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Math Help - Check solution: find the probability density function of Y=logX

  1. #1
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    Check solution: find the probability density function of Y=logX

    I think I have a good answer to this one, but I'd like to make sure it's correct.

    If X has exponential distribution with parameter \lambda, find the probability density function of Y=\log X.

    Here is what I did.
    F_Y(y) = P \{ Y \leq y \} = P \{\log X \leq y \} = P \{ X \leq e^y \} = \int_{0}^{e^y} \lambda e^{- \lambda x} dx
    Let u=\lambda x, du=\lambda dx
    \int_{0}^{e^y} \lambda e^{- \lambda x} dx = \int_{0}^{\lambda e^y} e^{-u} du = -e_0^{\lambda e^y} = -e^{-x} \Big |_0^{e^y} = -e^{-e^y}  + 1 = 1 - e^{-e^y}
    So,
    \frac {d F_{|X|} (x) } {dx} = p(x) = e^x e^{-e^x}

    Does this look solid?
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  2. #2
    MHF Contributor harish21's Avatar
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    You have to differentiate the cdf in order to get the pdf..

    since, F_{Y}(y)= P \{ X \leq e^y \} = F_{X}(e^Y)

    \therefore f_{Y}(y)= \dfrac{d}{dy}F_{X}(e^y) ={e^y \;.\; f_{X}(e^y)}=..........
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  3. #3
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    Quote Originally Posted by harish21 View Post
    You have to differentiate the cdf in order to get the pdf..

    since, F_{Y}(y)= P \{ X \leq e^y \} = F_{X}(e^Y)

    \therefore f_{Y}(y)= \dfrac{d}{dy}F_{X}(e^y) ={e^y \;.\; f_{X}(e^y)}=..........
    Thanks for that. Just to be absolutely sure, the final answer would be this, right?
    f_Y(y)=\frac{dF_X(e^y)}{dx}=e^y\cdot f_X(e^y)=e^y e^{-e^y}
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  4. #4
    MHF Contributor harish21's Avatar
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    for an exponential distribution:

     f_{X}(x)=\lambda \; e^{-\lambda x}

    so,
    f_{X}(e^y) = ....

    and f_{Y}(y)=...
    Last edited by harish21; November 9th 2010 at 08:49 AM.
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  5. #5
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    Ah, I see. I was looking at solutions for which \lambda=1. I've revised my answer to this:

    f_Y(y)=F'_Y (y)=f_X(e^y)\cdot\lambda e^{-\lambda y}=\lambda e^{-\lambda (y+e^y)}

    Any mistakes I might have made?
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  6. #6
    MHF Contributor harish21's Avatar
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    f_Y(y)=F'_X (e^y)=e^{y} \; f_X(e^y) = e^y \; (\lambda \; e^{-\lambda e^{y}}) = \lambda \; e^{e^{y}-\lambda e^{y}}
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