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.