Sampling distribution from two independent Bernoulli populations

Let’s assume that we have samples of two independent Bernoulli random variables, Ber(θ1) and Ber(θ2).

How do we prove that (ˉX1ˉX2)(θ1θ2)θ1(1θ1)n1+θ2(1θ2)n2dN(0,1)?

Assume that n1n2.

Answer

Put
a=θ1(1θ1)n1, b=θ2(1θ2)n2,
A=(ˉX1θ1)/a,
B=(ˉX2θ2)/b. We have
AdN(0,1), BdN(0,1).
In terms of characteristic functions it means
ϕA(t)EeitAet2/2, ϕB(t)et2/2.
We want to prove that
D:=aa2+b2Aba2+b2BdN(0,1)

Since A and B are independent,
ϕD(t)=ϕA(aa2+b2t)ϕB(ba2+b2t)et2/2,
as we wish it to be.

This proof is incomplete. Here we need some estimates for uniform convergence of characteristic functions. However in the case under consideration we can do explicit calculations. Put p=θ1, m=n1.
ϕX1,1(t)=1+p(eit1),ϕˉX1(t)=(1+p(eit/m1))m,ϕˉX1θ1(t)=(1+p(eit/m1))meipt,ϕA(t)=(1+p(eit/mp(1p)1))meiptm/p(1p)=((1+p(eit/mp(1p)1))eipt/mp(1p))m=(1t22m+O(t3m3/2))m
as t3m3/20. Thus, for a fixed t,
ϕD(t)=(1a2t22(a2+b2)n1+O(n3/21))n1(1b2t22(a2+b2)n2+O(n3/22))n2et2/2
(even if a0 or b0), since |ey(1y/m)m|y2/2m  when  y/m<1/2 (see https://math.stackexchange.com/questions/2566469/uniform-bounds-for-1-y-nn-exp-y/ ).

Note that similar calculations may be done for arbitrary (not necessarily Bernoulli) distributions with finite second moments, using the expansion of characteristic function in terms of the first two moments.

Attribution
Source : Link , Question Author : An old man in the sea. , Answer Author : Viktor

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