From a Normalized Limit to a Population Forecast
A supercritical branching process has a natural growth scale: if the mean number of children per individual is \(m>1\), generation \(n\) is measured relative to \(m^n\). The normalized population can converge even when the unnormalized population becomes very large. That limit gives a practical way to compare populations and estimate probabilities of crossing thresholds, but it does not make every finite-time forecast an asymptotic one. In particular, the generation at which an observation is made matters.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((Z_n)_{n\ge0}\) be a Galton–Watson process with state space \(S=\{0,1,2,\ldots\}\), initial population \(Z_0=1\), and offspring variable \(\xi\). The offspring counts of all individuals are independent copies of \(\xi\). Write \(m=\mathbb{E}\xi\), and let \(\mathcal{F}_n\) contain the population and reproduction information through generation \(n\). Assume \(m>1\). The normalized population is \(W_n=Z_n/m^n\).
For an integer threshold \(K\), the event that the generation-\(n\) population exceeds \(K\) is exactly \(\{W_n>K/m^n\}\), with a strict inequality. If \(K\) is on the scale of \(m^n\), the threshold \(K/m^n\) can approach a positive constant. The limit law of \(W\) can then approximate the exceedance probability. A separate, exact forecast is available for the conditional mean at any observation time.
Proof. Given \(\mathcal{F}_t\), there are \(Z_t\) individuals in generation \(t\). Each has expected \(m^h\) descendants \(h\) generations later, by iterating the offspring mean: the expected number of descendants one generation later is \(m\), and conditional expectation at each subsequent generation multiplies the expectation by \(m\). Linearity of conditional expectation over the \(Z_t\) individuals gives \(Z_t m^h\). If \(Z_t=0\), there are no descendants and both sides are zero. \(\square\)
Worked Example: A Four-Generation Conditional Forecast
Suppose the offspring mean is \(m=2\), and the population observed at generation \(t=7\) is \(Z_7=3\). The conditional expected population four generations later is
This value is a conditional mean, not a claim that the population at generation 11 will be 48. Its direct check is \(48/3=16=2^4\), the four-generation mean multiplier. No large-\(t\) assumption was used.
When the Limit Gives a Growth Ratio
The normalized limit also gives a pathwise statement about growth over a fixed horizon, provided the observation generation itself tends to infinity. On survival, \(W>0\), so the population is eventually nonzero and can be used as a denominator.
Proof. On survival, \(W>0\) almost surely. Since \(W_t=Z_t/m^t\to W\), both \(W_{t+h}\) and \(W_t\) tend to the same positive value \(W\) as \(t\to\infty\), with \(h\) fixed. Thus \(W_{t+h}/W_t\to1\). For all sufficiently large \(t\), \(Z_t>0\), and the definitions give \(Z_{t+h}/Z_t=m^h(W_{t+h}/W_t)\). Taking the limit proves the claim. When \(h=0\), the ratio is exactly \(1=m^0\). \(\square\)
This result justifies \(Z_t m^h\) as a long-run, fixed-horizon growth approximation when \(t\) is large and the process survives. It does not justify deriving that ratio by dividing two limit approximations at a fixed \(t\) while \(h\to\infty\). At fixed \(t\), the denominator \(W_t\) need not be close to \(W\); the normalized limit only controls generations tending to infinity. For an arbitrary observation time, \(Z_t m^h\) remains valid as the conditional mean, but the limit theorem alone does not make it a pathwise approximation to the realized future population.
Worked Example: Same Mean, Different Long-Run Forecasts
Compare two processes with \(m=2\), each started from one individual. In the first, every individual has exactly two children. In the second, offspring have the geometric law \(\mathbb{P}(\xi=k)=\tfrac13(\tfrac23)^k\), \(k\ge0\). Its probabilities sum to \(1\) by the geometric series, and its mean is \((2/3)/(1/3)=2\). Both laws satisfy the Kesten–Stigum moment condition.
In the deterministic process \(Z_n=2^n\), so \(W_n=1\) and \(W=1\). In the geometric process the generating function is \(f(s)=1/(3-2s)\). Its extinction probability is the smaller root of \(q=f(q)\): solving \(q(3-2q)=1\) gives \(q=1/2\) or \(q=1\), hence \(q=1/2\).
For completeness, the geometric law also gives the limit distribution explicitly. For \(s<1\), set \(y(s)=1/(1-s)\). Direct substitution gives \(y(f(s))=1+y(s)/2\). Iterating this relation \(n\) times yields \(y(f^{\circ n}(s))=2(1-2^{-n})+2^{-n}y(s)\), where \(f^{\circ n}\) is the \(n\)-fold composition. Since \(\mathbb{E}[e^{-uW_n}]=f^{\circ n}(e^{-u/2^n})\), for \(u>0\) we have \(2^{-n}y(e^{-u/2^n})\to1/u\), and therefore \[ \mathbb{E}[e^{-uW}]=\lim_{n\to\infty}f^{\circ n}(e^{-u/2^n}) =1-\frac{1}{2+1/u}=\frac{u+1}{2u+1}. \] This is the Laplace transform of a distribution with mass \(1/2\) at zero and, conditional on being positive, an exponential distribution of rate \(1/2\): its transform is \(\tfrac12+\tfrac12\,\frac{1/2}{1/2+u}=\frac{u+1}{2u+1}\). By uniqueness of Laplace transforms, it is the law of \(W\).
Consider the strict threshold \(Z_n>2^n\), or equivalently \(W_n>1\). In the deterministic process this probability is exactly zero at every \(n\). In the geometric process the limiting probability is \(\mathbb{P}(W>1)=\tfrac12 e^{-1/2}\): the atom at zero contributes nothing, while the positive component has rate \(1/2\). Thus two laws with the same mean have different long-run threshold forecasts. The difference comes from the distribution of \(W\), not from the mean growth multiplier.
Scaled Thresholds and Exceedance Probabilities
Proof. On the event \(W\ne x\), almost sure convergence \(W_n\to W\) and ordinary convergence \(x_n\to x\) imply that the two comparisons \(W_n>x_n\) and \(W>x\) eventually agree. Indeed, if \(W>x\), the positive gap \(W-x\) eventually exceeds \(|W_n-W|+|x_n-x|\); if \(W<x\), the corresponding negative gap eventually keeps both comparisons false. The exceptional event \(\{W=x\}\) has probability zero by assumption. Hence the indicators \(\mathbf{1}_{\{W_n>x_n\}}\) converge almost surely to \(\mathbf{1}_{\{W>x\}}\). They are bounded by 1, so the bounded convergence theorem gives convergence of their expectations, which are the stated probabilities. Finally, \(Z_n>K_n\) is equivalent to \(W_n>K_n/m^n\), proving the branching-process version. \(\square\)
At finite \(n\), replacing \(W_n\) by \(W\) is an approximation, not an identity. The proposition explains why it becomes reliable for large generations when the scaled thresholds converge to a continuity point of the limit distribution. If \(W\) has an atom at \(x\), the proposition does not supply the limit at that threshold; extra information about how the finite-generation probabilities approach the atom would be needed.
Worked Example: A Large-Generation Threshold Estimate
Use the geometric offspring process from the previous example, and ask for the probability that generation 10 exceeds 512 individuals. Here \(m^{10}=2^{10}=1024\), so the normalized threshold is \(512/1024=1/2\). The limit law has no atom at \(1/2\), and its positive part has exponential rate \(1/2\). Therefore the large-generation estimate is
The factor \(1/2\) is the probability of being in the positive component; the exponential tail at \(1/2\) is \(e^{-(1/2)(1/2)}=e^{-1/4}\). The approximation is based on the generation-10 normalized population being close to its limit, not on an exact finite-generation identity. The strict exceedance threshold is respected: using \(>\) rather than \(\ge\) makes no difference to the limit here because \(W\) has no atom at \(1/2\).
Fixed Thresholds, Survival, and a Common Error
A fixed absolute threshold behaves differently from a threshold proportional to \(m^n\). If \(K\ge0\) is fixed, then \(K/m^n\to0\), and \(W\) has positive probability at zero whenever extinction is possible. Thus the scaled-threshold proposition cannot be applied at \(x=0\) in that case. The survival statement in the Kesten–Stigum theorem gives the appropriate conclusion directly: on extinction, \(Z_n\) eventually equals zero; on survival, \(W_n\to W>0\), so \(Z_n=m^nW_n\to\infty\). Consequently, for each fixed finite \(K\), \(\mathbb{P}(Z_n>K)\to\mathbb{P}(\text{survival})\). To see the probability limit, the indicators of \(\{Z_n>K\}\) converge almost surely to the survival indicator, and bounded convergence applies.
Common mistake. One might write \(Z_{t+h}\approx Wm^{t+h}\) and \(Z_t\approx Wm^t\), divide, and conclude \(Z_{t+h}\approx Z_tm^h\) for any fixed \(t\). The limit theorem does not guarantee that \(Z_t/m^t\) is close to \(W\) at that fixed time. For fixed \(t\), on survival the limit theorem instead gives \(Z_{t+h}/(Z_tm^h)=W_{t+h}/W_t\to W/W_t\) as \(h\to\infty\), provided \(Z_t>0\). This need not be 1. The correction is to use \(Z_tm^h\) as the exact conditional-mean forecast, or use the pathwise ratio proposition with \(t\to\infty\) and fixed \(h\).
In practice, the mean multiplier answers how the conditional expectation scales; the limit distribution answers how long-run outcomes vary around that scale. Forecasting a threshold requires specifying whether it is fixed or grows with \(m^n\), and whether the desired comparison is strict. A point mass in the limit law can affect a threshold estimate, so continuity at the scaled threshold is a substantive condition, not a formality.
Check Your Understanding
For each question, distinguish exact finite-generation identities from conclusions that require a large-generation limit.
- If \(Z_t=5\), \(m=3/2\), and \(h=2\), what is \(\mathbb{E}[Z_{t+h}\mid\mathcal{F}_t]\)?
- Why does the fixed-horizon ratio proposition require \(t\to\infty\) while \(h\) stays fixed?
- For thresholds \(K_n\) satisfying \(K_n/m^n\to x\), what condition on the distribution of \(W\) is needed for the stated exceedance limit?
- For a fixed finite \(K\), why does \(\mathbb{P}(Z_n>K)\) tend to the survival probability rather than follow directly from the scaled-threshold proposition at \(x=0\)?
- In the geometric mean-two example, calculate the limiting estimate for \(\mathbb{P}(Z_n>2^{n+1})\), taking care to use a strict threshold.