chapter 7 proof of Var(S)

[kylie jane]

So we have
S=X_1+X_2+...+X_n

E(S|N=n)=nE(X)
var(S|N=n) = nvar(X)
E(S)=E(E(S|N)) = E(NE(X))=E(N)E(E(X))=E(N)E(X)

then for var(S)
var(S) = E(var(S|N))+var(E(S|N))
= E(Nvar(X))+var(NE(X))
= E(N)var(X) +var(N)[E(X)]^2
How do we get that
var(NE(X))=var(N)[E(X)]^2

 

[Busy_Bee4422]

E(X) is treated as a constant.

Therefore you use the relationship

Var(bZ) = [b^2] * Var(Z)

 

[kylie jane]

I just realised - E(X) is a constant, so using the forumula
var(aX+b) = a^2var(X) you get the result! doh!