Results 1 to 6 of 6

Thread: Check solution: find the probability density function of Y=logX

  1. #1
    Member
    Joined
    Jan 2010
    Posts
    232

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

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

    Does this look solid?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    You have to differentiate the cdf in order to get the pdf..

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

    $\displaystyle \therefore f_{Y}(y)= \dfrac{d}{dy}F_{X}(e^y) ={e^y \;.\; f_{X}(e^y)}=..........$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2010
    Posts
    232
    Quote Originally Posted by harish21 View Post
    You have to differentiate the cdf in order to get the pdf..

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

    $\displaystyle \therefore f_{Y}(y)= \dfrac{d}{dy}F_{X}(e^y) ={e^y \;.\; f_{X}(e^y)}=..........$
    Thanks for that. Just to be absolutely sure, the final answer would be this, right?
    $\displaystyle f_Y(y)=\frac{dF_X(e^y)}{dx}=e^y\cdot f_X(e^y)=e^y e^{-e^y}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    for an exponential distribution:

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

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

    and $\displaystyle f_{Y}(y)=...$
    Last edited by harish21; Nov 9th 2010 at 07:49 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2010
    Posts
    232
    Ah, I see. I was looking at solutions for which $\displaystyle \lambda=1$. I've revised my answer to this:

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

    Any mistakes I might have made?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    $\displaystyle f_Y(y)=F'_X (e^y)=e^{y} \; f_X(e^y) = e^y \; (\lambda \; e^{-\lambda e^{y}}) = \lambda \; e^{e^{y}-\lambda e^{y}}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Nov 11th 2011, 05:06 AM
  2. help on to find probability density function!!!
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: Jul 2nd 2011, 10:46 PM
  3. find all real numbers x for logx=x/100
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: Apr 13th 2011, 10:53 AM
  4. Replies: 4
    Last Post: Oct 27th 2010, 05:41 AM
  5. Density function solution check
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Apr 4th 2008, 11:35 PM

Search Tags


/mathhelpforum @mathhelpforum