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This story yesterday (no need to follow the link to understand the question!)

http://www.cnn.com/2011/US/02/01/texas.oldest.person.dies/index.html?hpt=T2

reminds me that I've often wondered about the following:

Suppose you have a fixed mortality curve $f(x)$ expressing the probability of a person's remaining alive at age $x$. Suppose also, for simplicity, you have a steady state population, so a given fixed birth rate. How would one compute from $f(x)$, and the birth rate, the probability distribution $\rho$ governing the time intervals from one "world's oldest person dies" event to the next?

Actually though I think I could probably write down explicit integrals to answer my own question literally, but I don't expect they would say much as such. So I'm really looking for a softer answer that would explain what features of $f(x)$, what measures of the heaviness of its tail I guess, dominate the behavior of $\rho$ and how?

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  • $\begingroup$ Given the steady state of the population, the expected distribution of ages would be the same at eg midnight each day, so the probability of the death happening on any particular day would be the same. This would imply a poisson process. Or is this just a misconceived comment which has missed some subtlety? $\endgroup$ Feb 2, 2011 at 12:24
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    $\begingroup$ @Mark: to imply a Poisson process you would also need independence between the days. This doesn't hold in general here. A process can be stationary in time without being Poisson. $\endgroup$ Feb 2, 2011 at 13:28
  • $\begingroup$ @James - thanks. One problem with my intuition here is assuming a stable population and assuming that different people have identically distributed life chances are both unrealistic. So there are bulges - and life expectancy varies hugely between social groups and geographical locations. $\endgroup$ Feb 3, 2011 at 18:53
  • $\begingroup$ The reason that the variation in life expectancy is significant is that if there is a sub-population of - say - 10% of the world's population which has a much higher expectation of life than the rest, then the problem reduces to that population modulo a very small adjustment. $\endgroup$ Feb 3, 2011 at 18:57
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    $\begingroup$ @Bridge: that's an interesting process. $\endgroup$
    – Romeo
    Feb 4, 2011 at 0:39

1 Answer 1

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There was a discussion on this quite a long time ago at sci.mat. See: http://groups.google.com/group/sci.math/browse_thread/thread/531c39a2e5723df8/528137c752566266?lnk=gst&q=oldest+person#528137c752566266

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  • $\begingroup$ Thank you, Or Zuk, for this. That discussion trails off where my question begins. Of course as a mathematician I'm not so interested in real-world mortality table, but rather in the theoretical relationship between one distribution and the other. And the integrals in that discussion are basically what I had in mind...but analyzing what they say about how the distributions relate still seems hard! $\endgroup$ Feb 6, 2011 at 9:26

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