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A DQM score has mean zero and is unique

Statement

Assume the density-version convention for the path in A dominated differentiable-in-quadratic-mean submodel. If s∈L2(P) is a DQM score, then Ps=0, so s∈L02(P) (Mean-zero square-integrable functions). Moreover, any two scores satisfying the same DQM expansion agree P-almost surely. The normalization argument itself uses no choice beyond the density versions already supplied in the definition.

Facts & Assumptions

Given: A common-dominated probability path Pt through P=P0, with density versions pt and a DQM score s∈L2(P).

[F1]

DQM means ∥pt−p−(t/2)sp∥L2(μ)=o(t) as t↓0 (A dominated differentiable-in-quadratic-mean submodel).

[F2]

A Radon–Nikodym density pt represents Pt by Pt(E)=∫Ept dμ for every measurable E (A sigma-finite signed measure that is absolutely continuous with respect to a sigma-finite positive measure has a unique almost-everywhere density).

[F3]

The L2 pairing is finite and satisfies Cauchy–Schwarz (Cauchy-Schwarz inequality for L2).

[F4]

The Lebesgue integral is linear on L1(μ) (The Lebesgue integral is linear on L1(μ)).

Proof

technique · Expand the squared norm and use normalization of the path
1.1F1F2

Put q=p, qt=pt, and rt=qt−q−(t/2)sq. By [F1], ∥rt∥2=o(t), while [F2] gives ∥q∥22=∫p dμ=1 and ∥qt∥22=∫pt dμ=1.

1.2F1algebra

If s′ is another score with remainder rt′=oL2(t), subtracting its expansion from [F1] gives ∥(t/2)(s−s′)q∥2≤∥rt∥2+∥rt′∥2=o(t).

2.1step 1.1F3F4algebra

Set ut=(t/2)sq+rt. Since ∥sq∥22=∫s2p dμ=Ps2<∞, we have ∥ut∥2=O(t) and qt=q+ut. The products qrt and ut2 are integrable by [F3]. Pointwise expansion and linearity [F4] give 0=∫(q+ut)2dμ−∫q2dμ=t Ps+2∫qrt dμ+∥ut∥22. By [F3], ∣∫qrt dμ∣≤∥q∥2∥rt∥2=o(t) and ∥ut∥22=O(t2)=o(t). Dividing by t>0 and taking t↓0 gives Ps=0.

3.1step 1.2step 2.1algebra∎

Divide by t/2 and let t↓0. Then ∥(s−s′)q∥22=P[(s−s′)2]=0, so s=s′ as an element of L2(P). Together with step 2.1 this proves mean zero and uniqueness.

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