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.

Re: Expected Value of a Combination of Random Variables

Quote:

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

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.

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.

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).

Re: Expected Value of a Combination of Random Variables

Quote:

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