In this chapter we will consider a random model for population growth in the absence of spatial or any other resource constraints. So, consider a population of individuals which evolves according to the following rule: in every generation n = 0,1,2, . . ., each individual produces a random number of offspring in the next generation, independently of other individuals. The probability mass function for offspring is often called the offspring distribution and is given by
pi =P(number of offspring =i),
for i= 0,1,2, . . .. We will assume that p0 < 1 andp1 < 1 to eliminate the trivial cases. This model was introduced by F. Galton in the late 1800s to study the disappearance of family names;
in this case pi is the probability that a man has isons.
We will start with a single individual in generation 0 and generate the resulting random family tree. This tree is either finite (when some generation produces no offspring at all) or infinite — in the former case, we say that the branching processdies outand, in the latter case, that it survives.
We can look at this process as a Markov chain where Xn is the number of individuals in generationn. Let us start with the following observations:
• IfXn reaches 0, it stays there, so 0 is an absorbing state.
• Ifp0>0, P(Xn+1 = 0|Xn=k)>0, for all k.
• Therefore, by Proposition 13.5, all states other than 0 are transient if p0 >0; the popu-lation must either die out or increase to infinity. If p0 = 0, then the population cannot decrease and each generation increases with probability at least 1−p1, therefore it must increase to infinity.
It is possible to write down the transition probabilities for this chain, but they have a rather complicated explicit form, as
P(Xn+1 =i|Xn=k) = P(W1+W2+. . .+Wk=i),
whereW1, . . . , Wk are independent random variables, each with the offspring distribution. This suggests using moment generating functions, which we will indeed do. Recall that we are as-suming that X0 = 1.
Let
δn=P(Xn= 0)
be the probability that the population is extinct by generation (which we also think of as time)n.
The probabilityπ0 that the branching process dies out is, then, the limit of these probabilities:
π0 =P(the process dies out) =P(Xn = 0 for somen) = lim
n→∞P(Xn= 0) = lim
n→∞δn.
Note thatπ0 = 0 ifp0 = 0. Our main task will be to computeπ0 for general probabilitiespk. We start, however, with computing the expectation and variance of the population at generationn.
Let µand σ2 be the expectation and the variance of the offspring distribution, that is, µ=EX1 =
X∞
k=0
kpk, and
σ2 = Var(X1).
Let mn = E(Xn) and vn = Var(Xn). Now, Xn+1 is the sum of a random number Xn of independent random variables, each with the offspring distribution. Thus, we have by Theorem 11.1,
mn+1 =mnµ, and
vn+1=mnσ2+vnµ2.
Together with initial conditions m0 = 1,v0 = 0, the two recursive equations determine mn and vn. We can very quickly solve the first recursion to get mn=µn and so
vn+1=µnσ2+vnµ2.
When µ = 1, mn = 1 and vn = nσ2. When µ 6= 1, the recursion has the general solution vn=Aµn+Bµ2n. The constantA must satisfy
Aµn+1 =σ2µn+Aµn+2, so that,
A= σ2 µ(1−µ).
Fromv0 = 0 we getA+B = 0 and the solution is given in the next theorem.
Theorem 14.1. Expectation mn and variance vn of the nth generation count.
We have
mn=µn and
vn=
(σ2µn(1−µn)
µ(1−µ) if µ6= 1, nσ2 if µ= 1.
We can immediately conclude thatµ <1 impliesπ0= 1, as P(Xn6= 0) =P(Xn≥1)≤EXn=µn→0;
if the individuals have less than one offspring on average, the branching process dies out.
Now, let φ be the moment generating function of the offspring distribution. It is more convenient to replaceet in our original definition withs, so that
φ(s) =φX1(s) =E(sX1) =
∞
X
k=0
pksk.
In combinatorics, this would be exactly the generating function of the sequencepk. Then, the moment generating function of Xnis
φXn(s) =E[sXn] =
∞
X
k=0
P(Xn=k)sk.
We will assume that 0≤s≤1 and observe that, for suchs, this power series converges. Let us get a recursive equation forφXn by conditioning on the population count in generation n−1:
φXn(s) = E[sXn]
= X∞
k=0
E[sXn|Xn−1=k]P(Xn−1 =k)
= X∞
k=0
E[sW1+...+Wk]P(Xn−1 =k)
= X∞
k=0
(E(sW1)E(sW2)· · ·E(sWk)P(Xn−1 =k)
= X∞
k=0
φ(s)kP(Xn−1 =k)
= φXn−1(φ(s)).
So,φXn is the nth iterate ofφ,
φX2(s) =φ(φ(s)), φX3(s) =φ(φ(φ(s))), . . . and we can also write
φXn(s) =φ(φXn−1(s)).
Next, we take a closer look at the properties of φ. Clearly, φ(0) =p0 >0 and
φ(1) = X∞
k=0
pk= 1.
Moreover, for s >0,
φ′(s) = X∞
k=0
kpksk−1 >0,
soφis strictly increasing, with
φ′(1) =µ.
Finally,
φ′′(s) =
∞
X
k=1
k(k−1)pksk−2 ≥0, soφis also convex. The crucial observation is that
δn=φXn(0),
and so δn is obtained by starting at 0 and computing the nth iteration of φ. It is also clear thatδnis a nondecreasing sequence (because Xn−1 = 0 implies thatXn= 0). We now consider separately two cases:
• Assume that φ is always above the diagonal, that is, φ(s) ≥ s for all s ∈ [0,1]. This happens exactly whenµ=φ′(1)≤1. In this case, δn converges to 1, and so π0 = 1. This is shown in the right graph of the figure below.
• Now, assume that φ is not always above the diagonal, which happens when µ > 1. In this case, there exists exactly one s < 1 which solves s = φ(s). As δn converges to this solution, we conclude that π0 < 1 is the smallest solution to s =φ(s). This is shown in the left graph of the figure below.
1 1
1 1
δ2
φ φ
δ1 δ2 δ1
The following theorem is a summary of our findings.
Theorem 14.2. Probability that the branching process dies out.
If µ≤1, π0 = 1. If µ >1, thenπ0 is the smallest solution on[0,1] tos=φ(s).
Example 14.1. Assume that a branching process is started with X0 =k instead of X0 = 1.
How does this change the survival probability? The kindividuals all evolve independent family trees, so that the probability of eventual death is π0k. It also follows that
P(the process ever dies out|Xn=k) =π0k
for everyn.
Ifµis barely larger than 1, the probabilityπ0 of extinction is quite close to 1. In the context of family names, this means that the ones with already a large number of representatives in the population are at a distinct advantage, as the probability that they die out by chance is much lower than that of those with only a few representatives. Thus, common family names become ever more common, especially in societies that have used family names for a long time. The most famous example of this phenomenon is in Korea, where three family names (Kim, Lee, and Park in English transcriptions) account for about 45% of the population.
Example 14.2. Assume that
pk=pk(1−p), k= 0,1,2, . . . .
This means that the offspring distribution is Geometric(1−p) minus 1. Thus, µ= 1
1−p −1 = p 1−p
and, ifp≤ 12,π0= 1. Now suppose that p > 12. Then, we have to compute φ(s) =
X∞
k=0
skpk(1−p)
= 1−p 1−ps.
The equationφ(s) =s has two solutions,s= 1 ands= 1−pp. Thus, when p > 12, π0 = 1−p
p .
Example 14.3. Assume that the offspring distribution is Binomial(3,12). Computeπ0. As µ= 32 >1,π0 is given by
φ(s) = 1 8+ 3
8s+3 8s2+ 1
8s3 =s, with solutions s= 1, −√
5−2, and√
5−2. The one that lies in (0,1), √
5−2≈0.2361, is the probability π0.
Problems
1. For a branching process with offspring distribution given byp0 = 16, p1 = 12, p3= 13,determine (a) the expectation and variance ofX9, the population at generation 9, (b) the probability that the branching process dies by generation 3, but not by generation 2, and (c) the probability that
the process ever dies out. Then, assume that you start 5 independent copies of this branching process at the same time (equivalently, change X0 to 5), and (d) compute the probability that the process ever dies out.
2. Assume that the offspring distribution of a branching process is Poisson with parameter λ.
(a) Determine the expected combined population through generation 10. (b) Determine, with the aid of computer if necessary, the probability that the process ever dies out forλ= 12,λ= 1, and λ= 2.
3. Assume that the offspring distribution of a branching process is given by p1 =p2 =p3 = 13. Note that p0 = 0. Solve the following problem for a = 1,2,3. Let Yn be the proportion of individuals in generation n (out of the total number of Xn individuals) from families of size a. (A family consists of individuals that are offspring of the same parent from the previous generation.) Compute the limit of Yn asn→ ∞.
Solutions to problems
1. For (a), compute µ= 32,σ2 = 72 −94 = 54, and plug into the formula. Then compute φ(s) = 1
6+ 1 2s+1
3s3. For (b),
P(X3 = 0)−P(X2= 0) =φ(φ(φ(0)))−φ(φ(0))≈0.0462.
For (c), we solveφ(s) =s, 0 = 2s3−3s+ 1 = (s−1)(2s2+ 2s−1), and soπ0= √32−1 ≈0.3660.
For (d), the answer isπ05. 2. For (a),µ=λand
E(X0+X1+. . .+X10) =EX0+EX1+. . .+EX10= 1 +λ+· · ·+λ10= λ11−1 λ−1 , ifλ6= 1, and 11 ifλ= 1. For (b), ifλ≤1 then π0 = 1, but ifλ >1, thenπ0 is the solution for s∈(0,1) to
eλ(s−1) =s.
This equation cannot be solved analytically, but we can numerically obtain the solution forλ= 2 to getπ0≈0.2032.
3. Assuming Xn−1 = k, the number of families at time n is also k. Each of these has, in-dependently, a members with probability pa. If k is large — which it will be for large n, as the branching process cannot die out — then, with overwhelming probability, the number of children in such families is aboutapak, whileXnis about µk. Then, the proportion Yn is about
apa
µ , which works out to be 16, 13, and 12, fora= 1, 2, and 3.