The number of defects in a yard of fabric Y has a Poisson di

The number of defects in a yard of fabric, Y, has a Poisson distribution with parameter X. The parameter X is random with density f(x) = e-x/(0, infinity) (x). Find E(Y|X) Find E(Y) Find V(Y).

Solution

Y|X=x follows Poisson distribution with parameter x i.e. Y|X=x ~Poi(x) distribution.

a)E(Y|X=x)=x

b)here X~exp(1) i.e exponential distribution with parameter 1.

E(Y)=E(E(Y|X=x))=E(x)=1

c)V(Y)=V(E(Y|X=x))+E(V(Y|X=x))=V(X)+E(X)=1²+1=1+1=2