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A Gaussian maximum without independence

Example

Assume AC. Let σ>0, let G∼N(0,1), and set Zj=σG for every 1≤j≤m, where m≥1. These variables are perfectly dependent. For every t>0, P(max⁡1≤j≤m∣Zj∣≥t)≤2mexp⁡ ⁣(−t22σ2).

Facts & Assumptions

Given: AC, σ>0, m≥1, and a standard normal random variable G.

[A1]

Assume AC as required by the standard-normal law construction (The Axiom of Choice).

[F1]

The law N(0,1) has density ϕ(x)=e−x2/2/2π, and N(0,σ2) is the law of x↦σx applied to N(0,1) (Standard normal and normal laws).

[F2]

The standard normal density is positive and integrates to one (The standard normal density has total mass one).

[F3]

For a nonnegative measurable g, or an integrable real g, expectation of g(G) is its integral against the law of G (Change of variables for expectation). Integration against the density law satisfies ∫g dN(0,1)=∫g(x)ϕ(x) dx for nonnegative measurable g (Integrating against a density agrees with integrating the product); applying this to the positive and negative parts gives the signed formula once absolute integrability is established.

[F4]

One-dimensional substitution satisfies ∫φ(c)φ(d)h=∫cd(h∘φ)φ′ when h is continuous and the stated differentiability and integrability hypotheses hold (Substitution: if φ is differentiable on [c,d] with φ′ integrable and f is continuous on an interval containing φ([c,d]), then ∫φ(c)φ(d)f=∫cd(f∘φ) φ′).

[F5]

Increasing nonnegative measurable functions pass to their limit under the integral (Monotone convergence for the integral).

[F9]

A continuous real function is Borel measurable under AC (Continuous functions on Euclidean spaces are Borel measurable).

[F10]

A centered variable is sub-Gaussian with proxy σ2 when its moment generating function is bounded by eλ2σ2/2 for every real λ (Sub-Gaussian variables and vectors, and sub-exponential variables).

[F11]

For m possibly dependent centered sub-Gaussian variables with common proxy σ2, P(max⁡j∣Yj∣≥t)≤2me−t2/(2σ2) (Finite sub-Gaussian maxima and epsilon-net reduction).

[F12]

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

[F13]

Under Countable Choice, every bounded Riemann-integrable function on a nondegenerate compact interval has the same Lebesgue and Riemann integrals (A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral). AC in [A1] implies Countable Choice, so this applies to the bounded continuous integrands supplied by [F8].

Verification

1.1A1F1F2F3F4F5F7F8F13algebra

Write ϕ(x)=e−x2/2/2π. By [F7] and composition, u↦e−u is continuous and integrable on compact intervals by [F8]. The map x↦x2/2 is differentiable with derivative x by [F7]; the identity x is continuous because ∣x−y∣<η whenever ∣x−y∣<δ with δ=η, so it is integrable on [0,R] by [F8]. Thus [F4] applies. By [F7], (−e−u)′=e−u, so [F8] evaluates the resulting integral: ∫0Rxϕ(x) dx=(1−e−R2/2)/2π≤1/2π. The integrand xϕ(x) is continuous and bounded on every compact interval by [F7] and [F8], so [F13] identifies these Riemann integrals with Lebesgue integrals. Reflection x=−y in [F4], followed by [F13], gives ∫−R0(−x)ϕ(x) dx=∫0Rxϕ(x) dx since ϕ is even. Letting integer R increase to infinity, [F2], [F3] for nonnegative functions, and [F5] give E∣G∣=2∫0∞xϕ(x) dx<∞. Since ϕ is even, the positive and negative parts of xϕ(x) have equal integrals, so [F3] for integrable functions gives EG=0.

1.2F3F6F7F9algebra

Put a=λσ. The function x↦eax is continuous and Borel measurable by [F7] and [F9], so [F3] and completing the square, using [F6], give EeλσG=12π∫Reax−x2/2 dx=ea2/212π∫Re−(x−a)2/2 dx.

2.1F2F4F5F7F8F13step 1.2algebra

For every positive integer R, [F4] with translation by a gives ∫−RRe−(x−a)2/2 dx=∫−R−aR−ae−y2/2 dy. The two integrands are continuous and bounded on these compact intervals by [F7] and [F8]; [F13] therefore identifies both sides with their Lebesgue integrals. Both interval families increase to R, so [F5] passes these equal integrals to the whole-line limit; [F2] identifies that limit as 2π. Step 1.2 then yields EeλσG=ea2/2=eλ2σ2/2.

3.1F1F10F11F12step 1.1step 2.1∎

By [F12] and step 1.1, EZj=σEG=0; step 2.1 gives its moment generating function. Thus each Zj is centered sub-Gaussian with proxy σ2 by [F10]. Apply [F11] to the m variables. They are all equal, but the bound requires no independence.

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