Measure concentration

Markov's Inequality

Expectation of a non-negative random variable Z can be written (tail-sum formula) as

E[Z]=x=0P[Zx]dx

Since P[Zx] is monotonically nonincreasing, we have

a0,E[Z]x=0aP[Zx]dxx=0aP[Za]dx=aP[Za].

Rearranging this, we get Markov's inequality:

a0,P[Za]E[Z]a

Chebyshev's Inequality

Applying Markov's inequality on the random variable (ZE[Z])2, we get:

a0,P[|ZE[Z]|a]Var[Z]a2

This can be used to derive that given a sequence Z1,Z2,,Zm of i.i.d. random variables with E[Z]=μ and Var[Z]1, then for any δ(0,1) with probability at least 1δ, we have:

|1mi=1mZiμ|1δm

(polynomial bound for the estimation error of the mean)

Chernoff's bounds

For independent Bernoulli random variables Z1,Z2,,Zm where i, P[Zi=1]=pi, denoting p=i=1mpi and Z=i=1mZi, we can show that:

P[Z>(1+δ)p]exp(pδ22+2δ3)

and

P[Z<(1δ)p]exp(pδ22+2δ3)

Hoeffding's Inequality

For a sequence of i.i.d. random variables Z1,Z2,,Zm with E[Z]=μ and P[aZb]=1, we have for any ϵ>0,

P[|1mimZiμ|>ϵ]2exp(2mϵ2(ba)2)

Proof:
Let Xi=Ziμ and X¯=1mimXi=Z¯μ. Using Markov's inequality, we have that for any λ>0 and ϵ>0,

P[X¯ϵ]=P[eλX¯eλϵ]eλϵ E[eλX¯]

Hoeffding's lemma: For a random variable X bounded between a and b with mean zero, for every λ>0, we have E[eλX]exp(λ2(ba)28).

Using independence and Hoeffding's lemma, we can obtain:

E[eλX¯]=E[ieλXi/m]=iE[eλXi/m]P[X¯ϵ]eλϵieλ2(ba)2/8m2=eλϵ+λ2(ba)2/8m

The exponent λϵ+λ2(ba)2/8m is minimized for λ=4mϵ(ba)2 yielding:

P[X¯ϵ]exp(2mϵ2(ba)2)

For the other side, we can apply the same argument on the variable X¯ to obtain:

P[X¯ϵ]exp(2mϵ2(ba)2)

Combining the two cases yields the desired bound.


Links

Sources