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Rates for randomized coordinate gradient descent and importance sampling

Statement

Let f:Rd→R be convex and continuously differentiable, with a partition into n≥1 nonempty blocks and block-gradient Lipschitz constants Li>0 as in Coordinate-block smoothness descent model. Let Mi≥Li, choose probabilities pi>0 with ∑ipi=1, and set ρ=min⁡ipi/Mi. Assume f∗=min⁡xf(x) is attained at x∗ and the initial sublevel set has finite radius R=sup⁡{∥x−x∗∥2:f(x)≤f(x0)}<+∞. Run Randomized coordinate gradient descent from the initial point x0 with exact updates, no stopping predicate, and finite budget K∈N. For every integer k with 1≤k≤K, E[f(xk)−f∗]≤2R2ρk. If, in addition, f is globally μ-strongly convex in the sense that f(y)≥f(x)+⟨∇f(x),y−x⟩+μ2∥y−x∥22(x,y∈Rd) for some μ>0, then E[f(xk)−f∗]≤(1−μρ)k(f(x0)−f∗). The strong-convexity factor is in [0,1]. Uniform sampling with Mi=Lmax⁡:=max⁡iLi gives the factor 1−μ/(nLmax⁡) and the bound 2nLmax⁡R2/k. For nonuniform sampling, choosing Mi=Li and pi=Li/∑jLj maximizes min⁡ipi/Mi, giving ρ=1/∑iLi; its constants are not the uniform constants.

Facts & Assumptions

Given: The objective, block partition, smoothness constants, sampling probabilities, and finite-history expectation in the Statement.

[F1]

Randomized coordinate gradient descent: at a selected block i, the update is xi+=xi−Mi−1∇if(x) and the block draw has probability pi.

[F2]

Coordinate-block smoothness descent model: f(x+Uih)≤f(x)+⟨∇if(x),h⟩+(Li/2)∥h∥22.

[F3]

The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn: the product Euclidean norm satisfies ∥x∥22=∑i∥xBi∥22.

[F5]

Differentiable convex functions are characterized by the gradient inequality: convex differentiability gives f(y)≥f(x)+⟨∇f(x),y−x⟩.

[F6]

Fermat's theorem: an interior differentiable local extremum has zero gradient: an unconstrained differentiable local minimizer has zero gradient.

[F7]

Laws of finite sums and finite products: finite weighted sums preserve nonnegativity and obey the ordinary algebraic sum laws.

[F8]

The principle of mathematical induction: a base case and successor implication prove a claim for every natural number.

[A1]

f is convex, continuously differentiable, and satisfies the block-gradient Lipschitz hypotheses stated above.

[A2]

The initial sublevel radius R is finite, the algorithm has no stopping predicate and k≤K, and every expectation at a fixed k is the finite weighted sum over the k-step index histories, so no infinite product probability space is needed.

[A3]

For the linear rate only, f satisfies the displayed global μ-strong-convexity inequality and has a minimizer.

Proof

1.1A1F1F2algebra

Fix x and a block i. With h=−∇if(x)/Mi, [F2] gives f(x+Uih)≤f(x)−1Mi∥∇if(x)∥22+Li2Mi2∥∇if(x)∥22≤f(x)−12Mi∥∇if(x)∥22, because Mi≥Li.

1.2A3F2algebra

Under [A3], strong convexity at x and x∗ gives, for e=x−x∗, f(x)−f∗≤⟨∇f(x),e⟩−(μ/2)∥e∥22≤∥∇f(x)∥22/(2μ); the last inequality follows by expanding 0≤∥∇f(x)−μe∥22. Hence ∥∇f(x)∥22≥2μδ(x). Also μ≤Li≤Mi for each block, since the strong-convexity lower model and [F2]'s upper model applied to any nonzero displacement in block i bound the same function difference.

2.1A2F1F7step 1.1algebra

Conditional on the first k draws, the next draw has probabilities pi. By the finite-sum laws [F7], averaging step 1.1 over those finitely many outcomes yields E[f(xk+1)∣xk]≤f(xk)−12∑ipiMi∥∇if(xk)∥22≤f(xk)−ρ2∥∇f(xk)∥22. Taking the finite-history expectation gives the same inequality with all terms averaged.

3.1A1A2F3F4F5F7F8step 2.1algebra

Put δ(x)=f(x)−f∗. If R=0, then x0=x∗ because x0 belongs to the initial sublevel set, and [F6] gives ∇f(x∗)=0; all later updates leave x∗ fixed and all gaps vanish. Now suppose R>0. By [F5], δ(x)≤⟨∇f(x),x−x∗⟩. Since every exact step decreases f, each iterate lies in the initial sublevel set; thus [F4] and the definition of R give δ(xk)≤R∥∇f(xk)∥2. For each finite k-step history ω, write wω for its probability and zω=∥∇f(xωk)∥2; the weights are nonnegative and sum to 1. With zˉ=∑ωwωzω, [F7] gives 0≤∑ωwω(zω−zˉ)2=∑ωwωzω2−zˉ2. Hence E∥∇f(xk)∥22≥(E∥∇f(xk)∥2)2≥ϕk2/R2. Step 2.1 therefore implies ϕk+1≤ϕk−ρϕk2/(2R2).

4.1F6F7step 3.1algebra

If some ϕj=0, then f(xj)=f∗ for every history of positive probability, [F6] gives zero gradient at each such unconstrained minimizer, and all later gaps are zero. Otherwise all ϕj>0, and ϕj+1≤ϕj gives 1ϕj+1−1ϕj=ϕj−ϕj+1ϕjϕj+1≥ϕj−ϕj+1ϕj2≥ρ2R2. Summing for j=0,…,k−1 telescopes by [F7] to 1/ϕk≥kρ/(2R2), which is the stated O(1/k) bound.

5.1A2A3F7F8step 2.1step 1.2algebra∎

Substituting step 1.2 in step 2.1 gives E[δ(xk+1)∣xk]≤(1−μρ)δ(xk). Because 0<μ≤Mi and pi≤1, 0≤μρ≤1. Taking expectations and using induction [F8] proves the linear bound. For any probabilities, [F7] gives 1=∑ipi≥ρ∑iMi, so ρ≤1/∑iMi; equality holds for pi=Mi/∑jMj. Since Mi≥Li, this is at most 1/∑iLi, attained by Mi=Li and pi=Li/∑jLj. The uniform and importance-sampling specializations follow by substitution.

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