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Branching Processes: Extensions · Tutorial 371 of 994

Population Forecasts From the Supercritical Limit

Learn how the normalized limit translates into long-run growth forecasts and threshold probabilities, and when those approximations are justified.

Advanced 12 min read

What You'll Learn

  • State the assumptions and implications of the supercritical normalized population limit
  • Derive conditional-mean forecasts from the branching property
  • Justify forecasts over a fixed horizon when the observation generation grows
  • Convert scaled population thresholds into limiting exceedance probabilities
  • Compare long-run forecasts for offspring laws with the same mean
  • Recognize why fixed thresholds and fixed observation times require care

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 \(W_n=Z_n/m^n\).

Standard result (Kesten–Stigum limit theorem): If \(\mathbb{E}[\xi\log^+\xi]<\infty\), where \(\log^+x=\max(\log x,0)\) for \(x>0\) and \(\log^+0=0\), then \(W_n\) converges almost surely and in \(L^1\) to a random variable \(W\), with \(\mathbb{E}W=1\). Moreover, \(W>0\) on survival and \(W=0\) on eventual extinction, up to events of probability zero. This is the Kesten–Stigum theorem from the earlier material; its proof is not repeated here.

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.

Proposition (conditional-mean forecast): For every \(t,h\ge0\), \(\mathbb{E}[Z_{t+h}\mid\mathcal{F}_t]=Z_t m^h\). This is an exact identity, not a conclusion that requires \(t\) to be large.

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

$$ \mathbb{E}[Z_{11}\mid\mathcal{F}_7]=Z_7m^4=3\cdot2^4=3\cdot16=48. $$

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.

Proposition (fixed-horizon growth ratio): Under the Kesten–Stigum assumptions, for each fixed integer \(h\ge0\), on the survival event, \(Z_{t+h}/Z_t\to m^h\) almost surely as \(t\to\infty\).

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

Proposition (scaled-threshold limit): Suppose \(W_n\to W\) almost surely, \(x_n\to x\), and \(\mathbb{P}(W=x)=0\). Then \(\mathbb{P}(W_n>x_n)\to\mathbb{P}(W>x)\). In a branching process, if \(K_n\) are integer thresholds with \(K_n/m^n\to x\), this gives \(\mathbb{P}(Z_n>K_n)\to\mathbb{P}(W>x)\), under the same no-atom condition.

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

$$ \mathbb{P}(Z_{10}>512)\ \approx\ \mathbb{P}(W>1/2) =\frac12\exp\left(-\frac12\cdot\frac12\right) =\frac12 e^{-1/4}. $$

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.

  1. If \(Z_t=5\), \(m=3/2\), and \(h=2\), what is \(\mathbb{E}[Z_{t+h}\mid\mathcal{F}_t]\)?
  2. Why does the fixed-horizon ratio proposition require \(t\to\infty\) while \(h\) stays fixed?
  3. 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?
  4. 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 \(x=0\)?
  5. 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.