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Stochastic approximation and stochastic gradient methods — Examples

1 · Prerequisites

2 · Summary

These examples work through one streaming least-squares update, a bounded-noise scalar Robbins–Monro root, and a biased simulator gradient. Together they distinguish conditional unbiasedness, almost-sure convergence with square-summable but nonsummable step sizes, and the nonzero stationarity floor caused by persistent bias.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

One streaming least-squares SGD update

Example

Let Y be equally likely to be −1 or 1 on a filtered space with trivial F0 and F1=σ(Y). Set ℓ(x,Y)=12(x−Y)2 and define the streaming least-squares objective f(x)=E[ℓ(x,Y)]. From x0=0, use the one-sample gradient G0=∂xℓ(x0,Y) and step α0=12. Compute the mean gradient, first iterate, and objective gap.

Facts & Assumptions

Given: The two-point finite probability space, the loss and objective in the Example, initial point x0=0, and step α0=1/2.

[A1]

The Axiom of Choice: AC is used for the canonical conditional-expectation interface.

[F1]

Finite probability spaces, outcome weights, events, and event probabilities: each of the two outcomes Y=−1,1 has weight 1/2.

[F2]

Expectation of a nonnegative or integrable random variable: on this finite space, expectation is the weighted finite sum.

[F3]

Filtration-adapted stochastic-gradient oracles and Robbins–Monro updates: the sample gradient is the oracle value, and the SGD step is x1=x0−α0G0.

[F6]

Conditional variance: Var⁡(G0∣F0)=E[(G0−E[G0∣F0])2∣F0].

Verification

1.1A1F1F2F3F5F6givenalgebra

By [F1]–[F2], f(x)=14(x+1)2+14(x−1)2=12(x2+1). Its difference quotient at x is (f(x+h)−f(x))/h=x+h/2, so [F5] gives ∇f(x)=x. Likewise (ℓ(h,Y)−ℓ(0,Y))/h=h/2−Y, so G0=−Y. Its conditional mean given the trivial F0 is zero, equal to ∇f(0), and [F6] with [F1]–[F2] gives conditional variance E[Y2]=1.

2.1F3F4step 1.1algebra∎

The update in [F3] gives x1=Y/2. Since Y2=1, f(x1)=12(1/4+1)=5/8, whereas f(0)=1/2 is the minimum because f(x)−f(0)=x2/2≥0. Hence the realized objective gap after this one sample is exactly 1/8.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A bounded-noise Robbins–Monro scalar root

Example

Consider h(x)=x with root x∗=0, initial point x0=0, and a noise sequence ξk∈{−1,1} where each ξk is Fk+1-measurable and satisfies E[ξk∣Fk]=0 and E[ξk2∣Fk]=1. No independence across iterations is assumed. Use steps αk=14(k+2)−3/4 and oracle Gk=xk+ξk. Verify the strong-monotonicity, stability, and stepsize conditions for the almost-sure Robbins–Monro theorem.

Facts & Assumptions

Given: The one-dimensional filtered model and noise conditions in the Example.

[A1]

The Axiom of Choice: AC is used by the conditional-expectation and almost-sure convergence interfaces.

[F1]

Filtration-adapted stochastic-gradient oracles and Robbins–Monro updates: xk+1=xk−αkGk, and conditional means and moments are interpreted almost surely.

[F2]

The p-series for a real exponent p converges exactly when p is greater than one: ∑n≥1n−p diverges for p≤1 and converges for p>1.

[F3]

Robbins–Monro converges almost surely for a strongly monotone root: a strongly monotone drift with a root, uniform conditional second moment, and positive steps with divergent sum and square-summable squares gives almost-sure convergence.

[F4]
[F5]

Taking out what is known: a known integrable factor may be taken outside conditional expectation.

[F6]

Basic algebra and order properties of conditional expectation: conditional expectation is linear and fixes integrable measurable quantities.

[F7]

The principle of mathematical induction: a bound true at the first index and inherited by each successor holds at every natural index.

Verification

1.1A1F1F4F5F6F7F8givenalgebrainduction

Since 0<αk<1, [F7] and [F8] give ∣xk∣≤1: it holds at x0=0, and if ∣xk∣≤1 then ∣xk+1∣=∣(1−αk)xk−αkξk∣≤(1−αk)∣xk∣+αk∣ξk∣≤1. Taking out the Fk-measurable xk with [F5] and using [F6] with the stated conditional moments gives E[Gk∣Fk]=xk=h(xk) and E[Gk2∣Fk]=xk2+1≤2. Also (h(x)−h(y))(x−y)=(x−y)2, so h is strongly monotone with μ=1.

2.1A1F2F3step 1.1givenalgebra∎

By [F2], ∑kαk=14∑n≥2n−3/4=∞ and ∑kαk2=116∑n≥2n−3/2<∞. Each αk≤1/4≤1/(2μ). Thus [F3] applies with M2=2, proving xk→0 almost surely.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

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.