Offspring Distributions and Probability Generating Functions

Encoding an Offspring Law

In a Galton–Watson branching process, the offspring count of one individual is a nonnegative integer-valued random variable. Its probability generating function packages the probabilities of all possible offspring counts into a single function. For finite-support laws, this function is a polynomial: its coefficients give the probabilities, while its derivatives give useful summaries such as the mean and variance. These relationships are algebraic, but careful attention to factorials is essential.

Let \(X\) be an offspring count on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with state space \(\mathbb{N}_0=\{0,1,2,\ldots\}\). Write \(p_j=\mathbb{P}(X=j)\). The probabilities satisfy \(p_j\ge0\) and \(\sum_{j=0}^{\infty}p_j=1\). The probability generating function, abbreviated pgf, is

The definition immediately gives two useful endpoint identities. At \(s=1\), every term \(1^j\) equals \(1\), so \(f(1)=\sum_jp_j=1\). At \(s=0\), only the term for \(j=0\) remains, so \(f(0)=p_0\), the probability of no offspring. More generally, the coefficient of \(s^j\) in the pgf is the probability \(p_j\). Derivatives at zero recover those coefficients, but with a factorial multiplier.

Coefficients from Derivatives at Zero

Proof. Differentiate each monomial \(s^j\) exactly \(k\) times. If \(j<k\), its derivative is zero. If \(j=k\), its derivative is the constant \(k!\). If \(j>k\), its derivative is \(j(j-1)\cdots(j-k+1)s^{j-k}\), which is zero at \(s=0\) because \(j-k\ge1\). Therefore, in the finite sum defining \(f^{(k)}(0)\), all terms vanish except the term \(j=k\), which contributes \(k!p_k\). For \(k=0\), the same argument says \(f(0)=p_0\), since every positive power of zero vanishes. Dividing by \(k!\), which is nonzero, gives the coefficient formula. \(\square\)

For example, the coefficient of \(s^3\) is \(f'''(0)/3!\), not simply \(f'''(0)\). The denominator matters increasingly at higher orders. For a law with infinite support, the power series still determines its coefficients, and derivatives at zero can be used to extract them; derivatives at \(s=1\), however, require more care because the corresponding moments may be infinite.

Derivatives at One and Offspring Moments

For finite-support laws, every derivative exists at \(s=1\). Differentiating a monomial gives a factorial product. For \(r\ge1\), define the falling factorial \((j)_r=j(j-1)\cdots(j-r+1)\) when \(j\ge r\), and set \((j)_r=0\) when \(j<r\). Then

These are factorial moments, rather than ordinary powers of \(X\). In particular, \(f'(1)=\mathbb{E}[X]\), the mean offspring count. Since \(X^2=X(X-1)+X\), the second ordinary moment is \(f''(1)+f'(1)\). Subtracting the square of the mean gives the variance.

Proof. Write \(f(s)=\sum_{j=0}^{M}p_js^j\). For \(j\ge r\), the \(r\)th derivative of \(s^j\) is \(j(j-1)\cdots(j-r+1)s^{j-r}\); for \(j<r\), it is zero. Substituting \(s=1\) into the finite sum therefore gives \(f^{(r)}(1)=\sum_{j=r}^{M}j(j-1)\cdots(j-r+1)p_j\). By the definition of expectation for a discrete random variable, this sum is the expectation of the falling factorial. Taking \(r=1\) gives the mean, and taking \(r=2\) gives \(\mathbb{E}[X(X-1)]\). The identity \(X^2=X(X-1)+X\) holds for every integer value of \(X\), so taking expectations gives \(\mathbb{E}[X^2]=f''(1)+f'(1)\). Finally, \(\operatorname{Var}(X)=\mathbb{E}[X^2]-(\mathbb{E}[X])^2\), yielding the stated variance formula. \(\square\)

For finite support, the formulas are finite sums and do not require separate moment assumptions. For an infinite-support law, the pgf is still defined on \([0,1]\), but a derivative at \(1\) need not be finite. One should not infer a finite mean or variance merely from the existence of \(f(s)\) for \(s<1\).

Worked Finite-Support Examples

Reading a PGF Reliably

The pgf represents the entire offspring distribution, not just its mean. When the support is finite, its coefficients can be read off directly, and the derivatives at one provide moment summaries. These are different operations: derivatives at zero isolate individual probabilities, while derivatives at one combine probabilities across the support. A pgf can therefore be used both to check a proposed offspring law and to calculate quantities needed to describe it.

Common mistake: forgetting the factorial. In the first example, \(f'''(0)=3/2\). Setting \(p_3=3/2\) is wrong: it treats the third derivative as the coefficient itself and even gives a value greater than one. The correct calculation is \(p_3=f'''(0)/3!=(3/2)/6=1/4\). In general, the \(k\)th derivative at zero is \(k!p_k\), so coefficient recovery requires division by \(k!\).

A second useful check is normalization. For any proposed finite-support pgf, its coefficients must be nonnegative and sum to \(f(1)=1\). If either check fails, it cannot describe an offspring probability distribution. Likewise, when using the moment formulas, check the mean independently from \(\sum_j jp_j\) if the support is short enough to make that sum transparent.