Convexity and the Fixed-Point Geometry of Extinction

Why the Shape of the Generating Function Matters

For a Galton–Watson process, extinction probabilities are linked to fixed points of the offspring generating function. That equation is useful only if we understand how many solutions it can have and where they lie. Convexity supplies that geometric control: except for a special linear case, there can be at most two fixed points, and the behavior at the endpoints tells us when zero itself is one of them.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((Z_n)_{n\ge 0}\) be a discrete-time Galton–Watson process with state space \(S=\mathbb{N}_0\), started from one individual, so \(Z_0=1\). Each individual has an independent number \(\xi\in\mathbb{N}_0\) of offspring, with probabilities \(p_k=\mathbb{P}(\xi=k)\) satisfying \(p_k\ge 0\) and \(\sum_{k=0}^{\infty}p_k=1\). Its offspring generating function is \(f(s)=\mathbb{E}[s^\xi]=\sum_{k=0}^{\infty}p_k s^k\) for \(s\in[0,1]\). The extinction probability is \(q=\mathbb{P}(Z_n=0\text{ for some }n\ge 0)\). The earlier material on extinction probability as a fixed point establishes that \(q\) is the smallest solution of \(f(s)=s\) in \([0,1]\). Here we study the geometry of all such solutions.

Monotonicity and Convexity

For \(0\le s\le t\le1\), each power satisfies \(s^k\le t^k\), so summing with nonnegative weights gives \(f(s)\le f(t)\). Hence \(f\) is nondecreasing. For each integer \(k\ge0\), the function \(s\mapsto s^k\) is convex on \([0,1]\); the \(k=0\) and \(k=1\) terms are constant and linear, respectively. A nonnegative weighted sum of convex functions is convex, so \(f\) is convex. The series defines a continuous function on \([0,1]\): if \(s_r\to s\), then \(s_r^\xi\to s^\xi\) and \(0\le s_r^\xi\le1\), so bounded convergence gives \(f(s_r)\to f(s)\). Subtracting the linear function \(s\) preserves convexity, so \(g\) is convex too.

Proof. For \(k\ge2\), \(s\mapsto s^k\) is strictly convex on \([0,1]\). If \(\mathbb{P}(\xi\ge2)>0\), at least one such term has positive weight. For distinct \(x,y\in[0,1]\) and \(0<\theta<1\), that term satisfies \((\theta x+(1-\theta)y)^k<\theta x^k+(1-\theta)y^k\), while all other terms satisfy the corresponding non-strict convexity inequality. Summing proves strict convexity of \(f\); subtracting \(s\) proves strict convexity of \(g\).

If \(g\) had three distinct zeros \(a<b<c\), strict convexity would give \(g(b)<\frac{c-b}{c-a}g(a)+\frac{b-a}{c-a}g(c)=0\), contradicting \(g(b)=0\). Since \(g(1)=0\), there is at most one other zero. Also, \(g(0)=f(0)=p_0\). If \(p_0=0\), then \(g(0)=g(1)=0\), so these are two zeros; strict convexity gives \(g(s)<(1-s)g(0)+s g(1)=0\) for every \(0<s<1\), ruling out any others. If \(p_0>0\), zero is not a root, and there can be at most one root other than \(1\), necessarily in \((0,1)\). Finally, if \(\mathbb{P}(\xi\ge2)=0\), then \(p_0+p_1=1\) and \(f(s)=p_0+p_1s\). Thus \(g(s)=p_0(1-s)\), which has only the root \(1\) when \(p_0>0\), and is identically zero when \(p_0=0\). \(\square\)

The theorem counts roots but does not always decide whether an interior root exists. When \(p_0>0\) and the law gives positive probability to at least two offspring, the mean offspring number provides a useful test. Write \(m=\mathbb{E}[\xi]=\sum_{k=0}^{\infty}kp_k\), allowing \(m=\infty\).

Proof. For \(0\le s<1\), the finite geometric-sum identity gives \(\frac{1-f(s)}{1-s}=\sum_{k=0}^{\infty}p_k(1+s+\cdots+s^{k-1})\), where the \(k=0\) term is zero. As \(s\) increases to \(1\), each inner sum increases to \(k\). By the monotone convergence theorem, this ratio tends to \(m\), possibly infinity. Therefore \(\frac{g(s)}{1-s}=1-\frac{1-f(s)}{1-s}\) tends to \(1-m\). If \(m>1\), the ratio is negative for \(s\) sufficiently close to \(1\), so \(g(s)<0\) there. Since \(g(0)=p_0>0\) and \(g\) is continuous, the intermediate value theorem gives a zero in \((0,1)\); the theorem above makes it unique.

Now suppose \(m\le1\). For \(s<1\), convexity implies that the secant slope \(\frac{g(1)-g(s)}{1-s}\) is at most the left derivative of \(g\) at \(1\), which is \(m-1\). This secant-slope inequality follows from convexity: slopes of chords ending at \(1\) do not exceed their limiting value as the left endpoint approaches \(1\). The left derivative is \(m-1\), as also follows from the displayed ratio limit. Thus \(-g(s)/(1-s)\le m-1\le0\), so \(g(s)\ge0\). If \(g(s)=0\) for some \(s<1\), strict convexity and \(g(1)=0\) would give \(g(x)<0\) for every \(s<x<1\), contradicting \(g(x)\ge0\). Hence no such root exists. \(\square\)

Worked Fixed-Point Geometries

Reading the Graph Without Losing a Root

For a strictly convex generating function, the graph can meet the line \(y=s\) at no more than two points, and \(s=1\) is guaranteed to be one of them. The value at zero determines whether the other possible intersection can be the endpoint \(0\): \(f(0)=p_0\), so zero is fixed exactly when \(p_0=0\). When \(p_0>0\), any second intersection must be strictly between zero and one. The offspring mean offers a further test in that case, but it does not replace checking the endpoint assumptions.

Common mistake. For \(f(s)=\tfrac12s+\tfrac12s^2\), one might set \(f(s)=s\), divide both sides by \(s\), and obtain \(\tfrac12+\tfrac12s=1\), hence \(s=1\). The division is valid only when \(s\ne0\), so it discards the root \(s=0\). The correction is to rearrange without division: \(\tfrac12s(s-1)=0\), giving both roots. Always check \(s=0\) separately before dividing by \(s\).