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.

  • $\begingroup$ 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? $\endgroup$
    – lulu
    Commented Jul 2, 2022 at 18:00
  • $\begingroup$ 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. $\endgroup$
    – keepfrog
    Commented Jul 2, 2022 at 18:11
  • $\begingroup$ 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. $\endgroup$
    – lulu
    Commented Jul 2, 2022 at 18:35
  • $\begingroup$ 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 $\endgroup$
    – keepfrog
    Commented Jul 2, 2022 at 18:43
  • $\begingroup$ 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. $\endgroup$
    – lulu
    Commented Jul 2, 2022 at 18:53

1 Answer 1


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: $\exp\left(-\lambda \int_0^T p (t)\mathrm{d}t \right)$


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