# Poisson process: the probability of no arrival and no action at each time

Suppose that a customer arrives according to a Poisson process at a rate $$\lambda$$. The time is denoted by $$t \in \mathbb{R}_+$$. At each time, the probability that a customer has not arrived is $$1 - F(t)$$, where $$F=1-\exp(-\lambda t)$$. When a customer arrives, he buys a product w.p. $$p(t)$$ and does not buy w.p. $$p(t)$$. In any case, the customer exits instantly after the purchase decision is made. What is the probability that no customer made a purchase at $$T$$?

If $$p(t)=1, \forall t$$, then it is just a probability that no customer has arrived, so it is $$\exp(-\lambda T)$$. My hunch is that
$$\exp(-\lambda \int_0^T p(t))dt,$$ but I am not sure how to formally show this.

• Not sure this is clear. Specifically, it's not clear what $t$ is nor is it clear what "made a purchase at $T$" mean. How long does a customer stay? If they stay for a millionth of a second, presumably they don't buy anything, right?
– lulu
Commented Jul 2, 2022 at 18:00
• I added some explanation. I'm thinking about a situation in which each customer randomly arrives and buys or not w.p. $p(t)$, and then he is done. Thus, the probability that I want to derive is that at time $T$, (i) no one arrived or (ii) everyone who arrived did not make a purchase. Commented Jul 2, 2022 at 18:11
• I don't see how any of that is clearer than what you wrote before. In any case, all of my questions are still open.
– lulu
Commented Jul 2, 2022 at 18:35
• Thee customer does not stay. When arrived, he makes the purchase w.p. $p(t)$. Another way to put this is that the arrival consists of two stage: a customer arrives at each t according to Poisson and only a fraction $p(t)$ of them actually arrives. In this case, I'm interested in the probability that no customer has actually arrived Commented Jul 2, 2022 at 18:43
• The probability that a customer arrives at a specific time $t$ is $0$. You could try to look at an interval $[t,t+\Delta t]$ where arrival is poisson with a scaled mean. Perhaps that's more the sort of thing you had in mind.
– lulu
Commented Jul 2, 2022 at 18:53

Let $$\Big\{[t_{i-1},t_i),s_i\Big\}_{i=1}^n$$ be a uniform tagged partition of $$[0,T]$$ into $$n$$ subintervals of equal length $$\Delta t=\frac{T-0}{n}$$.
The number of customers who arrive on $$[t_{i-1},t_i)$$ is $$\text{Poisson}\left(\lambda\Delta t\right)$$, and these customers can be grouped into two distinct categories: those who purchase an item and those who do not. If $$n$$ is large we expect $$\lambda p(s_i)\Delta t$$ customers to purchase and item and $$\lambda (1-p(s_i))\Delta t$$ to leave the establishment empty handed.
In this way we see that our Poisson process on $$[t_{i-1},t_i)$$ splits, and the number of customers who purchase an item on $$[t_{i-1},t_i)$$ is approximately $$\text{Poisson}\left(\lambda p(s_i)\Delta t\right)$$.
Observe now that the total of customers who purchase an item on $$[0,T]$$ is approximately $$\text{Poisson}\left(\sum_{i=1}^n\lambda p(s_i)\Delta t\right)$$ which becomes $$\text{Poisson}\left(\lambda\int_0^T p(t)\mathrm{d}t\right)$$ after taking $$n$$ to $$+\infty$$.
So, the probability nobody purchased an item is exactly what you proposed: $$\exp\left(-\lambda \int_0^T p (t)\mathrm{d}t \right)$$