From Offspring Laws to a Mean Matrix
In a one-type branching process, a single expected offspring number summarizes the average contribution of each individual to the next generation. With several types, each type can produce different numbers of individuals of every type. A matrix records all these averages at once. Its powers then give expected type counts over multiple generations, without requiring the distribution of the entire population.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). A finite-type branching process has types \(1,\ldots,d\), where \(d\) is a positive integer, and generations \(n=0,1,2,\ldots\). Each individual of type \(i\) produces a random vector \(\xi^{(i)}=(\xi^{(i)}_1,\ldots,\xi^{(i)}_d)\) in \(\mathbb{N}_0^d\), where \(\xi^{(i)}_j\) is the number of type-\(j\) children. Assume that distinct individuals reproduce independently, and that each individual’s offspring vector has a distribution determined only by that individual’s type. Offspring vectors have finite means.
Let \(Z_n=(Z_{n,1},\ldots,Z_{n,d})\) be the row vector of type counts in generation \(n\). The state space of the population-count process is \(\mathbb{N}_0^d\). The matrix \(M\) is not generally a transition matrix: its entries are nonnegative expected counts, and its rows need not sum to one. A row sum instead gives the expected total number of children of a parent of that row’s type.
The Expected-Count Recursion
Let \((\mathcal{F}_n)_{n\ge0}\) be the filtration containing the population and reproduction information revealed through generation \(n\), but not the offspring vectors of the individuals in generation \(n\). Given \(\mathcal{F}_n\), the \(Z_{n,i}\) type-\(i\) individuals each have expected \(m_{ij}\) children of type \(j\). Conditional expectation and independence therefore give the following result.
Proof. For each type \(j\), the count \(Z_{n+1,j}\) is the sum of the type-\(j\) offspring counts from all individuals in generation \(n\). Conditional on \(\mathcal{F}_n\), each type-\(i\) individual has expected type-\(j\) offspring count \(m_{ij}\). Linearity of conditional expectation, applied to the finite population in generation \(n\), gives \(\mathbb{E}[Z_{n+1,j}\mid\mathcal{F}_n]=\sum_{i=1}^d Z_{n,i}m_{ij}\). In row-vector form this is \(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]=Z_nM\). Taking expectations yields \(\mathbb{E}[Z_{n+1}]=\mathbb{E}[Z_n]M\). The finite means and finite number of types make each component integrable at each step: if the components of \(Z_n\) have finite expectations, the displayed conditional-mean formula gives finite expectations for \(Z_{n+1}\). Induction from the assumed integrable \(Z_0\) proves this for all \(n\). Iterating the expectation recursion gives \(\mathbb{E}[Z_n]=\mathbb{E}[Z_0]M^n\); for deterministic \(Z_0=z_0\), this is \(z_0M^n\). \(\square\)
Proof. The total population is \(\sum_{j=1}^d Z_{n,j}=Z_n\mathbf{1}\). Taking expectations and using the theorem gives \(\mathbb{E}[Z_n\mathbf{1}]=\mathbb{E}[Z_n]\mathbf{1}=z_0M^n\mathbf{1}\). \(\square\)
The recursion concerns expectations, not the exact population on a particular sample path. Individuals produce random offspring, so an observed generation can be much larger or smaller than its expected count. Matrix multiplication predicts means; it does not remove the randomness of reproduction.
Worked Example: A Two-Type Model with Fixed Offspring Counts
Suppose every type-1 individual has exactly one type-1 child and two type-2 children. Every type-2 individual has exactly two type-1 children and no type-2 children. Start with one type-1 individual. Find the expected counts in generations one, two, and three.
The offspring vectors are fixed, so their expectations are the same vectors. The rows of the mean progeny matrix are therefore \((1,2)\) and \((2,0)\):
For generation one, multiply the initial row vector by \(M\): \(z_0M=(1\cdot1+0\cdot2,\;1\cdot2+0\cdot0)=(1,2)\). For generation two, multiply the resulting row vector by \(M\): \((1,2)M=(1\cdot1+2\cdot2,\;1\cdot2+2\cdot0)=(5,2)\). For generation three: \((5,2)M=(5\cdot1+2\cdot2,\;5\cdot2+2\cdot0)=(9,10)\).
Check: Each type-1 parent contributes three children in total, and each type-2 parent contributes two. The generation-two total is \(5+2=7\), which is also \(1\cdot3+2\cdot2=7\), computed from the generation-one counts. The generation-three total is \(9+10=19\), also \(5\cdot3+2\cdot2=19\). These checks agree with the offspring totals in the model.
Worked Example: A Stochastic Two-Type Offspring Law
For a type-1 parent, let the offspring vector be \((2,0)\) with probability \(\tfrac12\) and \((0,2)\) with probability \(\tfrac12\). For a type-2 parent, let it be \((1,0)\) with probability \(\tfrac14\) and \((0,2)\) with probability \(\tfrac34\). These probabilities sum to one for each parent type: \(\tfrac12+\tfrac12=1\) and \(\tfrac14+\tfrac34=1\). Starting with one type-2 individual, compute the expected counts in generations one and two.
For a type-1 parent, the expected type-1 count is \(\tfrac12\cdot2+\tfrac12\cdot0=1\), and the expected type-2 count is \(\tfrac12\cdot0+\tfrac12\cdot2=1\). For a type-2 parent, the corresponding expectations are \(\tfrac14\cdot1+\tfrac34\cdot0=\tfrac14\) and \(\tfrac14\cdot0+\tfrac34\cdot2=\tfrac32\). Thus
The generation-one expected counts are \(z_0M=(0\cdot1+1\cdot\tfrac14,\;0\cdot1+1\cdot\tfrac32)=(\tfrac14,\tfrac32)\). For generation two, multiply once more: \[ (\tfrac14,\tfrac32)M = (\tfrac14\cdot1+\tfrac32\cdot\tfrac14,\; \tfrac14\cdot1+\tfrac32\cdot\tfrac32) = (\tfrac14+\tfrac38,\;\tfrac14+\tfrac94) = (\tfrac58,\tfrac52). \]
Check: The expected total in generation one is \(\tfrac14+\tfrac32=\tfrac74\). A type-1 parent has expected total offspring \(1+1=2\), and a type-2 parent has expected total offspring \(\tfrac14+\tfrac32=\tfrac74\). Weighting these by the generation-one expected counts gives \(\tfrac14\cdot2+\tfrac32\cdot\tfrac74=\tfrac12+\tfrac{21}{8}=\tfrac{25}{8}\). The generation-two total from the computed vector is \(\tfrac58+\tfrac52=\tfrac58+\tfrac{20}{8}=\tfrac{25}{8}\), as required.
A Three-Type Calculation
Worked Example: Three Types and Two Generations
Consider three types. A type-1 parent has offspring vector \((2,0,0)\) with probability \(\tfrac12\) and \((0,1,1)\) with probability \(\tfrac12\). A type-2 parent always has offspring vector \((0,1,2)\). A type-3 parent has offspring vector \((1,0,0)\) with probability \(\tfrac13\) and \((0,3,0)\) with probability \(\tfrac23\). Each listed offspring law has probabilities summing to one. Start with one type-1 and one type-3 individual, so \(z_0=(1,0,1)\).
For a type-1 parent, the expected offspring vector is \((\tfrac12\cdot2+\tfrac12\cdot0,\;\tfrac12\cdot0+\tfrac12\cdot1,\;\tfrac12\cdot0+\tfrac12\cdot1)=(1,\tfrac12,\tfrac12)\). The type-2 parent’s expected vector is \((0,1,2)\), since its offspring are fixed. For a type-3 parent, it is \((\tfrac13\cdot1+\tfrac23\cdot0,\;\tfrac13\cdot0+\tfrac23\cdot3,\;\tfrac13\cdot0+\tfrac23\cdot0)=(\tfrac13,2,0)\). Consequently, the three rows of \(M\) are these vectors:
The expected generation-one counts are \[ z_0M=(1,0,1)M =(1+\tfrac13,\;\tfrac12+2,\;\tfrac12+0) =(\tfrac43,\tfrac52,\tfrac12). \] The expected generation-two counts are \[ (\tfrac43,\tfrac52,\tfrac12)M = (\tfrac43+\tfrac16,\;\tfrac23+\tfrac52+1,\;\tfrac23+5) =(\tfrac32,\tfrac{25}{6},\tfrac{17}{3}). \] For the first component, the contributions are \(\tfrac43\cdot1\), \(\tfrac52\cdot0\), and \(\tfrac12\cdot\tfrac13=\tfrac16\), summing to \(\tfrac32\). For the second, they are \(\tfrac43\cdot\tfrac12=\tfrac23\), \(\tfrac52\cdot1=\tfrac52\), and \(\tfrac12\cdot2=1\), summing to \(\tfrac{25}{6}\). For the third, they are \(\tfrac43\cdot\tfrac12=\tfrac23\), \(\tfrac52\cdot2=5\), and \(\tfrac12\cdot0=0\), summing to \(\tfrac{17}{3}\).
Check: Generation one has expected total \(\tfrac43+\tfrac52+\tfrac12=\tfrac{13}{3}\). The expected total offspring per parent type is the row sum of \(M\): type 1 gives \(1+\tfrac12+\tfrac12=2\), type 2 gives \(0+1+2=3\), and type 3 gives \(\tfrac13+2+0=\tfrac73\). Applying these totals to generation one gives \(\tfrac43\cdot2+\tfrac52\cdot3+\tfrac12\cdot\tfrac73=\tfrac83+\tfrac{15}{2}+\tfrac76=\tfrac{16+45+7}{6}=\tfrac{34}{3}\). The generation-two vector sums to \(\tfrac32+\tfrac{25}{6}+\tfrac{17}{3}=\tfrac{9+25+34}{6}=\tfrac{34}{3}\), confirming the total independently.
Orientation and Interpretation
The row-vector convention determines the order of multiplication: \(z_0M^n\), not \(M^nz_0\). The row index of \(M\) is the parent type, so a row of population counts weights the rows of \(M\) to produce the next expected population. This is different from a transition matrix, whose rows sum to one because they describe probability distributions. A mean progeny matrix has no such requirement; its row sums measure average family sizes.
Common mistake: transposing the mean matrix. In the stochastic two-type example, the correct generation-one vector from one type-2 parent is \((0,1)M=(\tfrac14,\tfrac32)\). A mistaken calculation using \(M^{\mathsf T}\) gives \((0,1)M^{\mathsf T}=(1,\tfrac32)\). The first component is wrong: the type-2 parent has a type-1 child only in the outcome \((1,0)\), which occurs with probability \(\tfrac14\), so its expected type-1 offspring count is \(\tfrac14\), not \(1\). Keep parent types in rows and child types in columns, then multiply the population row vector on the left.
Mean progeny matrices are useful for comparing how different types contribute to future expected counts and for organizing multigeneration forecasts into a single calculation. Those forecasts do not, by themselves, describe the variability of the population or determine its realized path. In particular, an expected count need not be an integer, as the examples show. The matrix calculation answers a specific question—what are the expected counts by type?—and should be interpreted as such.
Check Your Understanding
Use the parent-row, child-column convention and show the arithmetic in each calculation.
- A type-1 parent produces \((1,2)\) with probability \(\tfrac12\) and \((3,0)\) with probability \(\tfrac12\). Compute its row in the mean progeny matrix.
- For \(M=\begin{bmatrix}1&2\\0&\tfrac12\end{bmatrix}\) and \(z_0=(0,2)\), find the expected generation-one count vector.
- In the three-type example, verify the type-2 component of the expected generation-two vector by adding the contributions from each type in generation one.
- Explain why a row sum of a mean progeny matrix need not equal one, and state what that row sum represents.
- State the finite-mean condition used to ensure the expected-count recursion is finite at each generation.