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Math Help - P.D.F. with two variables! Help!

  1. #1
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    P.D.F. with two variables! Help!

    I am having some trouble with multiple variables when one is fixed, and vice versa. I was wondering if someone could please help me - I missed a day of class, and it seems like everything was taught while I was sick!


    Anyway, if someone could please explain the following problem, I would really appreciate it!

    Let the probability density function of variables X and Y equal

    f(x,y) = 1/2 when 0<x<y, 0<y<2 and zero elsewhere. Find the conditional means and variances for each variable when another is fixed and the correlation of X and Y.


    Thanks in advance!
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  2. #2
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    Quote Originally Posted by ScienceGeniusGirl View Post
    I am having some trouble with multiple variables when one is fixed, and vice versa. I was wondering if someone could please help me - I missed a day of class, and it seems like everything was taught while I was sick!


    Anyway, if someone could please explain the following problem, I would really appreciate it!

    Let the probability density function of variables X and Y equal

    f(x,y) = 1/2 when 0<x<y, 0<y<2 and zero elsewhere. Find the conditional means and variances for each variable when another is fixed and the correlation of X and Y.


    Thanks in advance!
    Definitions:

    Marginal density of Y: f(y) = \int_{-\infty}^{+\infty} f(x, y) \, dx.

    Marginal density of X: f(x) = \int_{-\infty}^{+\infty} f(x, y) \, dy.

    Conditional density of X given Y = y: f(x | y) = \frac{f(x, y)}{f(y)} for f(y) > 0 and 0 otherwise.

    Conditional density of Y given X = x: f(y | x) = \frac{f(x, y)}{f(x)} for f(x) > 0 and 0 otherwise.

    E(X | Y = y) = \int_{-\infty}^{+\infty} x \, f(x | y) \, dx.

    E(Y | X = x) = \int_{-\infty}^{+\infty} y \, f(y | x) \, dy.

    Cov(X, Y) = E(XY) - E(X) E(Y)  = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} x y \, f(x, y) \, dx \, dy - \left( \int_{-\infty}^{+\infty} x \, f(x) \, dx \right) \, \left( \int_{-\infty}^{+\infty} y \, f(y) \, dy \right) .



    Application of the definitions to the given joint pdf:


    Marginal density of Y: f(y) = \int_{x = 0}^{x = y} \frac{1}{2} \, dx = \frac{y}{2}, x < y < 2.

    Marginal density of X: f(x) = \int_{y=x}^{y=2} \frac{1}{2} \, dy = 1 - \frac{x}{2}, 0 < x < y.

    Conditional density of X given Y = y: f(x | y) = \frac{1/2}{\frac{y}{2}} = \frac{1}{y} for x < y < 2.

    Conditional density of Y given X = x: f(y | x) = \frac{1/2}{1 - \frac{x}{2}} = \frac{1}{2-x} for 0 < x < y.

    E(X | Y = y) = \int_{x=0}^{x=y}\frac{x}{y} \, dx = \frac{y}{2}.

    E(Y | X = x) = \int_{y=x}^{y=2} \frac{y}{2-x}\, dy = \frac{2+x}{2}.

    Cov(X, Y):

    E(XY) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} x y \, f(x, y) \, dx \, dy = \int_{y=0}^{y=2} \int_{x=0}^{x=y} \frac{xy}{2} \, dx \, dy = \int_{y=0}^{y=2} \frac{y^3}{4} \, dy = 1.

    The rest of the calculations are left for you to do.
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  3. #3
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    Cov(X, Y):

    E(XY) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} x y \, f(x, y) \, dx \, dy = \int_{y=0}^{y=2} \int_{x=0}^{x=y} \frac{xy}{2} \, dx \, dy = \int_{y=0}^{y=2} \frac{y^3}{4} \, dy = 1.

    The rest of the calculations are left for you to do.[/quote]


    I guess I don't understand how you got to this point - I understand that if the E(X,Y) is equal to one, then to be calculating the Cov(X,Y), I need to calculate the -E(x)E(y) part - but how does one do that?

    I vaguely follow the above posted math, but not strongly enough to calculate the rest of it?
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  4. #4
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    Mostly, how did you get from the left side of that equation to the right side?


    I'm sorry for all the trouble, I just really don't understand.
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  5. #5
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    Quote Originally Posted by ScienceGeniusGirl View Post
    Mostly, how did you get from the left side of that equation to the right side?


    I'm sorry for all the trouble, I just really don't understand.
    By definition: E(XY) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} x y \, f(x, y) \, dx \, dy


    Put in integral terminals that define the region of the xy-plane in which the joint pdf is non-zero: = \int_{y=0}^{y=2} \int_{x=0}^{x=y} \frac{xy}{2} \, dx \, dy

    To get these terminals, draw the lines y = x and y = 2. 0 < x < y and 0 < y < 2 define the region of the xy-plane that lies between the y-axis and the lines y = x and y = 2. This is the area the integral terminals define. You're expected to have experience with doing this sort of thing.


    Integrate with respect to x:

    = \int_{y=0}^{y=2} \left[\frac{x^2 y}{4}\right]_{0}^{y} \, dx \, dy

    (note that y is treated as a constant because the integration is with respect to x)


    = \int_{y=0}^{y=2} \frac{y^3}{4} \, dy


    Integrate with respect to y: \left[ \frac{y^4}{16}\right]_{0}^{2} = 1.
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  6. #6
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    Ok, so I did the calculations for E(x) and E(Y) - Or tried to!

    I got E(X) = -1/2
    and E(Y) = 1


    Therefore, my Cov(X,Y) = 1 - (-1/2)(1) = 1.5 ???


    Currently working on my correlation, based on this covariance
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  7. #7
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    Quote Originally Posted by ScienceGeniusGirl View Post
    Ok, so I did the calculations for E(x) and E(Y) - Or tried to!

    I got E(X) = -1/2 Mr F says: Since the pdf is non-zero only for 0<x<y, 0<y<2, this cannot possibly be correct. Surely 0 < E(X) < 2 .....!

    and E(Y) = 1 Mr F says: Incorrect.


    Therefore, my Cov(X,Y) = 1 - (-1/2)(1) = 1.5 ???


    Currently working on my correlation, based on this covariance
    How did you get those answers?

    E(X) = \int_{-\infty}^{+\infty} x \, f(x) \, dx = \int_{0}^{2} x \, \left(1 - \frac{x}{2} \right) \, dx = \frac{2}{3}.

    E(Y) = \int_{-\infty}^{+\infty} y \, f(y) \, dy = \int_{0}^{2} y \, \left(\frac{y}{2} \right) \, dy = \frac{4}{3}.
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