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Sub-exponential concentration for independent sums

Statement

Let X1,…,Xn be independent centered real random variables, with Xi sub-exponential with parameters (νi2,bi) as defined in Sub-Gaussian variables and vectors, and sub-exponential variables, where n≥1 and bi>0. Put v=∑i=1nνi2, b=max⁡1≤i≤nbi, and S=∑i=1nXi. For t>0, if v>0 then P(∣S∣≥t)≤2exp⁡ ⁣(−12min⁡ ⁣{t2v,tb}), and if v=0 then P(∣S∣≥t)=0.

Facts & Assumptions

Given: Independent centered real random variables Xi with sub-exponential parameters (νi2,bi), where νi2≥0 and bi>0.

[F1]

Sub-exponentiality means EeλXi≤eλ2νi2/2 whenever ∣λ∣≤1/bi (Sub-Gaussian variables and vectors, and sub-exponential variables).

[F2]

Independence means independence of the sigma-algebras generated by the variables (Independent random elements).

[F3]

For independent real variables and nonnegative Borel functions, the expectation of the product factors (Expectations factor over finite products of independent random variables).

[F4]

Markov's inequality gives P(Y≥r)≤E[Y]/r for Y≥0 and r>0 (Markov's inequality for random variables).

[F5]

The exponential is continuous and strictly increasing, positive, satisfies e−x=1/ex and ex+y=exey, and has e0=1; these imply ex+e−x−2=e−x(ex−1)2≥0, with equality only at x=0 (The real exponential function and the number e by a power series, The exponential addition formula exp⁡(x+y)=exp⁡(x)exp⁡(y), The exponential is positive and satisfies exp⁡(−x)=1/exp⁡(x), The exponential function is strictly increasing).

[F6]

A nonnegative integrable function with integral 0 vanishes almost everywhere (A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere).

[F7]

Probability is subadditive on finite unions (Basic identities for a probability measure).

[F8]
[F9]

Expectation is linear on integrable real random variables (Linearity, monotonicity, and the modulus bound for expectation).

[F10]

Finite sums of real-valued measurable functions are measurable (Arithmetic and lattice operations preserve measurability whenever they are defined).

Proof

1.1givenF1F5F6F7F9F10algebra

If v=0, then every νi2=0. Fix i and choose λi=1/(2bi)>0. By [F1], both EeλiXi and Ee−λiXi are at most 1. The nonnegative random variable Qi=eλiXi+e−λiXi−2 therefore has expectation at most 0 by [F9]. By [F5], Qi=0 only when Xi=0; [F6] gives Xi=0 almost surely. The finite sum S is consequently 0 almost surely by [F7].

1.2F1F2F3F5F8givenconstruct

Suppose v>0. By [F8], b=max⁡ibi exists and is positive. For each real θ, the map gθ(x)=eθx is continuous: scalar multiplication is continuous since ∣θx−θy∣=∣θ∣∣x−y∣, and the exponential is continuous by [F5]. Its inverse images of open sets are open, so the collection of Borel sets with Borel preimage is a sigma-algebra containing the opens; hence gθ is Borel by The Borel sigma-algebra of a topological space, For each Borel B, {gθ(Xi)∈B}=Xi−1(gθ−1(B)) is measurable. For 0≤λ≤1/b, [F1] applies to every Xi. By [F5], eλS=∏ieλXi; independence [F2] and factorization [F3] give EeλS=∏iEeλXi≤eλ2v/2.

2.1F4F5F10step 1.2algebra

For t>0 and 0<λ≤1/b, [F10] makes S a measurable real random variable, and S≥t implies eλS≥eλt>0 by [F5]. Markov [F4], step 1.2, and the reciprocal and addition formulas in [F5] yield P(S≥t)≤exp⁡(−λt+λ2v/2). Choose λ=min⁡{t/v,1/b}. If t≤v/b, the exponent is −t2/(2v). If t>v/b, it is −t/b+v/(2b2)≤−t/(2b). Hence P(S≥t)≤exp⁡[−12min⁡{t2/v,t/b}].

3.1F1F2F3F4F5F7F10step 1.1step 1.2step 2.1algebra∎

For 0<λ≤1/b, [F1] applies at parameter −λ to every original Xi. By [F5], e−λS=∏ie−λXi; the original variables are independent by [F2], the functions x↦e−λx are nonnegative Borel by step 1.2, and factorization [F3] gives Ee−λS≤eλ2v/2. Markov [F4] and the same optimizing choice as in step 2.1 yield the stated lower-tail bound. The finite union bound [F7] combines the tails; step 1.1 handles v=0. No independence assertion about the transformed variables is needed.

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