Results 1 to 6 of 6

Math Help - Joint Probability Function

  1. #1
    Junior Member
    Joined
    Oct 2007
    Posts
    55

    Joint Probability Function

    Let X and Y have joint probability density function

    fX,Y(x,y) = x+y, for 0 <=x<= 1, o<=y<=1
    0, otherwise

    How do i go about determining the corr(X,Y)?

    Could someone outline the steps please.. Any help would be appreciated..
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    One has {\rm Corr}(X,Y)=\frac{{\rm Cov}(X,Y)}{\sqrt{{\rm Var}(X){\rm Var}(Y)}}=\frac{E[XY]-E[X]E[Y]}{\sqrt{(E[X^2]-E[X]^2)(E[Y^2]-E[Y]^2)}}, right? So what you need to compute is E[XY], E[X], E[Y], E[X^2] and E[Y^2]. In fact, the pdf of (X,Y) is symmetric in x,y, so that X and Y have same distribution and the formula reduces to: {\rm Corr}(X,Y)=\frac{E[XY]-E[X]^2}{E[X^2]-E[X]^2}.
    To compute E[XY], just integrate xy times the pdf of (X,Y) (over the square 0\leq x,y\leq 1).
    You can do the same with the other ones (integrating x and x^2), but you may find it quicker to first determine the pdf of X. For that, you just have to integrate the pdf of (X,Y) with respect to the variable y, keeping x fixed.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Laurent View Post
    One has {\rm Corr}(X,Y)=\frac{{\rm Cov}(X,Y)}{\sqrt{{\rm Var}(X){\rm Var}(Y)}}=\frac{E[XY]-E[X]E[Y]}{\sqrt{(E[X^2]-E[X]^2)(E[Y^2]-E[Y]^2)}}, right? So what you need to compute is E[XY], E[X], E[Y], E[X^2] and E[Y^2]. In fact, the pdf of (X,Y) is symmetric in x,y, so that X and Y have same distribution and the formula reduces to: {\rm Corr}(X,Y)=\frac{E[XY]-E[X]^2}{E[X^2]-E[X]^2}.
    To compute E[XY], just integrate xy times the pdf of (X,Y) (over the square 0\leq x,y\leq 1).
    You can do the same with the other ones (integrating x and x^2), but you may find it quicker to first determine the pdf of X. For that, you just have to integrate the pdf of (X,Y) with respect to the variable y, keeping x fixed.
    I was just about to reply when I saw yours. I misread the question as find Cov(X, Y) ..... So it's lucky you replied first (although my reply would have got the OP half way ...... )
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Oct 2007
    Posts
    55
    Thanks so much, Ok this is what i got...


    E[XY] = 1/3

    E[X]^2 = (1/3 + y/2)^2 = y^2/4 + 2y/6 + 1/9

    E[X^2] = 1/4 + y/3

    E[Y] = 1/3 + x/2

    And as a final answer.. i gained..

    y^4/16 - 2y^3/24 - 13y^2/144 + 10y/216 + 10/324

    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by brd_7 View Post
    Thanks so much, Ok this is what i got...

    E[X]^2 = (1/3 + y/2)^2 = y^2/4 + 2y/6 + 1/9

    There's no y in the answer (what is y?). In order to find the average, you integrate y as well:
    E[X]=\int_0^1\int_0^1 x(x+y)dx\,dy. The same for X^2.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Oct 2007
    Posts
    55
    Oh right, that makes sense.. Im assuming you would do the same if i needed E(Y)?

    Well as a final answer i got -1/11..

    Thanks for all the help again.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Joint Probability Function
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: August 4th 2010, 03:45 AM
  2. Joint probability function
    Posted in the Statistics Forum
    Replies: 1
    Last Post: February 10th 2010, 04:08 PM
  3. Replies: 5
    Last Post: December 5th 2009, 11:30 PM
  4. Joint probability function
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: April 28th 2009, 10:58 PM
  5. Joint Probability Function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 5th 2009, 02:56 AM

Search Tags


/mathhelpforum @mathhelpforum