Expected Value of a Combination of Random Variables

• Aug 23rd 2011, 05:10 AM
joeyjoejoe
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.
• Aug 23rd 2011, 05:18 AM
CaptainBlack
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
• Aug 23rd 2011, 06:16 AM
joeyjoejoe
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.
• Aug 23rd 2011, 12:59 PM
JeffN12345
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...
• Aug 23rd 2011, 01:30 PM
joeyjoejoe
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.
• Aug 23rd 2011, 02:27 PM
JeffN12345
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).
• Aug 23rd 2011, 07:39 PM
CaptainBlack
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