Poisson process from Binomial random variables and nonstandard analysis

2012 October 17
by Daniel Lakeland

Yesterday in the discussion class for my undergrad students I mentioned how the poisson distribution arises from a limit of Binomial random variables, and I started to derive it but cut myself short because I hadn’t prepared any notes and got a little muddled in some of the middle steps. So anyway I thought i’d give it a careful treatment using nonstandard analysis here on the blog. It’s a pretty straightforward derivation but the argument is particularly easy using nonstandard analysis so that might be helpful for someone who is studying either probability or nonstandard analysis for the first time.

First, a poisson process is some process that occurs in time or space, where discrete events happen at unknown locations in a continuous axis. So for example in time yellow cars arrive at a nearby intersection, if they arrive according to a poisson process then there is some constant rate but the actual times when each one arrives is random and therefore the count of total arrivals after a certain time is a random discrete variable (0,1,2,3… but not $$\sqrt{\pi}$$ for example). Similarly, perhaps cracks in plastic crazy straws occur one per inch on average, but if you look at a 7 inch straw you aren’t always going to have 7 cracks… you have a poisson random variable of cracks.

Let’s imagine a very long plastic crazy straw blank (it’s straight for ease of measuring), and let’s break up this crazy straw into a large number of thin slices. We’ll pick $$N$$ slices and then $$dx=L/N$$ is the width of each slice. We’ll make the following approximation, for a given slice the probability of a crack in the slice is $$p(0) = 1-p$$, $$p(1)=p$$ and there is no probability for greater than one crack because the slice is too small to fit more than one crack. Furthermore, we’ll assume that when $$N$$ is large, and hence $$dx$$ is small, that $$p = r dx$$ with a rate $$r$$ in other words the probability of a crack is proportional to the length of the small slice for small enough slices.

Now what’s the probability of $$n$$ cracks in length $$L$$ ? Well there are $$N$$ (capital N) slices, and each slice represents a binomial random variable with either one crack or no cracks. So based on the binomial distribution:

\[{N\choose n}p^n(1-p)^{(N-n)}\] is the probability of $$n$$ cracks. Let’s rewrite it as:


Now let’s transfer into the realm of idealization and let the cracks be infinitesimally small within a continuous medium and let $$N$$ be a nonstandard number so that $$dx = L/N$$ is an infinitesimal. By one definition of $$e$$ based on continuously compounded interest $$e^r = (1+r/N)^N$$. Therefore:

\[(1-rdx)^{(L/dx-n)} = (1-\frac{rL}{(L/dx)})^{L/dx-n}\sim e^{-rL}\]

since $$(1-rdx)^{-n} \sim 1$$ because $$rdx$$ is infinitesimal and $$n$$ is a small standard integer.

Now, when $$N$$ is nonstandard $$\frac{N!}{(N-n)!} = N(N-1)(N-2)…(N-n) \sim N^n +O(N^{n-1})= (L/dx)^n + O((L/dx)^{n-1})$$. We use the “Big-O” notation because we’re going to find out that the thing inside the $O$ becomes an infinitesimal correction in the final result regardless of what its coefficient is, what we really care about is the size of its exponent (n-1). Plugging this into the equation so far:

\[p(n) \sim \frac{1}{n!}((L/dx)^n +O((L/dx)^{n-1}))(rdx)^ne^{-rL}\]

cancelling the $$dx^n$$ we get

\[p(n) \sim \frac{1}{n!}((rL)^n+O(dx))e^{-rL}\sim \frac{1}{n!}(rL)^ne^{-rL}\]

which is the Poisson distribution.


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