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