What if we consider X/(Y+1) instead? What would the variance of that be?
To get the variance, you need the expectation, and the expectation of the square. Since the variables are independent, you need and . I guess you know how to get and ( and since I know the variance is ). So let's consider the other ones.
; then let in the sum to see that it equals .
is more tricky. I'd let and compute , which we have computed above ; now integrate to get , and multiply by to get .
Last edited by mr fantastic; March 23rd 2010 at 07:25 PM.
Reason: Made the square brackets around fractions bigger (such a lovely reply deserved it).