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A biased simulator gradient has limiting squared gradient b2

Example

Let the probability space be the one-point space Ω={ω0} with P({ω0})=1 and the trivial filtration Fk={∅,Ω} for every k. Let f(x)=x2/2, so finf⁡=0 and ∇f(x)=x. A deterministic simulator returns the biased gradient Gk=xk+b for fixed b∈R. Starting at x0=0 with constant step α=1/4, compute the iterates and their average squared true gradient.

Facts & Assumptions

Given: The one-dimensional objective, biased oracle, initial value, and stepsize in the Example.

[A1]

The Axiom of Choice: AC is used by the conditional-expectation formulation of the bias-floor theorem.

[F1]

Filtration-adapted stochastic-gradient oracles and Robbins–Monro updates: the unconstrained update is xk+1=xk−αGk.

[F2]

Biased SGD has an explicit weighted stationarity floor: for an L-smooth objective and steps α≤1/(4L), bounded bias β and variance σ2 give an asymptotic average squared-gradient bound (5/2)β2+Lασ2.

[F5]

The principle of mathematical induction: a formula true at k=0 and preserved by the recurrence holds for every natural index.

[F7]

Basic algebra and order properties of conditional expectation: conditional expectation preserves real constants, so E[c∣Fk]=c for every c∈R.

[F8]

Conditional variance: conditional variance is the conditional expectation of the squared deviation from the conditional mean.

Verification

1.1F1F3F4F5givenalgebrainduction

With q=3/4, [F1] gives xk+1=qxk−b/4. [F5] applies to xk=−b(1−qk): it holds at k=0, and substitution gives qxk−b/4=−b(q(1−qk)+(1−q))=−b(1−qk+1). Thus [F3] implies xk→−b and ∥∇f(xk)∥22=xk2→b2.

2.1A1F2F4F6F7F8step 1.1givenalgebra∎

The difference quotient (f(x+h)−f(x))/h=x+h/2 tends to x, so [F6] gives ∇f(x)=x. For all x,y, ∣∇f(x)−∇f(y)∣=∣x−y∣=∥x−y∥2 by [F4], and the stated objective satisfies f(x)≥0. The recurrence in step 1.1 makes each xk deterministic; [F7] then gives conditional mean xk+b=∇f(xk)+b, and [F8] gives conditional variance 0. Also x0=0 is square-integrable. For every N≥1, 1N∑k=0N−1xk2=b2(1−2(1−qN)N(1−q)+1−q2NN(1−q2))⟶b2. The two fractions tend to zero because their numerators are bounded and their denominators grow linearly in N. With L=1, β=∣b∣, σ=0, and α=1/4=1/(4L), [F2] applies and gives the (valid, though nonsharp) upper limit (5/2)b2, while the exact average tends to b2.

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