So I'm doing practice questions for the actuarial exams, and this question came up:
If X, Y, and Z are independent identically distributed random variables each with mean 3, then what is E[(X+Y+Z)^2]?
The solution says it's 108, but I don't have a worked out solution to look at. Any ideas? Thanks in advance.
X,Y,Z are iid some distribution with mean M and variance V
E((X+Y+Z)^2)=E(x^2)+E(Y^2)+E(Z^2)+2(E(XY)+E(YZ)+E( ZX))
Var(X)=E(X^2)-E(X)^2=E(x^2)-M^2 => E(X^2)=V+M^2
Cov(X,Y)=E(XY)-E(X) E(Y)=E(XY)-M^2,
X,Y are independent => cov(X,Y)=0)=>E(XY)=M^2
=> E((X+Y+Z)^2)=3(V+M^2)+2(3(M^2))=3V+9(M^2)
M=3 => E((X+Y+Z)^2)=3V+81
was the variance also required to be 9 or did i miss something?
If the variance was not 9 then the solution would not be 108...
No, the variance has to be 9 as well, any distribution will do as long as the mean is 3 and the variance is 9.
This is the case for the Poisson with mean 3, but you do not need to specify or assume a distribution. If you were free to pick any distribution with mean 3 you could have chosen the normal with mean 3 and variance 102,345. Which would not work.
CB