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Math Help - distribution function of Z = XY

  1. #1
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    distribution function of Z = XY

    X and Y are independent uniform variables distributed between 0 and 1.

    Z = XY

    How to calculate the distribution function of Z?
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  2. #2
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    Hello,

    Their joint pdf, that is to say the pdf of (X,Y) is the product of their pdf, since they're independent.
    It equals 1, for 0<x<1 and 0<y<1.

    Then, use the Jacobian transformation that transforms (X,Y) into (XY,Y) (see here : http://www.mathhelpforum.com/math-he...ion-r-v-s.html for more information) to get the joint pdf of (XY,Y)

    Then integrate with respect to the second variable to get the pdf of XY
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  3. #3
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    how would i do it if i want to take the integral
    from 0 to 1 for dy and 0 to z/x for dx

    In other words, if I have the
    P(X <=1, Y <= u/x)
    = double integral of 0 to 1 dx and 0 to u/x dy

    how to solve this integral?
    Last edited by mr fantastic; August 4th 2009 at 04:16 PM. Reason: Merged posts
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  4. #4
    MHF Contributor matheagle's Avatar
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    I solved that a few months ago on this board.
    You may have to look through a few pages to find it.
    The actual problem was to find the density of

    W=(XY)^Z

    where X,Y and Z are iid U(0,1) rvs.
    It turns out that W is U(0,1) too.
    On the way to solving that I found the density of XY.
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  5. #5
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    Quote Originally Posted by vcarter3 View Post
    X and Y are independent uniform variables distributed between 0 and 1.

    Z = XY

    How to calculate the distribution function of Z?
    Quote Originally Posted by vcarter3 View Post
    how would i do it if i want to take the integral

    from 0 to 1 for dy and 0 to z/x for dx

    In other words, if I have the

    P(X <=1, Y <= u/x)

    = double integral of 0 to 1 dx and 0 to u/x dy

    how to solve this integral?

    The cdf of Z is

    F(z) = \Pr(Z < z) = \Pr(XY < z) = \Pr \left(Y < \frac{z}{X} \right)

    where 0 \leq z \leq 1

    = \int_{x = 0}^{x = z} \int_{y = 0}^{y = 1} \, dy \, dx + \int_{x = z}^{x = 1} \int_{y = 0}^{y = z/x} \, dy \, dx .

    The pdf of Z is f(z) = \frac{dF}{dz}
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  6. #6
    Super Member Random Variable's Avatar
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    Quote Originally Posted by mr fantastic View Post
    The cdf of Z is

    F(z) = \Pr(Z < z) = \Pr(XY < z) = \Pr \left(Y < \frac{z}{X} \right)

    where 0 \leq z \leq 1

    = \int_{x = 0}^{x = z} \int_{y = 0}^{y = 1} \, dy \, dx + \int_{x = z}^{x = 1} \int_{y = 0}^{y = z/x} \, dy \, dx .

    The pdf of Z is f(z) = \frac{dF}{dz}
    I'm having trouble understanding the limits.
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  7. #7
    MHF Contributor matheagle's Avatar
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    I got f_Z(z)=-\ln z on 0<z<1.

    And that is a valid density and I think that's what I got the last time.

    f_{XY}(x,y)=1 on 0<x<1, 0<y<1.

    Let Z=XY and W=X.

    The inverse relationship is W=W and Y=Z/W.

    So, f_{ZW}(z,w)=1/w on 0<w<1, 0<z<w.

    And f_{Z}(z)=\int_z^1 dw/w=-\ln z on 0<z<1.

    NEXT, let Q=Z^U wher U is U(0,1), indep of X and Y, and it turns out Q is U(0,1).
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  8. #8
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    Quote Originally Posted by Random Variable View Post
    I'm having trouble understanding the limits.
    Draw the square defined by x = 0, x = 1, y = 0 and y = 1. Draw the part of the curve y = z/x that lies inside the square. Integrate over the region of the square that's below this curve.

    The result is consistent with baldeagle's calculation.
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  9. #9
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by mr fantastic View Post
    The result is consistent with boldeagle's calculation.
    If that's the case, then you must be correct.
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  10. #10
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    Quote Originally Posted by mr fantastic View Post
    The cdf of Z is

    F(z) = \Pr(Z < z) = \Pr(XY < z) = \Pr \left(Y < \frac{z}{X} \right)

    where 0 \leq z \leq 1

    = \int_{x = 0}^{x = z} \int_{y = 0}^{y = 1} \, dy \, dx + \int_{x = z}^{x = 1} \int_{y = 0}^{y = z/x} \, dy \, dx .

    The pdf of Z is f(z) = \frac{dF}{dz}
    How did you know to choose z as the point to break the integral into two?
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  11. #11
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    Quote Originally Posted by matheagle View Post
    I got f_Z(z)=-\ln z on 0<z<1.

    And that is a valid density and I think that's what I got the last time.

    f_{XY}(x,y)=1 on 0<x<1, 0<y<1.

    Let Z=XY and W=X.

    The inverse relationship is W=W and Y=Z/W.

    So, f_{ZW}(z,w)=1/w on 0<w<1, 0<z<w.

    And f_{Z}(z)=\int_z^1 dw/w=-\ln z on 0<z<1.

    NEXT, let Q=Z^U wher U is U(0,1), indep of X and Y, and it turns out Q is U(0,1).
    Some questions:
    1) You mean X = W for the inverse yes?
    2) Why is w the upper bound for z in your joint density function f(z,w)?
    3) How did you get z and 1 for the limits of integration for fz
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  12. #12
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    Quote Originally Posted by vcarter3 View Post
    How did you know to choose z as the point to break the integral into two?
    What part of post #8 don't you understand?
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