23

Will You Be Alive 10 Years from Now?

23.1 THE PROBLEM

As one gets older the above question takes on ever-increasing interest. The only real answer to it is something like “Who knows?,” or “Maybe, but maybe not,” or “Yes, God willing.” Those are all a bit less than satisfying, of course; can we at least calculate the probability of being alive 10 years from now? The answer to that question is a definite yes, and you can do it for yourself with information you can get right off the Web. All you need is your age (which I’m sure you know) and the so-called life-expectancy table for your particular situation (race and gender, which I’m also sure you know).

A life-expectancy table gives, at every age, how many more years you can expect to live. (See, for example, the U.S. Social Security Administration’s website, which has a Life Expectancy Calculator indexed on gender and birthdate—what statisticians call the cohort.) Naturally, as you get older your expected number of years yet to live continually decreases, or at least that has been the universal experience of humanity: no exceptions have ever been observed. For example, a man alive now who is age 50 might have an expected years yet to live of 30 years, but if he makes it to age 51 he would then have an expected years yet to live of only 29.5 years. A table of this sort does not directly answer our opening question, but it does implicitly contain the answer, and what I’ll now show you is how to extract the probability you’ll be alive 10 (or any other number you wish) years from now from such a table.

23.2 THEORETICAL ANALYSIS

Let p(x) be the probability a person alive now is still alive x years from now (now is time zero), where p(0) = 1. (Our opening question is, “What is p(10)?”) Then, p(x + Δx) is the probability the person is alive x + Δx years from now, where Δx ≈ 0, and so p(x) − p(x + Δx) is the probability the person dies in the interval x to x + Δx years from now. So, as of now the person has x years left to live with probability p(x) − p(x + Δx), and the average or expected years left to live, φ, is found by integrating the product x [p(x) − p(x + Δx)] over all x. That is,

Since

then, as Δx → 0 we have

and so

In the well-known integration-by-parts formula,

let u = x (and so du = dx) and dv = dp (and so v = p). Then,

Or, since lim_{x → 0}xp(x) = 0 because there is some finite value of x (certainly less than 200 years!) when p(x) = 0, we have the expected years left to live from now as

If we move forward from now (x = 0) to some later time (x > 0) the expected years left to live will, more generally, be φ(x), calculated just as in the above integral except for the lower limit, which is the then new now, x. That is, with u as a dummy variable of integration,

Our first integral expression for φ is, in fact, φ(0).

Here’s what we have so far: a person alive now will be alive x years from now with probability p(x) and have a life expectancy then of φ(x), or will be dead x years from now with probability 1 − p(x) and have (obviously!) a life expectancy then of zero. So, the overall life expectancy is the average of these two possibilities, given by

p(x)φ(x) + [1 − p(x)]0 = p(x)φ(x).

But we know the life expectancy at age x is φ(x), and so

p(x)φ(x) = φ(x),

with the surprising solution p(x) = 1, surprising because p(x) = 1 says our person is alive for all x. He is immortal! Alas, were it but so. Math is wonderful, yes, but it isn’t the Fountain of Youth. This solution is mathematically correct, but it is also physical nonsense (just as, in high school algebra, when solving quadratic equations for the number of apples in a bag we would reject negative or complex solutions as physically unacceptable). Fortunately, there is another, far more realistic solution for p(x).

To find it, we write p(x)φ(x) = φ(x) in this way:

This is what mathematicians call an integral equation, with an unknown function appearing both inside and outside an integral. Such equations can be very tough to solve, often requiring powerful, advanced techniques, but we are in luck as there is a beautiful classical solution to this integral equation. What I’ll show you next is how to unravel the above integral equation to find p(x) as a function of φ(x) (the values of which are the entries in a life expectancy table).

Start by writing

Then, define the function P(u) as the indefinite integral of p(u), and so

(I’ve put this equation in bold because I’ll refer back to it in just a few steps.) Integrating both sides gives

Next, change variable to

g(u) = P(∞) − P(u),

and so, remembering that p(u) is the derivative of P(u),

and so

Thus,

Now, whatever P(∞) and P(0) are, they are constants, so let’s write c = P(∞) − P(0), and thus

Looking back at the bold equation, we see that P(∞) − P(x) = p(x)φ(x), and so

Since p(0) = 1, then

which says c = φ(0). So,

or, at last (and again in bold),

To see how to use this result in an actual calculation, suppose (just for sake of illustration) that we have the following life-expectancy table for a person of some age now and for the next 10 years:

Finally, we can answer our opening question: what is p(10), the probability this person will be alive 10 years from now? From the second bold equation we have

To evaluate the integral I’ll use a numerical technique called Simpson’s method (after the English mathematician Thomas Simpson [1710–1761], who wrote of it in 1743), which can be remarkably accurate. Here’s how it works. To estimate the value of , divide the integration interval (a, b) into an even number of equal subintervals of width h. Then (see any good calculus textbook) we have

In our problem a = 0, b = 10, h = 1 (since we have ten subintervals, each one year wide), and f(u) = 1/φ(u). We can do the numerical work in a systematic way by arranging the arithmetic as shown in the following table:

The sum of all the products in the rightmost column is 1.7937, and, as h/3 = 1/3, we have

and so

ln{p(10)} = 0.38595 − 0.5979 = −0.21195.

Thus, the probability our person will be alive 10 years from now is

p(10) = e^−0.21195 = 0.809.

Simple—when you see it.