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Math Help - derived probability density function

  1. #1
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    derived probability density function

    I have a problem about derive a pdf from two marginal pdfs:
    random variables x and y are distributed according to the joint probability density function (pdf):
    fxy(x,y) = 3x/2 , if 1 <= x <= y <= 2
    0 , otherwhise

    Random variable z is defined by z = y - x
    Determine the probablity density function fz(z)

    thanks.
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  2. #2
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    it's been a while since i've done this but i think you need to do a doubble integral over the domain specified on the function you have been given
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  3. #3
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    Quote Originally Posted by rbenito View Post
    I have a problem about derive a pdf from two marginal pdfs:
    random variables x and y are distributed according to the joint probability density function (pdf):
    fxy(x,y) = 3x/2 , if 1 <= x <= y <= 2
    0 , otherwhise

    Random variable z is defined by z = y - x
    Determine the probablity density function fz(z)

    thanks.
    For any value of z the function z = y-x defines a curve through xy space. The pdf of p(z) of z is the integral of the joint density over the area between the curve for z and z+\delta z divided by \delta z (rather the limit as \delta z goes to zero).

    Write the curves in the form: y=x+z, then we have:

    <br />
A= \delta z~ \int f(x,x+z) dx <br />

    So:

    <br />
p(z) = \int f(x,x+z) dx <br />

    If you can't complete this from here say so and I will finish it.

    RonL
    Last edited by CaptainBlack; September 28th 2007 at 09:56 PM.
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  4. #4
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    can you complete the integral?

    what are the limits?
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by rbenito View Post
    what are the limits?
    \mathbb{R},

    but since the support of x is [1,2], that will do, but you have to modify this because the function is non-zero only when 1<x<x+z<2

    RonL
    Last edited by CaptainBlack; September 30th 2007 at 11:01 AM.
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  6. #6
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    can you finish the integral?

    of the problem to obtain the derived pdf?
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    For any value of z the function z = y-x defines a curve through xy space. The pdf of p(z) of z is the integral of the joint density over the area between the curve for z and z+\delta z divided by \delta z (rather the limit as \delta z goes to zero).

    Write the curves in the form: y=x+z, then we have:

    <br />
A= \delta z~ \int f(x,x+z) dx <br />

    So:

    <br />
p(z) = \int f(x,x+z) dx <br />

    If you can't complete this from here say so and I will finish it.

    RonL
    If I've done this right:

    Now as z=y-x,\ p(z) is zero for z \not \in [0,1], and for z \in [0,1]:

    <br />
p(z) = \int_1^{2-z} f(x,x+z) dx = p(z) = \int_1^{2-z} \frac{3x}{2} dx = \frac{3}{4}~\left[(z-2)^2-1\right]<br />

    RonL
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  8. #8
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    what about the integral related to "y"?

    I've noticed that you integrated respect to "x", but as far as I know, if you want to derive a pdf from two random variables, the limits must be according to the marginal "x" and "y" pdfs.
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by rbenito View Post
    I've noticed that you integrated respect to "x", but as far as I know, if you want to derive a pdf from two random variables, the limits must be according to the marginal "x" and "y" pdfs.
    It is only a coincidence that it is x, what I actually did was integrate along the
    curve g(x,y)=z, it just so happened that x was a convenient parameterisation
    of the curve.

    RonL
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