# Math Help - Expected Value of a Combination of Random Variables

1. ## Expected Value of a Combination of Random Variables

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.

2. ## Re: Expected Value of a Combination of Random Variables

Originally Posted by joeyjoejoe
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.
Expand the square inside the Expectation, then distribute the Expectation over the summed terms:

CB

3. ## Re: Expected Value of a Combination of Random Variables

Thanks... but I don't see how you can compute E(X^2) without additional information. Unless I am missing something stupid. Which is likely.

4. 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...

5. ## Re: Expected Value of a Combination of Random Variables

Ah, so really any distribution would work so long as the mean is 3. Might as well pick Poisson. Thanks.

6. ## Re: Expected Value of a Combination of Random Variables

Poisson Distribution is a good choice as the variance must be 9 for the solution to be E((X+Y+Z)^2)=108 (which is the case for a Poisson distribution with mean 3).

7. ## Re: Expected Value of a Combination of Random Variables

Originally Posted by joeyjoejoe
Ah, so really any distribution would work so long as the mean is 3. Might as well pick Poisson. Thanks.
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