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Math Help - Help with Joint Probability Density Function

  1. #1
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    Help with Joint Probability Density Function

    Hi all,

    I am stuck with the problem below:

    The Joint pdf of random variables X and Y: f(x,y) = k/y , 0<x<y, 0<y<1 and f(x,y) = 0, otherwise.

    First of all, the question asked to draw a diagram showing the support of the joint pdf.
    Probably this is the part that i am unable to do, which leads to problems with later part of the question.

    The question asked to calculate k, I integrate with respect to x over the range 0 to y, and then integrate with respect to y over the range 0 to 1.
    The answer I get is k = 1, is it correct? Did i integrate the correct range for x and y?

    After that the question asked to find: P(0<=X<=1/4, 0<=Y<=1/3).
    For X, the range i used to integrate over function f is 0 to 1/4. However, for Y, i am not so sure about it, I guess it is from 1/4 to 1/3? Or 0 to 1/3 for Y?

    Any help regarding the problem would be appreciated.
    Thank you.
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  2. #2
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    Quote Originally Posted by Watts84 View Post
    Hi all,

    I am stuck with the problem below:

    The Joint pdf of random variables X and Y: f(x,y) = k/y , 0<x<y, 0<y<1 and f(x,y) = 0, otherwise.

    First of all, the question asked to draw a diagram showing the support of the joint pdf.
    Probably this is the part that i am unable to do, which leads to problems with later part of the question.

    The question asked to calculate k, I integrate with respect to x over the range 0 to y, and then integrate with respect to y over the range 0 to 1.
    The answer I get is k = 1, is it correct? Did i integrate the correct range for x and y?

    After that the question asked to find: P(0<=X<=1/4, 0<=Y<=1/3).
    For X, the range i used to integrate over function f is 0 to 1/4. However, for Y, i am not so sure about it, I guess it is from 1/4 to 1/3? Or 0 to 1/3 for Y?

    Any help regarding the problem would be appreciated.
    Thank you.
    The support is the triangular region defined by the lines y = 1 and y = x and the y-axis. Is that what you had?
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  3. #3
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    Yes, I did manage to get that... Thank you.
    In that case, the double integral range would be from 0<x<1, 0<y<1?
    If thats the case, it looked like integrating the whole square, instead of the right triangle?
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  4. #4
    MHF Contributor matheagle's Avatar
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    no, it's a triangle 0<x<y<1
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  5. #5
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    Quote Originally Posted by Watts84 View Post
    Yes, I did manage to get that... Thank you.
    In that case, the double integral range would be from 0<x<1, 0<y<1?
    If thats the case, it looked like integrating the whole square, instead of the right triangle?
    Have you been taught how to set up and solve double integrals over a region of the xy-plane? If not, then there's little point in attempting questions like this one until you.

    The integral you're expected to set up and solve is \int_{x = 0}^{x = 1} \int_{y = x}^{y = 1} f(x, y) \, dy \, dx. Then equate the result of this calculation to 1 in order to solve for k.
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  6. #6
    MHF Contributor matheagle's Avatar
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    I prfer dxdy, it has 0 as both lower bounds
    and the integration is a lot easier.

    \int_0^1\int_0^y f(x,y)dxdy
    Last edited by matheagle; November 28th 2009 at 06:39 PM.
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  7. #7
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    Quote Originally Posted by matheagle View Post
    I prfer dxdy, it has 0 as both lower bounds
    and the integration is a lot easier.

    \int_0^1\int_0^y f(x,y)dxdy

    This is just basic calc3
    I see now why baldeagles are an endangered species ..... Their lower bounds are zero.
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  8. #8
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    Yes, i have been taught how to do double integral over x-y plane. I also prefer the lower bounds to be set to zero too, much easier..Thanks Matheagle.
    Been quite confuse in deciding the range for integration.
    Thanks a lot of clarified my doubts.
    As for the last part: Finding P(0<=X<=1/4, 0<=Y<=1/3).
    The integration will look like this: <br />
\int_{x = 0}^{x = 1/4} \int_{y = x}^{y = 1/3} f(x, y) \, dy \, dx<br />
?
    If i wish to integrate by dxdy, range of x will be from 0 to 1/4. How about y? <br />
\int_{y = ?}^{y = ?} \int_{x = 0}^{x = 1/4} f(x, y) \, dx \, dy<br />
?
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  9. #9
    MHF Contributor matheagle's Avatar
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    dxdy needs two regions, dydx only one, draw it

    \int_0^{1/4}\int_x^{1/3} fdydx

    To do dxdy you need to split y into two parts (0,1/4) and (1/4,1/3)

     \int_0^{1/4}\int_0^y fdxdy+ \int_{1/4}^{1/3}\int_0^{1/4} fdxdy
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