In a city there are three kinds of subway lines: the red, green and orange lines. Subways on each line arrive at a station according to three independent Poisson processes. On average, there is one red train every 10 minutes, one green train every 15 minutes, and one orange train every 20 minutes.
(a) When you arrive at the station what is the probability that the 1st subway that arrives is for the green line?
(b) How long will you wait, on average, before some train arrives?
(c) You have been waiting 20 minutes for a red train and have watched three orange trains to go by. What is the expected additional time you will wait for your subway?
Are these Poisson-related problems?
Solution:
(a) $\frac{\frac{1}{15}}{ \frac{1}{10} + \frac{1}{15} + \frac{1}{20} } = \frac{4}{13}$
(b) $\frac{1}{ \frac{1}{10} + \frac{1}{15} + \frac{1}{20} } = \frac{60}{13}$
(c) $\frac{1}{\frac{1}{10}} = 10$
Are these correct solutions as I don't see any Poisson here?
Best Answer
Yes, these are Poisson processes; and the solutions are correct. The key thing to note is that summing independent Poisson processes gives a Poisson process. Thinking of Poisson processes as arrivals, this says that if you count arrivals from two independent Poisson processes, then you get a Poisson process. You can read more about Poisson processes in Section 1.4 of Perla Sousi's lecture notes.
The main result being used here is that interarrival times in a Poisson process are exponentially distributed, and hence have the memoryless property.
(a/b) The time to get a red is exponential with rate $\lambda_r = 1/10$, green with rate $\lambda_g = 1/15$ and orange with rate $\lambda_o = 1/20$; call these arrival times $E_r$, $E_g$ and $E_o$. The first arrival is then $\min\{E_r,E_g,E_o\}$, and minimum of exponentials is exponential with parameter $\lambda := \lambda_r+\lambda_g+\lambda_o$; the expected time is then $1/\lambda$.
For (a), you can just calculate directly $\Pr( \text{Exponential}(\lambda_g) < \text{Exponential}(\lambda_r+\lambda_o) )$.
You can also see this as a sort of symmetry argument. Suppose we're trying to calculate $\Pr( E_1 = \text{Exponential}(1) < E_2 = \text{Exponential}(2)$. Write $\text{Exponential}(2)$ as $\min\{E_{2,1} = \text{Exponential}(1),E_{2,1}' = \text{Exponential}(1)\}$. There are then three exponentials ($E_1$, $E_{2,1}$ and $E_{2,1}'$) each of rate 1; if the first is the smallest, then $E_1 < E_2$, while otherwise $E_2 > E_1$. In this set-up, clearly $\Pr(E_1 < E_2) = 1/3$, by symmetry.
For (c), just use the memoryless property and the fact that the Poisson processes are independent, so it doesn't matter that you've seen orange buses