Crump-Mode-Jagers Branching Process: A Numerical Approach

Abstract

The theory of Crump-Mode-Jagers branching processes gives us the expected future population as a solution of a renewal equation. In order to find the expected future population age structure a large amount of renewal equations must be solved. This paper presents a numerical approach for projecting the population age structure and solving these renewal equations based on the theory of General Branching Processes. It is shown that the Leslie matrix projection, widely used in demographics, is actually a special case of this method. Applying the continuous time theory of branching processes produces estimation error for the Leslie matrix projections that comes from discretization of time. The presented numerical method can also be used for solving renewal equations satisfying certain conditions of smoothness. It involves only simple matrix multiplications which results in a very fast calculation speed.

Appendix

Example 1

Forecasting the population age structure of Bulgaria.

In order to find the expected future population of Bulgaria we need to find models for the functions S(t) and μ(t) from empirical data. The data used for calculations in this paper are published by Eurostat. One knows the number of live births by mother’s age, the number of deaths by age and the number of people by age. The demographic theory uses these data to produce the life-tables (see Keyfitz [6]). The Chiang’s formula (see Chiang [3]) presents the relation between age-specific birth and death rates and the probabilities for birth and death by age. Using demographics we can find the distribution of the life length and the probability of a woman giving birth at a specific age from the demographic coefficients.

The empirical estimations of these age-specific probabilities suggest the assumption of smoothness for μ(t) and S(t) is appropriate in case of human populations. Using smoothing splines, we can find the functional model that generated the data (see Ramsay [10] and Boor [2]). The resulting functions for μ(t) and S(t) are presented in Figs. 10.1, 10.2, and 10.3.

Fig. 10.1

Point process density function μ′(t), Bulgaria 2013

Fig. 10.2

Expectation of point process μ(t), Bulgaria 2013

Fig. 10.3

Survival probability S(t), Bulgaria 2013

We know the initial age structure of the population N ₀(bh; (b + 1)h], i.e. the number of all women on a specific age at the beginning of 2013. Having modelled the life length of each individual and her point process from real data (published by Eurostat), we can substitute those in Corollary 10.2 and Theorem 10.2. On each step of the matrix multiplication, we obtain the expected future population age structure. The resulting total population count is presented in Fig. 10.4.

Fig. 10.4

Female population count

Example 2

Numerical solution to renewal equations.

Let us consider a Poisson renewal process, i.e. the inter-renewal time is exponentially distributed according to the distribution function F(x) = 1 − e ^{−λ x}. It is well-known that the number of renewals until time t follows Poisson distribution and the expected number of renewals until time t is given by $\mathbb{E}N(t) =\lambda t$. The function $U(t) = \mathbb{E}N(t) + 1 = 1 +\lambda t$ is called renewal function and it satisfies the following renewal equation.

$$\displaystyle{ U(t) = I(t) +\int \limits _{ 0}^{t}U(t - u)dF(u), }$$

(10.5)

where I(t) = 1 for t ≥ 0. If we consider the general equation

$$\displaystyle{ Z(t) = z(t) +\int \limits _{ 0}^{t}Z(t - u)dF(u), }$$

(10.6)

then (from renewal theory) its solution is Z(t) = (U ∗ z)(t). To see how the numerical method in Corollary 10.2 works we will use it for solving Eq. (10.5) and then Eq. (10.6), choosing for example z(t) = e ^−t.

From Eq. (10.5) it is obvious that S(t) = 1 for every t > 0 and

$$\displaystyle{ \begin{array}{rl} _{bh}\mu (h)& = \frac{\mu ((b+1)h)-\mu (bh)} {S(bh)} = e^{-\lambda bh}(1 - e^{-\lambda h}) \end{array} }$$

for every b. From _b μ(t) = [μ(t + b) −μ(b)]∕S(b) we can calculate the derivative _b μ′(t) = μ′(t + b)∕S(b). Then _bh μ′(0) = μ′(bh)∕S(bh) = μ′(bh) = λ e ^{−λ b h}.

In this case ω = +∞ but for the purpose of calculating the U(t) we can assume ω = t or we can assign some value that is greater than t to ω and the results will not change. However, it is more convenient for calculations to use the smallest possible ω = t so that the matrix A has the following form:

$$\displaystyle{ A = \left [\begin{array}{*{10}c} (1 - e^{-\lambda h})& \frac{(1-e^{-\lambda h})} {e^{-\lambda h}} & \ldots & \frac{(1-e^{-\lambda h})} {e^{-\lambda (t-2h)}} & \frac{(1-e^{-\lambda h})} {e^{-\lambda (t-h)}} \\ 1 & 0 &\ldots & 0 & 0\\ 0 & 1 &\ldots & 0 & 0\\ \vdots & \vdots &\ddots & \vdots & \vdots \\ 0 & 0 &\ldots & 1 & 0 \end{array} \right ]. }$$

Let us denote by $\hat{U } (t)$ the numerical solution of Eq. (10.5). It can be shown that $A^{k}\cdot \left [\begin{array}{*{10}c} 1&0&\ldots &0\\ \end{array} \right ]^{\intercal }\approx \left [\begin{array}{*{10}c} \lambda h&\ldots &\lambda h&1&0&\ldots &0\\ \end{array} \right ]^{\intercal }$ and $\hat{U }(kh) = [1]\cdot A^{k}\cdot \left [\begin{array}{*{10}c} 1&0&\ldots &0\\ \end{array} \right ]^{\intercal } = 1+(kh)\lambda +k\cdot O(h^{2}) = U(kh)+k\cdot O(h^{2})$ for every k = 1, …, t∕h, using Taylor expansion of e ^x. For k = t∕h, we obtain $U(t) = \hat{U } (t) + O(h)$.

Let us consider the second equation (10.6). Its theoretical solution is given by the Riemann-Stieltjes integral

$$\displaystyle{ \begin{array}{rl} Z(t)& =\int \limits _{ 0}^{t}e^{-(t-u)}d(1 +\lambda u) = e^{-t} +\lambda \int \limits _{ 0}^{t}e^{u-t}d(u - t) \\ & = e^{-t} +\lambda e^{u-t}\vert _{0}^{t} = e^{-t} +\lambda (1 - e^{-t}).\end{array} }$$

Note that the function U has a jump for t = 0: U(0) = 1 and U(t) = 0 for t < 0, so the term U(0) ⋅ z(t − 0) = e ^−t is added when calculating the Stieltjes integral.

In this particular case, we have S(t) = e ^−t, _bh S(h) = S((b + 1)h)∕S(bh) = e ^−h and _bh μ(h) = [μ((b + 1)h) −μ(bh)]∕S(bh) = e ^{−λ b h}(1 − e ^{−λ h})∕e ^−bh for every b. Then,

$$\displaystyle{ A = \left [\begin{array}{*{10}c} (1 - e^{-\lambda h})& \frac{e^{-\lambda h}(1-e^{-\lambda h})} {e^{-h}} & \ldots & \frac{e^{-\lambda (t-2h)}(1-e^{-\lambda h})} {e^{-(t-2h)}} & \frac{e^{-\lambda (t-h)}(1-e^{-\lambda h})} {e^{-(t-h)}} \\ e^{-h} & 0 &\ldots & 0 & 0 \\ 0 & e^{-h} &\ldots & 0 & 0\\ \vdots & \vdots &\ddots & \vdots & \vdots \\ 0 & 0 &\ldots & e^{-h} & 0 \end{array} \right ]. }$$

Since it can be shown by recursion that

$$\displaystyle{A^{k}\left [\begin{array}{*{10}c} 1&0&\ldots &0\\ \end{array} \right ]^{\intercal }\approx \left [\begin{array}{*{10}c} \lambda h&\ldots &\lambda h&(1 - kh)&0&\ldots &0\\ \end{array} \right ]^{\intercal },}$$

the numerical approximation of the function Z(t) is $\hat{Z} (kh) = [1]\cdot A^{k}\cdot \left [\begin{array}{*{10}c} 1&0&\ldots &0\\ \end{array} \right ]^{\intercal } = 1+(kh)\lambda -kh+k\cdot O(h^{2})$ for every k = 1, …, t∕h. The theoretical solution is Z(kh) = e ^−kh +λ(1 − e ^−kh) = 1 − kh +λ ⋅ kh + O(h ²) consequently it follows that $U(kh) = \hat{U } (kh) + k \cdot O(h^{2})$. When k = t∕h we get $U(t) = \hat{U } (t) + O(h)$. For a fixed t, we obtain the numerical solution of the renewal equation inside the interval [0, t] with estimation error O(h).