# Does the Poisson distribution arise from something deeper?

Is the Poisson distribution a mere approximation to the binomial distribution or a result of something more fundamental?

• I wouldn't say "more fundamental", exactly, but it is notable that the Poisson distribution also arises in the context of the Poisson stochastic process. A caveat: there is technically a way to cook up a "binomial process" which converges to the Poisson process. But somehow this "binomial process" comes across as artificial while the Poisson process comes across as natural. Meanwhile the reverse is true in the usual presentation of the two distributions.
– Ian
Oct 3, 2020 at 14:53
• @Ian when we approximate the binomial, we look at the limit of n , the number of events/trials. What is the analogy for that limit as it goes to infinity, as in, if we want to know how many phone calls arrive in one hour and we know the expected per hour, we use a Poisson distribution to model this but what does the n going to infinity represent when approximating this event from a binomial in the first place. Oct 3, 2020 at 14:58
• Consider the Poisson process with rate $\lambda$ on a fixed time interval $[0,T]$, the associated "binomial process" with at most $n$ jumps consists of $n$ subintervals $I_j=[(j-1)T/n,jT/n)$, and an event occurs in each subinterval with probability $\lambda T/n$. Obviously this is only consistent if $n \geq \lambda T$.
– Ian
Oct 3, 2020 at 15:52
• (It also only serves as a good approximation to the Poisson if $n \gg \lambda T$.)
– Ian
Oct 3, 2020 at 16:08
• I like the sub intervals analogy but it’s still quite not clear how the binomial is just not a mere approximation, so when we go from a binomial distribution to Poisson, we wash away the combinatorics of possible events and somewhat arrive at a distribution that takes in a parameter that is ‘global’ in nature, it takes the expected number of events within an interval for this expected number of events to occur. Oct 4, 2020 at 9:54

Suppose we have an extremely large number of atoms, $$n$$, each atom having a very small chance of radioactive decay in a second. Suppose that the large number and small individual probability end up giving a measurable expected number of decay events, $$v$$, in a second.
For all intents and purposes, the radioactive decay of one atom is independent of the decay of the other atoms. The Linearity of Expectation then says that the individual probability of decay in a second is $$\frac vn$$. The Binomial Theorem says the probability that $$k$$ decay events are observed in a second is $$\binom{n}{k}\left(\frac vn\right)^k\left(1-\frac vn\right)^{n-k}$$ Since $$\lim_{n\to\infty}\binom{n}{k}\left(\frac vn\right)^k\left(1-\frac vn\right)^{n-k}=\frac{v^k}{k!}e^{-v}$$ for large $$n$$, we get the Poisson Distribution.