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Math Help - Help with conditional probability of joint density functions.

  1. #1
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    Post Help with conditional probability of joint density functions.

    Let X and Y be continuous random variables with joint density function
    f(x,y)=\left\{\begin{array}{lr}(24xy)&0< x< 1;0< y<1-x\\0&otherwise\end{array}\right.

    How would you determine the correlation coefficient, say p, and then calculate the conditional probability of P [ Y < X | X= .5 ]

    thank you for your help times infinity!
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  2. #2
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    Cov(X,Y) = E(XY) - E(X)E(Y)

    \rho = \frac{Cov(X,Y)}{\sigma_X\sigma_Y}

    Figure out the expectation the joint distribution.

    E(XY) = \int_0^1\int_0^{1-x}xyf(x,y)dydx

    Figure out the conditional distributions of X and Y.

    f_X(x) = \int_0^{1-x}f(x,y)dy

    f_Y(y) = \int_0^1 f(x,y)dx

    Figure out the expectation of X and Y.

    E(X) = \int_0^1 xf_X(x)dx

    E(Y) = \int_0^{1-x} yf_Y(y)dy

    Figure out the variance of X and Y.

    \sigma_X^2 = E(X^2) - [E(X)]^2

    \sigma_Y^2 = E(Y^2) - [E(X)]^2

    Plug in the appropriate functions and now you have correlation.
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  3. #3
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    Quote Originally Posted by BERRY View Post
    Cov(X,Y) = E(XY) - E(X)E(Y)

    \rho = \frac{Cov(X,Y)}{\sigma_X\sigma_Y}

    Figure out the expectation the joint distribution.

    E(XY) = \int_0^1\int_0^{1-x}xyf(x,y)dydx

    Figure out the conditional distributions of X and Y.

    f_X(x) = \int_0^{1-x}f(x,y)dy

    f_Y(y) = \int_0^1 f(x,y)dx

    Figure out the expectation of X and Y.

    E(X) = \int_0^1 xf_X(x)dx

    E(Y) = \int_0^{1-x} yf_Y(y)dy

    Figure out the variance of X and Y.

    \sigma_X^2 = E(X^2) - [E(X)]^2

    \sigma_Y^2 = E(Y^2) - [E(X)]^2

    Plug in the appropriate functions and now you have correlation.
    one question...

    When you figure out the conditional distributions of X and Y.

    should it be

    f_Y(y) = \int_0^{1-x}f(x,y)dx
    not

    f_Y(y) = \int_0^1 f(x,y)dx ??


    and also,
    when you figure out the variance of X and Y.


    \sigma_Y^2 = E(Y^2) - [E(Y)]^2
    not
    \sigma_Y^2 = E(Y^2) - [E(X)]^2 right?
    Last edited by Statsnoob2718; November 17th 2009 at 09:41 AM.
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  4. #4
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    Quote Originally Posted by Statsnoob2718 View Post
    one question...

    When you figure out the conditional distributions of X and Y.

    should it be

    f_Y(y) = \int_0^{1-x}f(x,y)dx
    not

    f_Y(y) = \int_0^1 f(x,y)dx ??


    and also,
    when you figure out the variance of X and Y.


    \sigma_Y^2 = E(Y^2) - [E(Y)]^2
    not
    \sigma_Y^2 = E(Y^2) - [E(X)]^2 right?
    For the variance of Y, I mistyped it. You're correct.

    For the conditional of Y, you integrate over x (thus the dx) and x ranges from 0 to 1 so the original is correct.
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  5. #5
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    Oh, cool dude, yeah I just drew a picture of the bounds to confirm, you're right about the limits. Great work!
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  6. #6
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Statsnoob2718 View Post
    one question...

    f_Y(y) = \int_0^{1-x}f(x,y)dx
    not

    f_Y(y) = \int_0^1 f(x,y)dx ??

    Both are wrong

    A function of y cannot have x in it.
    The correct marginal is

    f_Y(y) = \int_0^{1-y}f(x,y)dx

    just like

    f_X(x) = \int_0^{1-x}f(x,y)dy
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  7. #7
    MHF Contributor matheagle's Avatar
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    Likewise the E(Y) is wrong here

    E(X) = \int_0^1 xf_X(x)dx

    E(Y) = \int_0^{1-x} yf_Y(y)dy.
    E(Y) = \int_0^{1} yf_Y(y)dy

    IF you want to obtain the marginal density.
    But it's better to look at this two dimensionally...

    E(Y) = \int_0^{1} yf_Y(y)dy =\int_0^{1} y\int_0^{1-y}f(x,y)dxdy=\int_0^{1} \int_0^{1-y}yf(x,y)dxdy

    Because you have the choice of which variable you can integrate first

    E(Y) = \int_0^{1} \int_0^{1-x}yf(x,y)dydx
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