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Math Help - Joint cumulative distribution function

  1. #1
    Member billym's Avatar
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    Joint cumulative distribution function

    I have the joint density function:

    <br />
f(x,y)=\left\{<br />
\begin{array}{lr}<br />
k(2x+y)&0\le x\le 1;0\le y\le 2\\<br />
0&otherwise<br />
\end{array}<br />
\right.<br />

    First, I had to find k. So I integrated for x with given limits and then for y with the given limits and ended up with k=1/4. Correct?

    Now I have to find the joint cumulative distribution function... stuck. Do I just put in k=1/4 into the density function and then find the derivative?
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  2. #2
    Member billym's Avatar
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    f_{X,Y}(x,y)=\frac{1}{4}(2x+y)

    F_{X,Y}(x,y)=\frac{1}{8}(2x^2y+xy^2)

    correct?
    Last edited by billym; October 27th 2009 at 11:46 AM.
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  3. #3
    Member billym's Avatar
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    Can somebody tell me if I am doing this at all right?

    Marginal densities:

    f_X(x)=\int_0^2 \frac{1}{4}x^2y+\frac{1}{8}xy^2 dy = \frac{1}{6}(3x^2+2x)

    f_Y(y)=\int_0^1 \frac{1}{4}x^2y+\frac{1}{8}xy^2 dy = \frac{1}{48}(4y+3y^2)

    Conditional density of Y given X = 1/4:

    f_{Y|X}(y|1/4)=\frac{f_{XY}(1/4,y)}{f_X(1/4)}=\frac{1}{22}(3y+6y^2)

    Since f_{Y|X}(y|x) \not = f_Y(y), X and Y are not independent.
    Last edited by billym; October 27th 2009 at 08:03 PM.
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  4. #4
    MHF Contributor matheagle's Avatar
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    X and Y are not independent by inspection
    f(x,y) cannot be factored into two functions, one of x and one of y.
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  5. #5
    Member billym's Avatar
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    are the marginal densities that I obtained correct?
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  6. #6
    MHF Contributor matheagle's Avatar
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    I'm busy grading right now
    I might have time later tonight to look it over.
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  7. #7
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    The joint cdf is defined as F_{XY}(x, y) = P(X \le x, Y \le y), so just use the joint pdf to calculate that probability like you would any other (e.g. if you were calculating P(X \le 3, Y \le 2) what would you do? Extend this method to arbitrary x and y, and you will have the joint cdf). The joint cdf that you posted is correct if you assume that  0 \le x \le 1, 0 \le y \le 2 , but that might not be the case. The cdf must be defined for everything in the plane.

    The marginal densities are wrong. You integrate the unwanted variable out using the joint pdf, not the joint cdf.
    Last edited by theodds; October 27th 2009 at 06:10 PM.
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  8. #8
    Member billym's Avatar
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    Ok well at least I'm nearly there (just started this today). Yeah I forgot to post 0 \le x \le 1, 0 \le y \le 2 at the beginning but they're there when I attempted the marginal densities. Thanks guys.
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  9. #9
    Member billym's Avatar
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    f_X(x)=\int_0^2 1/4(2x+Y) dy = 1/2+x

    f_Y(y)=\int_0^1 1/4(2x+Y) dx = 1/4+y/4

    f_{Y|X}(y|1/4)=\frac{f_{XY}(1/4,y)}{f_X(1/4)}=\frac{1/4(2x+y)}{1/2+x}=1/6+y/3

    f_{Y|X}(y|x) \not = f_Y(y) ; X and Y are not independent.
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  10. #10
    MHF Contributor matheagle's Avatar
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    Those two are the correct marginal densities (in your last post).
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