Discovering the Mixed Poisson Distribution

Years ago I tried, and failed, to give a short axiomatization that uniquely characterized Poisson distributions. However, through it, I did learn about mixed Poisson distributions. The following is a brief recounting of this:

(Mis)characterizing Poisson Distributions

(Incomplete) Axioms for a family of Poisson distributions, ${X_T}$, indexed by $T\ge 0$ with $P_T(n)$ denoting the probability $X_T= n$:

1. Bisection: $X_{T_1+T_2} = X_{T_1}+X_{T_2}$
2. Normalization: $\mathbb{E}[X_T] = T$.
3. Regularity: $P_T(n)$ is smooth in $T$

I thought this would uniquely characterize Poisson distributions with rate $r = 1$. Indexing our distributions by the time period $T$, we know that a distribution corresponding to $T = T_1+T_2$ is the sum of $X_{T_1} + X_{T_2}$ as this corresponds to dividing our period into two sub-periods. With the normalization and regularity axioms, I thought I could uniquely characterize $X_T$ as $T\rightarrow 0$ and thus determine $X_T$ for all $T$. I was wrong… Here is where I got though.

Let

\[G(T,z) = \sum_{n\ge 0} P_T(n)z^n\]

be the generating function for each $X_T$.¹ Bisection implies that $G(T_1+T_2,z) = G(T_1,z)G(T_2,z)$, and we can use this fact to compute the derivative $G’(T,z) = \frac{\partial G}{\partial T}(T,z)$:

\[\begin{align} G'(T,z) &= \lim_{\Delta T\rightarrow 0}\frac{G(T+\Delta T,z) - G(T,z)}{\Delta T}\\ &= G(T,z) \left( \lim_{\Delta T\rightarrow 0}\frac{G(\Delta T,z)-1}{\Delta T}\right)\\ &= G(T,z) G'(0,z)\\ G(T,z) &= \exp (G'(0,z)T) \end{align}\]

The last derivation uses the fact that $G(0,z) = 1$ and L’Hôpital’s rule. We then solve the exponential differential equation and get the constant in front of the exponential is $1$ by considering $X_0$. Let $\varphi (z) = G’(0,z)$. We know $G(T,1) = 1$, and $\frac{\partial G}{\partial z}(T, 1) = T$ since $X_T$ is a distribution with first moment $T$. This translates to two equations for $\varphi(z)$:

\[\begin{align} \varphi(1) &= 0\\ \varphi'(1) &= 1 \end{align}\]

So, what is $\varphi(z)$? Well, the simplest solution is just $\varphi(z) = z-1$. This solution does yield the Poisson family. $e^{T(z-1)}$ is the probability generating function for Poisson distributions. Do any other solutions work? Yes, one could take any smooth $\varphi(z)$ satisfying the equations above.

Lets try $\frac{1}{n}(z^n-1)$. What does this represent? This is basically the generating function for a rate $1/n$ Poisson process, except each event in the process has weight $n$. Suppose we had a Poisson process for the number of people arriving at a coffee shop in a given period and say its normalized to rate $1$. We can count the number of hands arriving at that store via multiplying the output by $2$. This would correspond to the generating function $e^{T(z^2-1)}$. Toes arriving would be characterized by $e^{T(z^{10}-1)}$.

If $\varphi(z)$ is analytic, and things converge nicely², we can decompose $\varphi(z)$ into

\[\varphi(z) = \sum_i r_i (z^i-1),\text{ with } \sum_i r_i/i = 1\]

With this, we can decompose the probability generating function into:

\[e^{T\varphi(z)} = \prod_i e^{r_i T (z^i-1)}\]

This expresses the mixed Poisson distribution as an infinite sum of rate $r_i$ order $i$ generalized Poisson distributions. Pretty neat.

But what about the original task of characterizing Poisson distributions?

Well, we can notice from the above that the solution corresponding to the Poisson distribution $\varphi(z) = z-1$ is the unique solution with no higher-order terms than the linear term. Is there a condition that can enforce this? It turns out, yes. You can add the condition that the variance of $X_T$ matches the expectation. But I’ll leave that for you to work out :)

Footnotes