Answer:
Step-by-step explanation:
Given that:
Let us consider an event is A and let the associated indicator random variable is IA
In same way, an event is B and the associated indicator random variable is IB.
Suppose that P(A) = P
P(B) = q, P(AnB) = r
Now, we find the variance of IA - IB in terms of p,q,r
Var(IA - IB) = var(IA) + var(IB) - 2cov(IA, IB)
Here, we find var(IA)
[tex]var(IA) = E(IA)^2 - [E(IA)]^2\\=I^2\dot P(A) + 0^2(I-P(A))-[1P(A)+0(I-P(A))]^2\\=P(A)+0-[P(A)+0]^2\\=P(A)-[P(A)]^2\\=P(A)[1-P(A)][/tex]
From the given data we have [tex]P(A)=P[/tex]
[tex]P[1-P]\\\therefore\, var(IA)=P[1-P]...(1)[/tex]
similarly [tex]var(IB)=q(1-q)....(2)[/tex]
[tex]cov(IA,\, IB)=E(IA\dot IB)-E(IA)E(IB)\\=1\times P(AnB)+[0times P(AnB)^c]-P\times q[/tex]
Frome the given data we have
[tex]P(AnB)=r\\1\times r +(0)-P\dot q\\=r-Pq[/tex]
[tex]cov(IA,\, IB)=r-Pq...(3)[/tex]
from 1,2,3 we have the variance of IA - IB is
[tex]var(IA-IB)=P(1-P)+q(1-q)-2(r-Pq)[/tex]
