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Convex SCAFFOLD control-lag recursion

Statement

Let fi:Rd→R be convex, differentiable and β-smooth, with f=N−1∑ifi minimized at x⋆. Run Option II SCAFFOLD with a uniform size-S client sample, 1≤S≤N, and conditionally unbiased local stochastic gradients of variance at most σ2. Assume the client controls are updated to averages of the K local stochastic gradients when selected, and define the control lag Cr and local drift Er as in the SCAFFOLD analysis-trajectory definition. Then, for every r≥1, Cr≤(1−SN)Cr−1+SN(4βE[f(xr−1)−f⋆]+2β2Er).

Facts & Assumptions

Given: The Option II control updates, local paths and convex equal-client objective in the Statement.

[F1]

SCAFFOLD with client and server control variates: a selected client replaces its control by the average of the local gradients used in the round; an unselected control persists.

[F2]

SCAFFOLD trajectories, client drift, control lag and tracking error: Cr is the expected squared error of the mean full gradient at a client's stored query anchors from ∇fi(x⋆), averaged over clients, and Er averages the squared displacements at the current gradient-query points yi,k−1r.

[F3]

Smooth convex gradient differences around a minimizer and Gradient of a finite sum is the finite sum of gradients: the average client gradient at the minimizer is zero and the population mean of squared client-gradient differences from x⋆ is at most 2β(f(x)−f⋆).

[F4]

Relaxed inequalities for squared Euclidean norms and The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn: squared norms are convex and satisfy ∥u+v∥22≤2∥u∥22+2∥v∥22.

[F5]

Expectation of a nonnegative or integrable random variable and Linearity, monotonicity, and the modulus bound for expectation: expectations preserve the nonnegative inequalities and finite weighted averages used below.

Proof

1.1F1F2F5givenalgebra

Set p=S/N. By [F1], a uniform sample retains a client's old query [given] anchors with probability 1−p and replaces them with the current local query points with probability p. The definition of Cr in [F2] and Jensen's inequality for the average of K vectors therefore give Cr≤(1−p)Cr−1+pNK∑i=1N∑k=1KE∥∇fi(yi,k−1r)−∇fi(x⋆)∥22.

2.1F2F3F4step 1.1givenalgebra∎

For each summand, decompose ∇fi(y)−∇fi(x⋆)=(∇fi(xr−1)−∇fi(x⋆))+(∇fi(y)−∇fi(xr−1)). By [F4], its squared norm is at most twice the squared norm of each term. Averaging the first term over clients and using [F3] gives at most 4βE[f(xr−1)−f⋆]; averaging the second over clients and local steps and using β-Lipschitzness gives at most 2β2Er. Average step 1.1 over clients and substitute these two bounds to obtain the recurrence.

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