11 13 Branching processes in discrete time

Question 1

look at notes

Answer 1

notes done?

Question 2

Branching processes in discrete time:Introduction
assumptions

Answer 2

Assume each individual of a population lives a fixed time (one generation) and produces at the end of its life a #offspring that is a RV with
some probability pₖ
#offspring for each is i.i.d
thus defines a branching process in discrete time.

All Markov processes are memoryless

Question 3

Branching processes in discrete time: offspring distribution

Answer 3

{pₖ}ₖ₌₀ ∞

collection of all pₖ

P {X = k} ≡ Probability{X = k} = pk, k = 0, 1, 2, . . .
where p_k = P {X = k}

Question 4

All Markov processes are memoryless

Answer 4

the number of individuals in generation n + 1, called
Zₙ₊₁, depend on Zₙ (number of individuals in generation n), but not on the past history of
the process Z0,Z1, · · · ,Zₙ₋₁(number of individuals in each generation 0 to n − 1).

Question 5

Branching processes in discrete time:

Zₙ - #individuals in the nth generation

The #individuals in the (n+1)th generation?

Answer 5

The #individuals in the (n+1)th generation written as the sum of Zₙ independent RVs:
Zₙ₊₁ = X₁+X₂+X₃+..+X_{Zₙ}

{Zₙ independent Rvs in this sum}

X_i are iid
X_i represent #offspring of the ith member of the nth generation, independent realisations

(Markov property is satisfied- this is also a RV)

Question 6

Branching processes in discrete time: Extinction of the population when?

Answer 6

if Z_n = 0 ⇒
Z_m = 0 for all m > n

if p₀≠ 0: one individual can have no offspring
extinction happens if all individuals in one gen have no offspring

Question 7

Branching processes in discrete time: The mean
assume Z_0 given

Answer 7

Z₀=1 define the mean number of offspring of one individual i of the population in a given generation as µ

µ ≡ E(Xᵢ) = E(X)
= Σ{ₖ₌₀,..∞} kpₖ = p₁ + 2p₂ + 3p₃ + · · ·

idd for every individual i of every generation n, thus
E(Xᵢ) = E(X) and
the mean number of offspring of one individual is µ

Question 8

Branching processes in discrete time: The mean number of individuals in generation n+1

Answer 8

E(Zₙ₊₁)
= E(ΣXᵢ) (for i=1,…,Zₙ)
= E(X₁) +…+E(X_{Zₙ})
= µ +…µ = µ E[Zₙ]
(Zₙ terms)

Zₙ is RV taking values k=0,1,….,E(Zₙ₊₁) also infinite sum of terms kµ with weights p(Zₙ=k) probability of k individuals in gen n having on average µ offspring:

E(Zₙ₊₁) = Σ{ₖ₌₀,..∞} kµP{Zₙ=k}
= µ E(Zₙ)

iterating:
E(Zₙ) = µE(Zₙ₋₁) = µµE(Zₙ₋₂)=…
=µⁿ⁻¹ µE(Z₀) = Z₀µⁿ = Z₀ exp(n lnu)

Question 9

Branching processes in discrete time: The mean number of individuals in generation n+1 used to show exponential function:

Answer 9

iterating
E(Zₙ) = µE(Zₙ₋₁) = µµE(Zₙ₋₂)=…
=µⁿ⁻¹ µE(Z₀) = Z₀µⁿ = Z₀ exp(n lnu)

Thus: mean number of individuals increases exponentially as a function
of n if µ > 1, and decreases exponentially with n if µ < 1

Question 10

Branching processes in discrete time: Extinction of the population looking into this

probability that the population is not extinct by generation n is

Answer 10

If Zₙ individuals in gen n then the probability that the pop goes extinct in gen n+1 is p₀^{Zₙ}
(Zₙ is an RV so hard to find)
However, if Zₙ=0 for some n then Zₘ=0 for all m>n
Thus the probability that the population goes extinct by generation n (i.e. at, or before, generation n) is
0 ≤ P { population is extinct by n} = P {Zn = 0}.

probability that the population is not extinct by generation n is

0 ≤ P { population is not extinct by n} =
P {Zₙ ≥ 1}
= P {Zₙ = 1}+P {Zₙ = 2}+P {Zₙ = 3}+· · · .

Question 11

Branching processes in discrete time: Comparing probabilities of extinction to E(Zₙ)

Answer 11

E(Zₙ)
= P{Zₙ=1}+2P{Zₙ=2}+3P{Zₙ=3}+…

Thus
probability that the population is not extinct by generation n is
0 ≤ P { population is not extinct by n } ≤ E(Zₙ).
We know that E(Zₙ) = Z₀µⁿ

Question 12

Branching processes in discrete time:Conditions

0 ≤ P { population is not extinct by n } ≤ E(Zₙ) = Z₀µⁿ

Answer 12

If µ < 1 POPULATION EXTINCT sooner or later/guaranteed

P {Zₙ ≥ 1} ≤ µⁿ
P { population not extinct } → 0 as n → ∞

If µ > 1 then extinction is not guaranteed, but is still possible

This means that in this case, the population can go extinct or keep growing with a certain probability. Some realisations of the population obeying the rules of the branching process go extinct, but some never do and keep growing.

Question 13

Branching processes in discrete time: notation the probability uₙ that the population consisting initially of a single individual Z₀ = 1 is extinct by generation n

Answer 13

uₙ ≡ P {Zₙ= 0|Z₀ = 1}.

Question 14

Branching processes in discrete time: finding the probability uₙ that the population consisting initially of a single individual Z₀ = 1 is extinct by generation n

uₙ ≡ P {Zₙ= 0|Z₀ = 1}.

Answer 14

u₀=0
u₁ =p₀
(if p₀=0 no pop extinction ever, so for this analisis assume 0<p₀<1)

uₙ prob of extinction by time for gen n and is an increasing function of n as it includes the possible extinction events at 1,2,..,n

Question 15

Branching processes in discrete time:
uₙ ≡ P {Zₙ= 0|Z₀ = 1}.
uₙ = φ(uₙ₋₁ ), with u₀ = 0,

how we found

Answer 15

k individuals when n = 0: probability that whole pop dies out before the nth gen is the prob that, independently k sub-families starting with one individual die out
P {Zₙ = 0|Z₀ = k} = uₙᵏ

if k individuals when n=1 we have
P {Zₙ = 0|Z₀ = k} =P {Zₙ₋₁ = 0|Z₀ = k} = uₙ₋₁ᵏ
If Z₀ =1
P {Z₁ = k|Z₀ = 1} = P {X = k} = pₖ

can then write expression of u_n in terms of u_{n-1} summing over all poss values of k which links to pgf of X

Question 16

Branching processes in discrete time: Example

Suppose that p0 = 0.2, p1 = 0.4 and p2 = 0.4.
Then φ(z) =
.

Answer 16

φ(z) = 0.2 + 0.4z + 0.4z^2
,
u1 = 0.2,
u2 = φ(u1) = φ(0.2) = 0.2 + 0.4 × 0.2 +0.4 × 0.2^2

for increasing n: 0,1,2,3,4,5
un
u0=0
u1= φ(u0)=0.296
u2=0.353…

Question 17

Branching processes in discrete time:

Is extinction the ultimate fate of the population?

Answer 17

*uₙ is an increasing function of n when p_0 > 0

*As n → ∞, uₙ approaches a limit u (with 0 ≤ u ≤ 1) which is the probability that extinction ever occurs,
and satisfies
u = φ(u)

the probability u that the pop extinction ever occurs is a fixed point! solutions s.t 0 ≤ u ≤ 1.
from pgf properties φ(1) = 1 is always a solution
u=1 always a fixed point

Question 18

Branching processes in discrete time:

Is extinction the ultimate fate of the population?

show u=1 is a unique fixed point

Answer 18

extinction certain
u = 1 is the unique fixed point of φ when φ’(1) ≤ 1
and it corresponds to population extinction with probability Hence,
population extinction is certain when φ’(1) ≤ 1.
diagram: φ(u) against u increases and meets line y=u

population extinction possible but not certain
when φ’(1) > 1 , there is another solution u∞= φ(u∞)
giving a fixed point
0 < u∞ < 1 (stable as |φ’(u∞)|<1) and u=1 unstable as because |φ’(1)| > 1)

diagram: φ(u) against u increases and meets line y=u twice, once at u∞ then at 1

Question 19

Extinction summary:

Answer 19

*The mean number of offspring from one individual is µ = φ’(1).

• If µ ≤ 1 then extinction is guaranteed: u_n → 1 as n → ∞.
  The entire population goes extinct, sooner or later.

*If µ > 1 then extinction is not guaranteed (but still possible):
un → u∞ as n → ∞,
where 0 < u∞ < 1 and satisfies u∞ = φ(u∞).

*In a fraction u∞ of all realisations of the branching process, the population goes extinct, and in the other realisations (a fraction 1 − u∞ of them) there is no extinction and the population grows unbounded

Question 20

Example. Suppose that p0 = 0.2, p1 = 0.4 and p2 = 0.4. Then, φ(z) =
extinction?

Answer 20

φ(z) = 0.2 + 0.4z +0.4z²
φ’(z) = 0.4 + 0.8z, and the solution to is
u = 0.2 + 0.4u + 0.4u²
0 = 0.2 − 0.6u + 0.4u²
⇒ two fixed points: u = 0.5, u = 1.
Here, µ = φ’(1) = 0.4 + 0.8 = 1.2 > 1, and hence u = 0.5 is stable and gives the extinction probability ⇒ probability that the population ever goes extinct is u∞ = 0.5

Question 21

Interested in more general branching processes in discrete time: what do we need for a complete understanding of a discrete-time branching process describing the number of individuals in a population?

Answer 21

pdf of each of the RVs Z_n

giving the number of individuals in the nth generation of the
population. These can be obtained from the probability generating functions for Zn,

Question 22

pgf for discrete time branching processes: Deriving the pgf for the #individuals Zₙ in each generation n
Z_0
Z_1

Answer 22

we use initial condition Z₀=1 assuming single individual in the population and assume each individual in all generations leave behind X offspring with pgf of X φ(z)

• If Z₀=1 : Z₁ has the same probability distribution as X pgf φ(z)
• pgf for each Zₙ:
  gₙ(z) = ΣP(Z ₙ=k)zᵏ = E(Z^{Zₙ})
  (k=0,…,∞)

*Z₀=1
g₀(z)=z or δₖ,₁zᵏ
where δₖ,₁ kronecker delta function

g₁(z) = φ(z) as same probabilities Z₁=k as X=k as its the initial individual

Question 23

δₖ,₁ kronecker delta function

Answer 23

δₖ,ₘ = 1 if k=m
=0 if k/= m

Question 24

pgf for discrete time branching processes: Deriving the pgf for the #individuals Zₙ in each generation n

Zₙ₊₁
related to Zₙ?

Answer 24

Zₙ₊₁ = Σ Xᵢ from i=1 to Z ₙ £
each parent i contributing to pop in Zₙ₊₁
RV sum of RVs
so pgf:
gₙ₊₁ (z) = E(z^{Zₙ₊₁})
= E(z^(£))
= Σ_{k=0,∞} P(Zₙ=k)E(z^{Σ {i=1,k}Xᵢ})
= Σ{k=0,∞} P(Zₙ=k)E(z^X₁)….E(z^Xₖ)
=Σ_{k=0,∞} P(Zₙ=k) (φ(z)ᵏ) = gₙ(φ(z))
** thus we have the recurrence relation**
gₙ₊₁(z)=gₙ(φ(z)) for n ≥ 0, with g₀(z) = z

Question 25

pgf for discrete time branching processes: Deriving the pgf for the #individuals Zₙ in each generation summary pgf for Z_n

Answer 25

g₁(z)=φ(z)
g₂(z)= g₁(φ(z)) = φ(φ(z))
….
gₙ(z)= φ(….φ(φ(z)…) (n-1 compositions)
= φ(gₙ₋₁)

gₙ₊₁(z)=φ(gₙ(z))
meaning we also have
gₙ₊₁(z) = φ(gₙ(z)) for n ≥ 0, with g₀(z) = z,

Question 26

pgf for discrete time branching processes: Deriving the pgf for the #individuals Zₙ in each generation WHEN Z_0=k>1

Answer 26

Each individual of the initial gen produces an independent family
Z_n in generation n is a sum of k identically distributed RVs

E(z^{Z_n} | Z₀=k)
=(E(z^{Zₙ}| Z₀=1))ᵏ
= (gₙ(z))ᵏ
(gₙ(z) is pgf of Zₙ when Z₀=1

This means that it
suffices to focus on the case with Z₀ = 1 to obtain gₙ(z) and then simply
use above to find the probability generating function of Zₙ if initially Z₀ = k>1

Question 27

Example. If the probabilities of having no children, one child and two children are
respectively p0, p1 and p2, then how many grandchildren can you have and with what
probability?

Answer 27

Let X be #children and let Z₂ be #grandchildren
g₂(z) be the pgf of Z₂.
assume p₀+ p1 + p₂ = 1,
with pₙ = 0 for n /∈ {0, 1, 2}.

pgf of X :** φ(z) = p₀ + p₁z + p₂z²**
pgf of Z₂:
g₂(z) = φ(φ(z))
= p₀ + p₁φ(z) + p₂φ(z)²
= p₀ + p₁(p₀ + p₁z + p₂z²)+ p₂(p₀ + p₁z + p₂z²)
=p₀ + p₀p₁ + p₀²p₂ + (p₁² + 2p₂p₀p₁)z²+ (2p₂²p₁)z³
+(p₂³)z⁴

defn of g₂(z) ≡
sum_k [ P {Z₂ = k} zᵏ

P {Z₂ = 0} = P { zero grandchildren } = p₀ + p₀p₁ + p₀²p₂
P {Z₂ = 1} =P{1 grandchild} = p₁² + 2p₂p₀p₁
P {Z₂ = 2} = P {2 grandchildren } = p₁p₂ + pp₁² + 2p₂²p₀
P{Z₂=3} =P{3 grandchildren } = 2p₂²p₁
P {Z₂ = 4} = P{ 4 grandchildren } = p₂³

{z }

Question 28

pgf for discrete time branching processes: summary part 2

Answer 28

We have obtained two equivalent forms for the recurrence relation giving the pgf of Zₙ₊₁, with
g₀(z) = z (Z₀ = 1):

gₙ₊₁(z) = gₙ(φ(z)) and gₙ₊₁(z) = φ(gₙ(z)).
It is also instructive to look at the second version:
gₙ₊₁(z) ≡ E(z^{Zₙ₊₁} ) = p₀ + p₁ E(z^{Zₙ₊₁}|Z₁ = 1) + p₂E(z^{Zₙ₊₁} |Z₁ = 2) + · · ·

= p₀ + p₁ E(z^{Zₙ}|Z₀ = 1) + p₂E(z^{Zₙ}|Z₀ = 2) + · · ·
= p₀ + p₁ gₙ(z) + p₂gₙ²(z) + · · ·
=Σₖ pₖgₙᵏ(z) ≡ φ(gₙ(z)).