Yule Process Moments And Growth

From the Population Law to Its Moments

The Yule process models a population in which each individual gives birth independently at a constant rate. Its population distribution at a fixed time is geometric when the process begins with one individual. That simple distribution gives more than a description of the probabilities: it gives exact formulas for the expected population size and its variability. Those formulas also clarify what exponential growth in expectation does—and does not—say about an individual population path.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with expectation \(\mathbb{E}\). Let \((X_t)_{t\ge0}\) be the Yule process with state space \(S=\{0,1,2,\ldots\}\), continuous time \(t\ge0\), and per-individual birth rate \(\lambda>0\). When the population is \(i\ge1\), the total birth rate is \(\lambda i\), and each birth increases the population by one. State \(0\) is absorbing. Take the filtration \((\mathcal{F}_t)_{t\ge0}\) to represent the information in the process up to time \(t\). We use the standard non-explosive Yule process. Write \(\mathbb{P}_m\) and \(\mathbb{E}_m\) when \(X_0=m\).

For one initial individual, the population law is geometric on the positive integers. In the parameterization used here, its probability of success is \(p=e^{-\lambda t}\): for \(t>0\) and \(k\ge1\), \(\mathbb{P}_1(X_t=k)=p(1-p)^{k-1}\). At \(t=0\), \(X_0=1\) with probability one. The support starts at one because the original individual remains in the population: the Yule process has births but no deaths.

Deriving the Geometric Moments

First take \(X_0=1\) and \(t>0\). Set \(p=e^{-\lambda t}\) and \(r=1-p\), so \(0<p<1\), \(0<r<1\), and \(\mathbb{P}_1(X_t=k)=p r^{k-1}\). The geometric-series identities needed below follow by differentiating \(\sum_{k=0}^{\infty}r^k=(1-r)^{-1}\) for \(0<r<1\). The series and its differentiated series converge in that range.

Proof. Since \(1-r=p\), differentiating the geometric series gives \(\sum_{k\ge1}k r^{k-1}=1/(1-r)^2=1/p^2\). Therefore $$ \mathbb{E}_1[X_t] =p\sum_{k\ge1} k r^{k-1} =p\frac{1}{p^2} =\frac{1}{p}. $$ Differentiating twice also gives \(\sum_{k\ge1} k(k-1)r^{k-1}=2r/(1-r)^3=2r/p^3\). Thus \(\mathbb{E}_1[X_t(X_t-1)]=2r/p^2\). Using \(X_t^2=X_t(X_t-1)+X_t\), we obtain $$ \operatorname{Var}_1(X_t) =\mathbb{E}_1[X_t(X_t-1)]+\mathbb{E}_1[X_t]-(\mathbb{E}_1[X_t])^2 =\frac{2r}{p^2}+\frac{1}{p}-\frac{1}{p^2} =\frac{r}{p^2}. $$ Substituting \(p=e^{-\lambda t}\) and \(r=1-e^{-\lambda t}\) yields \(\mathbb{E}_1[X_t]=e^{\lambda t}\) and \(\operatorname{Var}_1(X_t)=e^{\lambda t}(e^{\lambda t}-1)\). At \(t=0\), \(X_0=1\) deterministically, so its mean is \(1=e^0\) and its variance is \(0=e^0(e^0-1)\). \(\square\)

For \(m\ge1\), the branching construction gives another useful description. Start \(m\) independent family processes, each begun from one individual, and add their population sizes. Their sum has the Yule process law begun from \(m\): when the total population is \(i\), its \(i\) individuals each give birth at rate \(\lambda\), so the total birth rate is \(\lambda i\). Independence means expectations and variances add. Equivalently, the probability generating function, the function \(G(s)=\mathbb{E}_m[s^{X_t}]\) for \(0\le s\le1\), is the \(m\)th power of the one-family generating function.

Proof. Let \(Y_1,\ldots,Y_m\) denote the independent family sizes at time \(t\), so \(X_t\) has the distribution of \(\sum_{\ell=1}^m Y_\ell\), with each \(Y_\ell\) having the one-ancestor Yule law. Linearity of expectation gives \(\mathbb{E}_m[X_t]=\sum_{\ell=1}^m\mathbb{E}[Y_\ell]=m e^{\lambda t}\). Independence gives \(\operatorname{Var}_m(X_t)=\sum_{\ell=1}^m\operatorname{Var}(Y_\ell)=m e^{\lambda t}(e^{\lambda t}-1)\), since the covariance between distinct family sizes is zero. When \(t=0\), every family has size one, and these same formulas give mean \(m\) and variance zero. \(\square\)

The distribution for \(m\) initial individuals can also be written explicitly. For \(t>0\), it is the sum of \(m\) independent geometric variables with common success probability \(p=e^{-\lambda t}\). Its probability generating function is $$ G(s)=\left(\frac{ps}{1-(1-p)s}\right)^m, $$ and its probability at \(k\ge m\) is \(\binom{k-1}{m-1}p^m(1-p)^{k-m}\). This is a negative binomial distribution. The moment result above does not require summing this probability formula term by term; independence reduces it to the one-ancestor calculation.

Several Initial Individuals

When there is more than one initial individual, the mean and variance both scale linearly with the initial population \(m\), not quadratically. The mean adds across families. For the variance, independence is essential: it makes the variance of the sum equal to the sum of the family variances. The distribution’s support also shifts: starting from \(m\ge1\), the population cannot be less than \(m\).

Expected Growth and Relative Fluctuations

The mean \(m e^{\lambda t}\) grows exponentially in time, at rate \(\lambda\). This is growth of the expected population size across the process’s probability distribution; it does not say that \(X_t\) equals \(m e^{\lambda t}\), or even that an individual realization follows a smooth curve. The variance records the spread around the mean. For \(m\ge1\) and \(t\ge0\), the squared coefficient of variation—variance divided by the squared mean—is $$ \frac{\operatorname{Var}_m(X_t)}{(\mathbb{E}_m[X_t])^2} =\frac{m e^{\lambda t}(e^{\lambda t}-1)}{m^2e^{2\lambda t}} =\frac{1}{m}(1-e^{-\lambda t}). $$ For fixed \(m\ge1\) and \(\lambda>0\), this ratio increases from zero at \(t=0\) toward \(1/m\) as \(t\) grows. Thus the standard deviation becomes large in absolute terms as the mean grows, while the relative standard deviation, obtained by taking the square root of this ratio, remains bounded and approaches \(1/\sqrt{m}\). In particular, starting from one individual does not make relative variability disappear as time increases.

A Common Parameterization Pitfall

A common error is to use the variance formula for a geometric random variable on \(\{1,2,\ldots\}\) as though it were \((1-p)/p\). That expression belongs to a different geometric convention: the number of failures before the first success, supported on \(\{0,1,2,\ldots\}\). For the Yule population, the support begins at one and the law is \(\mathbb{P}_1(X_t=k)=p(1-p)^{k-1}\); its variance is \((1-p)/p^2\).

Wrong calculation: In the one-ancestor example with \(p=1/2\), using \((1-p)/p\) gives \((1/2)/(1/2)=1\). Correction: The support-\(\{1,2,\ldots\}\) formula gives \((1-p)/p^2=(1/2)/(1/4)=2\), consistent with the direct check \(4+2-2^2=2\). Check the support and the probability formula before selecting a geometric-moment identity.

The case \(t=0\) is also worth checking separately when using a geometric series: then \(p=1\) and \(r=0\), so the distribution is degenerate at \(1\), not a nondegenerate geometric distribution. The moment formulas remain valid at \(t=0\) by direct substitution or by observing \(X_0=1\) deterministically. If \(X_0=0\), the absorbing state gives \(X_t=0\) for all \(t\), so formulas for positive initial populations should not be applied without this separate case.