poisson process (merging process problem)

Question

can someone check whether my answers are right? Also, how do I find the variance of a 3 variable merging poisson process?

Passengers arrive at a train ticket booth according to a Poisson process with rate $\lambda=2$ vehicles per minute. Passengers arriving at the booth belong to classes 1, 2 or 3 with probabilities $\frac{1}{2}, \frac{1}{3}$ or $\frac{1}{6}$ respectively. The passenger pay ticket price of 1dollar, 2dollar or 5dollar depending on the classes 1, 2, or 3 their seat belong to.

1. Find the probability that exactly 1dollar is collected in a period of 2 minutes.
2. Find the probability that the first toll collected is 1dollar.
3. Find the probability that the waiting time between two passengers that pay 5dollar is more than 10 minutes.
4. Find the mean and variance of the amount in dollars collected in any given hour.

Can someone check whether my answers are right?

Let N1 represent the number of class 1 ticket – $\lambda1$=.5x 60=30

Let N2 represent the number of class 2 ticket – $\lambda2$=1/3 x 60=20

Let N3 represent the number of class 3 ticket – $\lambda3$=1/6 x 60 = 10

Let N represent the total number of train tickets

1. P(N1(1/30)=1, N2(1/30)=0, N3(1/30)=0)=$e^{-1}$$e^{-\frac{2}{3}}$$e^{-\frac{1}{3}}$=0.13533
2. $\frac{1}{2}$
3. $e^{-\frac{10}{6}}$=0.189
4. mean=(30 x 1)+(20 x 2)+(10 x 5)=120, how do i find the variance though?

Thank you.

Answer 1

I don't understand how you got those rates $\lambda_1 = 30, \lambda_2 = 20, \lambda_3 = 10$, or what they represent.

If the total number of arrivals to the booth is Poisson with rate $2$ per minute, this is equal to a rate of $4$ per $2$-minute intervals. Since Class $1$ represents half of these arrivals, $\lambda_1 = 4/2 = 2$ arrivals per $2$ minutes, and similarly, $\lambda_2 = 4/3$, and $\lambda_3 = 4/6 = 2/3$. Then $$\Pr[N_1 = 1, N_2 = 0, N_3 = 0] = e^{-2} \frac{2^1}{1!} e^{-4/3} e^{-2/3} = 2e^{-4} \approx 0.0366313.$$

Alternatively, you can solve the question by reasoning as follows. In order to collect exactly $1$ dollar in a $2$-minute period, the booth must have exactly one arrival, and given exactly one arrival, it must be Class $1$. Since we are told that $\lambda = 2$ for a one-minute period and the conditional probability that a passenger is Class $1$ given they arrived at the booth is $1/2$, it immediately follows that the desired probability is $$e^{-2\lambda} \frac{(2\lambda)^1}{1!} \cdot \frac{1}{2} = e^{-4} \frac{4}{2} = 2e^{-4}.$$

Your answer to Part 2 is correct, but you do not explain why. Try to think of why it is correct. How would you explain to someone who understands Poisson processes why $1/2$ is correct?

For Part 3, the rate of Class 3 arrivals is $\lambda/6 = 1/3$ per minute. So the interarrival time, as measured in minutes, is exponential with rate $1/3$, hence the probability the interarrival time exceeds $10$ minutes is $e^{-10/3} \approx 0.035674.$ You need to take into account that the overall arrival rate is $2$ per minute, not $1$ per minute.

For Part 4, the rate of arrivals per hour is $60 \lambda = 120$ per hour, of which the rates per class are $60, 40, 20$ per hour. The dollars collected is therefore the random variable $$T = N_1 + 2N_2 + 5N_3$$ where $$N_1 \sim \operatorname{Poisson}(60), \\ N_2 \sim \operatorname{Poisson}(40), \\ N_3 \sim \operatorname{Poisson}(20).$$

The mean is straightforward since linearity of expectation gives $$\operatorname{E}[T] = \operatorname{E}[N_1 + 2N_2 + 5N_3] = \operatorname{E}[N_1] + 2 \operatorname{E}[N_2] + 5 \operatorname{E}[N_3] = 60 + 2(40) + 5(20).$$ But your question about variance requires us to know that each of the $N_i$ are independent processes. Thus $$\operatorname{Var}[T] \overset{\text{ind}}{=} \operatorname{Var}[N_1] + 2^2 \operatorname{Var}[N_2] + 5^2 \operatorname{Var}[N_3].$$