World’s Oldest Man Dies

 

A Spaniard thought to be the oldest man in the world has died at the age of 114, his family said on Saturday.     Associated Press, 07 March 2004

 

Now and then the newspapers report the death of the world’s oldest person. How often does this occur? Judging from the newspapers, it doesn’t seem to be a particularly rare event. The death of the world’s oldest person was reported in March 2002, November 2003, December 2003, February 2004, and March 2004. The ages of the individuals were 115, 116, 114, 116, and 114 years. It’s hard to assess the accuracy of these reports, since we can’t really be sure if the world’s oldest person has been correctly identified at any given moment. Still it raises an interesting question about the expected rate at which the oldest member of a population fails.

 

Obviously the answer depends on the distribution of life spans. If one widget is produced every U seconds, and each widget has a life span of exactly L seconds, then each widget will be the oldest when it fails, so the world’s oldest widget will fail once every U seconds. On the other hand, if there is variability in the life spans, then many widgets will fail without ever having been the oldest widget.

 

In general, suppose one widget is produced each U seconds, and the life span of each widget is drawn from a given density distribution f(t). If the jth widget has a lifespan of Tj, then it will be the oldest surviving widget when it fails if and only if Tj−1 < Tj + 1U and Tj−2 < Tj + 2U and so on. In other words, the necessary and sufficient condition is that Tj−k < Tj + kU for k=1,2,3... The probability is therefore the infinite product of the probabilities of all these conditions. Integrating this over all possible values of Tj. gives the probability that any given widget will be the oldest widget when it fails.

Let F(T) denote the cumulative probability distribution, i.e., the integral of f(t) from t = 0 to T. The weight factor for the contribution of the incremental range around the time t is f(t)dt, so the probability that any given widget will become the world’s oldest is


To give an expected value of 1 for the number of “oldest widget failures”, we need 1/Poldest widgets, and since one widget is produced every U seconds, this implies that the mean time between the production of widgets destined to fail as the oldest widget is U/Poldest seconds.

 

If the life span density f(t) has the exponential distribution the result turns out to be particularly simple, at least in the rare event approximation, i.e., when λU is much less than 1. We have



Hence the probability that any given widget is destined to fail as the oldest widget is

 

 

By numerical integration we find that if Uλ is small compared with 1 then the value of this integral approaches Uλ. (An analytical proof of this is given below.) It follows that we need 1/(Uλ) widgets to get an expected value of 1 for the number of oldest widget failures, and since one widget is produced every U seconds, the mean time between oldest widget failures (for exponentially distributed life spans) is simply 1/λ, which of course is also the mean life span of the individual widgets. For example, if the average life span for widgets with exponentially distributed life spans is 100 years, then we would expect a failure of the oldest widget once every 100 years. (Needless to say, the fact that deaths of the world’s oldest person are much more frequent than this is due mainly to the fact that human life spans are not exponentially distributed.)

 

Incidentally, the integrand

 

 

in the expression for Poldest is a function of U and λ independently, even though the integral from t = 0 to infinity turns out to be strictly dependent on just the product Uλ. To illustrate, the plot below show the integrand functions for the cases (U=1, λ = 0.4) and (U = 2, λ = 0.2).

 

 

Another interesting feature of the integrand function I(t) is that, like the Riemann zeta function, it has no positive real roots and all the negative integers −1, −2, −3, etc. are roots. This similarity is perhaps not too surprising, considering that the zeta function has an infinite product representation

 

 

The behavior of I(t), interpreted as an analytic function for complex values of t, could also be considered. (For what complex values of t does the function I(t) vanish?)

 

If the value of Uλ is on the order of 1 or larger, then Poldest becomes progressively less than Uλ (since it obviously can never exceed 1). Approximate values of Poldest for various values of Uλ are tabulated below.

 

 

For large values of Uλ we can show that Poldest approaches

 

 

To analytically prove this, and also to prove that Poldest approaches Uλ for sufficiently small values of Uλ, we can expand the product in the function I(t) to give the sum of series

 

 

Collecting terms with the same multiplier of U in the exponents, we find that the coefficients are the numbers of partitions of the multiplier into distinct positive integers. Using the fact that the generating function for the number of partitions into k distinct parts is

 

 

the first three series can be evaluated as

 

 

 

Thus we have

 

 

If λU is large the higher-order terms become small very quickly, so up to the first order we can take just the first series and integrate to give

 

 

On the other hand, if λU is very small we can replace each expression of the form 1−e−kλU in the denominator of the previous equation with kλU (at least up to some value of k) to give

 

 

(It’s true that eventually kλU will become large for any finite value of λU, but by then the numerators will have become negligible.) Integrating this from t = 0 to infinity, we get

 

 

for sufficiently small values of λU.

 

Another way of evaluating the expression for Poldest is to set λ equal to 1 (since the integral depends only on λU) to give the simplified expression

 

 

Then we expand the product, integrate term by term, and collect terms to give the result

 

 

The coefficient of e−kU depends on the partitions of k. If we let pj(k) denote the number of partitions of k into j distinct positive integers (without regard to order), then the coefficient of e−kU is given by

 

 

For example, to find c10 we examine the partitions of 10 into distinct positive integers

 

 

Thus we have p1(10)=1, p2(10)=4, p3(10)=4, and p4(10)=1, so the coefficient is

 

 

The first twenty-eight coefficients are listed below:

 

 

For small values of U the function Poldest(U) approaches U, so setting U = 0 we should find that the sum of these coefficients (plus the leading unit coefficient) is zero. To show that this is indeed the case, consider all 2N subsets of the set of natural numbers from 1 to N. For example, the set {1, 2, 3} has the non-empty subsets {null}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}, each of which is a partition into distinct integers of some natural number. As N goes to infinity this encompasses more and more of the components of the ck coefficients (albeit not in the same order, so conditional convergence must still be considered). In general the number of subsets with k elements is the binomial coefficient C(N,k), so we can consider the sum (including the leading unit coefficient)

 

 

For example, with N = 6 we have

 

 

The fact that C(6,k)/(k+1) happens to be an integer for all k from 0 to 5 is related to the fact that 6 is a perfect number. This is also true for the next perfect number, i.e., C(28,k)/(k+1) is an integer for all k from 0 to 27. However, for all integers N we have sN = 1/(N+1). Therefore, letting N go to infinity, we see that the sum of all these subsets, which is the same as the sum of all the partitions into distinct integers of every natural number, divided by the appropriate denominator, is zero.

 

We also have the following interesting relation:

 

 

Making this substitution, we can express the probability Poldest for sufficiently small U as

 

 

consistent with the observation that Poldest approaches U for small U.

 

The 114 year old Spaniard who died in March of 2004 as the world’s oldest person is said to have attributed his longevity partly to sleeping 15 hours per day. (The mathematician Abraham de Moivre in his old age supposedly noticed himself sleeping 15 minutes longer each day, and by extrapolating this trend to the point when he would be sleeping 24 hours per day he predicted the date of his death, a prediction which reportedly turned out to be exactly correct.) Reports of the death of the world’s oldest person always remind me of the final lines of “King Lear”:

 

The weight of this sad time we must obey,

speak what we feel, not what we ought to say.

The oldest hath borne most. We who are young

will never see so much, nor live so long.

 

 

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