I'm having troubles translating the following process to continuous time.
Consider the time interval $t$, which we can split into $N$ subintervals of length $\Delta t= t/N$. Let the probability mass function of a small enough but finite interval $\Delta t$ be given by $Poisson(k; \lambda \Delta t)$.
Then, the probability of no events in that period is given by $\exp(-\lambda t/N)$. Consequently, the probability of no event during $t$ is given by
$$ P(0, t) = \exp(-\lambda t/N)^N = \exp(-\lambda t)$$
which means that the continuous time approximation of a Poisson event is again a Poisson event, at least in the $k=0$ case. I would like to see how this plays out for $k > 0$.
Given that an event is distributed by $Poisson(k, \lambda \Delta t)$ at any $\Delta t$, how does the distribution of events over $t$ look?
The $k=0$ case was a very neat knife-edge case, which I could solve through standard combinatorics logics. How would I proceed here?
If you are looking for a hint rather than a full answer, the way to proceed is to form a differential equation from the following relationship:
$P(k,t + \Delta t ) = P(k,t )(1-\Delta t \lambda ) + P(k-1,t )(\Delta t \lambda ) $
You will then see that the Poisson distribution satisfies that equation.
Or consider as the limit of a binomial distribution?