## Introducing the Product Integral

Of all the actuary's standard formulae derived from the life table, none is more important in survival modelling than:

${}_tp_x = \exp\left(-\int_0^t\mu_{s+s}ds\right).\qquad(1)$

Stephen covered the derivation of this in a previous blog, but I want to look more closely at the right-hand side of equation (1).� In particular, we can find an entirely different representation of $${}_tp_x$$ as a product integral, which leads to many insights in survival models.

Recall how the integral in equation (1) is constructed.� Choose a partition of the interval $$[0,t]$$, that is some sequence $$\Delta_1,\Delta_2,\ldots,\Delta_n$$ of non-overlapping sub-intervals that exactly cover the interval.� Define $$ds_i$$ to be the length of sub-interval $$\Delta_i$$ and choose a value $$\mu_i$$ contained in each $$\Delta_i$$.� Then let $$n\to\infty$$ in such a way that all the $$ds_i\to0^+$$ in a well-behaved way.� The result is that the following sum yields the Riemann integral in the limit:

$\lim_{n\to\infty}\sum_{i=1}^n\mu_ids_i=\int_0^t\mu_{x+s}ds.\qquad(2)$

Each term $$\mu_ids_i$$ in the sum in equation (2) has the usual intuitive interpretation as the probability of dying during the sub-interval $$\Delta_i$$ of length $$ds_i$$, assuming the life is alive at the start of the sub-interval (see equation (4) of Stephen's previous blog).� Equation (2) above lets us rewrite equation (1) as:

${}_tp_x = \exp\left(-\lim_{n\to\infty}\sum_{i=1}^n\mu_ids_i\right).\qquad(3)$

Now exchange the order of the exponential operator and the limit in equation (3) (please don't show this to a proper mathematician):

\begin{eqnarray*}{}_tp_x &= \lim_{n\to\infty}\exp\left(-\sum_{i=1}^n\mu_ids_i\right)\\&= \lim_{n\to\infty}\prod_{i=1}^n\exp\left(-\mu_ids_i\right).\end{eqnarray*}

Expanding each exponential term to first order gives the following:

${}_tp_x\approx\lim_{n\to\infty}\prod_{i=1}^n\left(1-\mu_ids_i\right).\qquad(4)$

The question is: does this approximation of products of survival probabilities (over the times $$ds_i$$) make mathematical sense, leading in the limit to an equality?� The answer is yes, and it defines the product integral as follows:

$\prod_0^t\left(1-\mu_{x+s}ds\right) = \lim_{n\to\infty}\prod_{i=1}^n\left(1-\mu_ids_i\right).\qquad(5)$

Equation (5) is the promised new representation of $${}_tp_x$$, namely:

${}_tp_x = \prod_0^t\left(1-\mu_{x+s}ds\right) = \exp\left(-\int_0^t\mu_{x+s}ds\right).\qquad(6)$

Readers who remember their first-year analysis courses may recall the following formula:

$\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n = e^x.\qquad(7)$

See pages 410�411 of Hardy (1908, reissued in 1992) for a nice proof.� This is really the heart of the product integral, which generalizes $$x$$ to a function $$f(x)$$, and generalizes the partitions $$\frac{1}{n}$$ to the $$ds_i$$.� For a wide range of functions $$f(x)$$ (but not all functions) it is true that:

$\prod_0^t\left(1-f(s)ds\right) = \exp\left(-\int_0^tf(s)ds\right).\qquad(8)$

At first sight, expressions (6) and (8) seem to be just mathematical curiosities.� However, in our forthcoming book we describe several non-parametric estimators that rely on just these results.� It turns out that the product integral is the natural way to explain the relationship between these estimators.

References

Hardy, G. H. (1908) A Course of Pure Mathematics, Cambridge University Press (reissued in 1992 in the Cambridge Mathematical Library series).

Macdonald, A. S., Richards, S. J. and Currie, I. D. (2018) Modelling Mortality with Actuarial Applications, Cambridge University Press (forthcoming). ### RECENT POSTS

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