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The maximum of independent standard normals has expectation of order at least the square root of log dimension

Statement

Assume AC. There is an integer $m_{0}$ such that for every $m \geq m_{0}$ and independent standard-normal random variables $G_{1}, \dots, G_{m}$ , $E \max_{1 \leq i \leq m} G_{i} \geq \frac{1}{4} \sqrt{\log m} .$

Statement references

Rigollet and Hütter, §1.4, maximal inequalities

Facts & Assumptions

Given: AC and independent standard-normal random variables $G_{1}, \dots, G_{m}$ .

[F1]

A standard normal has density $ϕ (x) = (2 π)^{- 1 / 2} e^{- x^{2} / 2}$ , which is decreasing for $x \geq 0$ (Standard normal and normal laws).

[A1]

AC supplies the finite independent Gaussian family (The Axiom of Choice).

Proof

1.1A1F1algebra

Put $t_{m} = \frac{1}{2} \sqrt{\log m}$ and $q_{m} = P (G_{1} \geq t_{m})$ . Since $ϕ$ decreases on $[0, \infty)$ , $q_{m} \geq ϕ (t_{m} + 1 / (t_{m} + 1)) / (t_{m} + 1) \geq e^{- t_{m}^{2} / 2 - t_{m} - 1 / 2} / (\sqrt{2 π} (t_{m} + 1))$ ; the last inequality follows from $t_{m} / (t_{m} + 1) \leq t_{m}$ and $1 / (2 (t_{m} + 1)^{2}) \leq 1 / 2$ . Hence $m q_{m} \to \infty$ .

2.1F1step 1.1givenalgebra∎

Let $M_{m} = \max_{i} G_{i}$ . Independence gives $P (M_{m} < t_{m}) = (1 - q_{m})^{m} \leq e^{- m q_{m}} \to 0$ and $E [G_{1} 1_{{M_{m} < t_{m}}}] = E [G_{1} 1_{{G_{1} < t_{m}}}] (1 - q_{m})^{m - 1} \geq - (1 - q_{m})^{m - 1} / \sqrt{2 π}$ . Splitting at $M_{m} = t_{m}$ yields $E M_{m} \geq t_{m} (1 - (1 - q_{m})^{m}) - (1 - q_{m})^{m - 1} / \sqrt{2 π} \geq t_{m} / 2 = \frac{1}{4} \sqrt{\log m}$ for all sufficiently large $m$ .

Source notes

The cited notes give maximal-inequality context. The stated lower bound is derived here from a direct Gaussian-tail lower estimate and independence.

Depends on

Used by

The unrestricted Wainwright signed-support threshold claim fails without active-dimension control False statement

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Sources

Philippe Rigollet and Jan-Christian Hütter, High-Dimensional Statistics: Lecture Notes, §1.4 (standard reference, not scraped)