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Math Help - Catch my mistake: verifying joint pdf.

  1. #1
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    Catch my mistake: verifying joint pdf.

    I spent so much time doing this integral yet I can't seem to get the right answer. Please help me find where I made a mistake. Thanks!

    The joint pdf of X and Y is given by:

    f_{X,Y}(x,y) = \frac{16}{15}(x^{2} + \frac{xy}{2}), 0<x<1, 0<y<x+1

    Verify that this is a legitimate joint pdf. So I try to integrate the pdf and see if it equals one.

    \frac{16}{15}\int_0^1 \int_0^{x+1}(x^{2} + \frac{xy}{2})dydx = \frac{16}{15}\int_0^1 [x^{2}y + \frac{xy^{2}}{4}]_0^{x+1}dx =  \frac{16}{15}\int_0^1 (x^3 + x^2 + \frac{x^3 + 2x^2 + x}{4})dx = \frac{16}{15}[\frac{x^4}{4} + \frac{x^3}{3} + \frac{x^4}{16} + \frac{x^2}{4} +  \frac{x^2}{8}]_0^1=\frac{49}{45}

    As you can see, the integral of the pdf does not equal one. But it has to since it asked me to verify that it is legitimate! Where is my mistake? Thanks.
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  2. #2
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    Quote Originally Posted by jfz23 View Post
    [snip]

    \frac{16}{15} \int_0^1 \left( x^3 + x^2 + \frac{x^3 + 2x^2 + x}{4} \right) \, dx

    = \frac{16}{15} \left[ \frac{x^4}{4} + \frac{x^3}{3} + \frac{x^4}{16} + \frac{x^2}{4} + \frac{x^2}{8}\right]_0^1 Mr F says: The mistake is in this line.

    As you can see, the integral of the pdf does not equal one. But it has to since it asked me to verify that it is legitimate! Where is my mistake? Thanks.
    ..
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  3. #3
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    Ahh I found it! Such a simple calculation error...
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