12step 2.1algebra

The value b2−a2=2ad+d2 lies between LN and UN, and their gap is 2d2/N. By [F7] this gap can be made arbitrarily small. The Lebesgue integral equals the Riemann integral by [A2] and [F12], while [F5] bounds that integral between LN and UN for every N. Hence ∫[a,b]2x dμ(x)=b2−a2.

4.1A2F0F1F3F8F9F11F13F14step 3.1algebra

Step 3.1 with a=0,b=1 gives Q1([0,1])=1, so P1:=Q1 is a probability measure by [F13]. Set P0:=μ and p0(x):=1; for each A∈A, [F14] gives ∫Ap0 dμ=μ(A), so p0 is a density of P0. By definition of Q1, p1(x):=2x is a density of P1. Thus they define the experiment [F0], and their likelihood ratio is R(x)=2x. The common measure μ is finite and hence sigma-finite by [F11]. With c=3/2 and γ=0, P0(R>c)=μ((3/4,1])=1/4 and P0(R=c)=μ({3/4})=0 by [F1]. Thus [F9] gives the rejection region x>3/4 and size 1/4.

4.2A2F1F3F8step 3.1algebra

The power is P1((3/4,1])=∫[3/4,1]2x dμ(x)=1−(3/4)2=7/16 by [F1], [F3], and step 3.1; the singleton boundary has zero measure.

5.1A1F10step 4.1step 4.2∎

Under AC [A1], the Neyman–Pearson lemma [F10] says no test with size at most 1/4 has larger power.

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