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

1. ## 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?

2. You have to differentiate the cdf in order to get the pdf..

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

$\displaystyle \therefore f_{Y}(y)= \dfrac{d}{dy}F_{X}(e^y) ={e^y \;.\; f_{X}(e^y)}=..........$

3. Originally Posted by harish21
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?
$\displaystyle f_Y(y)=\frac{dF_X(e^y)}{dx}=e^y\cdot f_X(e^y)=e^y e^{-e^y}$

4. for an exponential distribution:

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

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

and $\displaystyle f_{Y}(y)=...$

5. 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?

6. $\displaystyle f_Y(y)=F'_X (e^y)=e^{y} \; f_X(e^y) = e^y \; (\lambda \; e^{-\lambda e^{y}}) = \lambda \; e^{e^{y}-\lambda e^{y}}$