## Sunday, April 20, 2008

### Children and siblings

Suppose the fertility rate in a given society is R. If one makes a poll among the adult population asking for the number of siblings each person has, including themselves, the estimated number, let us call it S, will certainly not be R: for one, women without children are nobody's mothers, so they are not affecting the poll result, unlike in the calculation of R. The question arises, what is the relationship between R and S? As it turns out, S is not a simple function of the fertility rate, but it depends on the exact fertility distribution.

To simplify things, let us only deal with mother-to-children relationships, i.e. if we say that two persons are siblings we mean that they are born to the same mother. Let r(n) be the probability that an arbitrary woman has n children; the fertility rate R is then the expected value of the fertility distribution r(n), that is, R = ∑ n·r(n). We now define:

s(n) = P(an arbitrary individual has n siblings, including herself)

(P stands for "probability"). Let us calculate s(n). Suppose we have a group of M women who collectively will give birth to N individuals during their lifetimes. We have then:

N = ∑ n·r(nM = RM,

with the individuals being distributed as follows:

r(1)·M individuals with 1 sibling (counting themselves),
2·r(2)·M individuals with 2 siblings (counting themselves),
r(nM individuals with n siblings (counting themselves)...

So, s(n) is simply the number of individuals in the n-th group, divided by N:

s(n) = n·r(nM/N = n·r(n)/R.

And S, the expected value of s(n), is

S = ∑ n2·r(n)/R,

which expresses the relationship between r(n) and S that we were looking for. Using Cauchy's Inequality and the fact that ∑ r(n) = 1, we have:

R2 = (∑ n·r(n))2 = (∑ (n·√r(n)) · √r(n))2
≤ (∑ n2r(n))(∑ r(n)) = ∑ n2r(n),

hence S = ∑ n2·r(n)/RR for every possible r(n).

A particularly simple case arises when the fertility follows a Poisson distribution with r(n) = RneR/n!. Simple formula manipulations show that in this situation r(n) = s(n + 1); hence S = ∑n ≥ 1 n·s(n) = n ≥ 1 n·r(n − 1)= n ≥ 0 n·r(n) + ∑n ≥ 0 r(n) = R + 1. The figure depicts a Poisson-distributed r(n) with R = 2.1, along with its associated s(n). 