# 0.7 Chernoff's bound and hoeffding's inequality  (Page 2/2)

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Take

$Z=|\stackrel{^}{{R}_{n}\left(f\right)}-R\left(f\right)|\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}\text{and}\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}t=ϵ$
$\begin{array}{ccc}\hfill P\left(|\stackrel{^}{{R}_{n}}\left(f\right)-R\left(f\right)|\ge ϵ\right)& \le & \frac{E\left[|\stackrel{^}{{R}_{n}\left(f\right)}-R\left(f\right){|}^{2}\right]}{{ϵ}^{2}}\hfill \\ & \le & \frac{\text{var}\left({\stackrel{^}{R}}_{n}\left(f\right)\right)}{{ϵ}^{2}}\hfill \\ & =& \frac{{\sum }_{i=1}^{n}\text{var}\left(\frac{{L}_{i}}{n}\right)}{{ϵ}^{2}}\hfill \\ & =& \frac{\text{var}\left(\ell \left(X\right),Y\right)}{n{ϵ}^{2}}\hfill \\ & =& \frac{{\sigma }_{L}^{2}}{n{ϵ}^{2}}\hfill \end{array}.$

So, the probability goes to zero at a rate of at least ${n}^{-1}$ . However, it turns out that this is an extremely loose bound. Accordingto the Central Limit Theorem

$\stackrel{^}{{R}_{n}}\left(f\right)=\frac{1}{n}\sum _{i=1}^{n}{L}_{i}\to N\left(R,\left(f\right),,,\frac{{\sigma }_{L}^{2}}{n}\right)\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}\text{as}\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}n\to \infty$

in distribution. This suggests that for large values of n,

$P\left(|\stackrel{^}{{R}_{n}}\left(f\right)-R\left(f\right)|\ge ϵ\right)\approx O\left({e}^{-\frac{n{ϵ}^{2}}{2{\sigma }_{L}^{2}}}\right).$

That is, the Gaussian tail probability is tending to zero exponentially fast.

## Chernoff's bound

Note that for any nonnegative random variable $Z$ and $t>0$ ,

$P\left(Z\ge t\right)=P\left({e}^{sZ}\ge {e}^{st}\right)\le \frac{E\left[{e}^{sZ}\right]}{{e}^{st}},\phantom{\rule{4pt}{0ex}}\forall s>0\phantom{\rule{4.pt}{0ex}}\text{by}\phantom{\rule{4.pt}{0ex}}\text{Markov's}\phantom{\rule{4.pt}{0ex}}\text{inequality}.$

Chernoff's bound is based on finding the value of $s$ that minimizes the upper bound. If $Z$ is a sum of independent random variables. For example, say

$Z=\sum _{i=1}^{n}\left(\ell ,\left(f\left({X}_{i}\right),{Y}_{i}\right),-,R,\left(f\right)\right)=n\left({\stackrel{^}{R}}_{n},\left(f\right),-,R,\left(f\right)\right)$

then the bound becomes

$P\left(\sum _{i=1}^{n},\left({L}_{i}-E\left[{L}_{i}\right]\right),\ge ,t\right)\le {e}^{-st}E\left[{e}^{s{\sum }_{i=1}^{n}\left({L}_{i}-E\left[{L}_{i}\right]\right)}\right]\le {e}^{-st}\prod _{i=1}^{n}E\left[{e}^{s\left({L}_{i}-E\left[{L}_{i}\right]\right)}\right],\phantom{\rule{4.pt}{0ex}}\text{from}\phantom{\rule{4.pt}{0ex}}\text{independence.}$

Thus, the problem of finding a tight bound boils down to finding a good bound for $E\left[{s}^{s\left({L}_{i}-E\left[{L}_{i}\right]\right)}\right]$ . Chernoff ('52), first studied this situation for binary random variables. Then,Hoeffding ('63) derived a more general result for arbitrary bounded random variables.

Theorem

## Hoeffding's inequality

Let ${Z}_{1},{Z}_{2},...,Zn$ be independent bounded random variables such that ${Z}_{i}\in \left[{a}_{i},{b}_{i}\right]$ with probability 1. Let ${S}_{n}={\sum }_{i=1}^{n}{Z}_{i}$ . Then for any $t>0$ , we have

$P\left(|{S}_{n}-E\left[{S}_{n}\right]|\ge t\right)\le 2{e}^{-\frac{2{t}^{2}}{{\sum }_{i=1}^{n}{\left({b}_{i}-{a}_{i}\right)}^{2}}}.$

The key to proving Hoeffding's inequality is the following upper bound: if $Z$ is a random variable with $E\left[Z\right]=0$ and $a\le Z\le b,$ then

$E\left[{e}^{sZ}\right]\le {e}^{\frac{{s}^{2}{\left(b-a\right)}^{2}}{8}}.$

This upper bound is derived as follows. By the convexity of theexponential function,

${e}^{sz}\le \frac{z-a}{b-a}{e}^{sb}+\frac{b-z}{b-a}{e}^{sa},\phantom{\rule{4.pt}{0ex}}\text{for}\phantom{\rule{4.pt}{0ex}}a\le z\le b.$

Thus,

$\begin{array}{ccc}\hfill E\left[{e}^{sZ}\right]& \le & E\left[\frac{Z-a}{b-a}\right]{e}^{sb}+E\left[\frac{b-Z}{b-a}\right]{e}^{sa}\hfill \\ & =& \frac{b}{b-a}{e}^{sa}-\frac{a}{b-a}{e}^{sb}\phantom{\rule{4.pt}{0ex}}\text{,}\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}\text{since}\phantom{\rule{4.pt}{0ex}}E\left[Z\right]=0\hfill \\ & =& \left(1-\theta +\theta {e}^{s\left(b-a\right)}\right){e}^{-\theta s\left(b-a\right)}\phantom{\rule{4.pt}{0ex}}\text{,}\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}\text{where}\phantom{\rule{4.pt}{0ex}}\theta =\frac{-a}{b-a}\hfill \end{array}.$

Now let

$u=s\left(b-a\right)\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}\text{and}\phantom{\rule{4.pt}{0ex}}\text{define}\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}\phi \left(u\right)\equiv -\theta u+log\left(1-\theta +\theta {e}^{u}\right).$

Then we have

$E\left[{e}^{sZ}\right]\le \left(1-\theta +\theta {e}^{s\left(b-a\right)}\right){e}^{-\theta s\left(b-a\right)}={e}^{\phi \left(u\right)}.$

To minimize the upper bound let's express $\phi \left(u\right)$ in a Taylor's series with remainder :

$\phi \left(u\right)=\phi \left(0\right)+u{\phi }^{\text{'}}\left(0\right)+\frac{{u}^{2}}{2}{\phi }^{\text{'}\text{'}}\left(v\right)\phantom{\rule{4.pt}{0ex}}\text{for}\phantom{\rule{4.pt}{0ex}}\text{some}\phantom{\rule{4.pt}{0ex}}v\in \left[0,u\right]$
$\begin{array}{ccc}\hfill {\phi }^{\text{'}}\left(u\right)& =& -\theta +\frac{\theta {e}^{u}}{1-\theta +\theta {e}^{u}}⇒{\phi }^{\text{'}}\left(u\right)=0\hfill \\ \hfill {\phi }^{\text{'}\text{'}}\left(u\right)& =& \frac{\theta {e}^{u}}{1-\theta +\theta {e}^{u}}-\frac{{\left(\theta {e}^{u}\right)}^{2}}{{\left(1-\theta +\theta {e}^{u}\right)}^{2}}\hfill \\ & =& \frac{\theta {e}^{u}}{1-\theta +\theta {e}^{u}}\left(1-\frac{\theta {e}^{u}}{1-\theta +\theta {e}^{u}}\right)\hfill \\ & =& \rho \left(1-\rho \right)\hfill \end{array}.$

Now, ${\phi }^{\text{'}\text{'}}\left(u\right)$ is maximized by

$\rho =\frac{\theta {e}^{u}}{1-\theta +\theta {e}^{u}}=\frac{1}{2}⇒{\phi }^{\text{'}\text{'}}\left(u\right)\le \frac{1}{4}.$

So,

$\phi \left(u\right)\le \frac{{u}^{2}}{8}=\frac{{s}^{2}{\left(b-a\right)}^{2}}{8}$
$⇒E\left[{e}^{sZ}\right]\le {e}^{\frac{{s}^{2}{\left(b-a\right)}^{2}}{8}}.$

Now, we can apply this upper bound to derive Hoeffding's inequality.

$\begin{array}{ccc}\hfill P\left({S}_{n}-E\left[{S}_{n}\right]\ge t\right)& \le & {e}^{-st}\prod _{i=1}^{n}E\left[{e}^{s\left({L}_{i}-E\left[{L}_{i}\right]\right)}\right]\hfill \\ & \le & {e}^{-st}\prod _{i=1}^{n}{e}^{\frac{{s}^{2}{\left({b}_{i}-{a}_{i}\right)}^{2}}{8}}\hfill \\ & =& {e}^{-st}{e}^{{s}^{2}{\sum }_{i=1}^{n}\frac{{\left({b}_{i}-{a}_{i}\right)}^{2}}{8}}\hfill \\ & =& {e}^{\frac{-2{t}^{2}}{{\sum }_{i=1}^{n}{\left({b}_{i}-{a}_{i}\right)}^{2}}}\hfill \\ & & \text{by}\phantom{\rule{4.pt}{0ex}}\text{choosing}\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}s=\frac{4t}{{\sum }_{i=1}^{n}{\left({b}_{i}-{a}_{i}\right)}^{2}}\hfill \end{array}$

Similarly, $P\left(E\left[{S}_{n}\right]-{S}_{n}\ge t\right)\le {e}^{\frac{-2{t}^{2}}{{\sum }_{i=1}^{n}{\left({b}_{i}-{a}_{i}\right)}^{2}}}$ . This completes the proof of the Hoeffding's theorem.

## Application

Let ${Z}_{i}={1}_{f\left({X}_{i}\right)\ne {Y}_{i}}-R\left(f\right),$ as in the classification problem. Then for a fixed f, it follows fromHoeffding's inequality (i.e., Chernoff's bound in this special case) that

$\begin{array}{ccc}\hfill P\left(|\stackrel{^}{{R}_{n}}\left(f\right)-R\left(f\right)|\ge ϵ\right)& =& P\left(\frac{1}{n},|{S}_{n}-E\left[{S}_{n}\right]|,\ge ,ϵ\right)\hfill \\ & =& P\left(|{S}_{n}-E\left[{S}_{n}\right]|\ge nϵ\right)\hfill \\ & \le & 2{e}^{-\frac{2{\left(nϵ\right)}^{2}}{n}}\hfill \\ & =& 2{e}^{-2n{ϵ}^{2}}\hfill \end{array}.$

Now, we want a bound like this to hold uniformly for all $f\in \mathcal{F}$ . Assume that $\mathcal{F}$ is a finite collection of models and let $|\mathcal{F}|$ denote its cardinality. We would like to bound the probability that ${max}_{f\in \mathcal{F}}|\stackrel{^}{{R}_{n}}\left(f\right)-R\left(f\right)|\ge ϵ$ . Note that the event

$\left\{\underset{f\in \mathcal{F}}{max},|\stackrel{^}{{R}_{n}}\left(f\right)-R\left(f\right)|,\ge ,ϵ\right\}\phantom{\rule{4pt}{0ex}}\equiv \phantom{\rule{4pt}{0ex}}\left\{\bigcup _{f\in \mathcal{F}},|\stackrel{^}{{R}_{n}}\left(f\right)-R\left(f\right)|,\ge ,ϵ\right\}.$

Therefore

$\begin{array}{ccc}\hfill P\left(\underset{f\in \mathcal{F}}{max},|\stackrel{^}{{R}_{n}}\left(f\right)-R\left(f\right)|,\ge ,ϵ\right)& =& P\left(\bigcup _{f\in \mathcal{F}},|\stackrel{^}{{R}_{n}}\left(f\right)-R\left(f\right)|,\ge ,ϵ\right)\hfill \\ & \le & \sum _{f\in F}P\left(|\stackrel{^}{{R}_{n}}\left(f\right)-R\left(f\right)|\ge ϵ\right),\phantom{\rule{4.pt}{0ex}}\text{the}\phantom{\rule{4.pt}{0ex}}\text{}\mathit{\text{union}}\phantom{\rule{4.pt}{0ex}}\mathit{\text{of}}\phantom{\rule{4.pt}{0ex}}\mathit{\text{events}}\text{''}\phantom{\rule{4.pt}{0ex}}\text{bound}\hfill \\ & \le & 2|F|{e}^{-2n{ϵ}^{2}},\phantom{\rule{4.pt}{0ex}}\text{by}\phantom{\rule{4.pt}{0ex}}\text{Hoeffding's}\phantom{\rule{4.pt}{0ex}}\text{inequality.}\hfill \end{array}$

Thus, we have shown that with probability at least $1-2|F|{e}^{-2n{ϵ}^{2}}$ , $\forall f\in \mathcal{F}$

$|\stackrel{^}{{R}_{n}}\left(f\right)-R\left(f\right)|<ϵ.$

And accordingly, we can be reasonably confident in selecting $f$ from $\mathcal{F}$ based on the empirical risk function ${\stackrel{^}{R}}_{n}$ .

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