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One-sided concentration bound for independent median trials

Statement

Let R≥1 be an integer, let Z1,…,ZR be independent Bernoulli random variables on a finite probability space with E[Zi]≤p<1/2, and put S=∑i=1RZi. For every t≥0, Pr⁡(S−ES≥t)≤e−2t2/R. Consequently, let R be odd and let T1,…,TR be independent finite-valued real estimates on a finite probability space. If I⊆R is an interval and Pr⁡(Ti∈I)≥1−p for every i, then their median (the middle order statistic) lies in I except with probability at most exp⁡(−2R(1/2−p)2).

Facts & Assumptions

Given: A finite probability space, an integer R≥1, and an independent Bernoulli family Z1,…,ZR with means pi=EZi≤p<1/2. For the median consequence, also the independent finite-valued estimates T1,…,TR, the interval I, and Pr⁡(Ti∈I)≥1−p for each i.

[F1]

Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials: a Bernoulli variable takes values 0,1 with probabilities 1−pi,pi.

[F2]

Independent random elements: mutual independence means the generated event sigma-algebras factor on finite intersections.

[F3]

Expectation factors over a finite product of mutually independent random variables: expectations of finite products of independent finite-valued real variables factor.

[F4]

Markov's inequality on a finite probability space: for nonnegative W and c>0, Pr⁡(W≥c)≤E[W]/c.

[F7]

The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a): if a function is continuous on [a,b] and differentiable on (a,b), then its endpoint increment is its derivative at some interior point times b−a.

[F8]

Expectation of a real random variable on a finite probability space: the expectation of a finite-valued variable is the finite weighted sum of its values, including the Bernoulli formula.

[F9]

Linearity, monotonicity, and the modulus bound for expectation: expectation is linear for the finite-valued real random variables below.

[F10]

A function differentiable at c is continuous at c: a function differentiable at a point is continuous there.

[F11]

Expectation agrees with the published finite weighted sum identifies general expectations of finite random variables on this probability space with the finite weighted sums in [F8].

Proof

1.1givenF1F5F6F8algebra

For one Bernoulli variable Zi, [F1] and [F8] give EZi=pi. Define Mi(λ)=E[exp⁡(λ(Zi−pi))]=exp⁡(−λpi)(1−pi+piexp⁡(λ))>0 by its two possible values and [F5], and put Gi(λ)=ln⁡Mi(λ). Set qi(λ)=piexp⁡(λ)/(1−pi+piexp⁡(λ))∈[0,1]. Differentiating by [F6] gives Gi(0)=Gi′(0)=0 and Gi′′(λ)=qi(λ)(1−qi(λ))≤1/4, including pi=0 and pi=1.

2.1step 1.1F5F6F7F10algebra

At λ=0 the desired bound is equality. For λ>0, put Hi(t)=Gi(t)−t2/8 on [0,λ]. By step 1.1, Hi′′(t)≤0. The functions Hi and Hi′ are continuous by [F10], so the hypotheses of [F7] apply. For any 0<v≤λ, [F7] applied to Hi′ on [0,v] gives Hi′(v)−Hi′(0)=Hi′′(c)v≤0 for some c∈(0,v); since Hi′(0)=0, this shows Hi′(v)≤0. Applying [F7] to Hi on [0,λ] then gives Hi(λ)−Hi(0)=Hi′(d)λ≤0 for some d∈(0,λ). As Hi(0)=0, Gi(λ)≤λ2/8. Therefore E[exp⁡(λ(Zi−pi))]≤exp⁡(λ2/8) for every λ≥0.

3.1givenF2F3F5F8F9F11step 2.1algebra

By [F11], the general expectations in [F3] and [F9] agree with the finite weighted sums in [F8]. Thus [F8]–[F9] give ES=∑ipi and S−ES=∑i(Zi−pi). For λ≥0, put Wi=exp⁡(λ(Zi−pi)). Since each Zi is finite-valued, every event defined by Wi is a union of level sets of Zi, so σ(Wi)⊆σ(Zi); the independence in [F2] therefore makes the Wi independent. The exponential addition law [F5] and finite product factorization [F3] now give E[exp⁡(λ(S−ES))]=∏iE[Wi]≤exp⁡(Rλ2/8) by step 2.1.

4.1givenF4step 3.1algebra

For t>0, choose λ=4t/R>0 and apply [F4] to the nonnegative variable eλ(S−ES) at threshold eλt. Step 3.1 gives Pr⁡(S−ES≥t)≤e−λt+Rλ2/8=e−2t2/R. For t=0, the probability is at most one, equal to the stated bound.

5.1givenF1F2F8F9F11step 4.1algebra∎

Let Yi indicate the event Ti∉I. Each Yi is Bernoulli with mean at most p by [F1], [F8], and the interval-success hypothesis. Each Yi is a function of the finite-valued Ti, so every event defined by Yi is a union of level sets of Ti; hence σ(Yi)⊆σ(Ti) and [F2] makes the Yi independent. By [F11], [F9] applies to the same finite weighted-sum expectations as [F8], so ES=∑iEYi≤pR. If the odd median is outside I, at least (R+1)/2 estimates lie on one side of I and fail, so S=∑iYi≥R/2 and S−ES≥R(1/2−p). Apply step 4.1 with this t to get the stated median bound.

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