The Central Idea

Let

\[ Z_n \]

denote the population size in generation \(n\).

If each individual in generation \(n\) independently produces a random number of offspring, then

\[ \boxed{ Z_{n+1} = \sum_{i=1}^{Z_n}X_{n,i} }, \]

where \(X_{n,i}\) is the number of offspring produced by individual \(i\) in generation \(n\).

The Galton–Watson Model

The classical discrete-time branching process is the Galton–Watson process.

Its standard assumptions are:

These assumptions produce a time-homogeneous branching model.

The Offspring Distribution

Let \(X\) represent the number of offspring produced by one individual.

Write

\[ p_k=P(X=k), \qquad k=0,1,2,\ldots \]

with

\[ \boxed{ \sum_{k=0}^{\infty}p_k=1 }. \]

The entire branching process is driven by this one-generation reproduction law together with the starting population.

A Simple Offspring Example

Suppose one individual has

\[ P(X=0)=0.25, \] \[ P(X=1)=0.50, \]

and

\[ P(X=2)=0.25. \]

An individual may therefore produce no offspring, one offspring, or two offspring.

Generation Zero

The process begins with an initial population

\[ Z_0. \]

A common introductory case uses

\[ Z_0=1, \]

so the process begins with a single ancestor.

More generally, the process may begin with any nonnegative integer number of individuals.

From One Generation to the Next

If

\[ Z_n=r, \]

then the next generation is the sum of \(r\) independent offspring counts:

\[ Z_{n+1} = X_{n,1}+X_{n,2}+\cdots+X_{n,r}. \]

Thus, the population size determines how many random reproduction variables must be added.

A Sample Family Tree

Suppose

\[ Z_0=1. \]

If the initial individual has \(3\) offspring, then

\[ Z_1=3. \]

If those three individuals have \(2\), \(0\), and \(1\) offspring, respectively, then

\[ Z_2=2+0+1=3. \]

A later generation can be larger, smaller, or the same size as the one before it.

Branching Processes Are Random Sequences

The generation sizes form the random sequence

\[ Z_0,Z_1,Z_2,\ldots \]

Each \(Z_n\) is a random variable taking values in

\[ \{0,1,2,\ldots\}. \]

The sequence is therefore a discrete-time stochastic process.

Branching Processes Are Markov Chains

Under the Galton–Watson assumptions, the distribution of \(Z_{n+1}\) depends on the past only through the current population \(Z_n\).

Therefore,

\[ P(Z_{n+1}=j\mid Z_n=i,Z_{n-1},\ldots,Z_0) = P(Z_{n+1}=j\mid Z_n=i). \]

The generation-size process is a Markov chain on the nonnegative integers.

Zero Is an Absorbing State

If

\[ Z_n=0, \]

then there are no individuals available to reproduce.

Hence,

\[ Z_{n+1}=0, \]

and all later generations are also zero.

Thus,

\[ \boxed{0\text{ is an absorbing state}}. \]

Extinction

The branching process becomes extinct if it eventually reaches population zero.

The extinction event is

\[ \{\exists n:Z_n=0\}. \]

Once extinction occurs, it is permanent.

The Extinction Time

Define

\[ T_0 = \min\{n\ge0:Z_n=0\}. \]

This is the first generation in which the population is extinct.

If extinction never occurs, then \(T_0\) is treated as infinite.

Mean Number of Offspring

Let

\[ m=E[X]. \]

In terms of the offspring probabilities,

\[ \boxed{ m = \sum_{k=0}^{\infty}k\,p_k }. \]

This single quantity provides the first major classification of branching behavior.

Expected Next Generation

Conditional on the current population,

\[ E[Z_{n+1}\mid Z_n] = mZ_n. \]

Each of the \(Z_n\) current individuals contributes an average of \(m\) offspring.

Expected Population Size

Taking expectations gives

\[ E[Z_{n+1}] = mE[Z_n]. \]

Repeating this recursion yields

\[ \boxed{ E[Z_n] = Z_0m^n } \]

when \(Z_0\) is a fixed initial population.

Example: Expected Growth

Suppose

\[ Z_0=2 \]

and the mean offspring count is

\[ m=1.5. \]

Then

\[ E[Z_4] = 2(1.5)^4 = 10.125. \]

The expected generation size need not be an integer because it is an average over many possible population outcomes.

Subcritical Branching

The process is called subcritical when

\[ \boxed{m<1}. \]

The expected population size decreases geometrically:

\[ E[Z_n]=Z_0m^n\to0. \]

In the standard nondegenerate Galton–Watson model, extinction occurs with probability \(1\).

Critical Branching

The process is called critical when

\[ \boxed{m=1}. \]

Then

\[ E[Z_n]=Z_0. \]

A constant expected population does not mean the actual population remains stable. In a nondegenerate critical Galton–Watson process, extinction still occurs with probability \(1\).

The Deterministic Critical Exception

If every individual always produces exactly one offspring,

\[ P(X=1)=1, \]

then

\[ Z_n=Z_0 \]

for all \(n\).

This process has \(m=1\) but does not become extinct when \(Z_0>0\).

Therefore, extinction statements for the critical case require the usual nondegeneracy assumption.

Supercritical Branching

The process is called supercritical when

\[ \boxed{m>1}. \]

The expected population grows geometrically.

Even so, extinction can still occur because early generations may happen to produce too few descendants.

Expected Growth Does Not Guarantee Survival

Suppose the mean offspring count satisfies

\[ m>1. \]

Then

\[ E[Z_n] \]

increases with \(n\), but individual sample paths can still hit zero.

Mean growth and survival probability are related but distinct quantities.

Probability Generating Function

The offspring distribution can be summarized by its probability generating function:

\[ \boxed{ f(s) = E[s^X] = \sum_{k=0}^{\infty}p_ks^k }. \]

For \(0\le s\le1\), this function encodes the complete offspring distribution.

Generating Function Example

If

\[ P(X=0)=0.25,\qquad P(X=1)=0.50,\qquad P(X=2)=0.25, \]

then

\[ \boxed{ f(s) = 0.25+0.50s+0.25s^2 }. \]

The Mean from the Generating Function

Differentiate:

\[ f'(s) = \sum_{k=1}^{\infty}k p_k s^{k-1}. \]

Evaluating at \(s=1\) gives

\[ \boxed{ f'(1)=E[X]=m }. \]

Extinction Probability

Assume the process begins with one ancestor:

\[ Z_0=1. \]

Let

\[ q=P(\text{eventual extinction}). \]

Then \(q\) satisfies the fixed-point equation

\[ \boxed{f(q)=q}. \]

Why the Extinction Equation Works

Condition on the number \(X\) of children produced by the initial ancestor.

If the ancestor has \(k\) children, the entire process becomes extinct only if all \(k\) descendant lines eventually become extinct.

By independence, that probability is

\[ q^k. \]

Averaging over \(X\) gives

\[ q = \sum_{k=0}^{\infty}p_kq^k = f(q). \]

Which Fixed Point Is the Extinction Probability?

The equation

\[ f(s)=s \]

can have more than one solution in \([0,1]\).

The extinction probability is the

\[ \boxed{\text{smallest solution in }[0,1]}. \]

Why \(1\) Is Always a Fixed Point

Because the offspring probabilities sum to \(1\),

\[ f(1) = \sum_{k=0}^{\infty}p_k = 1. \]

Therefore,

\[ f(1)=1. \]

The question is whether there is a smaller fixed point.

Extinction in the Subcritical and Critical Cases

For a standard nondegenerate Galton–Watson process with

\[ m\le1, \]

the smallest fixed point is

\[ \boxed{q=1}. \]

Thus extinction occurs almost surely.

The deterministic case \(P(X=1)=1\) is the important critical exception.

Extinction in the Supercritical Case

In a supercritical process with a genuine possibility of extinction,

\[ m>1, \]

the fixed-point equation has a solution

\[ q<1. \]

Therefore,

\[ \boxed{ P(\text{survival})=1-q>0 }. \]

Example: Solving for Extinction

Suppose

\[ P(X=0)=\frac14 \]

and

\[ P(X=2)=\frac34. \]

Then

\[ f(s) = \frac14+\frac34s^2. \]

The mean is

\[ m = 0\left(\frac14\right) + 2\left(\frac34\right) = \frac32. \]

The process is supercritical.

Solving the Fixed-Point Equation

Solve

\[ \frac14+\frac34q^2=q. \]

Multiplying by \(4\),

\[ 1+3q^2=4q. \]

Rearranging,

\[ 3q^2-4q+1=0. \]

Factoring gives

\[ (3q-1)(q-1)=0. \]

The solutions are

\[ q=\frac13 \]

and

\[ q=1. \]

The extinction probability is the smaller one:

\[ \boxed{q=\frac13}. \]

Survival Probability

In the preceding example,

\[ P(\text{survival}) = 1-q = 1-\frac13 = \boxed{\frac23}. \]

Supercritical growth creates a positive chance of indefinite survival, but extinction remains possible.

Starting with Several Ancestors

Suppose

\[ Z_0=r. \]

Each initial ancestor generates an independent descendant tree.

If the extinction probability from one ancestor is \(q\), then extinction from \(r\) independent ancestors requires all \(r\) family lines to die out:

\[ \boxed{ P(\text{extinction}\mid Z_0=r)=q^r }. \]

Example: Several Initial Ancestors

If

\[ q=\frac13 \]

and

\[ Z_0=3, \]

then

\[ P(\text{extinction}) = \left(\frac13\right)^3 = \boxed{\frac1{27}}. \]

The Branching Property

A central structural feature is that descendants of different individuals evolve independently and according to the same law.

Conditional on

\[ Z_n=r, \]

the future population can be viewed as the sum of \(r\) independent copies of the process started from one individual.

This is called the branching property.

Generating Functions Across Generations

Let

\[ F_n(s)=E[s^{Z_n}] \]

when \(Z_0=1\).

The branching property gives

\[ \boxed{ F_{n+1}(s)=F_n(f(s)) }. \]

Equivalently, \(F_n\) is obtained by repeatedly composing \(f\) with itself.

Repeated Composition

Starting from one ancestor,

\[ F_1(s)=f(s), \] \[ F_2(s)=f(f(s)), \]

and more generally

\[ \boxed{ F_n(s)=f^{\circ n}(s) }, \]

where \(f^{\circ n}\) means \(n\)-fold composition, not an ordinary power.

Probability of Extinction by Generation \(n\)

Because

\[ 0^{Z_n} = \begin{cases} 1,&Z_n=0,\\ 0,&Z_n>0, \end{cases} \]

we have

\[ P(Z_n=0) = F_n(0). \]

Therefore,

\[ \boxed{ P(Z_n=0) = f^{\circ n}(0) }. \]

Approaching the Eventual Extinction Probability

The probabilities

\[ P(Z_n=0) \]

increase with \(n\), because once extinction occurs it persists.

Their limit is the eventual extinction probability:

\[ \boxed{ q = \lim_{n\to\infty}f^{\circ n}(0) }. \]

Variance of the Offspring Distribution

Let

\[ \sigma^2=\operatorname{Var}(X). \]

Conditional on \(Z_n\),

\[ \operatorname{Var}(Z_{n+1}\mid Z_n) = \sigma^2 Z_n. \]

More individuals create more independent sources of reproductive variability.

Variance of Generation Size

For \(Z_0=1\), finite offspring variance \(\sigma^2\), and \(m\ne1\),

\[ \boxed{ \operatorname{Var}(Z_n) = \sigma^2 m^{n-1} \frac{m^n-1}{m-1} }. \]

For the critical case \(m=1\),

\[ \boxed{ \operatorname{Var}(Z_n) = n\sigma^2 }. \]

Why Critical Processes Can Still Be Highly Variable

In a critical process,

\[ E[Z_n]=1 \]

when \(Z_0=1\), but

\[ \operatorname{Var}(Z_n)=n\sigma^2. \]

The expected size stays constant while uncertainty grows.

This is another example of why a mean alone does not describe the behavior of a stochastic process.

Population Size Can Be Zero or Very Large

Branching distributions are often strongly skewed.

Many sample paths may already be extinct while a smaller set of surviving paths have become large.

The expectation averages across both types of outcomes.

Conditional Growth Given Survival

In a supercritical process, unconditional expectation includes extinct paths whose population is zero.

Consequently, the population size among surviving lineages can be much larger than the unconditional mean suggests.

Conditioning on survival changes the distribution and must be stated explicitly.

Deterministic Branching as a Special Case

If every individual produces exactly \(r\) offspring,

\[ P(X=r)=1, \]

then

\[ Z_n=Z_0r^n. \]

The model is formally a branching process, but there is no reproductive randomness.

Binary Branching

A common simplified model allows only

\[ X=0 \]

or

\[ X=2. \]

This produces a population in which individuals either leave no descendants or split into two.

Binary branching appears in idealized cell-division and particle models.

Poisson Offspring Distribution

Another common model uses

\[ X\sim\operatorname{Poisson}(\lambda). \]

Then

\[ m=\lambda \]

and

\[ \sigma^2=\lambda. \]

The generating function is

\[ \boxed{ f(s)=e^{\lambda(s-1)} }. \]

Criticality for Poisson Offspring

In the Poisson offspring model:

The mean offspring count directly determines the criticality class.

Branching Processes and Epidemic Intuition

In the early stages of an epidemic, when susceptible depletion and repeated contacts can be neglected, one infected individual may generate a random number of secondary cases.

A branching process can approximate the resulting transmission tree.

The approximation becomes less accurate when population structure, immunity, intervention, or competition for susceptible individuals becomes important.

Branching Processes and Family Lines

Branching models were historically developed to study the survival of family names across generations.

The same mathematics now appears in population biology, genetics, epidemiology, nuclear chain reactions, algorithms, and network cascades.

Branching Processes and Reliability Trees

A branching structure can also represent cascading failures or propagating events.

One failure may trigger a random number of secondary failures, each of which can trigger further events.

Independence and identical-distribution assumptions must be assessed carefully in such applications.

Branching Processes and Algorithms

Recursive algorithms can generate random numbers of subproblems.

The number of active tasks at each recursion depth can sometimes be modeled by a branching process.

Extinction then corresponds to termination of all branches.

Simulation of a Branching Process

A branching process can be simulated generation by generation.

  1. Choose the initial population \(Z_0\).
  2. For each current individual, sample an offspring count.
  3. Add the offspring counts to obtain \(Z_{n+1}\).
  4. If \(Z_{n+1}=0\), stop because extinction has occurred.
  5. Otherwise, repeat for the next generation.

What Simulation Can Estimate

Repeated branching simulations can estimate:

Estimating Extinction by Simulation

If \(M\) independent simulated branching processes are run and \(E\) of them become extinct within a sufficiently long observation horizon, then

\[ \widehat q = \frac{E}{M} \]

provides an empirical estimate of extinction probability.

A finite simulation horizon can miss extinctions that would occur later.

Simulation Horizon Matters

A process that has not become extinct by generation \(N\) has not necessarily survived forever.

Therefore, simulation estimates of eventual extinction require either long horizons, theoretical correction, or both.

Model Assumption: Independent Reproduction

The Galton–Watson model assumes that one individual's offspring count does not directly alter another individual's offspring count.

Competition, shared resources, common environmental shocks, or coordinated behavior can violate this assumption.

Model Assumption: Identical Reproduction Law

Standard branching theory assumes all individuals share the same offspring distribution.

Real populations can differ by age, type, location, health, or environment.

More advanced multitype and varying-environment branching processes relax this assumption.

Model Assumption: No Population Feedback

In a basic Galton–Watson process, reproduction does not depend on the current population size.

Density dependence, resource limitation, and carrying capacity require different models.

Common Error: Confusing Individuals with Generations

The random variable \(X\) describes the offspring count of one individual.

The random variable \(Z_n\) describes the total population in generation \(n\).

These quantities play different roles.

Common Error: Writing \(Z_{n+1}=mZ_n\)

The relation

\[ E[Z_{n+1}\mid Z_n]=mZ_n \]

is an expectation statement.

The actual next generation is random:

\[ Z_{n+1} = \sum_{i=1}^{Z_n}X_{n,i}. \]

Replacing the random process by its mean removes the central uncertainty.

Common Error: Assuming \(m>1\) Means Certain Survival

Supercritical branching gives positive survival probability, not guaranteed survival.

If there is a chance that individuals produce no descendants, the process can still become extinct.

Common Error: Assuming \(E[Z_n]=1\) Means Stability

In a nondegenerate critical process,

\[ E[Z_n]=1 \]

when \(Z_0=1\), yet extinction occurs with probability \(1\).

Rare large surviving populations can balance many extinct paths in the expectation.

Common Error: Choosing the Wrong Fixed Point

Because

\[ f(1)=1, \]

the equation

\[ f(q)=q \]

always has \(q=1\) as a solution.

In a supercritical process, the extinction probability is the smallest fixed point in \([0,1]\), which may be less than \(1\).

Common Error: Confusing Composition with Powers

The expression

\[ f^{\circ n}(s) \]

means repeated function composition:

\[ f(f(\cdots f(s)\cdots)). \]

It does not mean

\[ [f(s)]^n. \]

Common Error: Treating a Finite-Horizon Survivor as Immortal

Observing

\[ Z_N>0 \]

means only that the process has survived through generation \(N\).

Extinction may still occur in a later generation.

A Reliable Procedure for Expected Generation Size

  1. Identify the offspring distribution.
  2. Compute \(m=E[X]\).
  3. Identify the starting population \(Z_0\).
  4. Use \(E[Z_n]=Z_0m^n\).
  5. Interpret the result as an average, not as a guaranteed population size.

A Reliable Procedure for Extinction Probability

  1. Write the offspring probabilities \(p_k\).
  2. Construct \(f(s)=\sum_kp_ks^k\).
  3. Compute the mean \(m=f'(1)\) as an initial classification check.
  4. Solve \(f(q)=q\).
  5. Keep only solutions in \([0,1]\).
  6. Select the smallest such solution.
  7. If there are \(r\) initial ancestors, use \(q^r\).

A Reliable Procedure for Branching Simulation

  1. Set the current population equal to \(Z_0\).
  2. Generate one offspring count for each current individual.
  3. Sum those counts to obtain the next generation.
  4. Record the generation size.
  5. Stop at extinction or at a chosen simulation horizon.
  6. Repeat the entire simulation many times for empirical probabilities.
  7. Compare simulated estimates with theoretical values when available.

Example: Full Branching Workflow

Suppose

\[ P(X=0)=0.2,\qquad P(X=1)=0.3,\qquad P(X=2)=0.5. \]

The mean offspring count is

\[ m = 0(0.2)+1(0.3)+2(0.5) = 1.3. \]

Therefore, the process is supercritical.

Its generating function is

\[ f(s) = 0.2+0.3s+0.5s^2. \]

Solve

\[ 0.2+0.3q+0.5q^2=q. \]

Rearranging,

\[ 0.5q^2-0.7q+0.2=0. \]

Multiplying by \(10\),

\[ 5q^2-7q+2=0. \]

Factoring,

\[ (5q-2)(q-1)=0. \]

Therefore,

\[ q=0.4 \]

or

\[ q=1. \]

The extinction probability is

\[ \boxed{q=0.4}, \]

so the survival probability is

\[ \boxed{0.6}. \]

Practice 1: Compute the Mean Offspring Count

Suppose

\[ P(X=0)=0.4,\qquad P(X=1)=0.4,\qquad P(X=2)=0.2. \]

Find \(m=E[X]\).

Practice 2: Classify the Process

Using the offspring distribution from Practice 1, classify the branching process as subcritical, critical, or supercritical.

Practice 3: Expected Generation Size

A branching process has

\[ m=1.2 \]

and

\[ Z_0=5. \]

Find \(E[Z_3]\).

Practice 4: Construct a Generating Function

Suppose

\[ P(X=0)=0.5,\qquad P(X=2)=0.5. \]

Write \(f(s)\).

Practice 5: Find an Extinction Probability

For the offspring distribution

\[ P(X=0)=0.25,\qquad P(X=2)=0.75, \]

find the eventual extinction probability starting from one ancestor.

Practice 6: Several Initial Ancestors

If the one-ancestor extinction probability is

\[ q=0.4, \]

find the extinction probability when

\[ Z_0=2. \]

Practice 7: Extinction by Generation \(n\)

For a process beginning with one ancestor, what does

\[ f^{\circ 4}(0) \]

represent?

Practice 8: Critical Variance

A critical branching process has

\[ m=1,\qquad \sigma^2=2, \]

and starts from one ancestor.

Find

\[ \operatorname{Var}(Z_6). \]

Practice 9: Identify the Absorbing State

What state is absorbing in the generation-size Markov chain of a standard branching process, and why?

Practice 10: Identify the Error

A student says:

“Because a supercritical branching process has \(m>1\), its population must grow forever.”

Explain the error.

Practice Answers

Practice 1

\[ m = 0(0.4)+1(0.4)+2(0.2) = 0.8. \]

Therefore,

\[ \boxed{m=0.8}. \]

Practice 2

Since

\[ m=0.8<1, \]

the process is

\[ \boxed{\text{subcritical}}. \]

Practice 3

\[ E[Z_3] = 5(1.2)^3 = 5(1.728) = \boxed{8.64}. \]

Practice 4

The probability generating function is

\[ \boxed{ f(s)=0.5+0.5s^2 }. \]

Practice 5

The generating function is

\[ f(s)=\frac14+\frac34s^2. \]

Solving

\[ f(q)=q \]

gives

\[ q=\frac13 \]

and \(q=1\). The smaller solution is the extinction probability:

\[ \boxed{q=\frac13}. \]

Practice 6

Independent initial family lines must both become extinct:

\[ q^2 = (0.4)^2 = \boxed{0.16}. \]

Practice 7

It represents

\[ \boxed{P(Z_4=0)}, \]

the probability that the process is extinct by generation \(4\).

Practice 8

For a critical process,

\[ \operatorname{Var}(Z_n)=n\sigma^2. \]

Therefore,

\[ \operatorname{Var}(Z_6) = 6(2) = \boxed{12}. \]

Practice 9

The absorbing state is

\[ \boxed{0}. \]

Once no individuals remain, there is nobody left to produce offspring, so all later generations also have size \(0\).

Practice 10

The condition

\[ m>1 \]

describes average reproductive growth. It does not prevent a particular sample path from producing too few descendants and reaching population zero.

A supercritical process typically has

\[ 0 so both extinction and indefinite survival can have positive probability.

What to Remember

A Galton–Watson branching process evolves through

\[ \boxed{ Z_{n+1} = \sum_{i=1}^{Z_n}X_{n,i} }. \]

The central modeling principle is: each current individual independently creates a random number of future individuals, and those local reproduction events compound across generations.

Where We Go Next

Branching processes evolve in discrete generations. Another fundamental stochastic model focuses instead on random event arrivals occurring through continuous time.

The next lesson, Poisson Processes, develops event counts, independent increments, arrival rates, waiting times, and the connection between Poisson counts and exponential interarrival times.