From small steps to big results
In survival-model work there is a fundamental relationship between the \(t\)-year survival probability from age \(x\), \({}_tp_x\), and the force of mortality, \(\mu_x\):
\[{}_tp_x = \exp\left(-\int_0^t\mu_{x+s}ds\right).\qquad(1)\]
Where does this relationship come from? We start by extending the survival time by an amount, \(h\), and look at the \((t+h)\)-year survival probability:
\[{}_{t+h}p_x = {}_tp_x.{}_hp_{x+t}\qquad(2)\]
which is simply to say that in order to survive \((t+h)\) years, you first need to survive \(t\) years and then you need to survive a further \(h\) years. Of course, surviving \(h\) years is the same as not dying in \(h\) years, so equation (2) can be written thus:
\[{}_{t+h}p_x = {}_tp_x.(1-{}_hq_{x+t}).\qquad(3)\]
If the period \(h\) is small enough, we can express the probability of dying, \({}_hq_{x+t}\), in terms of the force of mortality, \(\mu_{x+t}\):
\[{}_hq_{x+t} = h.\mu_{x+t}+o(h)\qquad(4)\]
where the function \(o(h)\) collects second- and higher-order powers of \(h\) and, crucially, is such that:
\[\lim_{h\to0^+}\frac{o(h)}{h} = 0\qquad(5)\]
i.e. \(o(h)\) tends to zero faster than \(h\) does. If we substitute equation (4) into equation (3) and re-arrange we get the following:
\[\frac{{}_{t+h}p_x-{}_tp_x}{h} = -{}_tp_x\mu_{x+t} + \frac{o(h)}{h}.\qquad(6)\]
We can now let \(h\to0^+\) and make use of equation (5):
\[\lim_{h\to0^+}\frac{{}_{t+h}p_x-{}_tp_x}{h} = -{}_tp_x\mu_{x+t}.\qquad(7)\]
The left-hand side of equation (7) is the definition of the first partial derivative of \({}_tp_x\) with respect to \(t\), so we have an ordinary differential equation (ODE) of degree 1 and order 1:
\[\frac{\partial}{\partial t}{}_tp_x = -{}_tp_x\mu_{x+t}.\qquad(8)\]
We are nearly there, as the solution to equation (8) is:
\[{}_tp_x = \exp\left(-\int_0^t\mu_{x+s}ds\right)+C\qquad(9)\]
where \(C\) is the constant of integration. However, we also have a boundary condition: since the probability of dying in a time interval of length zero is zero, \({}_0p_x=1\). From this we know that \(C=0\) in equation (9) and thus we have the result in equation (1) at the start of this posting.
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Occupational hazard
Everything points to Poisson
One recurring theme in our forthcoming book, Modelling Mortality with Actuarial Applications, is the all-pervading role of likelihoods that suggest the lurking presence of a Poisson distribution. A popular assumption in modelling hazard rates is that the number of deaths observed at any given age is a Poisson random variable, so perhaps that might explain it?
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