设随机变量X和Y相互独立,且P{X=1}=P{Y=1}=P>0,P{X=0}=P{Y=0}=1-P>0,令Z=1（X+Y为偶数）Z=0（X+Y为奇数）,要使X与Z相互独立,则p为?

Question

设随机变量X和Y相互独立,且P{X=1}=P{Y=1}=P>0,P{X=0}=P{Y=0}=1-P>0,令Z=1（X+Y为偶数） Z=0 （X+Y为奇数）,要使X与Z相互独立,则p为?

Answer 1

P(Z=1) = P(X+Y为偶数) = P(X=0,Y=0) + P(X=1,Y=1) = P(X=0)P(Y=0) + P(X=1)P(Y=1) = (1-p)^2 + p^2,
P(Z=0) = P(X+Y为奇数) = P(X=0,Y=1) + P(X=1,Y=0) = P(X=0)P(Y=1) + P(X=1)P(Y=0) = (1-p)*p + p*(1-p) = 2*p*(1-p),
P(X=0,Z=1) = P(X=0,X+Y为偶数) = P(X=0,Y=0) = (1-p)^2,要独立需要= P(X=0)*P(Z=1) = (1-p)*[(1-p)^2+p^2],so 1-p = (1-p)^2 + p^2 = 1-2*p+2*(p^2),so 2*(p^2)-p=0,so p = 1/2.