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A bounded-noise Robbins–Monro scalar root

Example

Consider $h (x) = x$ with root $x^{*} = 0$ , initial point $x_{0} = 0$ , and a noise sequence $ξ_{k} \in {- 1, 1}$ where each $ξ_{k}$ is $F_{k + 1}$ -measurable and satisfies $E [ξ_{k} ∣ F_{k}] = 0$ and $E [ξ_{k}^{2} ∣ F_{k}] = 1$ . No independence across iterations is assumed. Use steps $α_{k} = \frac{1}{4} (k + 2)^{- 3 / 4}$ and oracle $G_{k} = x_{k} + ξ_{k}$ . 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: $x_{k + 1} = x_{k} - α_{k} G_{k}$ , 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: $\sum_{n \geq 1} n^{- p}$ diverges for $p \leq 1$ 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]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : in dimension one, $∥ x ∥_{2} = ∣ x ∣$ .

[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.

[F8]

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: the Euclidean norm satisfies the triangle inequality.

Verification

1.1A1F1F4F5F6F7F8givenalgebrainduction

Since $0 < α_{k} < 1$ , [F7] and [F8] give $∣ x_{k} ∣ \leq 1$ : it holds at $x_{0} = 0$ , and if $∣ x_{k} ∣ \leq 1$ then $∣ x_{k + 1} ∣ = ∣ (1 - α_{k}) x_{k} - α_{k} ξ_{k} ∣ \leq (1 - α_{k}) ∣ x_{k} ∣ + α_{k} ∣ ξ_{k} ∣ \leq 1.$ Taking out the $F_{k}$ -measurable $x_{k}$ with [F5] and using [F6] with the stated conditional moments gives $E [G_{k} ∣ F_{k}] = x_{k} = h (x_{k})$ and $E [G_{k}^{2} ∣ F_{k}] = x_{k}^{2} + 1 \leq 2$ . Also $(h (x) - h (y)) (x - y) = (x - y)^{2}$ , so $h$ is strongly monotone with $μ = 1$ .

2.1A1F2F3step 1.1givenalgebra∎

By [F2], $\sum_{k} α_{k} = \frac{1}{4} \sum_{n \geq 2} n^{- 3 / 4} = \infty$ and $\sum_{k} α_{k}^{2} = \frac{1}{16} \sum_{n \geq 2} n^{- 3 / 2} < \infty$ . Each $α_{k} \leq 1 / 4 \leq 1 / (2 μ)$ . Thus [F3] applies with $M^{2} = 2$ , proving $x_{k} \to 0$ almost surely.

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