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Normal location score, information, and attainment

Example

Assume AC. Let X∼N(μ,σ2), where μ∈R is the parameter and σ2>0 is fixed. With Lebesgue measure as dominator, the density is pμ(x)=12πσ2exp⁡ ⁣(−(x−μ)22σ2). The estimator T=X is unbiased for ψ(μ)=μ, and it attains the scalar Cramér–Rao bound.

Facts & Assumptions

Given: AC; σ2>0; the normal location model and density displayed above; and T(X)=X.

[A1]

The Axiom of Choice: AC supplies the normal probability law and the arbitrary-index density representatives used by the general score definition.

[F1]

Standard normal and normal laws: N(0,1) has density ϕ(z)=e−z2/2/2π, and N(μ,σ2) is the law of μ+σZ for Z∼N(0,1).

[F2]

The standard normal density has total mass one: the standard normal density ϕ(z)=e−z2/2/2π integrates to 1. The required second moment is established locally below by compact-interval integration. Also Gaussian even moments for Brownian increments gives EZ2=1: instantiate its increment as Xt=Z minus the constant Xs=0, with labels s=0,t=1 and m=1; no Brownian process is required.

[F3]

Change of variables for expectation, Integrating against a density agrees with integrating the product, and The Lebesgue integral is linear on L1(μ): expectations under a density are density integrals, affine normal laws are computed from X=μ+σZ, and expectation is linear on integrable variables.

[F4]

Score and Fisher information: the score is the derivative of the log density and scalar information is its expected square.

[F5]

Mean-zero score and the information identity: the stated regularity hypotheses imply mean-zero score and I=−E[ℓ′′].

[F6]

Moments, variance, and covariance on a probability space: variance and covariance are centered second moments.

[F7]

Scalar Cramér–Rao lower bound: under the score-covariance identity, the scalar variance bound has equality exactly when the centered estimator is (ψ′/I) times the score.

[F11]

Monotone convergence for the integral: increasing sequences of nonnegative measurable functions pass to the limit under the integral.

[F12]

A linear map T of Rn sends Lebesgue measurable sets to Lebesgue measurable sets, with λn(T[E])=∣det⁡T∣ λn(E) when T is invertible and T[E] Lebesgue null when it is not: under Countable Choice, for a>0 and Lebesgue measurable E⊆R, λ(aE)=aλ(E). The assumed AC supplies this premise. For nonnegative measurable f, correspondence of simple minorants under the bijection z↦az and the defining supremum for the nonnegative integral give ∫f(az) dz=a−1∫f(y) dy. Likewise Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation and the same simple-minorant argument give ∫f(x−b) dx=∫f(y) dy for every real b.

Verification

technique · direct
1.1A1F1F9givenalgebra

Let Z∼N(0,1) have density ϕ, and put G(z):=zϕ(z). By [F9], ϕ′(z)=−zϕ(z) and G′(z)=ϕ(z)−z2ϕ(z). These functions are smooth.

2.1F3F10step 1.1algebra

For each integer n≥1, G′ is continuous on [−n,n], hence integrable there by [F10]. Applying the second FTC and then the Riemann–Lebesgue integral identity in [F10] gives ∫[−n,n](ϕ−z2ϕ) dλ=G(n)−G(−n)=2nϕ(n). Linearity [F3] therefore yields ∫[−n,n]z2ϕ dλ=∫[−n,n]ϕ dλ−2nϕ(n).

3.1F2F3F9F11step 2.1algebra

By the power-series representation in [F9], en2/2≥(n2/2)2/2!=n4/8, so 0≤nϕ(n)≤8/(2πn3)→0. Also ϕ1[−n,n]↑ϕ and z2ϕ1[−n,n]↑z2ϕ. Monotone convergence [F11], normalization [F2], and the compact calculation in step 2.1 give EZ2=∫z2ϕ dλ=lim⁡n∫[−n,n]z2ϕ dλ=1<∞, also the second moment supplied in [F2].

4.1F1F3F4step 3.1givenalgebra

Differentiating the displayed normal density gives ∂ηpη=(x−η)pη/σ2, ∂ηηpη=((x−η)2/σ4−1/σ2)pη, Sμ=(X−μ)/σ2, and ℓμ′′=−1/σ2. The support is all of R, and step 3.1 with X=μ+σZ gives EμSμ2=1/σ2<∞.

5.1F1F3F5F12step 3.1step 4.1givenalgebra

Fix μ0∈R and δ>0, put y=x−μ0, and take ∣η−μ0∣<δ. Then ∣x−η∣≤∣y∣+δ and (x−η)2≥y2/2−δ2, so both first and second density derivatives are bounded by C(1+y2)e−y2/(4σ2) for a finite C independent of η,x. First, write the displayed density as pμ(x)=σ−1ϕ((x−μ)/σ). For every Borel set A, the scaling and translation identities [F12] with a=σ and b=μ give ∫Apμ(x) dx=∫1A(μ+σz)ϕ(z) dz=P(μ+σZ∈A), so pμ is indeed a density of the specified law N(μ,σ2). To justify the envelope transformation, put a=2σ>0 and q(y)=(4πσ2)−1/2e−y2/(4σ2). For every Borel set A, [F12] applied to f=1Aq gives ∫Aq(y) dy=a∫1A(az)q(az) dz=∫1A(az)ϕ(z) dz, since aq(az)=ϕ(z). By [F1] and [F3], the last integral equals P(aZ∈A); thus q is indeed a density of the affine law Y=aZ∼N(0,2σ2). The envelope is a constant multiple of (1+y2)q(y). By [F3] and step 3.1, its integral in y is finite, since E(1+Y2)=1+a2EZ2=1+2σ2. Translation invariance in [F12] transfers this finite integral from y to x=y+μ0, proving integrability of the displayed envelope on the original data space. Thus the fixed-support, continuous-derivative, envelope, and score-moment hypotheses of [F5] hold, so EμSμ=0 and I(μ)=−Eμℓμ′′=1/σ2.

6.1F1F3F6step 3.1step 5.1algebra

From X−μ=σ2Sμ and step 5.1, EμX=μ. Also Eμ[(X−μ)2]=σ2EZ2=σ2, so T=X has variance σ2.

7.1F4F6F7F8step 5.1step 6.1algebra

The pointwise identity X−μ=σ2Sμ and [F6] give Cov⁡μ(X,Sμ)=σ2I(μ)=1. On R, ψ(μ)=μ has difference quotient 1 at every x≠μ, so [F8] gives ψ′(μ)=1; the score-covariance identity required by [F7] holds.

8.1F6F7step 6.1step 7.1algebra∎

The bound [F7] is Var⁡μ(X)≥1/I(μ)=σ2, and step 6.1 gives equality. Also X−μ=σ2Sμ=(ψ′(μ)/I(μ))Sμ, so the stated equality condition holds.

Depends on

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