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

Question

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?

This might be useful: https://math.stackexchange.com/questions/115022/pdf-of-the-difference-of-two-exponentially-distributed-random-variables — Locally unskillful, Feb 15 '19 at 17:36
Use convolution theorem $f_Z(z) = \int_{0}^{\infty} f_X(x) f_Y(x-z)\ \text {dx}.$ — little o, Feb 15 '19 at 17:44

score 3 · Accepted Answer · answered Feb 15 '19 at 17:38

3

Since $X,\,Y$ are iids, $X-Y$ has a symmetric distribution, so the cdf satisfies $F_Z(-t)=1-F_Z(t)$. For $t<0$, $F_Z(t)=\frac{1}{2}\exp t$.

answered Feb 15 '19 at 17:38

J.G.

115,835

score 3 · Answer 2 · answered Feb 15 '19 at 17:44

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 $$ \int\limits_0^\infty e^{-x}\,dx\int\limits_{x-t}^\infty e^{-y}\,dy=\frac{1}{2}e^t. $$

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

2 Answers2