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Binary zero-one loss risks are the testing error probabilities

Statement

Let Θ0,Θ1 be disjoint subsets of a parameter set Θ in a statistical experiment. Let K be a randomized test on the binary action space {0,1}, with 1 meaning reject, and write φ(x)=K(x,{1}). Define the zero-one error loss by L(θ,a)=1Θ0(θ)a+1Θ1(θ)(1−a), so L is zero when θ∉Θ0∪Θ1. For βφ(θ)=Eθ[φ(X)] and the risk R(θ,K) of Loss functions and risk, R(θ,K)=βφ(θ)(θ∈Θ0),R(θ,K)=1−βφ(θ)(θ∈Θ1).

Facts & Assumptions

Given: A statistical experiment, disjoint hypotheses Θ0,Θ1, and a binary randomized test K with test function φ.

[F1]

Randomized tests: the action space is {0,1} with its full sigma-algebra, each K(x,⋅) is a probability measure, and φ(x)=K(x,{1}).

[F2]

Loss functions and risk: if each action section of L is measurable and nonnegative, then R(θ,K)=∫[∫L(θ,a)K(x,da)]Pθ(dx).

[F3]

Nonnegative simple measurable functions, The integral of a nonnegative simple function, and The nonnegative integral agrees with the simple integral on simple functions: the integral of an indicator of a measurable singleton under K(x,⋅) is the K(x,⋅)-measure of that singleton.

[F4]

Expectation of a nonnegative or integrable random variable: expectation under Pθ is integration with respect to Pθ.

[F5]

Monotonicity and nonnegative homogeneity of the nonnegative integral: if 0≤f≤g, then ∫f dPθ≤∫g dPθ.

[F6]

Integrable real and complex functions, and their integrals: a real measurable function is integrable when the integral of its absolute value is finite.

[F7]

The Lebesgue integral is linear on L1(μ): the integral is linear on integrable functions.

[F8]

Statistical experiments and Probability measures and probability spaces: each Pθ is a probability measure with Pθ(X)=1.

Proof

technique · direct
1.1givenF1F2

For fixed θ, the action section a↦L(θ,a) is one of a, 1−a, or 0. Each is nonnegative and measurable on the full sigma-algebra of {0,1}, so [F2] defines its risk.

2.1F1F3step 1.1algebra

On Θ0, the section is a=1{1}(a), whose integral under K(x,⋅) is K(x,{1})=φ(x) by [F1] and [F3]. On Θ1, the section is 1−a=1{0}(a), whose integral is K(x,{0})=1−φ(x) because K(x,⋅) is a probability measure.

3.1F2F4step 2.1

Substituting these inner integrals into [F2] and using [F4] gives R(θ,K)=Eθφ(X) on Θ0 and R(θ,K)=Eθ[1−φ(X)] on Θ1.

4.1F1F5F6F7F8step 3.1algebra∎

Since 0≤φ≤1 and 0≤1−φ≤1, [F5] and [F8] show that each has integral at most 1. Thus their absolute-value integrals are finite, so they are integrable by [F6]. Linearity [F7] and [F8] give Eθ[1−φ]=1−Eθ[φ]. The two risk identities now have the asserted form.

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