Results 1 to 5 of 5

Math Help - Conditional Expectation

  1. #1
    Member
    Joined
    Sep 2008
    Posts
    79

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    It seems you only need E(300-2X)=300-2(50)

    Y=aX+b is always a uniform rv, if X is.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by uberbandgeek6 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2008
    Posts
    79
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by uberbandgeek6 View Post
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: July 22nd 2011, 01:39 AM
  2. Conditional Expectation
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: March 29th 2011, 05:01 PM
  3. Conditional Expectation
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: January 29th 2010, 08:34 PM
  4. Expectation & Conditional Expectation
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: February 1st 2009, 10:42 AM
  5. Conditional Expectation
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: December 9th 2007, 08:22 PM

Search Tags


/mathhelpforum @mathhelpforum