# $X \sim \exp(1)$, $Y \sim \exp(1)$ Independent what is the CDF of $Z=X-Y$?

Let $$X \sim \exp(1)$$, $$Y \sim \exp(1)$$ $$Z=X-Y$$.

$$X,Y$$ are independent.

What is the distribution of Z?

For $$t\geq0$$, I simply calculated that the old fashioned way.

$$F_Y(t)=P(Z\leq t)=P(X-Y\leq t)=\int_{0}^{\infty}\int_{0}^{t+y}e^{-x}e^{-y}dxdy=1-\frac{1}{2}e^{-t}$$

What can I do about the second case? $$t<0$$?

If the joint PDF of $$X$$ and $$Y$$ is non-negative, how can I integrate in this region?

Since $$X,\,Y$$ are iids, $$X-Y$$ has a symmetric distribution, so the cdf satisfies $$F_Z(-t)=1-F_Z(t)$$. For $$t<0$$, $$F_Z(t)=\frac{1}{2}\exp t$$.
The joint distribution of $$(X,Y)$$ is identically zero on the second, third and fourth quadrants. If $$t\le 0$$, your probability is an integral over an angle, $$P\{X-Y\le t\}=P\{X\ge 0\,\,\,\textrm{and} \,\,Y\ge X-t\},$$ which equals $$\int\limits_0^\infty e^{-x}\,dx\int\limits_{x-t}^\infty e^{-y}\,dy=\frac{1}{2}e^t.$$