guide:B846f441d7

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label{Ch-TB} We start this chapter by explaining a general technique to obtain bounds for [math]\P\bigl[Y_1+\cdots+Y_d\geqslant a\bigr][/math] where the [math]Y_i[/math] are independent random variables. A bound that arises from this ‘recipe’ is often referred to as a Chernoff bound. The basic idea is given in the following Lemma which establishes the ‘generic’ Chernoff bound.

Lemma

Let [math](\Omega,\Sigma,\P)[/math] be a probability space and let [math]Y_1,\dots,Y_n\colon\Omega\rightarrow\mathbb{R}[/math] be mutually independent random variables. Then

[[math]] \P\bigl[Y_1+\cdots+Y_n\geqslant a\bigr]\leqslant \inf_{t \gt 0}\exp(-ta)\prod_{i=1}^n\E(\exp(tY_i)) [[/math]]

holds for every [math]a \gt 0[/math].

Show Proof

For fixed [math]t \gt 0[/math] the function [math]\exp(t\cdot\shortminus)[/math] is increasing. We put [math]Y=Y_1+\cdots+Y_n[/math] and get

[[math]] \P[Y\geqslant a] = \P[\exp(tY)\geqslant \exp(ta)]\leqslant\frac{\E(\exp(tY))}{\exp(ta)} [[/math]]

by applying the Markov bound. Now we take the infimum over [math]t \gt 0[/math] and employ the independence to get

[[math]] \P[Y\geqslant a]\leqslant \inf_{t \gt 0}\exp(-ta)\E(\exp(t\Bigsum{i=1}{n} tY_i))=\inf_{t \gt 0}\exp(-ta)\prod_{i=1}^n\E(\exp(tY_i)) [[/math]]

as desired.

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The classical Chernoff bound recipe is now to compute or estimate the right hand side by exploiting additional information on the [math]Y_i[/math] that is given in concrete situations. Readers unfamiliar with Chernoff bounds can find a classical example on how the recipe works in Problem where the [math]Y_i[/math] are independent Bernoulli random variables. In Theorem below the additional knowledge comes from the assumption that the moments of each [math]Y_i[/math] grow at most like factorial.

Theorem

Let [math](\Omega,\Sigma,\P)[/math] be a probability space and let [math]Y_1,\dots,Y_d\colon\Omega\rightarrow\mathbb{R}[/math] be mutually independent random variables with expectation zero and [math]|\E(Y_i^k)|\leqslant k!/2[/math] for [math]i=1,\dots,d[/math] and [math]k\geqslant2[/math]. Then we have

[[math]] \P\bigl[|Y_1+\cdots+Y_d|\geqslant a\bigr]\leqslant 2\exp\bigl(-C\min\bigl(\medfrac{a^2}{d},a\bigr)\bigr) [[/math]]

for every [math]a \gt 0[/math] and with [math]C=1/4[/math].

Show Proof

Let [math]Y[/math] be one of the random variables [math]\pm Y_1,\dots,\pm Y_d[/math] and [math]0 \lt t\leqslant 1/2[/math]. Then we estimate

[[math]] \begin{align*} \E(\exp(tY)) & = \bigl|\E\bigl(1+tY +\Bigsum{k=2}{\infty}\medfrac{(tY)^k}{k!}\bigr)\bigr| \leqslant 1 + \Bigsum{k=2}{\infty}\medfrac{t^k|\E(Y^k)|}{k!}\\ & \leqslant 1 + \Bigsum{k=2}{\infty}\medfrac{t^k}{2} = 1 + \smallfrac{1}{2}\bigl(\medfrac{1}{1-t} - (t+1)\bigr)\\ & = 1 + \medfrac{1}{2}\medfrac{t^2}{1-t}\leqslant 1+t^2\leqslant \exp(t^2) \end{align*} [[/math]]

where the last equality can easily be checked by means of calculus. Now we use Lemma and get

[[math]] \begin{align*} \P\bigl[Y_1+\cdots+Y_d\geqslant a\bigr] & \leqslant \inf_{t \gt 0}\exp(-ta)\prod_{i=1}^d\E(\exp(tY_i))\\ & \leqslant \inf_{0 \lt t\leqslant1/2}\exp(-ta)\prod_{i=1}^d\exp(t^2)\\ & \leqslant \inf_{0 \lt t\leqslant1/2}\exp(-ta+dt^2). \end{align*} [[/math]]

By solving [math]\frac{\dd}{\dd t}(-ta+dt^2)=-a+2dt=0[/math] for [math]t[/math] and considering the constraint [math]t\leqslant1/2[/math] this leads to [math]t=\min(\frac{a}{2d},\frac{1}{2})[/math], and further to

[[math]] -ta+dt^2 \leqslant -\medfrac{1}{2}ta -\medfrac{1}{2}ta + dt\medfrac{a}{2d}=-\medfrac{1}{2}ta \leqslant -\medfrac{1}{2}\min\bigl(\medfrac{a}{2d},\medfrac{1}{2}\bigr)a = -\min\bigl(\medfrac{a^2}{4d},\medfrac{a}{4}\bigr). [[/math]]

This implies

[[math]] \P\bigl[Y_1+\cdots+Y_d\geqslant a\bigr] \leqslant \exp\Bigl(-\min\bigl(\medfrac{a^2}{4d},\medfrac{a}{4}\bigr)\Bigr) [[/math]]

and similiarly we get

[[math]] \P\bigl[Y_1+\cdots+Y_d\leqslant -a\bigr] = \P\bigl[-Y_1-\cdots-Y_d\geqslant a\bigr] \leqslant \exp\Bigl(-\min\bigl(\medfrac{a^2}{4d},\medfrac{a}{4}\bigr)\Bigr). [[/math]]

Consequently,

[[math]] \P\bigl[\bigl|Y_1+\cdots+Y_d\bigr|\geqslant a\bigr] \leqslant 2\exp\Bigl(-\min\bigl(\medfrac{a^2}{4d},\medfrac{a}{4}\bigr)\Bigr) [[/math]]

as desired.

■

\section*{Problems}

General references

Wegner, Sven-Ake (2024). "Lecture Notes on High-Dimensional Data". arXiv:2101.05841 [math.FA].