Extinction Probabilities in Multitype Branching

Extinction as a Coupled Probability

In a multitype branching process, one individual can produce children of several types. The future of a type 1 individual can therefore depend on whether its descendants include type 2 individuals, whose own reproduction law may be different. Extinction probabilities are consequently coupled: they must be solved together, rather than one type at a time.

Consider a process in discrete generations on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. There are two types, and $Z_n=(Z_{n,1},Z_{n,2})\in S=\mathbb{N}_0^2$ records the numbers of individuals of each type in generation $n$. Each individual of type $i$ independently produces an offspring vector $(\xi_{i1},\xi_{i2})\in\mathbb{N}_0^2$, with the same distribution for all individuals of type $i$. Reproduction is independent across individuals and generations. Let $\mathcal{F}_n$ contain the population and reproduction information through generation $n$. We write $\mathbb{P}_i$ for the law with one initial individual of type $i$, and define extinction as the event that some generation has no individuals of either type.

Definition: The offspring generating function for type $i$ is $F_i(s_1,s_2)=\mathbb{E}[s_1^{\xi_{i1}}s_2^{\xi_{i2}}]$, for $(s_1,s_2)\in[0,1]^2$. The type-specific extinction probability is $q_i=\mathbb{P}_i(\text{the process eventually has no individuals})$. The vector $q=(q_1,q_2)$ lies in the unit cube $[0,1]^2$.

The joint offspring distribution matters. For example, a type 1 individual that has either two type 2 children or no children has generating function $a+(1-a)s_2^2$, if the probability of no children is $a$. The square records that both type 2 descendants must have extinct lineages for the entire family to become extinct.

The Fixed-Point Equations and Their Meaning

Theorem: For a two-type branching process with finite offspring counts almost surely and independent reproduction as specified above, the extinction vector satisfies $$ q_1=F_1(q_1,q_2),\qquad q_2=F_2(q_1,q_2). $$ Moreover, $q$ is the smallest solution of these equations in $[0,1]^2$, in the coordinatewise order: if $x\in[0,1]^2$ also satisfies $x_i=F_i(x_1,x_2)$ for $i=1,2$, then $q_i\le x_i$ for both types.

Conditioning on the first generation explains the equations. A type $i$ ancestor has $\xi_{i1}$ type 1 children and $\xi_{i2}$ type 2 children. Given these counts, the child lineages reproduce independently. The probability that every lineage dies out is therefore $q_1^{\xi_{i1}}q_2^{\xi_{i2}}$; taking its expectation gives $F_i(q_1,q_2)$. The proof below also establishes that this extinction vector is the smallest fixed point, a fact needed when the equations have multiple solutions.

Proof. For $n\ge0$, let $q_i^{(n)}=\mathbb{P}_i(Z_n=(0,0))$, the probability that generation $n$ is empty from one type $i$ ancestor. At $n=0$, the initial ancestor is present, so $q_i^{(0)}=0$. The empty population remains empty in every later generation. Conditioning on the first ancestor's offspring, then using independence of the child lineages, gives $$ q_i^{(n+1)}=\mathbb{E}\!\left[(q_1^{(n)})^{\xi_{i1}}(q_2^{(n)})^{\xi_{i2}}\right] =F_i(q_1^{(n)},q_2^{(n)}). $$ Here a factor with exponent zero is $1$, including when its base is zero. Each $F_i$ is nondecreasing in each coordinate on $[0,1]^2$, because every term $s_1^{a}s_2^{b}$ has nonnegative integer exponents and is nondecreasing there. Since $q^{(0)}=(0,0)\le q^{(1)}$, induction gives $q^{(n)}\le q^{(n+1)}$ coordinatewise. Thus $q^{(n)}$ has a coordinatewise limit $r\in[0,1]^2$. The events $\{Z_n=(0,0)\}$ increase with $n$, and their union is precisely eventual extinction. Hence $r_i=q_i$. For each fixed offspring vector, the powers in the expectation defining $F_i(q^{(n)})$ converge to those defining $F_i(q)$, and the products are bounded by $1$. Bounded convergence therefore gives $F_i(q)=\lim_n F_i(q^{(n)})=\lim_n q_i^{(n+1)}=q_i$. This proves the fixed-point equations. Finally, let $x\in[0,1]^2$ be any fixed point. We have $q^{(0)}\le x$. If $q^{(n)}\le x$, coordinatewise monotonicity gives $q^{(n+1)}=F(q^{(n)})\le F(x)=x$, where $F=(F_1,F_2)$. Induction and passage to the limit give $q\le x$, proving the smallest-solution assertion. $\square$

The theorem does not say that every solution is an extinction vector. In particular, $(1,1)$ is always a fixed point, since an offspring generating function equals $1$ at $(1,1)$. In a model that has a smaller fixed point, choosing $(1,1)$ would incorrectly conclude that extinction is certain. The smallest-solution condition resolves that ambiguity.

Solving Two-Type Systems

Worked Example: Symmetric Cross-Type Reproduction

A type 1 individual has no children with probability $\tfrac14$, and has two type 2 children with probability $\tfrac34$. A type 2 individual has no children with probability $\tfrac14$, and has two type 1 children with probability $\tfrac34$. Find both extinction probabilities.

Each offspring law has total probability $\tfrac14+\tfrac34=1$. The generating functions and fixed-point equations are $$ F_1(s_1,s_2)=\tfrac14+\tfrac34s_2^2,\qquad F_2(s_1,s_2)=\tfrac14+\tfrac34s_1^2, \qquad x=\tfrac14+\tfrac34y^2,\quad y=\tfrac14+\tfrac34x^2. $$ Subtracting the equations gives $x-y=-\tfrac34(x-y)(x+y)$, or $(x-y)(1+\tfrac34(x+y))=0$. Since $x,y\in[0,1]$, the second factor is positive, so $x=y$. Substitution gives $x=\tfrac14+\tfrac34x^2$, equivalently $3x^2-4x+1=0=(3x-1)(x-1)$. The fixed points are $x=y=\tfrac13$ and $x=y=1$; the smallest is the extinction vector.

Check: Substituting $(\tfrac13,\tfrac13)$ into either right-hand side gives $\tfrac14+\tfrac34(\tfrac13)^2=\tfrac14+\tfrac1{12}=\tfrac13$. Substituting $(1,1)$ gives $\tfrac14+\tfrac34=1$. Both are fixed points, and $\tfrac13<1$ in each coordinate, so the selected solution is $q=(\tfrac13,\tfrac13)$.

Worked Example: Unequal Extinction Probabilities

A type 1 individual has no children with probability $\tfrac14$, and one type 2 child with probability $\tfrac34$. A type 2 individual has no children with probability $\tfrac14$, and two type 1 children with probability $\tfrac34$. Solve for the extinction vector.

Both offspring laws sum to $\tfrac14+\tfrac34=1$. Their generating functions give $$ x=\tfrac14+\tfrac34y,\qquad y=\tfrac14+\tfrac34x^2. $$ The first equation yields $y=(4x-1)/3$. Substituting this into the second equation gives $(4x-1)/3=\tfrac14+\tfrac34x^2$. Multiplying by $12$ gives $16x-4=3+9x^2$, so $9x^2-16x+7=0$. Factoring, $(x-1)(9x-7)=0$, and thus $x=1$ or $x=\tfrac79$. The corresponding values of $y$ are $1$ and $(4\cdot\tfrac79-1)/3=\tfrac{19}{27}$. Both solutions lie in the unit square, and $(\tfrac79,\tfrac{19}{27})$ is coordinatewise smaller. Therefore $q=(\tfrac79,\tfrac{19}{27})$.

Check: For type 1, $\tfrac14+\tfrac34\cdot\tfrac{19}{27}=\tfrac9{36}+\tfrac{19}{36}=\tfrac79$. For type 2, $\tfrac14+\tfrac34(\tfrac79)^2=\tfrac{81}{324}+\tfrac{147}{324}=\tfrac{228}{324}=\tfrac{19}{27}$. The candidate satisfies both equations. The other solution $(1,1)$ also satisfies them, and the smallest-fixed-point theorem identifies which one is extinction.

The second example illustrates why type-specific probabilities should not be replaced by a single scalar without justification. Type 1 offspring depend on the type 2 extinction probability, while type 2 offspring depend on the type 1 probability. The two equations express that dependence directly.

Worked Example: Joint Offspring and an Initial Mixed Population

For an individual of either type, suppose the offspring vector is $(0,0)$ with probability $\tfrac13$ and $(1,1)$ with probability $\tfrac23$. Find the extinction probability from one ancestor of each type, and from an initial population containing one of each type.

Each type's probabilities sum to $\tfrac13+\tfrac23=1$, and both generating functions are $F_i(s_1,s_2)=\tfrac13+\tfrac23s_1s_2$. The equations are $x=\tfrac13+\tfrac23xy$ and $y=\tfrac13+\tfrac23xy$. Their right-hand sides are identical, so every fixed point has $x=y$. Writing $x=y$ gives $x=\tfrac13+\tfrac23x^2$, or $2x^2-3x+1=0=(2x-1)(x-1)$. The solutions are $(\tfrac12,\tfrac12)$ and $(1,1)$; the smallest is $q=(\tfrac12,\tfrac12)$.

With one ancestor of each type, the two initial individuals' descendant families reproduce independently. The probability that both families become extinct is $q_1q_2=\tfrac12\cdot\tfrac12=\tfrac14$. Check: Substitution gives $\tfrac13+\tfrac23\cdot\tfrac12\cdot\tfrac12=\tfrac13+\tfrac16=\tfrac12$ in each fixed-point equation; multiplying the two type-specific probabilities gives $\tfrac14$ for the mixed initial population.

Result: If the initial population has $z_1$ type 1 individuals and $z_2$ type 2 individuals, its extinction probability is $q_1^{z_1}q_2^{z_2}$. This includes an empty initial population, whose extinction probability is $1$.

Proof. Each initial individual starts a descendant family. By the independence of reproduction across individuals, these families are independent, and the probability that a family from a type $i$ individual becomes extinct is $q_i$. The whole population becomes extinct exactly when every one of its $z_1+z_2$ families becomes extinct. Multiplying their probabilities gives $q_1^{z_1}q_2^{z_2}$. If $z_1=z_2=0$, the event already holds; the empty product is $1$, in agreement with the formula. $\square$

Common Mistakes and a Reliable Procedure

A frequent error is to treat offspring counts of different types as independent when the model specifies a joint offspring vector. In the third example, the type 1 individual produces either zero of both types or one of each type. Its generating function is $\tfrac13+\tfrac23s_1s_2$, not $(\tfrac13+\tfrac23s_1)(\tfrac13+\tfrac23s_2)$. The latter expression describes a different model in which the two offspring counts are independent Bernoulli variables. At $(s_1,s_2)=(\tfrac12,\tfrac12)$, the correct generating function equals $\tfrac13+\tfrac16=\tfrac12$; the incorrect product equals $(\tfrac13+\tfrac13)^2=\tfrac49$. Thus using the product would not even preserve the fixed-point equation for the stated offspring law.

Another error is to solve the equations and select the largest root, or simply select $(1,1)$. The fixed-point equations alone can have multiple solutions in the unit cube. The extinction vector is the smallest one, as proved above. A sound calculation therefore has three stages: derive each generating function from the joint offspring probabilities, solve the coupled equations within $[0,1]^2$, and compare all admissible fixed points coordinatewise. Substitution back into both equations is an essential check.

Check Your Understanding

For each question, write the relevant generating function or equations before calculating.

A type 1 individual has no children with probability $\tfrac12$ and two type 2 children with probability $\tfrac12$. Write its offspring generating function and the type 1 extinction equation.
Why does the extinction vector satisfy $q_i=F_i(q_1,q_2)$, rather than an equation involving only $F_i(q_i,q_i)$ in general?
In the unequal-extinction example, substitute $q=(\tfrac79,\tfrac{19}{27})$ into both equations and verify each coordinate.
Explain why $(1,1)$ is always a fixed point, and why that fact does not imply certain extinction.
If the initial population has two type 1 individuals and one type 2 individual, express its extinction probability using $q_1$ and $q_2$.

Extinction Probabilities in Multitype Branching

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