# Adding beans to a pot

I’ve been reading through Crispin Gardiner’s Handbook on stochastic processes’. It’s a really dense but helpful book, designed sort of as a reference as opposed to a teaching tool. There are plenty of calculations worth going through, but the very first one I wanted to redo is using generating functions to derive the Poisson process, that’s the probability distribution generated from a process that has a mean rate of happening in a given time interval. The events must be independent. I’ll frame the process in terms of adding a single bean $n=1$ to a pot with mean chance $\lambda$ in a given time interval $\Delta t$.

Thus, the probability we consider is a function of time and number of beans $p(n,t)$. The probability that the number of beans is increased by 1 in the time interval $t+\Delta t$ is $p(n\to n+1,t+\Delta t) = \lambda\Delta t$

That means the probability that $n$ stays constant in the time interval is made up of the addition of two probabilities: the complement of the probability of having $n$ and increasing the number, and the probability that we orginally had $n-1$ beans and one was added. Together then $p(n,t+\Delta t) = p(n,t)[1 - \lambda\Delta t] + p(n-1,t)\lambda\Delta t$

we can expand and factor and then as $\Delta t\to 0$ we have a partial differential equation (PDE) for the probability: $\frac{1}{\lambda}\frac{\partial p}{\partial t} = p(n-1,t)-p(n,t)$

We can solve this partial differential equation, for example, by using the method of generating functions. We define a probability generating function (pgf) as a power law sum: $G(s,t) = \sum_{n=0}^{\infty} s^n p(n,t)$

The reason for calling it a generating function is intuitive. Taking the first derivative with respect to $s$ and evaluating at $s=0$ gives us $\left.\partial_s G(s,t)\right|_{s=0} = \sum_{n=1}^{\infty} n s^{n-1} p(n,t) = 1*0^0*p(1,t) + 2*0^1*p(2,t) + 3*0^2*p(3,t) + \ldots$

where we had to readjust the sum to avoid division by zero. It is clear then that $\left.\partial_s G(s,t)\right|_{s=0} = p(1,t)$

so taking derivatives of the pgf and evaluating at 0 gives us the probability of having 1 bean over time. Likewise the second derivative evaluated at zero gives us the probability of having 2 beans $\left.\partial_s^2 G(s,t)\right|_{s=0} = \sum_{n=2}^{\infty} n(n-1) s^{n-2} p(n,t) = 2*1*p(2,t)$

so that the general formula for the probability of having $k$ beans is $p(k,t) = \frac{1}{k!} \left. \partial_s^k G(s,t)\right|_{s=0}$

and this function generates’ the probabilities.

Now, we can return to solving the PDE. Writing the time derivative of the pgf $\partial_t G(s,t) = \sum_{n=0}^{\infty} s^n \partial_t p(n,t)$

we now substitute in the time rate-change of the probability from the PDE we are interested in solving $\partial_t G(s,t) = \lambda \sum_{n=0}^{\infty} s^n [p(n-1,t)-p(n,t)]$

the right hand side sum can be written out using the notation $p(s,t)=p_s$ $\partial_t G(s,t) = \lambda[s^1p_0-s^1p_1+S^2p_1-s^2p_2+\ldots]$

and grouping terms $\partial_t G(s,t) = \lambda[sp_0 + p_1s(s-1) +p_2s^2(s-1) +\ldots]$

or $\partial_t G(s,t) = \lambda(s-1)[\frac{s}{s-1}p_0 + p_1s +p_2s^2 +\ldots]$

so that for $s\gg1$ this is approximately $\partial_t G(s,t) \approx \lambda(s-1)[p_0 + p_1s +p_2s^2 +\ldots] = \lambda(s-1)G(s,t)$

thus the solution is $G(s,t) = e^{\lambda(s-1)t} G(0,t)$

If we assume that $n=0$ at $t=0$, we have $p(0,0)=1$ and $p(n\geq 1,0)=0$. That means that $G(s,0)=1$. We can write $G(s,t) = e^{\lambda(s-1)t} = e^{-\lambda t}\left(e^{\lambda t}\right)^s$

and expand in powers of $s$ to obtain $G(s,t) = e^{-\lambda t}[1+s\ln e^{\lambda t} + \frac{1}{2!}s^2(\ln e^{\lambda t})^2 + \frac{1}{3!}s^3(\ln e^{\lambda t})^3 + \ldots]$ $G(s,t) = \sum s^k \,\frac{e^{-\lambda t} (\lambda t)^k}{k!}$

so that we may identify $p(k,t)$ obtaining the solution to the Poisson process for the number of beans in a jar if they are added randomly with a mean number in a time interval to be $p(n,t) = \frac{e^{-\lambda t} (\lambda t)^n }{n!}$

The form of this probability distribution might be familiar.