# How to formally apply the exponential distribution's memoryless property?

Lilian arrives at a bank to find a Joe being served at the counter. The time of service is distributed $\mathrm{Exp}\left(\lambda\right)$. How is Lilian's waiting time distributed?

Suggested solution. Denote by $Y$ Joe's service time and by $X$ the time Joe had already been served when Lilian arrived at the bank. We know that $P\left(0\leq X\leq Y\right)=1$. With this terminology Lilian's waiting time $Z$ is given by $Z=Y-X$. We're interested in $Z$'s distribution.

It can be shown by the exponential distribution's memoryless property that for all $0\leq a, b$, $P\left(\left.Z\leq a+b\right|X\geq a\right)=P\left(Y\leq b\right)$. Does this mean that $Z\sim Y$?

-
Your statement about conditional probabilities is not true. What is true by the memoryless property is that (assuming $Z$ is independent of $X$, given the fact that $X \le Y$) $Z \sim Y$. – Robert Israel Jan 27 '13 at 5:59
@RobertIsrael: Dear Robert, thanks very much for your reply. I apologize for my late reaction. I failed to notice your comment before. – Evan Aad Feb 10 '13 at 10:58