Results 1 to 4 of 4

Math Help - Probability proof question

  1. #1
    Member
    Joined
    Oct 2006
    Posts
    84

    Probability proof question

    I"m not really sure how to go about this problem, so any help would be greatly appreciated!

    Let X and Y be independent random variables with the same distribution, taking values 0 and 1 with equal probability. Show that

    E((X+Y)(|X-Y|)) = E(X+Y)E(|X+Y|), but that X+Y and |X-Y| are not independent.

    (Here, E is the expectation)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by clockingly View Post
    I"m not really sure how to go about this problem, so any help would be greatly appreciated!

    Let X and Y be independent random variables with the same distribution, taking values 0 and 1 with equal probability. Show that

    E((X+Y)(|X-Y|)) = E(X+Y)E(|X+Y|), but that X+Y and |X-Y| are not independent.

    (Here, E is the expectation)
    X+Y can take the values 0, 1, 2, with probability 1/4, 1/2, 1/4

    |X-Y| can take the values 0, 1, with probabilty 1/2, 1/2

    Also the RV ((X+Y), |X-Y|) can take values (0,0), (0,1), (1,0), (1,1), (2,0), (2,1),
    with probabilities 1/4, 0, 0, 1/2, 1/4, 0.

    So now you can just evaluate E((X+Y)(|X-Y|)) and E(X+Y)E(|X+Y|) to show that
    they are equal

    If X+Y and |X-Y| are independent then their joint distribution:

    P((X+Y)=a, |X-Y|=b)=P((X+Y)=a)P(|X-Y|=b)

    and you have sufficient information to check this

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2006
    Posts
    84
    Thanks! I just realized that I made a mistake in typing the problem. Sorry about that. It's actually show that E((X+Y)(|X-Y|)) = E(X+Y)E(|X-Y|) but I'm assuming the values would be the same.

    Okay, so using this information, I'm getting that:
    E(X+Y) = 0*1/4 + 1*1/2 + 2*1/4 = 1
    and
    E(|X-Y|) = 0*1/2 + 1*1/2 = 1/2

    Multiplying these two expectations gives 1/2, which is E(X+Y)E(|X-Y|).

    However, how would one find (X+Y) and (|X-Y|) individually and then compute the expectation? I guess I'm a little unclear about what values to use for this because there are coordinates involved.
    Last edited by clockingly; October 8th 2007 at 05:52 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2006
    Posts
    84
    Sorry, you can disregard my previous question. I just figured it out!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Probability Proof
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: January 30th 2011, 02:06 PM
  2. Probability-measure theory proof question
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: March 24th 2010, 05:53 PM
  3. Probability proof?
    Posted in the Advanced Statistics Forum
    Replies: 7
    Last Post: October 1st 2009, 08:55 AM
  4. Probability Models proof question
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: March 8th 2009, 08:53 PM
  5. Probability Proof Question
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: February 9th 2009, 02:00 PM

Search Tags


/mathhelpforum @mathhelpforum