So if we were to say $X(t)$ is a Poisson process with rate $λ$, I'm trying to understand this idea:
If we fix $n$ then we would expect the time between the $n^{th}$ arrival and the $(n+1)^{th}$ arrival to be equal to $1/λ$, right?
However, if we were to fix $t$, what is the expected time between the previous arrival before $t$ and the next arrival after $t$?
Who can help me? Why should the answer be bigger than $1/\lambda$?
Many thanks.
On
This is simply an application of the independent increments property of Poisson processes, which says:
$\{X_{t-s}\} \perp \{X_{s}\}, \text{where } s < t$
In plain English: the process after time $s$ is independent from everything that happened before $s$. We can treat it like a new process starting at time $s$ regardless of how long since the previous arrival. This intuitive hurdle is actually central to the Poisson process. No matter how long you've been waiting for an arrival, the expected remaining waiting time (assuming no arrivals happen in between) after 1 unit of time and after 100 units of time is exactly the same: $1/\lambda$.
As you know, the time between the $n$ and $n+1^{th}$ arrival has distribution $\operatorname {Exp}(\mathrm{rate} = \lambda),$ so you are right that the expected time between the two is $\cfrac 1\lambda$. If we fix the time $t$, the amount of time until the next arrival is still distributed $\operatorname {Exp}(\mathrm{rate} = \lambda),$ so the expected time until the next arrival is still $\cfrac 1\lambda$. Unless we can be certain that the most recent arrival is exactly at $t$, then the expected time between the two arrivals will be greater than $\cfrac 1\lambda$, and of course we cannot be certain of that.