How can I find the probability distribution of $Z=e^{XY}$ given that $X$ and $Y$ are independent and have a uniform probability distribution in $[0,1]$?
My work up to now:
The probability density functions of $X$ and $Y$ are $f_{X}(x)=f_{Y}(y)=1$, thus, $f_{X,Y}(x,y)=1$. I also know that $Z$ can take values from $1$ to $e$.
I tried to find the probability that $Z$ will be less than a certain value z integrating $f_{X,Y}(x,y)$ over the corresponding area: $$ P(Z<z)=F_{Z}(z)=\ln(z)+\int_{\ln(z)}^{1}\int_{0}^{\ln(z)/x}\,dy\,dx $$ I got the limits of the integral from a plot of $y=\ln(z)/x$ vs $x$. From this i get: $$ F_{Z}(z)=\ln(z)[1-\ln(z)] $$
Here is the problem, if I replace $z=e$ in $F_{Z}(z)$ I should get $1$, since the probability of $Z$ being smaller or equal to $e$ is $1$, right? So there is something wrong with my solution, but i just can't find it. Help!
My mistake was evaluating $\ln(x)$ in $x=\ln(z)$ as $\ln(z)$ instead of $\ln(\ln(z))$ in the integral. Here is the correct procedure:
$F_{Z}(z)=\ln(z)+\int_{\ln(z)}^{1}\int_{0}^{\ln(z)/x}dydx$
$= \ln(z)+\int_{\ln(z)}^{1}\frac{\ln(z)}{x} dx$
$= \ln(z)+\ln(z)\left(\ln(1)-\ln(\ln(z))\right)$
$= \ln(z)-\ln(z)\ln(\ln(z))$
Replacing $z=e$ I should get $F_{Z}(e)=1$
$F_{Z}(e)=\ln(e)-\ln(e)\ln(\ln(e)) = 1 - 1\ln(1) = 1 - 0 = 1$