CDF of Joint Cumulative Distribution Function For X, Y two independent Random Variables with identical distribution
$$X,Y ~f(t)  = \lambda e^{-\lambda t}, \text{ for } t\in (0,+\infty)$$
I want to find the cdf of $$Z = \vert X-Y\vert$$
I use $$F_Z(t) = P(-t<X-Y<t) = \int^{+\infty}_{x=-\infty}\int^{x+t}_{x-t} \lambda e^{-\lambda x}\lambda e^{-\lambda y}dydx$$
$$ = \frac{1}{2}(e^{\lambda t}-e^{-\lambda t})$$
which should be clearly wrong as it is a monotonically increasing function.
Can anyone tell me where is the problem?
 A: One thing is that the lower bound of the first integral must be $0$, but I think you had interpreted it as zero since otherwise the integral would diverge. The second thing is that the lower bound of the second integral is also incorrect, since the density of $Y$ is zero if $y$ is negative. The last thing is that a CDF can be increasing so that is certainly not a problem. The problem you should worry about is that the function you got goes to infinity if $t\to\infty$. 
Okay, you do it correctly like this 
\begin{align} 
F_Z(t) =P(-t<Y-X<t) =\int^\infty_0 \int^{x+t}_{\max\{x -t, 0\}} \lambda e^{-\lambda x} \lambda e^{-\lambda y} \, dy\, dx 
\end{align} 
Since the density of $Y$ is zero if $y$ is negative. The  calculations is left for you. 
A: $X,Y$ are only supported over positive real values.   Integration boundaries must include this restriction.
$$\begin{split}F_Z(t) &= \mathsf P(0<X<t, 0<Y<X+t)+\mathsf P(t\leqslant X, X-t\leqslant Y<X+t)\\[2ex] &= \left(\int_0^t\int_0^{x+t} \lambda^2 e^{-\lambda(x+y)}~\mathsf d y~\mathsf dx+\int_t^\infty\int_{x-t}^{x+t} \lambda^2 e^{-\lambda(x+y)}~\mathsf d y~\mathsf dx\right)\mathbf 1_{t\in(0;\infty)}\\[2ex]&~~\vdots\end{split}$$
