Chebyshev.

Two point sampling.

Chernoff Bound

Intro

Theorem:

Pr[X > (1 + $\displaystyle \delta$)$\displaystyle \mu$] < $\displaystyle \left[\vphantom{
\frac{e^\delta}{(1+\delta)^{(1+\delta)}}}\right.$$\displaystyle {\frac{e^\delta}{(1+\delta)^{(1+\delta)}}}$$\displaystyle \left.\vphantom{
\frac{e^\delta}{(1+\delta)^{(1+\delta)}}}\right]^{\mu}_{}$.

Proof.

Pr[X > (1 + $\displaystyle \delta$)$\displaystyle \mu$]

=

Pr[exp(tX) > exp(t(1 + $\displaystyle \delta$)$\displaystyle \mu$)]

 

 

<

$\displaystyle {\frac{E[\exp(tX)]}{\exp(t(1+\delta)\mu)}}$

 

E[exp(tX)]

=

$\displaystyle \prod$E[exp(tXi)]

 

E[exp(tXi)]

=

piet + (1 - pi)

 

 

=

1 + pi(et - 1)

 

 

$\displaystyle \le$

exp(pi(et - 1))

 

¬…       So overall bound is

$\displaystyle {\frac{\exp((e^t-1)\mu)}{\exp(t(1+\delta)\mu)}}$

True for any t. Plug in t = ln(1 + $ \delta$).

¬…       This in turn less than e- $\scriptstyle \mu$$\scriptstyle \delta^{2}$/4 for $ \delta$< 2e - 1. (Less than 2- (1 + $\scriptstyle \delta$)$\scriptstyle \mu$ for larger $ \delta$).

¬…       By same argument on exp(- tX),

Pr[X < (1 - $\displaystyle \delta$)$\displaystyle \mu$] < $\displaystyle \left[\vphantom{ \frac{e^{-\delta}}{(1-\delta)^{(1-\delta)}}}\right.$$\displaystyle {\frac{e^{-\delta}}{(1-\delta)^{(1-\delta)}}}$$\displaystyle \left.\vphantom{ \frac{e^{-\delta}}{(1-\delta)^{(1-\delta)}}}\right]^{\mu}_{}$

bound e- $\scriptstyle \delta^{2}$/2.

Summary, Probability of deviation by relative error $ \delta$< 1 is at most e- $\scriptstyle \delta^{2}$$\scriptstyle \mu$/3 in each direction. Large $ \delta$gives 2- (1 + $\scriptstyle \delta$)$\scriptstyle \mu$

Remark: bound applies to any vars distributed in range [0, 1].

Basic applications:

Zillions of Chernoff applications; will see next time.

¬…       Basic applications:

o      Set balancing. minimize max bias. 4$ \sqrt{n\ln n}$.

o      cn log n balls in c bins. max matches average.