Number of Dead People vs Living

arX

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6 Aug

2014

DID THE EVER DEAD OUTNUMBER THE LIVING AND

WHEN? A BIRTH-AND-DEATH APPROACH

JEAN AVAN, NICOLAS GROSJEAN AND THIERRY HUILLET

Abstract. This paper is an attempt to formalize analytically the questionraised in World Population Explained: Do Dead People Outnumber Living,Or Vice Versa? Huffington Post, [7]. We start developing simple determin-

istic Malthusian growth models of the problem (with birth and death rateseither constant or time-dependent) before running into both linear birth anddeath Markov chain models and age-structured models.

Keywords: population growth; Malthusian; constant vs time-dependentbirth/death rates; time-inhomogeneous Markov chain; age-structured models.

Corresponding author.

1. Introduction

Starting from a question raised in (of all places) Huffington Post, asking whetherthe total number of dead people ever outgrew the total number of living people ata certain time, and when the crossover (if any) occurred, leads us to analyze somesimplified models of population growth based on mean-field approximations, i.e.dealing with the notion of mean birth/death rates.

We shall successively consider a simple deterministic growth model for the wholepopulation (Section 2); a linear birth/death Markov chain model for the growthof the whole population (Section 3); a deterministic age-structured model (Lotka-McKendrick-von Foerster) in Section 4, for which we shall explicitly (as far aspossible) derive general solutions for the population sizes (in all 3 cases), the ex-tinction probability, time to extinction and joint law for the number of people everborn and ever dead (for the Markovian process). Some explicitly solvable cases willbe considered.

2. Simple birth and death models for population growth(deterministic)

We get first interested in the simple deterministic evolution of a population whoseindividuals are potentially immortal, non-interacting and living in a bath withunlimited resources.

2.1. Constant rates (Malthusian model). Let the mean size of some populationat time t = 0 be x (0) = x0 and let x (t) be its mean size at time t.

1

2 JEAN AVAN, NICOLAS GROSJEAN AND THIERRY HUILLET

Let (b, d) be the birth and death rates per individual, assumed constant in thefirst place. With . = d/dt, consider the evolution

(1).x (t) = (b d)x (t) =: .xb (t) .xd (t) ,

where.x (t) is the growth rate at time t of the population, whereas

.xb (t) = bx (t)

and.xd (t) = dx (t) are the birth and death rates at time t of this population (the

rates at which newborns and dead people are created, respectively).

Integrating, we have x (t) = xb (t)xd (t), with x (0) = xb (0) . The quantities xb (t)and xd (t) are the mean number of individuals ever born and ever dead betweentimes 0 and t and therefore x (t) is the current population size (xb (t) includes theinitial individuals). Obviously, with := b d

x (t) = x (0) et

and therefore (if b 6= d)

xb (t) = x (0) +b

b d (x (t) x (0))

xd (t) =d

b d (x (t) x (0)) .

Under this model, given that some individual is alive at time t, its lifetime d obeysP (d > t+ | d > t) = P (d > ) = ed , independent of t (the memory-lessproperty of the exponential distribution). The individuals are immortal in thatthere is no upper bound for their lifetime although very long lifetimes are unlikelyto occur, i.e. with exponentially small probabilities.

Similarly, its reproduction time b obeys P (b > t+ | b > t) = P ( b > ) =eb . After each reproduction time, each individual is replaced by two individualsby binary splitting. Then 1d and

1b are the mean lifetime and reproduction

times.

We assume the supercritical condition: = b d > 0 so that x (t) grows expo-nentially fast. Then, as t

(2)xb (t)

x (t) b

b d andxd (t)

x (t) d

b d .

Remark: The cases b = d or b d < 0 correspond to the critical and sub-critical situations, respectively. In the first case, the population size x (t) remainsconstant (while xb (t) and xd (t) grow linearly) whereas in the second case, the pop-ulation size goes to 0 exponentially fast. Unless otherwise specified we shall stickto the supercritical case in the sequel.

Suppose we ask the questions:- how many people have ever lived in the past and

- do the ever dead outnumber the living and when?

The answer to the first question is xb (t).

The answer to the second question is positive iff b > d andd

bd> 1 or else

if d < b < 2d. Then the time when the overshooting occurred is t defined by

A BIRTH-AND-DEATH APPROACH 3

xd(t)x(t)

= 1. Thus

t = 1log

(2 b

d

)> 0,

which is independent of x (0). At time t, the number of people who have ever livedin the past is thus

xb (t) = x (0)

(1 +

bb d

(et 1)) = x (0) 2d

2d b > x (0) .

Note also that the population size at t is

x (t) = x (0) et = x (0)

d2d b > x (0) ,

so twice less than xb (t) consistently with x (t) = xb (t) xd (t) and xd (t) =x (t) .

Suppose that at some (large) terminal time tf we know:

- xb (tf ) the number of people who ever lived before time tf .

- the initial population size x (0) at time t = 0.

- the current population size x (tf ) .

Then

= b d = 1tf

log

(x (tf )

x (0)

)and

bb d

xb (tf )

x (tf )

and both b and d are known. As a result,

t = 1log

(2 b

d

) tf log (1 + x (tf ) / (xb (tf ) 2x (tf )))

log (x (tf ) /x (0))

can be estimated from the known data.

Example:

The number of people who ever lived on earth is estimated to xb (tf ) = 105.109.

Suppose 10.000 years ago the initial population was of 106 units. Then x (0) = 106

and tf = 104. The current world population is around x (tf ) = 7.10

9. Then

b 15.103, d 14.103 and t 75 years.These rough estimates based on most simple Malthus growth models are very sen-sitive to the initial condition.

2.2. Time-dependent rates. Because the assumption of constant birth and deathrates is not a realistic hypothesis, we now allow time-dependent rates, so both band d are now assumed to depend on current time t. We shall assume that b (t)is bounded above by +b < and below by b > 0 and similarly for d (t) withupper and lower bounds d . Unless specified otherwise, we assume supercriticality

throughout, that is b >

d . We shall also assume that both b (t) and d (t) arenonincreasing functions of t.1

Let b (t) = t0 dsb (s) , d (t) =

t0 dsd (s) be primitives of b (t) and d (t) .

1Under this hypothesis, both the expected lifetime and the mean reproduction time of eachindividual increase as time passes by, reflecting an improvement of the general survival conditions.


Under this model, given some individual is alive at time t, its lifetime d obeysP (d > t+ | d > t) = e(d(t+)d(t)) so P (d dt | d > t) = d (t) dt.Similarly, its reproduction time b obeys P ( b > t+ | b > t) = e(b(t+)b(t))so P ( b dt | b > t) = b (t) dt. After each reproduction time, each individualis again replaced by two individuals, giving birth to a new individual by binarysplitting.

With d/dt, we thus have to consider the new evolution

(3).x (t) = (b (t) d (t))x (t) =: .xb (t) .xd (t) ,

where.x (t) is the growth rate at time t of the population, whereas

.xb (t) = b (t) x (t)

and.xd (t) = d (t)x (t) are again the birth and death rates at time t of this popu-

lation.

We immediately have x (t) = xb (t)xd (t), with x (0) = xb (0) . The quantities xb (t)and xd (t) are the mean number of individuals ever born and ever dead betweentimes 0 and t (xb (t) including the initial individuals) and therefore x (t) is thecurrent population size. Clearly, with (t) = b (t) d (t)

x (t) = x (0) e(t)

and therefore

xb (t) = x (0)

(1 +

t0

dsb (s) e(s)

)xd (t) = x (0)

t0

dsd (s) e(s).

Note x (t) = x (0)(1 +

t0 ds (b (s) d (s)) e(s)

)We assume that (t) is non-decreasing with t so that x (t) grows and also that,with b >

d > 0

b (t) b > 0 and d (t) d > 0 as t.Then, as t

(4)xb (t)

x (t)=

1 + t0 dsb (s) e

(s)

1 + t0 ds (b (s) d (s)) e(s)

b

b dand

xd (t)

x (t)=

t0 dsd (s) e

(s)

1 + t0ds (b (s) d (s)) e(s)

d

b d.

Indeed, by LHospital rule, if f(t)g(t) t c F (t)G(t) t c where (F,G) are the primi-

tives of (f, g) .Applying LHospital rule to f = b (t) e(t) and g = (b (t) d (t)) e(t)

gives the result. Suppose b >

d . Wheneverd

b

d

> 1 or else b < 2

d , there

is a unique t such thatxd(t)x(t)

= 1.

Remark: Assume b >

d > 0. We have the detailed balance equations:

d

dt

[xb (t)xd (t)

]=

[b (t) b (t)d (t) d (t)

] [xb (t)xd (t)

], with

[xb (0)xd (0)

]=

[xb (0)0

].


In vector form, this is also.x (t) = B (t)x (t), x (0) where x (t) 0 (componentwise).

The matrix B (t) has eigenvalues (0, b (t) d (t)) , with (0, b (t) d (t)) (0, b d

)as t . 1 (t) = b (t) d (t) is its dominant eigenvalue. Let-

ting X (t) be a 2 2 matrix with X (0) = I, and.

X (t) = B (t)X (t) , we havex (t) = X (t)x (0) and

et

01(s)dsX (t) xy

xyas t,

with B ()x = x, yB () = y, = b d := 1 () the dominant eigenvalueof B (). The left and right eigenvectors of B () associated to are found to bex =

(b ,

d

)and y = (1,1) . The latter convergence result holds because, with

y independent of t, yB (t) = 1 (t)y for all t > 0 and B (t)1 = 0 (1 is in the

kernel of B (t) for all t), see [2]. We are thus led to

et

01(s)dsX (t) = e(t)X (t) 1

b d

[b bd d

]as t

and e(t)x (t) x (0)b d

[bd

]as t,

which is consistent with the previous analysis making use of LHospital rule, recall-ing x (t) = xb (t) xd (t) = x (0) e(t). Note

1

t

t0

1 (s) ds as t.

Examples:

(i) (homographic rates) With b > a > 0 and d > c > 0, assume

b (t) =at+ b

t+ 1and d (t) =

ct+ d

t+ 1

Thus b = a, +b = b and

d = c, +d = d. Then

b (t) = at+ (b a) log (t+ 1) and d (t) = ct+ (d c) log (t+ 1)and with = (b a) (d c)

(t) = (a c) t+ log (t+ 1) and x (t) = x (0) (t+ 1) e(ac)t.Note 0, depending on b d. Thus

xb (t) = x (0)

(1 +

t0

ds (as+ b) (s+ 1)1

e(ac)s)

xd (t) = x (0)

t0

ds (cs+ d) (s+ 1)1

e(ac)s

and xd(t)x(t) tc

ac . When a > c, log x (t) /t a c > 0 and the population growsat exponential rate.

If in addition a < 2c, there is a unique t such that xd (t) = x (t) : t0

ds (cs+ d) (s+ 1)1

e(ac)s = (t + 1)e(ac)t .


(ii) Gompertz (a critical case b = a = c =

d ): We briefly illustrate here on anexample that population growth under criticality also has a rich structure. Let

b (t) = a+ (b a) et and d (t) = c+ (d c) et.Assume > 0, a = c and b d > . Then

b (t) = at+b a

(1 et) and b (t) = ct+ d c

(1 et)

(t) = b (t) d (t) = b d

(1 et) .

Because b = a = c =

d , xd (t) and x (t) will never meet.

We have x (t) = x (0) e(t) = x (0) e(bd)(1et)/ x (0) e(bd)/ as t : due

to bd > , after an initial early time of exponential growth, the population size sta-bilizes to a limit. This model bears some resemblance with the time-homogeneouslogistic (or Verhulst) growth model with alternation of fast and slow growth andan inflection point between the two regimes.

3. Birth and death Markov chain (stochasticity)

The deterministic dynamics discussed for (x (t) , xb (t) , xd (t)) arise as the meanvalues of some continuous-time stochastic integral-valued birth and death Markovchain for (N (t) , Nb (t) , Nd (t)) which we would like now to discuss; Nb (t) , Nd (t)are now respectively the number of people who ever lived and died at time t tothe origin, while N (t) = Nb (t) Nd (t) is the number of people currently aliveat t. Due to stochasticity, some new effects are expected to pop in, especially thepossibility of extinction, even in the supercritical regime.

3.1. The current population size N (t).

3.1.1. Probability generating function (pgf) of N (t). Let N (t) {0, 1, 2, ...} withN (0)

d pi0 for some given initial probability distribution pi0. Define the transitionmatrix of a (linear) birth and death Markov chain [9] as

(5) P (N (t+ dt) = n+ 1 | N (t) = n) = b (t)ndt+ o (dt)P (N (t+ dt) = n 1 | N (t) = n) = d (t)ndt+ o (dt) ,

with state {0} therefore absorbing. Let t (z) = EE(zN(t) | N (0)) =: E (zN(t)) be

the pgf of N (t) averaged over N (0) . Let 0 (z) = E(zN(0)

)be the pgf of N (0) .

Then, defining

g (t, z) = (b (t) d (t)) (z 1) + b (t) (z 1)2 ,t (z) obeys the PDE (see [1])

tt (z) = g (t, z)zt (z) , with t=0 (z) = 0 (z)

or equivalently, solves the non-linear Bernoulli ODE problem.t (z) = g (t, t (z)) , with t=0 (z) = z and t (z) = 0

(1t (z)

).

Setting (t) = e(t) and (t) = t0dsb (s) e

(t)(s), both going to as t ,assuming

b (t) b > 0 and d (t) d > 0 as t


and b 6= d , we easily get the solution

(6) t (z) = 0

(1 ( (t) (t)) (z 1)

1 (t) (z 1)),

where inside 0 () we recognize t (z) as an homographic pgf, solution to the ODEproblem. We choose for initial condition a Bernoulli thinned geometric randomvariable with homographic pgf2

0 (z) =q (q0 + p0z)

1 p (q0 + p0z) =q + p0q (z 1)q p0p (z 1) ;

this considerably simplifies (6), since the composition of two homographic pgfs isagain an homographic pgf. With this initial condition, we finally get

t (z) =q (q (t) p0q (t)) (z 1)q (q (t) + p0p (t)) (z 1) .

With x (0) = 0 (1) = p0/q > 0 (with the derivative with respect to z), averaging

over N (0), we obtain

x (t) = E (N (t)) = EE (N (t) | N (0)) = t (1) = 0 (1) (t) = x (0) e(t)Var (N (t)) =

t (1) +

t (1) t (1)2 = x (0)2 (2p 1) (t)2 + x (0) (t) (1 + 2 (t)) .Remark: For pgfs of the homographic form (6), it is easy to see that, averagingover N (0), for n 13

P (N (t) = n) = EP (N (t) = n | N (0))= [zn]t (z) = C (t)

n1(t (0)C (t) +D (t)) ,

where, with t (0) = P (N (t) = 1) =

0 (0) (t) (1 + (t))2

C (t) := 1

t (0)

1 t (0)and D (t) :=

t (0)

1 t (0) t (0) .

3.1.2. The extinction probability and time till extinction. The extinction probabilityreads

P (N (t) = 0) = t (0) = 1p0 (t)

q + q (t) + p0p (t).

Because P (N (t) = 0) is also P ( e t) where e is the extinction time (the evente = corresponds to non-extinction), we obtain

P (e > t) =p0 (t)

q + q (t) + p0p (t).

Whenever b >

d > 0 (supercritical regime), (t) / (t) > 1 (as t )because

(t) / (t) =

t0

dsb (s) e(s) >

t0

ds (s) e(s) = 1 e(t) 1.

2E(zN(0)

):= 0 (z) = G (B (z)) is the composition of a geometric pgf G (z) =

(qz) / (1 pz) (p + q = 1) with a Bernoulli pgf B (z) = q0 + p0z (p0 + q0 = 1). So

N (0)d=

G

i=1Bi where the Bis are iid Bernoulli, independent of G with E (N (0)) = p0/q

and Var(N (0)) = p0 (q0q + p0p) /q2 = q0E (N (0)) + pE (N (0))2 .

3[zn]t (z) is the coefficient of zn in the series expansion of t (z) .


As it can be checked, were both b (t) and d (t) be identified with their limitingvalues b and

d , then: (t) / (t) 1/ (1 ) = b /(b d

)where :=

d /

b (< 1 in the supercritical regime).

Thus, with = 1/ (1 ) > 1

(7) P ( e > t) P (e =) = p0q+ p0p

as t.

The extinction probability reads

e = P ( e 0 : Nd (t) Nb (t))the first time the ever dead outnumber the ever living. In the randomized setup,even in the supercritical regime of mean exponential growth for x (t) = xb (t)xd (t),there is a no-luck possibility that this event occurs and this corresponds to globalextinction of the population at time e.

These results answer the somehow related question on if and when the ever dead(Nd (t)) outnumber the ever living (Nb (t)) in the supercritical linear birth and deathMarkov chain model; but so far the question on whether the ever dead (Nd (t))outnumber the living (N (t) = Nb (t)Nd (t)) has not been addressed.

3.2. The joint law of Nb (t) and Nd (t). In order to access to a comparative studybetween Nd (t) and N (t) := Nb (t)Nd (t), we need to first compute the joint pgfof Nb (t) and Nd (t).

3.2.1. Joint pgf of Nb (t) and Nd (t). Define Nb (t) and Nd (t) as the number ofpeople ever born and ever dead till time t. We wish now to compute the jointlaw of Nb (t) and Nd (t), keeping in mind N (t) = Nb (t) Nd (t) and for someN (0) = Nb (0) .

With n = nb nd, the joint transition probabilities are obtained as(8) P (Nb (t+ dt) = nb + 1, Nd (t+ dt) = nd | N (t) = n) = b (t)ndt+ o (dt)

P (Nb (t+ dt) = nb, Nd (t+ dt) = nd + 1 | N (t) = n) = d (t)ndt+ o (dt)allowing to derive the evolution equation of EP (Nb (t) = nb, Nd (t) = nd | N (0)).Introducing the joint pgf of Nb (t) and Nd (t) as t (zb, zd) := E

(zNb(t)b z

Nd(t)d

), we

obtain its time evolution as

(9) tt = (b (t) (zb 1) + d (t) (zd 1)) (zbzb zdzd)t,with initial condition 0 (zb, zd) = 0 (zb), noting Nd (0) = 0.


3.2.2. Correlations. Consistently, we have xb (t) = zbt (1, 1) and xd (t) = zdt (1, 1) .Taking the derivative of (9) with respect to zb and also with respect to zd and eval-uating the results at zb = zd = 1 we are lead to

.xb (t) = b (t) (xb (t) xd (t)) =

b (t)x (t), xb (0) = x (0) and to.xd (t) = d (t)x (t), xd (0) = 0. With

E(Nb (t)

2):= 2zbzbt (1, 1) , E

(Nd (t)

2):= 2zdzdt (1, 1) , E (Nb (t)Nd (t)) :=

2zbzdt (1, 1)

we also have

d

dt

E(Nb (t)

2)

E(Nd (t)

2)

E (Nb (t)Nd (t))

= 2b (t) 0 2b (t)0 2d (t) 2d (t)

d (t) b (t) b (t) d (t)

E(Nb (t)

2)

E(Nd (t)

2)

E (Nb (t)Nd (t))

,

with initial condition

E(Nb (0)

2)

E(Nd (0)

2)

E (Nb (0)Nd (0))

= E

(N (0)2

)00

. In vector form,this is also

.x (t) = C (t)x (t), x (0) where x (t) 0 (componentwise).

The matrix C (t) tends to a limit C () as t.It has eigenvalues (0, b (t) d (t) , 2 (b (t) d (t))) , with

(0, b (t) d (t) , 2 (b (t) d (t)))(0, b d , 2

(b d

)),

as t . We identify now 1 (t) = 2 (b (t) d (t)) as its dominant eigenvalue(the one for which limt

1t

t0 (s) ds is largest). Letting X (t) be a 3 3 matrix

with X (0) = I and.

X (t) = C (t)X (t) , we have x (t) = X (t)x (0) and

et

01(s)dsX (t) xy

xyas t,

with C ()x = x, yC () = y, = 2 (b d ) := 1 () the dominanteigenvalue of C (). The left and right eigenvectors of C () associated to arefound to be x =

((b)2,(d)2, b

d

)and y = (1, 1,2) . The latter conver-

gence result holds because, with y independent of t, yC (t) = 1 (t)y for all t > 0

and C (t) 1 = 0 (1 is in the kernel of C (t) for all t).

Indeed,

X (t) = x (t)y solves.

X (t) = C (t)X (t)

and from Theorem 1 in [10], assuming4 .b (t) dt t) whereb > a > 0 and P (T > t) is the tail probability distribution of any positive random variable T.


where Cb,d = E(Nb (0)

2) (b

d

)/(b d

)2, Cb = E (Nb (0))

b /(b d

)and Cd = E (Nb (0))

d /(b d

)so Cb,dCbCd =Var(Nb (0))

(b

d

)/(b d

)2>

0, translating an asymptotic positive correlation of (Nb (t) , Nd (t)) . Note

1

t

t0

1 (s) ds as t.

3.2.3. Joint pgf revisited. Introducing the change of variables from (zb, zd) to

zb/zd and zbzd, the right-hand-side of (9) only contains . The

evolution equation for in the new variables indeed reads

tt =(b (t)

2 (b (t) + d (t)) + d (t) )t, 0 (, ) = 0 () .

t is therefore obtained as a function of a suitable combination of and time(solvable by the method of characteristics), and separately of in a way determinedby initial conditions.

Fixing to a certain value, one may then seek for an homographic ansatz

(11) t (, ) = 0

(a (t) + b (t)

c (t) + d (t),

)= 0

(a (t) + b (t)

c (t) + d (t)

),

with a (0) = d (0) = 1 and b (0) = c (0) = 0.

Plugging this ansatz into the evolution equation for and denoting respectively

(t) = b (t) , (t) = (b (t) + d (t)) , (t) = d (t) ,we need to solve

.

(t) = (t) (t)2+ (t) (t)+ (t) with initial condition (0) =

and the guess (t) = a(t)+b(t)c(t)+d(t) . We get.ac a .c = (ad bc)

.

bc b .c+ .ad a.

d = (ad bc).

bd b.

d = (ad bc) .Fixing the CP 3 gauge choice of any homographic form a+bc+d (invariant under

a, b, c, d a, b, c, d) by fixing ad bc = 1 yields

(12)d

dt

[a cb d

]=

[/2 /2

] [a cb d

],

to be solved as the ordered exp-integral[a cb d

](t) =exp

t0

[/2 /2

](s) ds.

Although this system can rarely be solved explicitly by quadrature for all t, this in

principle solves t (, ) hence t (zb, zd) when back to the original variables.

Remark: We note from the above differential equation (12) and the expression of(, , ) in the transition matrix that, as functions of

(a, b/, c/, d) are functions of 2,

or else that (besides t)

(13) a/c and d/b are functions of 2.


3.2.4. Asymptotics. In matrix form, (12) reads.

X (t) = A (t)X (t) , X (0) = I

where A (t) =

[/2 /2

]has zero trace and X (t) =

[a cb d

](t). It follows

that det(X) is a conserved quantity.

Note that, due to || 1, det (A (t)) = (b (t) + d (t))2 /4 + 2b (t)d (t) (b (t) d (t))2 /4 < 0: For all t, the eigenvalues of A (t) are real with oppositesign (t) = 12

(b (t) + d (t))

2 42b (t)d (t). With

A1 =

[ 1/2 0 1/2

]and A2 =

[ 1/2 0 1/2

],

we have A (t) = b (t)A1 + d (t)A2 with [A1, A2] 6= 0; In fact A1, A2 generate thefull su(2) algebra hence it is consistent to define

(14) X (t) = exp (1 (t)L1 + 2 (t)L2 + 3 (t)L3) ,

for some i (t) obeying a Wei-Norman type non-linear differential equation [8] andLi Pauli matrices. For recent further developments, see [12].

With i = 1, 2, let xi (t) = X (t)xi (0) with x1 (0) = (1, 0) x1 (t) = (a, b) and

x2 (0) = (0, 1) x2 (t) = (c, d) .

We set

A (t) = A () + A (t)A () = b A1 + d A2A (t) =

(b (t) b

)A1 +

(d (t) d

)A2.

Of course, the matrix A () has two real opposite eigenvalues

= 1

2

(b +

d

)2 42b dwith (t) as t.From Theorem 1 in [10], assuming

.b (t) dt


(a (T ) + b (T )) / (c (T ) + d (T )). This common limit exists and, from the above

scaling argument (13), it can be written in the form say h(2)/, for some (un-

known) function h encoding the finite size effects till time T ; it depends on and

2. Note that, consistently, both a (t) /b (t) and c (t) /d (t) also have a common limit(angular azimut of both x1,x2) which is

+ (1) /+ (2) =(b ) / (+ + (b + d ) /2) = u (2) /,

where u(2)=(2+

(b +

d

))/(2d

).

So the limit structure of X (t) is known and as t(16) t (, ) 0

(h(2)).

The dependence on has disappeared in the limit. Coming back to the originalvariables, recalling

zb/zd and zbzd, as t yields the limiting shape

of t (zb, zd) . We get

t (zb, zd) = E(zNb(t)b z

Nd(t)d

)t E

(zNb()b z

Nd()d

)= 0 (h (zbzd)) .

3.2.5. A particular solvable case. It turns out that in several case there is anexplicit solution to (12), illustrating the general conclusions just discussed. Let usgive an example.

Explicit solution. Let > 0 be a fixed number. When d (t) = b (t) , (12)can be solved explicitly while observing[

/2 /2

]= b (t)

[ (1+)2 (1+)2

]and diagonalizing the latter matrix. With b (t) =

t0 b (s) ds and putting

(2)=

1

2

(1 + )

2 42,we indeed get ad bc = 1 where

a = (1 + )2(2) sinh ( (2)b (t))+ cosh ( (2)b (t))

b =

(2) sinh ( (2)b (t))

c = (2) sinh ( (2)b (t))

d =(1 + )

2(2) sinh ( (2)b (t))+ cosh ( (2)b (t)) .

so that

t (, ) = 0

(a + b

c + d,

)= 0

(a + b

c + d

).

Recalling = zb and 2 = zbzd, coming back to the original variables, we get

t (zb, zd) = E(zNb(t)b z

Nd(t)d

)= 0

(At (zbzd) zb +Bt (zbzd) zbzdCt (zbzd) zb +Dt (zbzd)

)


where

At (zbzd) = 1 + 2 (zbzd)

sinh ( (zbzd) b (t)) + cosh ( (zbzd) b (t))

Bt (zbzd) =

(zbzd)sinh ( (zbzd) b (t))

Ct (zbzd) = 1 (zbzd)

sinh ( (zbzd) b (t))

Dt (zbzd) =1 +

2 (zbzd)sinh ( (zbzd) b (t)) + cosh ( (zbzd) b (t))

each depend on t and zbzd.

t asymptotics.This fully solvable case enables us to obtain exact asymptotics for the generatingfunction t (zb, zd) in the supercritical case < 1. From the explicit formulae for(a, b, c, d) we get:

limt

a (t)

c (t)=

1 + 2 (2)2

=2

1 + + 2(2) = lim

t

b (t)

d (t)

identified indeed by definition of (2). This illustrates the fact that the vectors

xi (t), i = 1, 2 become parallel up to exp (2b (t)) when t gets large. Note thatone still have of course ad bc = 1 (ad) / (bc) 1 = 1/ (bc) = O (exp (2b (t))).Indeed, it appears that a/c b/d = 1/ (cd) = O (exp (2b (t))) 0 (withoutbeing strictly = 0). Hence we get that the generating function yields

t (zb, zd) 0(1 + 2

2

), =

1

2

(1 + )

2 4zbzd.

Dependence on has disappeared. Observing that as t(17) t (zb, zd) = E

(zNb(t)b z

Nd(t)d

) eE

((zbzd)

Nd() | extinct),

because either extinction occurred (with probability e) andNb () = Nd ()


Note that, would one consider a subcritical case > 1, then h (1) = 1 leading toe = 0 (h (1)) = 1 (almost sure extinction) with E

(zNd()

)= 0 (h (z)) char-

acterizing the law of Nb () = Nd () < at extinction. Note E (Nd ()) =E (Nb (0))

1 .

In the critical case = 1, h (1) = 1 (e = 1) and E(zNd()

)= 0

(11 z)

with E (Nd ()) =.

Remark: This solvable case includes the situation where both b (t) and d (t)are independent of time. It suffices to particularize the latter formula while settingb (t) = bt.

If zb = z, zd = z1,

t(z, z1

)= E

(zN(t)

)= 0

(At (1) z +Bt (1)

Ct (1) z +Dt (1)

)which is the previous homographic expression for E

(zN(t)

)in the particular case

when d (t) = b (t).

If zb = z, zd = 1 or zb = 1, zd = z, we get the marginals

t (z, 1) = E(zNb(t)

)= 0

(zAt (z) +Bt (z)

Ct (z) z +Dt (z)

)or

t (1, z) = E(zNd(t)

)= 0

(At (z) +Bt (z) z

Ct (z) +Dt (z)

).

The processes (Nb (t) , Nd (t)) both are monotone non-decreasing.

If zb = z1, zd = z

2, we are interested in the Laurent series

(19) t (z) := E(z2Nd(t)Nb(t)

)= t

(z1, z2

)= 0

(At (z) z

1 +Bt (z) z

Ct (z) z1 +Dt (z)

).

The quantity t (z) = t(z1, z2

)turns out to be useful in the understanding

of the statistical properties of the first time that (t) := 2Nd (t) Nb (t) hits 0,starting from (0) = Nb (0) = N (0) < 0. We now address this point.

3.2.6. First time Nd (t) N (t) = Nb (t)Nd (t) ((t) 0). The process (t)takes values in Z. It moves up by 2 units when a death event occurs and downby 1 unit when a birth event occurs. Starting from (0) = Nb (0) < 0, theprocess (t) will enter the nonnegative region (t) 0 for the first time at somerandom crossing time (0) where the dependence on the initial condition has beenemphasized.

There are two types of crossing events:

- type-1: (t) enters the nonnegative region (t) 0 as a result of (t) = 1followed instantaneously by a death event leading to (t+) = +1.

- type-2: (t) enters the nonnegative region (t) 0 as a result of (t) = 2followed instantaneously by a death event leading to (t+) = 0.

Consider the process 1 (t) which is (t) conditioned on t = 0 being a type-1 firstcrossing time ( (0) = 1 and (0+) = +1). Let 1 (t) be the rate at which some


type-1 crossing occurs at t for 1. We let

1 (s) :=

0

est1 (t) dt

be the Laplace-Stieltjes transform (LST) of 1 (t) .

Let 1 denote the (random) time between 2 consecutive type-1 crossing times for1. By standard renewal arguments [5]

(20) 1 (s) = 1/(1 f1 (s)

),

where f1 (s) is the LST of the law of 1.

Similarly, consider the process 2 (t) which is (t) conditioned on t = 0 being atype-2 first crossing time ( (0) = 2 and (0+) = 0). Let 2 (t) be the rateat which some type-2 crossing occurs at t for 2. Let 2 denote the random time

between 2 consecutive type-2 crossing times for 2. Then 2 (s) = 1/(1 f2 (s)

)where 1 (s) is the LST of 2 (t) and f2 (s) the LST of the law of 2.

Clearly, between any two consecutive crossings of any type, there can be none, oneor more crossings of the other type.

The laws of the first return times (1, 2) of (1,2) are thus known once the rates(1 (t) , 2 (t)) are.

Let now 1,(0) (t) (respectively 2,(0) (t)) be the rate at which some type-1 (re-spectively type-2) crossing occurs at t when is now simply conditioned on any (0) < 0.

Let also 1,(0) (respectively 2,(0)) be the first time that some type-1 (respectivelytype-2) crossing occurs for given (0) < 0.

Again by renewal arguments for this now delayed renewal process:

(21) 1,(0) (s) = f1,(0) (s) /(1 f1 (s)

)and

2,(0) (s) = f2,(0) (s) /(1 f2 (s)

)where 1,(0) (s) (respectively 2,(0) (s)) is the LST of 1,(0) (t) (respectively

2,(0) (t)) and f1,(0) (s) (respectively f2,(0) (s)) is the LST of the law of 1,(0)(respectively 2,(0)). Finally, we clearly have

(22) (0) = inf(1,(0), 2,(0)

).

The knowledge of all these quantities (in particular the law of (0)) is thereforeconditioned on the knowledge of the rates. Conditioned on the initial condition (0) < 0, we have for instance

(23) 1,(0) (t) dt =[z1

]t (z) d (t) E(0) (Nd (t) + 1 | (t) = 1) dt

2,(0) (t) dt =[z2

]t (z) d (t) E(0) (Nd (t) + 2 | (t) = 2)dt,

corresponding respectively, through a jump of size 2 of (t), to the death events (t+ dt) = 1 given (t) = 1 and (t+ dt) = 0 given (t) = 2. In (23),t (z) = t

(z1, z2

)is given by (19).


The term[z1

]t (z) (respectively

[z2

]t (z)) is the probability that (t) = 1

(respectively (t) = 2).The term E (Nd (t) + 1 | (t) = 1) (respectively E (Nd (t) + 2 | (t) = 2)) isthe expected value of N (t) = Nb (t)Nd (t) given (t) = 1 (respectively of N (t)given (t) = 2).These last two informations can be extracted from the joint probability gener-ating function of (Nd (t) ,(t) := 2Nd (t)Nb (t) | (0)) which is known fromt (zb, zd) = E

(zNb(t)b z

Nd(t)d

).

Therefore, the law of (0) follows in principle, although the actual computationaltask left remains huge.

4. Age-structured models

We finally address a different point of view of the birth/death problems, once againdeterministic but now based on age-structured models (see e.g. [4], [3], [6] and [13]).

Let = (a, t) be the density of individuals of age a at time t. Then obeys

(24) t+ a = d (a, t) ,with boundary conditions

(a, 0) = 0 (a) and (0, t) =

b (a, t) (a, t) da.

The quantities b (a, t) and d (a, t) are the birth and death rates at age and time(a, t) and they are the inputs of the model together with 0 (a) . The equation (24)with its boundary conditions constitute the Lotka-McKendrick-Von Foerster model,[11].

The total population mean size at t is

x (t) =

(a, t) da.

With d/dt, integrating (24) with respect to age and observing (, t) = 0, we

get

.x (t) =

b (a, t) (a, t)da

d (a, t) (a, t) da =:

.xb (t) .xd (t)

where.xb (t) and

.xd (t) are the birth and death rates at time t. We note that

.xb (t) = (0, t), the rate at which new individuals (of age a = 0) are injected in thesystem.

Integrating, we have x (t) = xb (t)xd (t), with x (0) = xb (0) in view of d (a, 0) =0. xb (t) and xd (t) are the mean number of individuals ever born and ever deadbetween times 0 and t (xb (t) including, as before, the initial individuals). The an-swer to the question of whether and when the ever dead outnumbered the livingis transferred to the existence of the time t when some overshooting occurred,

defined by xd(t)x(t) = 1. As in Section 2, this requires the computation of both xd (t)

and x (t) and we now address this question.


With a t = min {a, t}, let

(25) d (a, t) =

at0

d (a s, t s) ds.

Integration of (24), using the method of characteristics yields its solution as

(26) (a, t) = 0 (a t) ed(a,t) if t < a

=.xb (t a) e

d(a,t) if t > a.

Recall 0 (a) = (a, 0) and.xb (t) = (0, t) as from the boundary conditions of (24).

We shall distinguish 3 cases.

4.1. Case: b (a, t) and d (a, t) independent of age a5. Suppose b (a, t) =

b (t) and d (a, t) = d (t) for some given time-dependent rate functions b (t) , d (t) .Then

.xb (t) = b (t)x (t) and

.xd (t) = d (t)x (t) . Supposing x (0) =

0 (a) da a.We thus obtain

xb (t) = x (0) +

t0

b (s)x (s) ds = x (0)

(1 +

t0

b (s) eb(s)d(s)ds

)xd (t) =

t0

d (s)x (s) ds = x (0)

t0

d (s) eb(s)d(s)ds

and for the age-structure

(a, t) = 0 (a t) ed(t) if t < a=

.xb (t a) e(d(t)b(ta)) if t > a,

which is completely determined because.xb (t) = x (0)d (t) e

b(t)d(t) is.

4.2. Case: b (a, t) and d (a, t) independent of time t6. Suppose b (a, t) =

b (a) and d (a, t) = d (a) for some given age-dependent rate functions b (a) , d (a) .Then

d (a, t) =

at0

d (a s) ds= d (a) d (a t) if t < a= d (a) if t > a.

5Not very realistic since it assumes that individuals may reproduce at any age and are poten-tially immortal eg no cutoff to b (a) for a < acrit or a > acrit, and d (a) does not increase witha.

6which amounts to assuming that no modification of the demographical dynamics ever occurs.


We thus obtain

(a, t) = 0 (a t) e(d(a)d(at)) if t < a=

.xb (t a) ed(a) if t > a

which is completely determined as well once.xb (t) = (0, t) is. We have

.xb (t) =

b (a) (a, t) da

=

t0

b (a).xb (t a) ed(a)da+

t

b (a) 0 (a t) e(d(a)d(at))da,

the sum of two convolution terms. Taking the Laplace transforms, with

(z) :=

0

ezab (a) ed(a)da and (z) :=

0

eza0 (a) ed(a)da,

we get the Laplace-transform of.xb (t) as

x.b (z) = x.b (z) (z) + (z) (z) ,

leading to

x.b (z) = (z) (z)

1 (z) .

Inverting the Laplace transform yields.xb (t). The solution of (24) is completely

determined from (26).

We also have

x (t) =

(a, t) da =

t0

.xb (t a) ed(a)da+

t

0 (a t) e(d(a)d(at))da

the sum of two convolution terms. Taking the Laplace transforms, with

0 (z) :=

0

ezaed(a)da,

we get

x (z) = x.b (z) 0 (z) + (z) 0 (z) .

So

(27) x (z) =0 (z) (z)

1 (z) =0 (z)

(z)x.b (z) .

Inverting this Laplace transform yields x (t).

4.3. The complete case. This is the realistic case when both b (a, t) and d (a, t)fully depend on (a, t) . Looking at

.x (t) =

b (a, t) (a, t)da

d (a, t) (a, t) da =:

.xb (t) .xd (t)

suggests to introduce the mean birth and death rates (averaging over age)

b (t) =

b (a, t) (a, t) da

(a, t) daand d (t) =

d (a, t) (a, t) da

(a, t)da.


Then, with x (t) = (a, t) da

.x (t) = (b (t) d (t))x (t) =:

.xb (t) .xd (t) .

Sox (t) = x (0) e

b(t)

d(t),

where b ,

d are the primitives of

b ,

d and

.xb (t) = x (0)

b (t) eb(t)

d(t),

.xd (t) = x (0)

d (t) eb(t)

d(t).

We are back to a case where b ,

d are independent of age although we know thatthere is an age dependency for both b (a, t) and d (a, t).

Suppose we are given (b (t) ,

d (t)) in the first place. Then.xb (t),

.xd (t) and x (t)

are known from Subsection 4.1. Furthermore (a, t) = (a, t) with

(a, t) = 0 (a t) e

d(t) if t < a

=.xb (t a) e(

d(t)

b(ta)) if t > a

is the known age-dependent density.

So we need to solve the following inverse problem: find the unknown functionsb (a, t) and d (a, t) such that

b (a, t) (a, t) da =

.xb (t) and

d (a, t)

(a, t) da =.xd (t)

which is a linear problem (not in the convolution class). Then (xb (t) , xd (t) , x (t))will constitute a solution consistent with (24) and the rates b (a, t) , d (a, t) .

We briefly sketch another point of view:

Suppose we were given d (a, t) and.xb (t) = (0, t) in the first place.

Then (a, t) is completely known from Equations (25) and (26), together withx (t) =

(a, t) da and

.xd (t) =

d (a, t) (a, t) da. Also xb (t) is known from

xb (0) = x (0) = (a, 0) da.

Now.xb (t) =

b (a, t) (a, t) da where now b (a, t) has to be determined. We thus

need to solve the inverse problem

.xb (t) =

t0

b (a, t).xb (t a) e

d(a,t)da+

t

b (a, t) 0 (a t) ed(a,t)da,

where the unknown function is b (a, t) . Suppose this inverse problem was solved.Then (xb (t) , xd (t) , x (t)) constitute a solution consistent with (24) and the ratesb (a, t) , d (a, t) .

Acknowledgments: T.H. acknowledges partial support from the labex MME-DII(Mode`les Mathematiques et Economiques de la Dynamique, de l Incertitude et desInteractions), ANR11-LBX-0023-01.


References

[1] Bharucha-Reid, A. T. Elements of the theory of Markov processes and their applications.Reprint of the 1960 original. Dover Publications, Inc., Mineola, NY, 1997. xii+468 pp.

[2] Cohen, J. E. Ergodic theorems in demography. Bull. Amer. Math. Soc. (N.S.) 1 (1979), no.2, 275-295.

[3] Cushing, J. M. An introduction to structured population dynamics. CBMS-NSF RegionalConference Series in Applied Mathematics, 71. Society for Industrial and Applied Mathe-matics (SIAM), Philadelphia, PA, 1998. xiv+193 pp.

[4] Charlesworth, B. Evolution in age-structured populations. Cambridge Studies in Mathemat-ical Biology, 1. Cambridge University Press, Cambridge-New York, 1980. xiii+300 pp.

[5] Feller, W. An introduction to probability theory and its applications. Vol. II. John Wiley &Sons, Inc., New York-London-Sydney 1966 xviii+636 pp.

[6] Hoppensteadt, F. C. Mathematical methods of population biology. Courant Institute of Math-ematical Sciences, New York University, New York, 1977. v+167 pp.

[7] Howard J. World Population Explained: Do Dead People Out-number Living, Or Vice Versa? Huffington Post, available athttp://www.huffingtonpost.com/2012/11/07/world-population-explained n 2058511.html.

[8] Huillet, T.; Monin, A.; Salut, G. Minimal realizations of the matrix transition Lie group forbilinear control systems: explicit results. Systems Control Lett. 9 (1987), no. 3, 267-274.

[9] Karlin, S.; McGregor, J. Linear growth birth and death processes. J. Math. Mech., (1958),no 7, 643-662.

[10] N. Levinson. The asymptotic nature of solutions of linear systems of differential equations.Duke Math. J. (1948), no 15, 111-126.

[11] McKendrick, A.G. Applications of mathematics to medical problems, Proc. Edinb. Math.Soc. (1926), no 44, 98-130.

[12] Ohkubo. J. Karlin-McGregor-like formula in a simple time-inhomogeneous birth-death pro-cess. Preprint: arXiv:1404.2703, (2014).

[13] Webb, G. F. Theory of Age Dependent Population Dynamics, Marcel Dekker, New York,1985.

CNRS, UMR-8089 and University of Cergy-Pontoise, 2, rue Adolphe Chauvin F-95302,Cergy-Pontoise, Cedex, FRANCE, E-mail(s): [email protected], [email protected],[email protected]

1. Introduction2. Simple birth and death models for population growth (deterministic)2.1. Constant rates (Malthusian model)2.2. Time-dependent rates

3. Birth and death Markov chain (stochasticity)3.1. The current population size N( t) 3.2. The joint law of Nb( t) and Nd( t)

4. Age-structured models4.1. Case: 0=x"0115b( a,t) and 0=x"0115d( a,t) independent of age aNot very realistic since it assumes that individuals may reproduce at any age and are potentially immortal eg no cutoff to 0=x"0115b( a) for aacrit, and 0=x"0115d( a) does not increase with a.4.2. Case: 0=x"0115b( a,t) and 0=x"0115d( a,t) independent of time twhich amounts to assuming that no modification of the demographical dynamics ever occurs.4.3. The complete case

References

Number of Dead People vs Living

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