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Thread: Joint Probability Function

  1. #1
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    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..
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  2. #2
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    One has $\displaystyle {\rm Corr}(X,Y)=\frac{{\rm Cov}(X,Y)}{\sqrt{{\rm Var}(X){\rm Var}(Y)}}=\frac{E[XY]-E[X]E[Y]}{\sqrt{(E[X^2]-E[X]^2)(E[Y^2]-E[Y]^2)}}$, right? So what you need to compute is $\displaystyle E[XY]$, $\displaystyle E[X]$, $\displaystyle E[Y]$, $\displaystyle E[X^2]$ and $\displaystyle E[Y^2]$. In fact, the pdf of $\displaystyle (X,Y)$ is symmetric in $\displaystyle x,y$, so that $\displaystyle X$ and $\displaystyle Y$ have same distribution and the formula reduces to: $\displaystyle {\rm Corr}(X,Y)=\frac{E[XY]-E[X]^2}{E[X^2]-E[X]^2}$.
    To compute $\displaystyle E[XY]$, just integrate $\displaystyle xy$ times the pdf of $\displaystyle (X,Y)$ (over the square $\displaystyle 0\leq x,y\leq 1$).
    You can do the same with the other ones (integrating $\displaystyle x$ and $\displaystyle x^2$), but you may find it quicker to first determine the pdf of $\displaystyle X$. For that, you just have to integrate the pdf of $\displaystyle (X,Y)$ with respect to the variable $\displaystyle y$, keeping $\displaystyle x$ fixed.
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  3. #3
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    Quote Originally Posted by Laurent View Post
    One has $\displaystyle {\rm Corr}(X,Y)=\frac{{\rm Cov}(X,Y)}{\sqrt{{\rm Var}(X){\rm Var}(Y)}}=\frac{E[XY]-E[X]E[Y]}{\sqrt{(E[X^2]-E[X]^2)(E[Y^2]-E[Y]^2)}}$, right? So what you need to compute is $\displaystyle E[XY]$, $\displaystyle E[X]$, $\displaystyle E[Y]$, $\displaystyle E[X^2]$ and $\displaystyle E[Y^2]$. In fact, the pdf of $\displaystyle (X,Y)$ is symmetric in $\displaystyle x,y$, so that $\displaystyle X$ and $\displaystyle Y$ have same distribution and the formula reduces to: $\displaystyle {\rm Corr}(X,Y)=\frac{E[XY]-E[X]^2}{E[X^2]-E[X]^2}$.
    To compute $\displaystyle E[XY]$, just integrate $\displaystyle xy$ times the pdf of $\displaystyle (X,Y)$ (over the square $\displaystyle 0\leq x,y\leq 1$).
    You can do the same with the other ones (integrating $\displaystyle x$ and $\displaystyle x^2$), but you may find it quicker to first determine the pdf of $\displaystyle X$. For that, you just have to integrate the pdf of $\displaystyle (X,Y)$ with respect to the variable $\displaystyle y$, keeping $\displaystyle x$ 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 ...... )
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  4. #4
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    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

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  5. #5
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    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 $\displaystyle y$ in the answer (what is $\displaystyle y$?). In order to find the average, you integrate $\displaystyle y$ as well:
    $\displaystyle E[X]=\int_0^1\int_0^1 x(x+y)dx\,dy$. The same for $\displaystyle X^2$.
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  6. #6
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    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.
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