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Math Help - Density Function

  1. #1
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    Density Function

    Let x and y be identically distributed variables have density function f(u)=\frac{e^{\frac{-u^2}{2}}}{(2\pi)^\frac{1}{2}}

    find the density function of z=x^2+y^2

    i have an answer given as g(z)=.5e^{\frac{-z}{2}} for z\geq 0

    but I cannot seem to get that answer, thanks in advance
    Last edited by emathinstruction; February 16th 2010 at 07:10 AM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    you're saying that X and Y are st normals
    However the constant is square root of 2 pi.
    not 2 pi squared.
    Then X^2 is a chi-square with one degree of freedom
    and so is Y^2 and their sum by independence is a chi-square with two degrees of freedom
    Last edited by matheagle; February 16th 2010 at 07:13 AM.
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  3. #3
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    yeah, i am well aware of that, but when I tried to get an answer i ended up with a root z in there that shouldnt be
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  4. #4
    MHF Contributor matheagle's Avatar
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    \frac{e^{\frac{-u^2}{2}}}{(2\pi)^2} is not a density

    However \frac{e^{\frac{-u^2}{2}}}{(2\pi)^{1/2}} is
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  5. #5
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    yes, that was a typo on my part sorry, but can you get this problem started for me, I was trying to use the distribution function method
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  6. #6
    MHF Contributor matheagle's Avatar
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    as I said

    Then X^2 is a chi-square with one degree of freedom
    and so is Y^2 and their sum by independence is a chi-square with two degrees of freedom
    Use MGFs to add the two rvs.
    Chi-squares are gammas.
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  7. #7
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    okay, thanks

    is it invalid to say z=2y^2, and then try to use the distribution function method?
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  8. #8
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by emathinstruction View Post
    okay, thanks

    is it invalid to say z=2y^2, and then try to use the distribution function method?
    there is no 2 infront
    let W=Y^2
    Squaring a st normal is very simple.
    That can be found in most books, probably yours.
    and it's been done here a dozen times...
    Last edited by matheagle; February 16th 2010 at 08:16 AM.
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