If $X$ and $Y$ are independent and identically distributed RVs with pdfs $f_X(x)=e^{-x}U(x)$ and $f_Y(y)=e^{-y}U(y)$, find the PDF of $Z$ if:
a. $Z=X-Y$
b. $Z=\min(X,Y)/\max(X,Y)$
$U$ represents the unit step function.
For the first part, I tried looking at the range of values that $y$ could be, and I got $y\geq x-z$ since $x-y \leq z$ for $F_Z(z)$. I set up my CDF integral to be
$$\int_0^\infty\int_0^{z+y}e^{-x-y} dx dy$$
and got $F_Z(z)= 1-e^{-z}/2$. I differentiated this to get my PDF, which is $e^{-z}/2$. This answer is incorrect, and I'm not sure what I did wrong.
For the second part, I again tried looking at the range of values that $y$ could be, and I got $y \leq zx$ where $0 \leq z \leq 1$. I set up my CDF integral to be
$$\int_0^\infty \int_0^{y/z}e^{-x-y} dx dy$$
and got $F_Z(z)= 1-\frac {z} {z+1}$. I differentiated this to get $\frac {-1} {(z+1)^2}$. This answer is also incorrect. I'm not sure what to do on either of these two problems at this point.
Best Answer
Just expanding drhab's answer.
For the first case you should consider when, for any given $z\in \Bbb R$, when $Y > X-z$. When $z>0$ you have the situation depicted below.
Thus, for $z>0$, \begin{eqnarray} F_Z(z) &=& \int_0^{+\infty}\int_0^{z+y} e^{-x-y} dx dy=\\ &=&1-\frac12 e^{-z}. \end{eqnarray}
For $z<0$ you can look at the figure below.
Then you can compute, in such interval, e.g. as \begin{eqnarray} F_Z(z) &=& \int_{0}^{+\infty}\int_{x-z}^{+\infty} e^{-x-y} dy dx=\\ &=&\frac12 e^{z}. \end{eqnarray}
Differentiation yields $$\boxed{f_Z(z) = \frac12 e^{-|z|}}.$$
As for the second question, consider that, if $X>Y$ you need to consider the event $Y<zX$; whereas, if $X<Y$, your event becomes $Y>\frac1{z}X$.
It is thus immediate to conclude that, for $z>1$, $$F_Z(z) = 1.$$
For $0<z<1$, consider then the shaded area in the figure below.
Therefore, for $0<z<1$, \begin{eqnarray} F_Z(z) &=&1- \int_0^{+\infty} \int_{zx}^{\frac{x}{z}} e^{-x-y}dy dx=\\ &=&\frac{2z}{1+z}. \end{eqnarray}
And, finally, by differentiation $$\boxed{f_Z(z) = \frac2{(1+z)^2}}, \ \ 0<z<1.$$