# Conditional Expectation

• Oct 24th 2010, 10:51 AM
uberbandgeek6
Conditional Expectation
Suppose that a company will incur a loss X that is uniformly distributed between 0 and 100. The company will pay a bonus to its employees that is uniformly distributed from 0 to 300−2X. What is the expected value of the bonus paid?

Since X~Uniform(0,100), the pdf would be f(x) = 1/100, 0 < x < 100. For Y = 300 - 2X, I used the change of variables formula to get f(y) = 1/200. I assume this problem must have something to do with the conditional mean, E(Y|X=x), but for that I would need a joint pdf f(x,y) and X = some value, which I'm not sure how I'd find. Where should I go from here?
• Oct 24th 2010, 11:49 PM
matheagle
It seems you only need E(300-2X)=300-2(50)

Y=aX+b is always a uniform rv, if X is.
• Oct 25th 2010, 05:49 AM
CaptainBlack
Quote:

Originally Posted by uberbandgeek6
Suppose that a company will incur a loss X that is uniformly distributed between 0 and 100. The company will pay a bonus to its employees that is uniformly distributed from 0 to 300−2X. What is the expected value of the bonus paid?

Since X~Uniform(0,100), the pdf would be f(x) = 1/100, 0 < x < 100. For Y = 300 - 2X, I used the change of variables formula to get f(y) = 1/200. I assume this problem must have something to do with the conditional mean, E(Y|X=x), but for that I would need a joint pdf f(x,y) and X = some value, which I'm not sure how I'd find. Where should I go from here?

From first principles the expected bonus is E(E(bonus|x))

(which is what the eagke is getting at)

CB
• Oct 28th 2010, 07:20 AM
uberbandgeek6
I understand what you are saying, but I was thinking about this some more. Since the bonus (Y) is uniformly distributed between 0 and 300-2X, the pdf of Y|X should be f(y|x) = 1 / (300-2x). Then the expected value of Y|X should be E(Y|X) = (300-2x)/2 = 150 - x. So if I know the expected value of X is 50, couldn't I just plug it in to get E(Y|X=50) = 150 - 50 = 100?
• Oct 28th 2010, 08:58 AM
CaptainBlack
Quote:

Originally Posted by uberbandgeek6
I understand what you are saying, but I was thinking about this some more. Since the bonus (Y) is uniformly distributed between 0 and 300-2X, the pdf of Y|X should be f(y|x) = 1 / (300-2x). Then the expected value of Y|X should be E(Y|X) = (300-2x)/2 = 150 - x. So if I know the expected value of X is 50, couldn't I just plug it in to get E(Y|X=50) = 150 - 50 = 100?

It will work in this case because E(Y|X) is linear in x, but it won't work in general.

CB