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 
_{} 


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 lU is much less than 1. We have 



_{} 

By numerical integration we find that if Ul is small compared with 1 then the value of this integral approaches Ul. (An analytical proof of this is given below.) It follows that we need 1/(Ul) 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/l, 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 P_{oldest} is a function of U and l independently, even though the integral from t = 0 to infinity turns out to be strictly dependent on just the product Ul. To illustrate, the plot below show the integrand functions for the cases (U=1, l = 0.4) and (U = 2, l = 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 Ul is on the order of 1 or larger, then P_{oldest} becomes progressively less than Ul (since it obviously can never exceed 1). Approximate values of P_{oldest} for various values of Ul are tabulated below. 


For large values of Ul we can show that P_{oldest} approaches 

_{} 

To analytically prove this, and also to prove that P_{oldest} approaches Ul for sufficiently small values of Ul, 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 lU is large the higherorder 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 lU is very small we can replace each expression of the form 1e^{k}^{lU} in the denominator of the previous equation with klU (at least up to some value of k) to give 

_{} 

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

_{} 

for sufficiently small values of lU. 

Another way of evaluating the expression for P_{oldest} is to set l equal to 1 (since the integral depends only on lU) 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 p_{j}(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 c_{10} we examine the partitions of 10 into distinct positive integers 

_{} 

Thus we have p_{1}(10)=1, p_{2}(10)=4, p_{3}(10)=4, and p_{4}(10)=1, so the coefficient is 

_{} 

The first twentyeight coefficients are listed below: 


For small values of U the function P_{oldest}(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 2^{N} subsets of the set of natural numbers from 1 to N. For example, the set {1, 2, 3} has the nonempty 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 c_{k} 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 s_{N} = 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 P_{oldest} for sufficiently small U as 

_{} 

consistent with the observation that P_{oldest} 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 turned out to be exactly correct.) In some ways Shakespeare’s “King Lear” strikes me as being about the death of the world’s oldest person… 

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. 
