## 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?Ā 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.Ā 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.Ā 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.Ā However, we also have a boundary condition: since the probability of dying in a time interval of length zero is zero, $${}_0p_x=1$$.Ā 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. ### RECENT POSTS

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##### Model types in Longevitas
LongevitasĀusers can choose between seventeen types of survival model (μx) and seven types of GLM (qx). In addition there are a further seven extensions of the GLM models for qx to span multi-year data without violation of the independence assumption. Longevitas also offers non-parametric analysis, including Kaplan-Meier survival curves and traditional A/E comparisons against standard tables.