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University of Texas at Austin
HW Assignment 5
Problem 5.1. Let {Zn}n∈N 0 be a branching process with the offspring distribution given by
p 0 = 0. 6 , p 1 = 0. 3 , p 2 = 0. 05 , p 3 = 0. 05 , pn = 0, n > 3. (1) Compute the pmf of Z 2. (2) Find the extinction probability.
(Note: use Mathematica; the command Expand may be useful. The numbers are not chosen so that the results are nice. In fact, they are not!)
Solution:
(1) The generating function of Z 2 is P (P (s)), where P (s) = 0.6 + 0. 3 s + 0. 05 s^2 + 0. 05 s^3 , is the generating function of the offspring distribution. Mathematica gives In[43]:= P@s_D^ =^ 0.6^ +^ 0.3^ s^ +^ 0.05^ s ^ 2^ +^ 0.05^ s ^ 3;
In[47]:= Expand@P@P@sDDD
Out[47]= 0.8088^ +^ 0.1242 s^ +^ 0.0333 s^2 +^ 0.02625 s^3 +^ 0.005225 s^4 + 0.0014875 s^5 + 0.00058125 s^6 + 0.00013125 s^7 + 0.00001875 s^8 + 6.25 ´ 10 -^6 s^9 Therefore, the pmf of Z 2 is p 0 = 0. 8088 , p 1 = 0. 1242 , p 2 = 0. 0333 , p 3 = 0. 02652 , p 4 = 0. 05225 , p 5 = 0. 0014875 , p 6 = 0. 00058125 , p 7 = 0. 00013125 , p 8 = 0. 00001875 , p 9 = 6. 25 × 10 −^6 , pn = 0, n > 9. (2) The extinction probability is a solution of the extinction equation P (s) = s, s ∈ [0, 1]. We use Mathematica again: In[48]:= Solve@P@sD^ ^ s, sD Out[48]= 88 s ® -4.60555<, 8 s ® 1.<, 8 s ® 2.60555<< The smallest (and only) solution in [0, 1] is s = 1, so we conclude that the extinction probability is equal to 1.
Problem 5.2. What happens when the population starts from k individuals instead of one? Let {Zn}n∈N 0 be a stochastic process constucted in the same manner as a simple branching process:
Zn+1 =
∑^ Zn
k=
Zn,k, n ∈ N 0 ,
where {Zn,k}n,k≥ 1 is a collection of independent random variables, all having the offspring distribution with pmf {pn}n∈N 0. The only diffrence is that Z 0 = k, where k ∈ N is fixed and non-random, instead of Z 0 = 1. Like always, let P be the generating function of {pn}n∈N 0.
(1) What is the generating function of Z 1? (2) What is the generating funcion of Z 2? Of Zn, for a general n ∈ N? (3) What does the extinction equation look like? How is the extinction probability for the classical case k = 1 related to the extinction probability for a general k ∈ N?
Solution:
(1) We can write Z 1 as a sum of k independent random variables
Z 1 =
∑^ k
j=
Z 0 ,j ,
and each Z 0 ,j has a pmf with generating function P. Therefore (since sums of independent random variables correspond to products in the world of generating functions) we have PZ 1 = [P (s)]k. (2) For n = 2, we have
Z 2 =
∑^ Z^1
j=
Z 1 ,j ,
so PZ 2 (s) = PZ 1 (P (s))] = [P (P (s))]k.. Similarly, PZn (s) = [P (P (... P (s)... ) ︸ ︷︷ ︸ n Ps
]k.
(3) The same ideas as in the case k = 1 give us P[E] = lim n PZn (0) = lim n
[P (P (... P (0)... )
n Ps
]k.
Therefore, (P[E])^1 /k^ = lim n P (P (... P (s)... ) ︸ ︷︷ ︸ n Ps
The limit on the right-hand side above is nothing but the extinction probability for the case k = 1. Therefore, P[E]^1 /k^ satisfies the extinction equation s = P (s), and so, the extinction equation in this case is s^1 /k^ = P (s^1 /k). By the equation above, solutions of this equation are merely kth^ powers of solutions to the classical extinction equation s = P (s). In particular, the extinction probability for general k is the kth^ power of the extinction probability for k = 1. If you think about it, it makes a lot of sense: each individual in the first generation starts his/her own population and the extinction of the whole process amounts to extinctions of k independent populations with offspring distribution given by P. The extinction probability of each of them is P[E], so the extinction probability for the whole process must be P[E]k.
Problem 5.3. In a branching process, the offspring distribution is given by its characteristic function P (s). Define T = inf{n ∈ N : Zn = 0} to be the extinction time of the population (note that it may be possible that T = +∞).
(1) Write an expression for P[T = n], for n ≥ 1, in terms of the function P (and its iterates P (m), m > 1). (2) Suppose that P (s) = p + qs (^) (5.1)
where p, q ∈ (0, 1). Compute P[T = n], n ∈ N 0 , in this case. (3) (Optional, but may appear on the exam!) Enhance the standard branching process model with P given by (5.1), by assuming that Z˜ 0 = i for some i ∈ N, i.e., allow for more than one individual in the initial generation and assume that they are all identical. Again, denote the total number of individuals in the nth^ generation of this enhanced branching process by Z˜n for every n ∈ N, and define the stopping time T˜ = inf{n ∈ N : Z˜n = 0}. Find P[ T˜ = n] for any n ∈ N for this enhanced branching process.
Solution:
After multiplying both sides by P[B] > 0, the equality above turns out the be equivalent to
P[C ∩ B] =
∑^ ∞
k=
P[C ∩ B ∩ Ak],
which it true by additivity of probability and the fact that C ∩ B = ∪∞ k=1C ∩ B ∩ Ak, and that the union is disjoint.
