Why Reducibility Changes the Analysis
In a multitype branching process, individuals of different types can have different offspring laws. If every type can eventually produce every other type, the mean progeny matrix is irreducible and a single strongly connected system describes reproduction. If some reproductive pathways are one-way, however, the matrix is reducible: one class of types may feed another without any return path. The classes that can be reached from the initial population then matter as much as the matrix’s overall growth rate.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \(S=\{1,\ldots,d\}\) be a finite type set, and let \(Z_n=(Z_n(1),\ldots,Z_n(d))\) be the row vector of type counts in generation \(n\), for discrete time \(n\geq0\). Each individual reproduces independently, with offspring law depending only on its type. Write \(\xi_{ij}\) for the number of type-\(j\) children of a type-\(i\) individual, and assume these offspring counts have finite means \(m_{ij}=\mathbb{E}[\xi_{ij}]\). The mean progeny matrix is \(M=(m_{ij})\): rows index parent types, columns index child types. Conditioning on generation \(n\) gives \(\mathbb{E}[Z_{n+1}]=\mathbb{E}[Z_n]M\), and therefore \(\mathbb{E}[Z_n]=\mathbb{E}[Z_0]M^n\).
Collapse each communicating class to a single vertex. The resulting graph has no directed cycles: a cycle among distinct classes would make those classes communicate and hence be one class. Thus its classes can be ordered so every edge goes from an earlier class to a later one. After applying the same type permutation to rows and columns, \(M\) is block upper triangular. Its diagonal blocks describe reproduction within classes; off-diagonal blocks describe one-way production between classes.
Proof. Order the strongly connected classes as above, so \(M\) is block upper triangular. For a block upper triangular matrix, the determinant of \(xI-M\) is the product of the determinants of \(xI-C\) over its diagonal blocks \(C\). This follows by the determinant formula for block triangular matrices, applied successively to the blocks. Therefore the characteristic polynomial of \(M\) is the product of the characteristic polynomials of those blocks, proving the eigenvalue assertion and the claim about the spectral radius.
For the growth claim, restrict to the set \(R\) of types reachable from the initial type. No type outside \(R\) can be produced from a type in \(R\), so the expected population started there is governed by the submatrix \(M_R\). Its diagonal blocks are exactly the reachable classes. The standard Perron–Frobenius theorem for each irreducible nonnegative block gives its exponential growth rate; the standard Gelfand formula for finite matrices identifies the exponential rate of powers of \(M_R\) with its spectral radius. For a lower bound from a particular reachable class, take a directed path from the initial type to that class. The path has fixed finite length and positive expected contribution. Thereafter, the population within the class has the class’s Perron–Frobenius growth rate. Taking the largest rate over reachable classes gives the stated rate. If every reachable class has spectral radius zero, the graph of reachable types has no positive-weight cycle, so its powers eventually vanish and there is no positive exponential growth rate. \(\square\)
The Perron–Frobenius and Gelfand results used here are standard results from finite-dimensional matrix theory; their proofs are not needed for the block decomposition. In particular, the spectral radius of the full matrix describes the largest rate present anywhere in the system, not necessarily the rate visible from every initial type.
Worked Example: A One-Way Chain of Type Classes
Consider three types with mean progeny matrix \(M=\begin{bmatrix}\tfrac32&\tfrac12&0\\0&\tfrac12&2\\0&0&\tfrac32\end{bmatrix}\). One possible offspring specification is: type 1 has a type-1 count distributed as \(\operatorname{Binomial}(3,\tfrac12)\) and an independent type-2 count distributed as \(\operatorname{Bernoulli}(\tfrac12)\); type 2 has a type-2 count distributed as \(\operatorname{Bernoulli}(\tfrac12)\) and exactly two type-3 children; type 3 has a type-3 count distributed as \(\operatorname{Binomial}(3,\tfrac12)\). All unspecified offspring counts are zero. These laws give exactly the displayed matrix.
The graph has edges \(1\to2\) and \(2\to3\), and each type is its own communicating class. The diagonal blocks are the scalars \(\tfrac32,\tfrac12,\tfrac32\), so the full spectral radius is \(\tfrac32\). Starting with one type-1 individual gives \(Z_0=(1,0,0)\). The first two expectations are
For the second multiplication, the type-1 coordinate is \(\tfrac32\cdot\tfrac32=\tfrac94\); the type-2 coordinate is \(\tfrac32\cdot\tfrac12+\tfrac12\cdot\tfrac12=1\); and the type-3 coordinate is \(\tfrac12\cdot2=1\). The class with rate \(\tfrac12\) is present in the pathway but is not the largest rate. The accessible type-3 class has rate \(\tfrac32\), so the expected population’s exponential rate from type 1 is \(\tfrac32\).
Check: multiplying the displayed \(\mathbb{E}[Z_1]\) by \(M\) gives \((\tfrac94,1,1)\), as reported. The entries are means, not transition probabilities; their rows are not required to sum to one.
Extinction Equations Follow the Dependency Graph
For each type \(i\), define its offspring probability generating function by \(F_i(s_1,\ldots,s_d)=\mathbb{E}[\prod_{j=1}^d s_j^{\xi_{ij}}]\) for \(s_j\in[0,1]\), using the convention \(s^0=1\), including at \(s=0\). The vector \(q=(q_1,\ldots,q_d)\), where \(q_i\) is the probability of eventual extinction starting from one type-\(i\) individual, satisfies \(q_i=F_i(q_1,\ldots,q_d)\). For a finite population, extinction means that some generation is empty; once empty, all later generations are empty.
Proof. Let \(q^{(0)}=(0,\ldots,0)\) and \(q^{(n+1)}=F(q^{(n)})\). By induction, \(q_i^{(n)}\) is the probability that generation \(n\) is empty when the process starts from one type-\(i\) individual: the claim holds at \(n=0\) for the event that the initial individual is absent only if interpreted as extinction by generation zero, so instead set \(q^{(0)}=0\) as the probability of having no descendants at generation zero, and use the recursion for extinction by generation \(n+1\). Equivalently, \(q^{(n)}\) is the probability that generation \(n\) is empty, with \(q^{(0)}=0\); the recursion follows by conditioning on the first individual’s offspring and independence of their descendant processes. The events that generation \(n\) is empty increase with \(n\), since an empty generation remains empty. Thus \(q^{(n)}\) increases to the eventual extinction probability \(q\). Generating functions are continuous on \([0,1]^d\), so taking limits in \(q^{(n+1)}=F(q^{(n)})\) gives \(q=F(q)\).
If \(r\in[0,1]^d\) is any fixed point, then \(q^{(0)}\leq r\) coordinatewise. Each \(F_i\) is nondecreasing in every coordinate, because it is a sum of nonnegative monomials, so induction gives \(q^{(n)}\leq r\) for every \(n\). Taking limits yields \(q\leq r\), proving minimality. Finally, a type’s generating function depends only on its own class and classes reachable by its outgoing edges. The class graph is a finite directed acyclic graph, so downstream classes can be evaluated first. Once their coordinates are fixed, iteration of the class’s generating-function map from the zero vector gives its smallest fixed point by the same monotonicity argument. \(\square\)
The zero vector in this iteration represents the initial lower bound for finding a fixed point; it does not assert that the initial individual is absent. Eventual extinction is the increasing limit of the probabilities of an empty generation, which is why it gives the smallest fixed point rather than an arbitrary solution of the equations.
Worked Example: A Critical Class Feeding a Supercritical Class
There are two types, \(A\) and \(B\). A type-\(A\) individual has a type-\(A\) count distributed as \(\operatorname{Binomial}(2,\tfrac12)\) and, independently, a type-\(B\) count distributed as \(\operatorname{Bernoulli}(\tfrac12)\). A type-\(B\) individual has a type-\(B\) count distributed as \(\operatorname{Binomial}(2,\tfrac34)\). The means are \(m_{AA}=1\), \(m_{AB}=\tfrac12\), \(m_{BA}=0\), and \(m_{BB}=\tfrac32\), so \(M=\begin{bmatrix}1&\tfrac12\\0&\tfrac32\end{bmatrix}\). The classes are \(\{A\}\) and \(\{B\}\), with a one-way edge from \(A\) to \(B\).
First solve for type \(B\). Its generating function is \(F_B(s_B)=(\tfrac14+\tfrac34s_B)^2\), so its extinction probability satisfies
The roots are \(\tfrac19\) and \(1\). The smaller root is \(q_B=\tfrac19\). Now \(F_A(s_A,s_B)=(\tfrac{1+s_A}{2})^2(\tfrac{1+s_B}{2})\). Substituting \(q_B=\tfrac19\), the equation for \(q_A\) is \(q_A=\tfrac5{36}(1+q_A)^2\). Rearranging gives \(5q_A^2-26q_A+5=0\), whose roots are \(\tfrac15\) and \(5\). Only \(\tfrac15\) lies in \([0,1]\), so \(q_A=\tfrac15\).
Check: \(F_B(\tfrac19)=(\tfrac14+\tfrac1{12})^2=(\tfrac13)^2=\tfrac19\). Also \(F_A(\tfrac15,\tfrac19)=(\tfrac35)^2(\tfrac59)=\tfrac15\). Both equations hold, and the selected roots are the smallest admissible fixed points.
Reachability Matters for Growth
A class can dominate the full matrix while making no contribution to a process started elsewhere. The following calculation shows why a growth claim should specify its initial type as well as the matrix.
Worked Example: A Supercritical Class That Cannot Be Reached
Let \(M=\begin{bmatrix}\tfrac12&\tfrac12&0\\0&\tfrac32&0\\0&0&2\end{bmatrix}\). For example, type 1 produces independent type-1 and type-2 counts, each distributed as \(\operatorname{Bernoulli}(\tfrac12)\); type 2 produces a type-2 count distributed as \(\operatorname{Binomial}(2,\tfrac34)\); and type 3 produces exactly two type-3 children. These laws give the stated means. The classes are singletons, and type 3 is not reachable from type 1. The full spectral radius is \(2\), but the largest reachable class rate from type 1 is \(\tfrac32\).
Start from one type-1 individual. Write \(\mathbb{E}[Z_n]=(x_n,y_n,z_n)\). The matrix recursion gives \(x_{n+1}=\tfrac12x_n\), \(y_{n+1}=\tfrac12x_n+\tfrac32y_n\), and \(z_{n+1}=2z_n\), with \((x_0,y_0,z_0)=(1,0,0)\). Thus \(x_n=(\tfrac12)^n\) and \(z_n=0\). Iterating the recurrence for \(y_n\) gives
For the equality, the sum is a finite geometric sum: its ratio is \(\tfrac13\), so it equals \(\tfrac12(\tfrac32)^{n-1}\big(1-(\tfrac13)^n\big)/(1-\tfrac13)\), which simplifies to the displayed expression. Hence the expected total is \(\tfrac12(\tfrac32)^n+\tfrac12(\tfrac12)^n\), whose \(n\)th root tends to \(\tfrac32\).
Check: At \(n=1\), the formula gives \(x_1=\tfrac12\), \(y_1=\tfrac12\), and \(z_1=0\), matching \((1,0,0)M\). Substitution into the recurrence gives \(y_{n+1}=\tfrac12(\tfrac12)^n+\tfrac32\cdot\tfrac12((\tfrac32)^n-(\tfrac12)^n)=\tfrac12((\tfrac32)^{n+1}-(\tfrac12)^{n+1})\).
Common Mistake: Applying the Full-Matrix Rate to Every Type
A tempting but incorrect calculation is to see the spectral radius \(2\) in the preceding example and conclude that a population started from type 1 grows at rate \(2\). This treats the type-3 block as if it were reachable. But the type-1 row has zero type-3 mean, and the type-2 row also has zero type-3 mean. Consequently \(z_n=0\) for every \(n\) when \(z_0=0\), and the actual expected total has rate \(\tfrac32\). The correction is to first identify the classes reachable from the initial type, then compare their block rates.
Extinction is also not determined by the spectral radius of the full matrix alone for each starting type. In the two-type example, the type-\(A\) diagonal block has mean \(1\), but \(A\) can produce \(B\), whose extinction probability is \(\tfrac19\). The full extinction equation gives \(q_A=\tfrac15\), not \(1\). For a reducible system, the equations and their dependency order retain information about these one-way reproductive pathways that a single global rate cannot express.
Check Your Understanding
For each question, specify the relevant classes or fixed-point equations before calculating.
- In the first example, list the directed edges and the diagonal block rates. Which blocks are reachable from type 2?
- Why does solving the type-\(B\) extinction equation before the type-\(A\) equation work in the two-type example?
- For the matrix in the inaccessible-class example, what is the expected type-3 count at generation \(n\) when the initial population is one type-1 individual? Explain using the graph.
- Verify directly that \(\tfrac15=\tfrac5{36}(1+\tfrac15)^2\).
- Can a class have the largest spectral radius in the full matrix and still have no effect on expected growth from a specified initial type? Give the relevant reachability condition.