LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generated

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Linear and quadratic Gaussian statistics have exponential tails

Statement

Assume AC. Let $n \geq 1$ be an integer. Let $Z_{1}, \dots, Z_{n}$ be independent $N (0, 1)$ variables, let $c_{1}, \dots, c_{n}, δ_{1}, \dots, δ_{n} \in R$ , and let $0 < σ \leq b$ . Put $∥ c ∥_{n}^{2} = n^{- 1} \sum_{i} c_{i}^{2}$ and $∥ δ ∥_{n}^{2} = n^{- 1} \sum_{i} δ_{i}^{2}$ . For $t > 0$ , $\Pr ⁣ (∣ n^{- 1} \sum_{i} c_{i} σ Z_{i} ∣ \geq t) \leq 2 \exp ⁣ (- \frac{n t^{2}}{2 σ^{2} ∥ c ∥_{n}^{2}}),$ with zero probability when $∥ c ∥_{n} = 0$ . If $∥ δ ∥_{n} \leq 1$ and $0 < t \leq b^{2} + 2$ , then $\Pr ⁣ (∣ n^{- 1} \sum_{i} (δ_{i} + σ Z_{i})^{2} - (σ^{2} + ∥ δ ∥_{n}^{2}) ∣ \geq t) \leq 2 \exp ⁣ (- \frac{n t^{2}}{4 V_{b}}), V_{b} = 2 b^{4} + 4 b^{2} .$

Facts & Assumptions

Given: The independent standard-normal variables and the stated deterministic coefficients.

[A1]

Assume AC for the standard-normal law Standard normal and normal laws.

[F1]

Expectations factor over finite products of independent random variables: expectations of integrable products of functions of independent variables factor.

[F2]

A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not gives positive linear scaling of Lebesgue measure, and Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation gives translation invariance. Applying these set identities first to nonnegative simple functions and then using Monotone convergence for the integral gives $\int_{R} f (a z + h) d z = a^{- 1} \int_{R} f (y) d y$ for nonnegative Borel $f$ , $a > 0$ , and $h \in R$ .

[F3]

Markov's inequality for random variables: for a nonnegative random variable $X$ and $a > 0$ , $\Pr (X \geq a) \leq E X / a$ .

[F4]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t: for $x > 0$ , $\log x = \int_{1}^{x} d u / u$ .

Proof

1.1A1F1F2F3givenalgebra

By Standard normal and normal laws, $Z$ has density $e^{- z^{2} / 2} / \sqrt{2 π}$ . Completing the square and translating the integration variable by $u$ using [F2] gives $E e^{u Z} = e^{u^{2} / 2}$ for every real $u$ . Independence and [F1] therefore give the moment generating function of $n^{- 1} \sum_{i} c_{i} σ Z_{i}$ as $\exp (λ^{2} σ^{2} \sum_{i} c_{i}^{2} / (2 n^{2}))$ . Exponential Markov [F3] with $λ = n^{2} t / (σ^{2} \sum_{i} c_{i}^{2})$ bounds each one-sided tail by $\exp (- n t^{2} / (2 σ^{2} ∥ c ∥_{n}^{2}))$ ; if all $c_{i} = 0$ , the statistic is identically zero.

1.2A1F1F2F4givenalgebra

For $∣ λ ∣ \leq 1 / (4 b^{2})$ , completing the square in the same density integral gives $E \exp {λ [(δ_{i} + σ Z_{i})^{2} - (δ_{i}^{2} + σ^{2})]} = (1 - 2 λ σ^{2})^{- 1 / 2} \exp ⁣ (- λ σ^{2} + \frac{2 λ^{2} σ^{2} δ_{i}^{2}}{1 - 2 λ σ^{2}}) .$ Here $∣ 2 λ σ^{2} ∣ \leq 1 / 2$ . By [F4], for $∣ x ∣ \leq 1 / 2$ , $- x - \log (1 - x) \leq x^{2}$ : for $x \geq 0$ integrate $u / (1 - u) \leq 2 u$ on $[0, x]$ , and for $x < 0$ integrate $u / (1 + u) \leq u$ on $[0, ∣ x ∣]$ . Taking logarithms in the display, summing over $i$ , and using [F1] therefore bounds the centered-sum log moment generating function above by $λ^{2} (2 n b^{4} + 4 b^{2} \sum_{i} δ_{i}^{2}) \leq n V_{b} λ^{2}$ .

2.1F3step 1.2givenalgebra∎

For $0 < t \leq b^{2} + 2 = V_{b} / (2 b^{2})$ , choose $λ = t / (2 V_{b})$ , which satisfies $0 < λ \leq 1 / (4 b^{2})$ . Exponential Markov [F3] and step 1.2 give the upper tail $\exp (- n t^{2} / (4 V_{b}))$ ; applying the same argument to the negative centered sum gives the lower tail. Adding the two bounds proves the claimed quadratic-statistic inequality.

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Sources

A. W. van der Vaart and J. H. van Zanten, Rates of contraction of posterior distributions based on Gaussian process priors, §3.3 (standard reference, not scraped)
Subhashis Ghosal and Aad W. van der Vaart, Convergence rates of posterior distributions for noniid observations, §3 (standard reference, not scraped)