Question

A cable car starts off with n riders. The times between successive stops of the car are independent exponential random variables with rate λ. At each stop one rider gets off. This takes no time, and no additional riders get on. After a rider gets off the car, he or she walks home. Independently of all else, the walk takes an exponential time with rate μ. (a) What is the distribution of the time at which the last rider departs the car?

Answer 1

Answer:

The distribution is $\frac{\lambda^{n}e^{- \lambda t}t^{n - 1}}{(n - 1)!}$

Solution:

As per the question:

Total no. of riders = n

Now, suppose the $T_{i}$ is the time between the departure of the rider i - 1 and i from the cable car.

where

$T_{i}$ = independent exponential random variable whose rate is $\lambda$

The general form is given by:

$T_{i} = \lambda e^{- lambda}$

(a) Now, the time distribution of the last rider is given as the sum total of the time of each rider:

$S_{n} = T_{1} + T_{2} + ........ + T_{n}$

$S_{n} = \sum_{i}^{n} T_{n}$

Now, the sum of the exponential random variable with $\lambda$ with rate $\lambda$ is given by:

$S_{n} = f(t:n, \lamda) = \frac{\lambda^{n}e^{- \lambda t}t^{n - 1}}{(n - 1)!}$