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Conditional variance of independent and without-replacement minibatches

Statement

Let n,d≥1, let (Ω,F,(Fk)k≥0,P) be a discrete filtered probability space, and fix k. Let fi:Rd→R be continuously differentiable, let x:Ω→Rd be Fk-measurable, and suppose each gi=∇fi(x) is square-integrable. Define F=n−1∑ifi, gˉ=n−1∑igi=∇F(x) by Gradient of a finite sum is the finite sum of gradients, and

σ2(x)=1n∑i=1n∥gi−gˉ∥22.

For an integer b≥1, draw I1,…,Ib as mutually independent uniform indices from {1,…,n}, with the whole index vector independent of Fk, and set Gb=b−1∑j=1bgIj. Then

E[Gb∣Fk]=gˉ,E[∥Gb−gˉ∥22∣Fk]=σ2(x)b.

For sampling without replacement, let S be a uniform subset of {1,…,n} of size b, independent of Fk, and set GS=b−1∑i∈Sgi. If n>1 and 1≤b≤n, then

E[GS∣Fk]=gˉ,E[∥GS−gˉ∥22∣Fk]=n−bb(n−1)σ2(x).

When n=1, the without-replacement case requires b=1; both estimators equal gˉ and both errors are zero.

Facts & Assumptions

Given: The discrete filtered probability space Filtration and filtered probability space, the time k, the Fk-measurable point and square-integrable gradients, and the two sampling laws in the Statement. Vector conditional expectations mean coordinatewise scalar conditional expectations.

[A1]

The Axiom of Choice: assume AC for conditional-expectation classes.

[F1]

Filtration-adapted stochastic-gradient oracles and Robbins–Monro updates: stochastic-gradient conditional expectations are taken with respect to the current filtration, and AC is assumed for their existence.

[F2]

Filtration and filtered probability space: Fk is a sigma-algebra and Fk⊆F.

[F3]

Conditional expectation as an ae class: under AC, conditional expectations of integrable random variables exist as almost-sure classes.

[F4]

Conditioning a known variable and an independent variable: an integrable random variable independent of the conditioning sigma-algebra has conditional expectation equal to its mean.

[F5]

Taking out what is known: an integrable factor measurable in the conditioning sigma-algebra can be taken outside conditional expectation.

[F6]

Basic algebra and order properties of conditional expectation: conditional expectation is linear and preserves finite integrable sums.

[F7]

The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn: ∥z∥22=⟨z,z⟩, and the inner product is bilinear.

[F8]

Gradient of a finite sum is the finite sum of gradients: the gradient of F is the average of the component gradients.

[F9]

Finite sums and finite products, by recursion: a finite average is the finite recursively defined sum scaled by 1/n.

[F10]

Laws of finite sums and finite products: finite sums are additive, obey scaling, and permit termwise algebra.

Proof

Given: The objects and sampling assumptions in the Statement.

1.1A1F1F2F3F4F5F6F7F8F9F10givenalgebra

For every j,i, independence and [F4] give E[1{Ij=i}∣Fk]=1/n. Since gi is Fk-measurable and square-integrable, [F1]–[F3], [F5], and [F6] allow gIj=∑i1{Ij=i}gi to be conditioned coordinatewise and termwise, yielding E[gIj∣Fk]=gˉ. For j≠ℓ and every A∈Fk, joint independence gives P(A,Ij=r,Iℓ=s)=P(A)/n2; hence E[⟨gIj−gˉ,gIℓ−gˉ⟩∣Fk]=n−2∑r,s⟨gr−gˉ,gs−gˉ⟩=⟨n−1∑r(gr−gˉ),n−1∑s(gs−gˉ)⟩=0, since each centered finite sum vanishes. Expanding ∥Gb−gˉ∥22 leaves b diagonal terms, each with conditional expectation σ2(x), so division by b2 proves the with-replacement identities.

1.2A1F1F2F4F6F10givenalgebra

Assume n>1. Let Zi=1{i∈S}. Relabeling the elements of the uniform size-b subsets shows that E[Zi∣Fk] has one common value p for all i, and that E[ZiZj∣Fk] has one common value r for all ordered pairs i≠j; independence of S from Fk makes these conditional values constant. There are n singleton indices and n(n−1) ordered distinct pairs, by choosing the first and then the second index. Every such subset satisfies ∑iZi=b and ∑i≠jZiZj=b(b−1), since each selected index has exactly b−1 other selected indices. Taking conditional expectations and using finite-sum linearity [F6, F10] gives np=b and n(n−1)r=b(b−1). Thus p=b/n and r=b(b−1)/(n(n−1)), including b=1 where r=0.

2.1A1F1F4F5F6F7F10step 1.2givenalgebra∎

The first inclusion probability in step 1.2, with [F5] and [F6], gives E[GS∣Fk]=b−1∑i(b/n)gi=gˉ. Put ai=gi−gˉ; since GS−gˉ=b−1∑iZiai, [F5]–[F7] give E[∥GS−gˉ∥22∣Fk]=b−2[(b/n)∑i∥ai∥22+(b(b−1)/(n(n−1)))∑i≠j⟨ai,aj⟩]. The cross sum is −∑i∥ai∥22 by expanding ⟨∑iai,∑jaj⟩=0, so the result is (n−b)/(b(n−1))σ2(x). If n=1, the only draw and subset contain the sole component, so both errors are zero.

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