What is your expected waiting time if limousine inter-departure times follow an exponential distribution? Limousines depart from the railway station to the airport from the early morning till late at night. The limousines leave from the railway station with independent inter-departure times that are exponentially distributed with an expected value of $20$ minutes. Suppose you plan to arrive at the railway station at three o’clock in the afternoon. What are the expected value and the standard deviation of your waiting time at the railway station until a limousine leaves for the airport?


Shouldn't the expected waiting time be $20$ minutes? Since it's an exponential distribution then for $t\to$  time to the departure, $f(t) = \frac{1}{20}e^{-\frac{1}{20}t}$. The expected value for this distribution would be $20$. 

Is this correct?
This is a special case of the so called "stationary excess distribution".

The CDF of the time you have to wait is given by $F(t) = \frac{1}{20} \int_0^t [1-F(s)] ds$ where $F(s)$ is the CDF of an Exponential RV with mean 20. 

An alternative way to do this is to use the fact that you're dealing with a poisson process, and if you start a poisson process at any time, it is still a poisson process with the same parameters, so the distribution to the next arrival is still Exponential with mean 20.
This is not exactly correct. If the number of limousines run every day is a constant $n$ = 60*(LateAtNight-YearlyMorning)/20,  and then the answer will be

$$20 * (n/(n+1)*(1+1/(n+1)/(n+2)) + 60*(24-(LateatNight-YearlyMorning))*pow((ArrivalTime-YearlyMorning)/(LateatNight-YearlyMorning),n)$$

As such, for a station opened from 5 AM till 1 AM and you arrived at 3PM, you have to wait ~ 19' 40''.