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Thread: Find the CDF for Z

  1. #1
    Yan is offline
    May 2008

    Find the CDF for Z

    Suppose X and Y are independent random variables, where Y is exponentially distributed with mean 1, and X has density function
    fX(x) = 2x*exp{-x^2}
    (x >= 0):
    Find the CDF for Z = X^2 + Y .

    I don't know what is CDF stand for.
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  2. #2
    MHF Contributor matheagle's Avatar
    Feb 2009
    That's easy.
    I can give you the density instead of the CDF, the cdf is easy too.

    Let $\displaystyle W=X^2$, then the density of W is...

    $\displaystyle f_W(w)=f_X(x){1\over 2} w^{-1/2}=e^{-w}$ when w>0 and zero otherwise.

    Hence $\displaystyle W\sim Exp(1)=\Gamma(1,1)$ as well as X.

    Since they are independent the sum is a $\displaystyle \Gamma(2,1)$, which gives you the density.

    The cumulative distribution function is $\displaystyle F_Z(z)=P(Z\le z)=\int_{-\infty}^zf_Z(t)dt$.
    Last edited by matheagle; May 27th 2009 at 10:30 PM.
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