Poisson distribution formula How is the Poisson distribution formula obtained? according to the theory it represents, for instance, the number of cars that go through some fixed route during a certain time. But again, how is the formula of the Poisson distribution derived? 

Suppose:


*

*The probability of an event occurring in a time interval of length $\Delta t$ is $\lambda \Delta t$, where $\Delta t$ is a small parameter (held fixed for the moment).

*Events in disjoint intervals are independent

*Only one event can occur in each interval.


The first two assumptions are natural; the third is a bit artificial, and I merely make it to simplify the calculations. With some modification to the other assumptions, this third assumption can be dropped.
In this case the number of events occurring in a time interval of length $n \Delta t$ is Binomial(n,$\lambda \Delta t$). In particular, the probability that there are exactly $k$ events is $\frac{n!}{k! (n-k)!} \lambda^k (\Delta t)^k (1-\lambda \Delta t)^{n-k}$
Now we want to fix $k$ and send $n \to \infty$ with $\Delta t = 1/n$. In the process we will get the number of events in a time interval of length $1$ in a Poisson process with intensity $\lambda$. In other words we will get $P(X=k)$ when $X$ is Poisson($\lambda$) distributed.
Thus the issue is to calculate
$$L=\lim_{n \to \infty} \frac{n!}{k! (n-k)!} \lambda^k n^{-k} (1-\lambda/n)^{n-k}.$$
We will write this as four factors:
$$L=\lim_{n \to \infty} \frac{\lambda^k}{k!} \frac{n!}{(n-k)! n^k} (1-\lambda/n)^{-k} (1-\lambda/n)^n.$$
The first factor is already independent of $n$. The second and third factors will converge to $1$. The last factor will converge to $e^{-\lambda}$. So we will recover the familiar formula for the Poisson distribution.
