From: israel@math.ubc.ca (Robert Israel) Newsgroups: sci.math Subject: Re: Death of the oldest person Date: 6 Jan 1998 19:37:20 GMT In article <1998Jan5.122140.8938@leeds.ac.uk>, eclrh@sun.leeds.ac.uk (Robert Hill) writes: |> This set me wondering on the question: in a country of known total |> population, how often would one expect the event "death of the oldest |> person in the country" to occur? What mathematics would one use |> to estimate this? What plausible assumptions would one have to |> make in order to arrive at a ball-park estimate (in the style of a |> physicist) if one did not have access to very detailed demographic data? Suppose births take place as a Poisson process with constant intensity b, and the lifetime of each individual has distribution F (i.e. F(x) is the probability of death by age x). Let A(t) be the age of the oldest person alive at time t. Let H(y) = int_y^infinity dx (1-F(x)). In particular H(0) is the life expectancy at birth. Then it can be shown that P(A(t) <= y) = exp(-b H(y)) (which is the probability that everyone born before time t-y is dead at time t). Note that there is a positive, but presumably very small probability exp(-b H(0)) that nobody is alive at time t. Given that the oldest person alive at time t has age y, the probability that this person survives to time t+x is (1-F(y+x))/(1-F(y)). On the other hand, if nobody is alive at time t we have to wait an expected time 1/b for someone to be born and then H(0) for that person to die. Thus the expected time from now until the next death of an oldest person is exp(-b H(0))(1/b + H(0)) + int_0^infinity dy (exp(-b H(y))' int_0^infinity dx (1-F(y+x))/(1-F(y)) = exp(-b H(0))(1/b + H(0)) + int_0^infinity dy int_0^infinity dx b (1 - F(x+y)) exp(-b H(y)) = exp(-b H(0))(1/b + H(0)) + b int_0^infinity dy H(y) exp(-b H(y)) Robert Israel israel@math.ubc.ca Department of Mathematics (604) 822-3629 University of British Columbia fax 822-6074 Vancouver, BC, Canada V6T 1Z2 ============================================================================== From: israel@math.ubc.ca (Robert Israel) Newsgroups: sci.math Subject: Re: Death of the oldest person Date: 7 Jan 1998 02:01:49 GMT In article <68u15g$ffd$1@nntp.ucs.ubc.ca>, israel@math.ubc.ca (Robert Israel) writes: |> In article <1998Jan5.122140.8938@leeds.ac.uk>, eclrh@sun.leeds.ac.uk (Robert Hill) writes: |> |> This set me wondering on the question: in a country of known total |> |> population, how often would one expect the event "death of the oldest |> |> person in the country" to occur? What mathematics would one use |> |> to estimate this? What plausible assumptions would one have to |> |> make in order to arrive at a ball-park estimate (in the style of a |> |> physicist) if one did not have access to very detailed demographic data? |> Suppose births take place as a Poisson process with constant intensity b, |> and the lifetime of each individual has distribution F (i.e. F(x) is the |> probability of death by age x). Let A(t) be the age of the oldest person |> alive at time t. Let H(y) = int_y^infinity dx (1-F(x)). In particular |> H(0) is the life expectancy at birth. |> Then it can be shown that |> P(A(t) <= y) = exp(-b H(y)) |> Thus the expected time from now until the next death of an oldest person |> is ... |> exp(-b H(0))(1/b + H(0)) + b int_0^infinity dy H(y) exp(-b H(y)) I didn't quite answer the question that was asked, if by "how often" you mean the expected interval from one death-of-oldest-person to the next. Note that by symmetry the answer I gave is equal to the expected time from the last death-of-oldest-person to now, or half the expected length of the interval containing "now", but because of the "inspection paradox" the current interval tends to be longer than a typical interval. Assume that the lifetime distribution F has a density f = F'. Given that the age of the currently oldest person is y > 0, the probability of a death-of-oldest person between now and now+h is f(y)/(1-F(y)) h + O(h^2). On the other hand, if there is nobody currently alive the probability is O(h^2) (because it requires both a birth and a death to occur). So the (unconditional) probability of a death-of-oldest-person in this time interval is h int_0^infinity dy f(y)/(1-F(y)) (exp(-b H(y))' + O(h^2) = h b int_0^infinity dy f(y) exp(-b H(y)) + O(h^2) The expected number of such deaths in a given time interval of length T is then T b int_0^infinity dy f(y) exp(-b H(y)) and thus the expected interval should be 1/(b int_0^infinity dy f(y) exp(-b H(y))) I should be careful here: the deaths-of-oldest-person are not a Markov process, so you can't just use the Elementary Renewal Theorem, but I expect that for any "reasonable" lifetime distribution the process has good enough ergodic properties that this will work. Note also that the expected number N of individuals alive at any given time is related to the birth intensity b and the life expectancy H(0) by N = b H(0). Robert Israel israel@math.ubc.ca Department of Mathematics (604) 822-3629 University of British Columbia fax 822-6074 Vancouver, BC, Canada V6T 1Z2 ============================================================================== From: "Rose Baker" Newsgroups: sci.math Subject: Re: Death of the oldest person Date: Wed, 7 Jan 1998 10:55:30 -0000 Robert Israel wrote in message <68unmd$3or$1@nntp.ucs.ubc.ca>... >The expected number of such deaths in a given time interval of length T is then > T b int_0^infinity dy f(y) exp(-b H(y)) >and thus the expected interval should be > 1/(b int_0^infinity dy f(y) exp(-b H(y))) >I should be careful here: the deaths-of-oldest-person are not a Markov process, >so you can't just use the Elementary Renewal Theorem, but I expect that for any >"reasonable" lifetime distribution the process has good enough ergodic >properties that this will work. > >Note also that the expected number N of individuals alive at any given time >is related to the birth intensity b and the life expectancy H(0) by N = b H(0). > I agree with this later mailing. Robert has said everything I would have said if I had been faster off the mark! I'd just add that he has assumed a steady-state population, and that it is possible to redo the calculation with age distribution in the population not derived from the survival distribution for individuals. These latter were published in 1991 and are effectively Gompertz curves (exponentially increasing hazard of death) in the UK. It would be possible to crank out an approximation to the integral, using the method of steepest descents. To get the thing more accurate, we would need to include the higher male mortality. Pity (for this purpose) that lifespan is not exponentially distributed, when the problem becomes much simpler---everyone has an equal probability of dying in any time interval, and the gap between deaths of the oldest is (I think from doing it in my head) just the average lifespan. All this would make a nice student project! Rose