Near-Critical Survival Probability From a Taylor Expansion

Why Near-Critical Survival Is Small

In a supercritical Galton–Watson process, the mean number of offspring exceeds one, so survival has positive probability under the usual nondegeneracy condition. When that mean is only slightly above one, survival is typically rare. The offspring generating function gives a way to quantify how rare: its local curvature near one determines the leading survival probability.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, and let $(Z_n)_{n\ge 0}$ be a Galton–Watson process in discrete time with state space $S=\{0,1,2,\ldots\}$. Each individual independently has $\xi$ offspring, where $\xi$ is a nonnegative integer-valued random variable. The process starts with $Z_0=1$, and $Z_{n+1}$ is the total number of offspring of the $Z_n$ individuals. State $0$ is absorbing. Write $p_k=\mathbb{P}(\xi=k)$, $m=\mathbb{E}\xi$, and $G(s)=\mathbb{E}[s^\xi]=\sum_{k\ge0}p_ks^k$ for $0\le s\le1$. The extinction probability is $q=\mathbb{P}(Z_n=0\text{ for some }n\ge0)$, and the survival probability is $\rho=1-q$.

Definition: A near-critical family here is a collection of offspring laws indexed by $\varepsilon>0$, with means $m_\varepsilon=1+\varepsilon$. For each law write $G_\varepsilon$ for its generating function, $q_\varepsilon$ for its extinction probability, and $\rho_\varepsilon=1-q_\varepsilon$ for its survival probability.

The result below makes precise what it means for the family to approach a nondegenerate critical law. In addition to approaching critical mean, the laws must not shift a significant amount of their offspring variation into rarer and rarer large values.

Theorem (Near-critical survival approximation): Suppose that, as $\varepsilon\downarrow0$, the offspring probabilities satisfy $p_{\varepsilon,k}\to p_{0,k}$ for every $k\ge0$, and that $\sup_{\varepsilon}\mathbb{E}[\xi_\varepsilon^3]<\infty$. Suppose also that the limiting law has mean $1$, variance $\sigma^2$ with $0<\sigma^2<\infty$, and $p_{0,0}>0$. For sufficiently small $\varepsilon>0$, the family is supercritical with positive survival probability, and $$ \lim_{\varepsilon\downarrow0}\frac{\rho_\varepsilon}{\varepsilon}=\frac{2}{\sigma^2}. $$

The coefficient involves the variance of the limiting critical law, not a variance evaluated at an unrelated fixed parameter. The third-moment bound is a convenient sufficient condition for the Taylor remainder to be controlled uniformly across the family.

Fixed Points and the Expansion

First, the survival probability satisfies a fixed-point equation. The extinction probability from one individual is $q_\varepsilon$. Conditioning on that individual’s offspring gives $q_\varepsilon=\mathbb{E}[q_\varepsilon^{\xi_\varepsilon}]=G_\varepsilon(q_\varepsilon)$. To see why this identifies the extinction probability, let $q_{\varepsilon,n}=\mathbb{P}(Z_n=0)$. Then $q_{\varepsilon,0}=0$, and independence of the offspring lines gives $q_{\varepsilon,n+1}=G_\varepsilon(q_{\varepsilon,n})$. These probabilities increase to $q_\varepsilon$: extinction by generation $n$ increases to eventual extinction. Since $G_\varepsilon$ is continuous on $[0,1]$, taking limits gives the fixed-point equation. Moreover, this limit is the smallest fixed point in $[0,1]$: if $r=G_\varepsilon(r)$, then induction gives $q_{\varepsilon,n}\le r$ for every $n$, starting from $0\le r$.

For small enough $\varepsilon$, $p_{\varepsilon,0}>0$, since it converges to $p_{0,0}>0$. Also, the mean exceeds one, so some offspring value is at least two with positive probability. Thus $G_\varepsilon(s)-s$ is strictly convex on $0<s<1$, is positive at $s=0$, and has derivative $m_\varepsilon-1=\varepsilon>0$ at $s=1$. It is negative just to the left of $1$, so it has a root below $1$. The smallest fixed point is therefore below $1$, and $\rho_\varepsilon>0$.

Substitute $q_\varepsilon=1-\rho_\varepsilon$ into the fixed-point equation. The Taylor expansion at $1$ has linear coefficient $m_\varepsilon$ and quadratic coefficient $\mathbb{E}[\xi_\varepsilon(\xi_\varepsilon-1)]/2$. We now justify the expansion and the limiting steps used in the theorem.

Proof. Let $a_\varepsilon=\mathbb{E}[\xi_\varepsilon(\xi_\varepsilon-1)]$. For each integer $k\ge0$, Taylor’s formula applied to $s^k$ at $s=1$ gives, for $0\le r\le1$, $$ (1-r)^k=1-kr+\frac{k(k-1)}{2}r^2+R_k(r), \qquad |R_k(r)|\le \frac{k^3}{6}r^3. $$ For $k=0,1,2$ the remainder is zero. For $k\ge3$, the third derivative on $[0,1]$ is bounded by $k(k-1)(k-2)\le k^3$, which gives the displayed remainder bound. Taking expectations and using the uniform third-moment bound yields $$ G_\varepsilon(1-r)=1-(1+\varepsilon)r+\frac{a_\varepsilon}{2}r^2+R_\varepsilon(r), \qquad |R_\varepsilon(r)|\le \frac{M}{6}r^3, $$ where $M=\sup_\varepsilon\mathbb{E}[\xi_\varepsilon^3]<\infty$.

We need both $a_\varepsilon\to\sigma^2$ and $\rho_\varepsilon\to0$. The third-moment bound makes the second moments uniformly integrable: for $K\ge1$, $$ \mathbb{E}\!\left[\xi_\varepsilon^2\mathbf{1}_{\{\xi_\varepsilon>K\}}\right] \le \frac{\mathbb{E}[\xi_\varepsilon^3]}{K}\le\frac{M}{K}. $$ Pointwise convergence of the offspring probabilities, followed by truncating the sums at $K$ and then letting $K\to\infty$, therefore gives convergence of first and second moments. In particular, $a_\varepsilon=\mathbb{E}[\xi_\varepsilon^2]-m_\varepsilon\to \mathbb{E}[\xi_0^2]-1=\sigma^2$.

The generating functions converge uniformly on $[0,1]$. Indeed, after truncating each generating-function sum at $K$, its finite part converges uniformly because each coefficient converges; the remaining part is bounded by $\mathbb{P}(\xi_\varepsilon>K)\le M/K^3$, uniformly in $s\in[0,1]$. The limiting critical generating function $G_0$ has no fixed point below $1$: strict convexity and $G_0'(1)=1$ imply $G_0(s)>G_0(1)+G_0'(1)(s-1)=s$ for $0\le s<1$. Strict convexity follows from positive variance, which for a mean-one nonnegative integer-valued law implies positive probability of an offspring value at least two.

Now take any sequence $\varepsilon\downarrow0$. The numbers $\rho_\varepsilon\in[0,1]$ have a convergent subsequence, say with limit $r$. Uniform convergence of $G_\varepsilon$ and the fixed-point equation imply $G_0(1-r)=1-r$. The limiting critical generating function has only the fixed point $1$, so $r=0$. Every convergent subsequence has this limit; hence $\rho_\varepsilon\to0$.

Finally, set $r=\rho_\varepsilon>0$ in the expansion and use $G_\varepsilon(1-r)=1-r$. Cancelling the constant terms and dividing by $r$ gives $$ \varepsilon=\frac{a_\varepsilon}{2}\rho_\varepsilon+ \frac{R_\varepsilon(\rho_\varepsilon)}{\rho_\varepsilon}. $$ The absolute value of the last term is at most $M\rho_\varepsilon^2/6$, which tends to zero. Since $a_\varepsilon\to\sigma^2>0$, it follows that $\varepsilon/\rho_\varepsilon\to\sigma^2/2$, or equivalently $\rho_\varepsilon/\varepsilon\to2/\sigma^2$. $\square$

Worked Calculations

Worked Example: Offspring Counts Zero or Two

For $0<\varepsilon<1$, let $\mathbb{P}(\xi_\varepsilon=2)=(1+\varepsilon)/2$ and $\mathbb{P}(\xi_\varepsilon=0)=(1-\varepsilon)/2$. These probabilities sum to $(1+\varepsilon+1-\varepsilon)/2=1$. The mean is $2(1+\varepsilon)/2=1+\varepsilon$. At $\varepsilon=0$, the limiting law has variance $\mathbb{E}\xi_0^2-(\mathbb{E}\xi_0)^2=1-1=1$.

The generating function is $G_\varepsilon(s)=(1-\varepsilon)/2+(1+\varepsilon)s^2/2$. The fixed-point equation gives $$ (1+\varepsilon)q_\varepsilon^2-2q_\varepsilon+1-\varepsilon=0 =(q_\varepsilon-1)\big((1+\varepsilon)q_\varepsilon-(1-\varepsilon)\big). $$ The root below $1$ is $q_\varepsilon=(1-\varepsilon)/(1+\varepsilon)$, so $$ \rho_\varepsilon=1-q_\varepsilon=\frac{2\varepsilon}{1+\varepsilon}. $$ At $\varepsilon=1/4$, the probabilities are $5/8$ for two offspring and $3/8$ for zero; $q_\varepsilon=(3/8)/(5/8)=3/5$, and $\rho_\varepsilon=2/5$. The near-critical leading term is $2\varepsilon/\sigma^2=2(1/4)/1=1/2$. The exact value $2/5$ is below that leading term, and the ratio $\rho_\varepsilon/\varepsilon=2/(1+\varepsilon)$ tends to $2=2/\sigma^2$ as required.

Worked Example: Offspring Counts Zero or Three

Let $\mathbb{P}(\xi_\varepsilon=3)=(1+\varepsilon)/3$ and $\mathbb{P}(\xi_\varepsilon=0)=(2-\varepsilon)/3$, for $0<\varepsilon<2$. The probabilities sum to $(1+\varepsilon+2-\varepsilon)/3=1$, and the mean is $3(1+\varepsilon)/3=1+\varepsilon$. The limiting critical law has second moment $9(1/3)=3$, so its variance is $3-1^2=2$.

Write $p=(1+\varepsilon)/3$. The fixed-point equation is $q_\varepsilon=1-p+pq_\varepsilon^3$, or $$ p q_\varepsilon^3-q_\varepsilon+1-p =(q_\varepsilon-1)\big(pq_\varepsilon^2+pq_\varepsilon+p-1\big)=0. $$ The quadratic factor has roots $\big(-1\pm\sqrt{4/p-3}\big)/2$. The root in $[0,1)$ uses the plus sign, so $\rho_\varepsilon=(3-\sqrt{4/p-3})/2$. For $\varepsilon=1/5$, $p=2/5$, $4/p-3=10-3=7$, and $\rho_\varepsilon=(3-\sqrt7)/2$. Substitution checks the fixed point: the chosen $q_\varepsilon=(\sqrt7-1)/2$ solves the quadratic factor, so the factored equation is zero. The leading approximation is $2\varepsilon/\sigma^2=2(1/5)/2=1/5$; the exact expression has ratio $\rho_\varepsilon/\varepsilon\to1=2/\sigma^2$ as $\varepsilon\downarrow0$.

Worked Example: A Three-Point Offspring Law

For $0<\varepsilon<1/4$, assign probabilities $p_{\varepsilon,0}=1/4-\varepsilon$, $p_{\varepsilon,1}=1/2+\varepsilon$, and $p_{\varepsilon,2}=1/4$. Their sum is $1/4-\varepsilon+1/2+\varepsilon+1/4=1$, and the mean is $0p_{\varepsilon,0}+p_{\varepsilon,1}+2p_{\varepsilon,2}=1+\varepsilon$. At criticality the variance is $1/2-1^2=1/2$.

Here $G_\varepsilon(s)=1/4-\varepsilon+(1/2+\varepsilon)s+s^2/4$. Subtracting $s$ and factoring gives $$ G_\varepsilon(s)-s=\frac14(s-1)(s-(1-4\varepsilon)). $$ Thus the extinction root below $1$ is $q_\varepsilon=1-4\varepsilon$, which lies in $(0,1)$ under the stated condition, and $\rho_\varepsilon=4\varepsilon$. At $\varepsilon=1/8$, the probabilities are $1/8,5/8,1/4$, summing to $1$; the roots are $1$ and $1/2$, so survival is $1/2$. The leading approximation is $2(1/8)/(1/2)=1/2$, exactly the same here. In general, $4\varepsilon/\varepsilon=4=2/(1/2)$, verifying the predicted coefficient.

What the Approximation Does—and Does Not—Say

The approximation is a leading-order statement as $\varepsilon\downarrow0$. It does not claim that $\rho_\varepsilon=2\varepsilon/\sigma^2$ exactly at a fixed positive distance from criticality. The first example shows why: at $\varepsilon=1/4$, the leading expression is $1/2$, whereas the exact survival probability is $2/5$.

Common mistake. In the Taylor expansion the quadratic coefficient is $\mathbb{E}[\xi_\varepsilon(\xi_\varepsilon-1)]/2$, not $\operatorname{Var}(\xi_\varepsilon)/2$. For the zero-or-two law at $\varepsilon=1/4$, $\operatorname{Var}(\xi_\varepsilon)=1-(1/4)^2=15/16$, while $\mathbb{E}[\xi_\varepsilon(\xi_\varepsilon-1)]=2(5/8)=5/4$. Replacing the quadratic coefficient by half the variance would give the finite-parameter expression $2\varepsilon/\operatorname{Var}(\xi_\varepsilon)= (1/2)/(15/16)=8/15$, not the exact $2/5$. The correction is to use the factorial second moment in the expansion. In the near-critical limit, that moment tends to $\sigma^2$, because $m_\varepsilon\to1$ and $\mathbb{E}[\xi_\varepsilon(\xi_\varepsilon-1)]=\operatorname{Var}(\xi_\varepsilon)+m_\varepsilon^2-m_\varepsilon$.

Finite positive variance for each member of a family, by itself, does not guarantee a uniform Taylor approximation as the mean changes. A sequence of laws could put progressively smaller probability on progressively larger offspring counts, while each individual law still has finite moments. The uniform moment condition in the theorem rules out this uncontrolled movement of mass and makes the remainder bound valid at the shrinking survival probability. More refined conditions can replace the third-moment bound, but they must still control the second-order expansion uniformly.

The practical interpretation is that, under regular approach to a nondegenerate critical offspring law, a small increase $\varepsilon$ in mean produces survival probability approximately $2\varepsilon/\sigma^2$. Greater variance in the limiting critical offspring count lowers the leading survival probability: variation creates more chance of unusually small families, including extinction, even when the mean is just above one.

Check Your Understanding

Use the fixed-point equation and the limiting variance to justify each answer.

For the zero-or-two family, find the exact extinction and survival probabilities at $\varepsilon=1/5$, and compare survival with its leading approximation.
Explain why the Taylor coefficient of $r^2$ in $G_\varepsilon(1-r)$ is $\mathbb{E}[\xi_\varepsilon(\xi_\varepsilon-1)]/2$.
For a limiting critical law with variance $\sigma^2=3$, what is the predicted limit of $\rho_\varepsilon/\varepsilon$, assuming the theorem’s hypotheses?
Why does convergence of the survival probabilities to zero matter when dividing the fixed-point equation by $\rho_\varepsilon$?
Name one role played by the uniform third-moment bound in the proof.

Near-Critical Survival Probability From a Taylor Expansion

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