When is your Poisson model not a Poisson model?

The short answer for mortality work is that your Poisson model is never truly Poisson. The longer answer is that the true distribution has a similar likelihood, so you will get the same answer from treating it like Poisson.  Your model is pseudo-Poisson, but not actually Poisson.

Assume that you have \(M\) identical lives and you observe \(D\) deaths from a total time lived \(E^c\).  A common assumption is that \(D\) has a Poisson distribution:

\[D\sim{\rm Poisson}(E^c\mu),\qquad(1)\]

where \(\mu\) is the mortality hazard that we typically want to estimate.  However, the Poisson probability function is:

\[\Pr(D=d)=\frac{e^{-E^c\mu}(E^c\mu)^d}{d!}, \quad d=\{0, 1, 2, \ldots\},\qquad (2)\]

which means that there is a positive probability of having \(d>M\) deaths.  A model that allows more deaths than there are lives is, to put it politely, poorly specified.  Despite this, we will press on because enlightenment is closer than it seems.  The likelihood that one would use for inference under the Poisson model is the probability function in equation (2) using the data:

\[L^{\rm Poisson}=\frac{e^{-E^c\mu}(E^c\mu)^D}{D!}.\qquad(3)\]

Multiplicative factors that don't involve parameters have no impact on inference, so in practice we use the following simplification:

\[L^{\rm Poisson}\propto e^{-E^c\mu}\mu^D.\qquad(4)\]

We have seen that \(D\) cannot have a Poisson distribution, since this permits more deaths than the \(M\) lives under observation.  So what is the correct model?  At the time Macdonald et al (2018) was published, Angus wrote about how Poisson-like likelihoods kept cropping up.  "Poisson-like" sounds close to Poisson, so let's pursue this.  In a recent blog I showed how the most fundamental unit of mortality modelling is a survival model for individual lives.  We assume that life \(i\) is observed for \(t_i\) years, and the indicator \(d_i\) takes the value 1 if the life is dead at time \(t_i\), and zero otherwise.  The likelihood for this individual, \(L_i\), is then:

\[L_i = \exp\left(-\int_0^{t_i} \mu_s ds\right)\mu_s^{d_i}.\qquad(5)\]

If the mortality hazard is constant, then \(\mu_s=\mu\) and the likelihood in equation (5) simplifies to:

\[L_i = e^{-t_i\mu}\mu^{d_i}.\qquad(6)\]

If you compare equation (6) with equation (4), you get an idea of where this is going.  If we observe \(M\) identical individuals, then the combined likelihood, \(L^{\rm not\ Poisson}\), is the product of equation (6) for all \(M\) lives:

\[L^{\rm not\ Poisson}=\prod_{i=1}^M e^{-t_i\mu}\mu^{d_i}.\qquad(7)\]

If we write \(E^c = \sum_i t_i\) and \(D=\sum_i d_i\), then equation (7) can be simplified to:

\[L^{\rm not\ Poisson}=e^{-E^c\mu}\mu^D,\qquad(8)\]

which will produce the same inference for \(\mu\) as equation (4).  Thus, we have a likelihood for a model that is not Poisson - equation (8) - but has the same likelihood for a Poisson model in equation (4) (apart from a factor that doesn't matter).  In Macdonald & Richards (2024) we refer to the model in equation (8) as pseudo-Poisson - it is not a Poisson model, but inference for \(\mu\) can proceed as if it were.  This is handy, as it means we can use existing machinery for Poisson estimation, even though the number of deaths does not have a Poisson distribution.  An important example for actuaries is the use of Poisson GLMs for fitting mortality-projection models.