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For an unbiased stochastic gradient $H$ at $X$ with finite conditional second moment, expand $H=\nabla f(X)+(H-\nabla f(X))$ in the squared Euclidean norm. The cross term has zero conditional expectation, which proves the displayed identity. It separates the squared true gradient from the total conditional variance of its random estimate.

Target: gradient-descent.bigb, under Stochastic gradient descent.

A sigma-algebra $\mathcal{G}$ on the same underlying set as $\mathcal{F}$, with $\mathcal{G}\subseteq \mathcal{F}$. In probability theory it represents information retained from the original probability space and is a possible conditioning collection for a conditional expectation.

If a finite group $\Gamma$ acts by measure-preserving maps on a probability space, let $\mathcal{I}$ be its invariant sigma-algebra. For an integrable random variable $Z$, the displayed group average is $\mathcal{I}$-measurable. For $A\in \mathcal{I}$, changing variables under each group element gives ${\int }_{A}Z\circ T_{g}d\mathbb{P}={\int }_{A}Zd\mathbb{P}$. The average therefore satisfies the defining integral identity for conditional expectation. For permutations of $N$ exchangeable coordinates, averaging a product attached to $d$ distinct indices yields the sum over all ordered distinct $d$-tuples divided by the falling factorial $(N{)}_d=N!/(N-d)!$.

For two strictly positive zero-rate stock martingales with bounded volatilities and the same initial value, proportional terminal prices must be equal because their expectations coincide. Conditional expectation then makes the whole paths equal. Comparing their stochastic integrals by the Itô isometry gives equality of the volatility coefficients almost everywhere in time and probability. Perfect positive correlation alone does not provide the required proportionality.

For one integrable random variable $Y$, the family $Z_{\mathcal{G}}=\mathbb{E}[Y\mid \mathcal{G}]$, over arbitrary sigma-algebras $\mathcal{G}$, has uniform integrability. On $A=\{|Z_{\mathcal{G}}|>K\}$, the defining property of conditional expectation gives $\mathbb{E}[|Z_{\mathcal{G}}|1_A]\le \mathbb{E}[|Y|1_A]$, while $\mathbb{P}(A)\le \mathbb{E}|Y|/K$ by the Markov inequality. The integrable random variable $Y$ has uniformly small absolute integrals over events of sufficiently small probability.

Write $X_{-i}$ for all inputs except $X_i$. Under the Efron–Stein inequality hypotheses, variance tensorization is the equivalent conditional variance formConditioning freezes the other inputs. Then $Z,Z_i^{\prime }$ are independent with the same conditional probability distribution, givingTaking expected values explains the factor $1/2$.^[ref]

For Bernoulli input $X_i$ with parameter $p_i$, let ${\Delta }_{i}f$ be the value of $f$ with coordinate $i$ set to one minus its value with that coordinate set to zero, keeping $X_{-i}$ fixed. The two-point conditional variance isFor the maximum example, ${\Delta }_{1}f=1-X_2$ and ${\Delta }_{2}f=1-X_1$. The two averaged conditional variances are $0.3(0.7)(0.4)=0.084$ and $0.6(0.4)(0.7)=0.168$, precisely half the respective resampling terms. In the dependent example $X_1=X_2=B$, each other coordinate already determines $W$, so both conditional variances vanish; variance tensorization also fails without independence.

For a short induction proof, separate the last input using the law of total variance. Apply the smaller-dimensional statement to the conditional expectation; Jensen's inequality for the square bounds its coordinate contributions by the original ones. One input gives equality, starting the induction.^[ref]