Suppose that the times at which customers enter a store can be modelled as a Poisson process. Over a 10-minute period, 8 customers entered the store.
How many customers would you expect to see enter the store in a two hours period?
What is the probability that no customers enter the store in a two minutes period?
What is the probability of waiting an additional 3 minutes for a customer to enter the store, if the last customer entering the store came n minutes ago?
I don't think I understand the Poisson process properly. Would the intensity, $\lambda = 0.8$, since we expect (theoretically) 0.8 customers every minute?
From that, I assume that the answer to (1) is $\lambda t = (0.8)(120) = 96$.
For (2), I think want $P(N_2 = 0)$, where $N_2 \sim PP(1.6)$. So I get $P(N_2 = 0) = \frac{1.6^0 e^{-1.6}}{0!} = 0.201$. Am I on the right track?
I am completely lost on (3). Could anyone perhaps provide me a tip to get started? Thanks!
Best Answer
To find the intensity, recall as you write, that the average number of customers in a time interval $[t_1,t_2]$ is $\lambda(t_2-t_1)$. Thus, if $t$ is given in minutes, $$ 10\lambda=8, $$ from which you deduce your answer $\lambda=0.8\,\mathrm{min}^{-1}$.
Your answer is correct.
Your answer is correct, since $N_2\sim\mathrm{Poiss}(2\lambda)$ (I'm not sure what your notation $PP(\cdot)$ is, it should reference the Poisson distribution, not the Poisson process).
Expanding on Henry's tip, recall that the increments of the Poisson process are independent and exponentially distributed. So it is asked to compute $$ \mathbb P\left(X\ge n+3\mid X\ge n\right), $$ where $X\sim\mathrm{Exp}(\lambda)$. Do you see how to use the memorylessness property of the Poisson process here? (Or more specifically, the memorylessness of the exponential distribution.)