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Suppose a fixed-step size stochastic gradient descent iteration satisfiesand $f$ has a global minimizer $x^{\ast }$. Taking expectations and forming a telescoping sum proves the displayed bound with ${\Delta }_0=f(X^0)-f(x^{\ast })$ and deterministic $X^0$. The Cauchy-Schwarz inequality then bounds the averaged, or the best, expected gradient Euclidean norm by $\sqrt{M+2L{\Delta }_0/k}$. This is a guarantee of stationarity in the indicated senses, rather than a last-iterate convergence assertion. The additive $M$ is a stochastic gradient noise floor.

Target: real-analysis.bigb, under Strongly convex function.

There is a summation-index misprint. On the native PDF, paper page 3, equation (4) has $X^k$ inside a sum indexed by $j$. The intended telescoping estimate has $X^j$ inside that sum; this is the estimate proved below and used in part (d). The index mismatch is present in the original PDF, rather than being introduced by the transcription.

Take the expectation of the supplied conditional expectation bound (3), using the tower property:Rearranging and forming a telescoping sum givesBecause $x^{\ast }$ is a global minimizer, $f(X^k)\ge f(x^{\ast })$ for every realisation. Hence the corrected equation (4) isAll expectations at a finite iteration are finite here: a deterministic starting point and finitely many choices of the sampled index produce only finitely many possible iterates. The bound controls accumulated squared gradients, with an additive contribution from the stochastic gradient noise floor.

Write ${\Delta }_0=f(x^0)-f(x^{\ast })$. Dividing the corrected estimate from part (c) by $k/(2L)$ gives the averaged stationarity bound for stochastic gradient descentThe Cauchy-Schwarz inequality, first within each expectation and then over the $k$ indices, givesThusEquivalently, if $J$ is chosen independently of the run with the discrete uniform distribution on $\{0,\dots ,k-1\}$, then $\mathbb{E}\Vert \nabla f(X^J){\Vert }^2\le M+2L{\Delta }_0/k$. The same bound for the expected Euclidean norm follows by the Cauchy-Schwarz inequality.

The guarantee is averaged or best-iterate stationarity. It does not assert that every last iterate converges, or that the limiting Euclidean norm equals $\sqrt{M}$. The squared gradient bound has excess $O(k^{-1})$ above $M$; the displayed expected Euclidean norm bound is within $O(k^{-1/2})$ of $\sqrt{M}$, including the zero-noise case. For fixed $M>0$, it is sharper:For example, to ensure ${\min }_{j<k}\mathbb{E}\Vert \nabla f(X^j)\Vert \le \sqrt{M}+\epsilon$ with $\epsilon >0$, it suffices that $k\ge 2L{\Delta }_0/(2\sqrt{M}\epsilon +{\epsilon }^2)$. The big O notation here concerns these upper bounds, rather than an asserted asymptotic equality for the stochastic gradient noise floor.

The printed distance needs a square. The native PDF, paper page 3, prints $\Vert x^0-x^{\ast }\Vert$ in the final bound, but the general conclusion from (5) contains $\Vert x^0-x^{\ast }{\Vert }^2$. We first prove the corrected statement and then show that the printed version can fail.

Take the expectation in (5), use the tower property, and sum over $j=0,\dots ,k-1$. The squared Euclidean distances form a telescoping sum:Discard the final nonnegative squared Euclidean distance. The smallest of $k$ numbers is at most their arithmetic mean, soThis is an $O(k^{-1})$ bound above the stochastic gradient noise floor $M/L$ for the best expected objective function value. By Jensen's inequality, it also bounds $\mathbb{E}f({\bar{X}}^k)-f(x^{\ast })$ for the averaged iterate ${\bar{X}}^k=k^{-1}{\sum }_{j=0}^{k-1}{X}^{j+1}$, because $f$ is a convex function.

To test the printed version, take one component function, $f_1(x)=f(x)=x^2/4$, with $L=1$, $x^{\ast }=0$ and $x^0=10$. This convex quadratic function has $\nabla f(x)=x/2$, hence it has a $1$-Lipschitz gradient. The stochastic gradient descent update is deterministic, its conditional variance is zero, and the assumed positive bound $M=1/10$ is valid. At $k=1$, the new iterate is $X^1=5$, soThus the printed inequality is false under its stated assumptions. The squared Euclidean distance in the corrected bound resolves the defect.