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A regularized logistic finite sum with an exact minimizer

Statement

Consider n=14 one-dimensional examples, seven with feature ai=1 and label yi=1, and seven with ai=1 and yi=−1. With unit ℓ2 regularization, define

fi(w)=log⁡(1+exp⁡(−yiw))+12w2,F(w)=114∑i=114fi(w). Here log⁡ denotes the natural logarithm.

Then each component is convex with 5/4-Lipschitz gradient, F is 1-strongly convex, and its unique minimizer is w∗=0. An SVRG epoch from snapshot w~=1 with η=2/25 and m=28 therefore satisfies

E[F(wJ)−F(0)]≤181224 [F(1)−F(0)],F(1)−F(0)=12log⁡(2+e+e−1)+12−log⁡2.

Facts & Assumptions

Given: The finite data set and objective in the Statement.

[F1]

Finite-sum SVRG: SVRG returns an independent uniform inner iterate and uses a full gradient plus two component gradients per inner update.

[F2]

Expected linear contraction of an SVRG epoch: if all components are convex, continuously differentiable and L-smooth, the average is μ-strongly convex with a stationary point x∗, n≥10L/μ, η=1/(10L), and m=2n, then from a deterministic snapshot with the theorem's independent uniform component draws and output index the expected gap relative to x∗ contracts by its stated q.

[F3]

Gradient of a finite sum is the finite sum of gradients: the gradient of the finite average is the average of component gradients.

[F4]

Strong convexity for an extended-real objective: μ-strong convexity is the strong-Jensen inequality with decrement (μ/2)θ(1−θ)∣x−y∣2.

[F5]

Lipschitz map, α-Hölder map for rational 0<α≤1, and contraction: an L-Lipschitz gradient obeys ∣fi′(u)−fi′(v)∣≤L∣u−v∣ in one dimension.

[F8]

The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a): the derivative of a composition is the derivative of the outer function applied to the derivative of the inner function.

[F11]

Multivariable Taylor formula with a Lagrange remainder along a line segment: a C2 scalar function on a segment has a second-order Taylor formula with Hessian remainder at an intermediate point.

[F12]
[F13]

The real exponential function and the number e by a power series: exp⁡(w) is the real exponential and e=exp⁡(1).

[F14]

Finite sums and finite products, by recursion: F is the finite average of its fourteen component losses.

[F15]

The natural logarithm as the inverse of the exponential function: for x>0, log⁡x is the unique real y with exp⁡(y)=x.

[F16]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: log⁡ is strictly increasing on (0,∞) and log⁡(ab)=log⁡a+log⁡b for a,b>0.

[F17]

The exponential is positive and satisfies exp⁡(−x)=1/exp⁡(x): exp⁡(w)>0 and exp⁡(−w)=1/exp⁡(w).

[F18]

Convex and strictly convex functions on Euclidean convex sets: convexity is the segment inequality for function values.

[F19]

Sign rules for products and monotonicity of multiplication: multiplication by a positive denominator preserves order. Squares of nonzero elements are positive makes a nonzero real square positive, and Multiplication by zero: 0⋅a=0 gives square zero in the zero case; thus every real square is nonnegative.

[F20]

Laws of finite sums and finite products: finite sums are additive and obey scaling.

[A1]

The Axiom of Choice: assume AC for the conditional-expectation interfaces inherited from the SVRG theorem [F2].

Proof

Given: The data, loss, and regularizer in the Statement.

1.1F6F7F8F9F13F17F19givenalgebra

Write r=exp⁡(w)>0 by [F13] and [F17]. By [F6]–[F9], f+′(w)=−1/(1+r)+w, f−′(w)=r/(1+r)+w, and f+′′(w)=f−′′(w)=1+r/(1+r)2. Since (1+r)2−4r=(r−1)2≥0 by [F19], 1≤fi′′(w)≤5/4 for every component and w.

2.1F4F5F10F11F12F14F18F20step 1.1givenalgebra

Apply [F10] to fi′ and [F12]; step 1.1 gives ∣fi′(u)−fi′(v)∣≤(5/4)∣u−v∣, so every component is 5/4-smooth. For any u,v, [F11] gives fi(v)≥fi(u)+fi′(u)(v−u)+(v−u)2/2. For 0≤θ≤1, put z=(1−θ)u+θv, apply this inequality from z to u,v, weight by 1−θ,θ, and add; the gradient terms cancel and (1−θ)(u−z)2+θ(v−z)2=θ(1−θ)(v−u)2. Hence (1−θ)fi(u)+θfi(v)≥fi(z)+12θ(1−θ)(v−u)2. Since the subtracted term is nonnegative, this gives ordinary convexity by [F18] and verifies the strong Jensen inequality in [F4]. Averaging these inequalities with [F20] proves that F is convex and 1-strongly convex.

3.1A1F1F2F3F13F14F15F16F17F20step 1.1step 2.1givenalgebra∎

At w=0, the two component derivatives in step 1.1 are −1/2 and 1/2, so [F3], [F14] and [F20] give F′(0)=0. Averaging the lower Taylor inequality from step 2.1 gives F(w)≥F(0)+F′(0)w+w2/2=F(0)+w2/2, so 0 is the unique minimizer. Substitution gives F(0)=log⁡2. The product law [F16] and e=exp⁡(1) from [F13], with e−1=1/e from [F17], give F(1)−F(0)=12log⁡(2+e+e−1)+12−log⁡2. Since n=14≥10(5/4)/1, η=2/25=1/(10L), and m=28=2n, [F1] and [F2] give the claimed SVRG guarantee with q=25L4μn+14=125224+56224=181224.

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