The Yule Process Population Distribution

From Birth Rates to Population Probabilities

A Yule process models a population in which each individual gives birth independently at the same constant rate, with no deaths. Because the total birth rate increases with the population, its transition rates depend on the current state. The forward equations record how probability flows between successive population sizes. Solving those equations from a one-individual initial state gives a geometric distribution at each positive time.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, and let $(X_t)_{t\ge0}$ be a continuous-time Markov chain with state space $S=\{1,2,3,\ldots\}$, initial state $X_0=1$, and natural filtration $\mathcal{F}_t=\sigma(X_s:0\le s\le t)$. Fix a per-capita birth rate $\lambda>0$. In state $n$, the total birth rate is $n\lambda$: the process waits an exponential time with rate $n\lambda$, then moves to $n+1$. There are no other transitions. Write $p_n(t)=\mathbb{P}(X_t=n)$.

Definition: The Yule process with per-capita birth rate $\lambda>0$, started from one individual, is the pure-birth process with transition rate $n\lambda$ from $n$ to $n+1$, for every $n\ge1$. Its population probabilities are $p_n(t)=\mathbb{P}(X_t=n)$.

We will use the waiting-time construction to ensure that the process does not make infinitely many births in a finite time. Let $E_n$ be the independent holding time spent in state $n$, so $E_n$ is exponential with rate $n\lambda$. For any positive integer $m$, independence gives

\mathbb{E}\!\left[\exp\!\left(-\sum_{n=1}^{m}E_n\right)\right] =\prod_{n=1}^{m}\frac{n\lambda}{n\lambda+1}.

For $x>0$, $\log(1+x)\ge x/(1+x)$, as follows by differentiating the difference and noting it is zero at $x=0$. Therefore, the logarithm of the product is $-\sum_{n=1}^{m}\log(1+1/(n\lambda))$, which tends to $-\infty$, since $\sum_{n\ge1}1/(n\lambda+1)$ diverges. The product tends to zero. By bounded convergence, $\mathbb{E}[\exp(-\sum_{n\ge1}E_n)]=0$, where $\exp(-\infty)=0$. Since this exponential is nonnegative, it is zero almost surely; hence $\sum_{n\ge1}E_n=\infty$ almost surely. Thus infinitely many births cannot occur in finite time.

The Forward Recurrences

The process can enter state $1$ only by starting there, and it leaves state $1$ at rate $\lambda$. To enter state $n\ge2$, it must move from $n-1$ to $n$, at rate $(n-1)\lambda$; it leaves $n$ at rate $n\lambda$. The resulting forward recurrences are

p_1'(t)=-\lambda p_1(t),\qquad p_n'(t)=(n-1)\lambda p_{n-1}(t)-n\lambda p_n(t),\quad n\ge2,

with initial conditions $p_1(0)=1$ and $p_n(0)=0$ for $n\ge2$. One way to justify each recurrence directly is to consider the last entry into state $n$. For $n\ge2$, entry occurs from $n-1$ at rate $(n-1)\lambda$, and after entry the process must remain in $n$ until time $t$. Thus

p_n(t)=\int_0^t (n-1)\lambda p_{n-1}(s)e^{-n\lambda(t-s)}\,ds.

There is no contribution from an initial state $n$, because $X_0=1$ and $n\ge2$. Differentiating this integral gives the stated recurrence. For $n=1$, the probability of no birth by time $t$ is $e^{-\lambda t}$, which gives its equation and initial value.

Theorem (Yule population distribution): For $\lambda>0$, $t>0$, and $n\ge1$, the population probabilities are $$ p_n(t)=e^{-\lambda t}\bigl(1-e^{-\lambda t}\bigr)^{n-1}. $$ In particular, the population at time $t$ has a geometric distribution on $\{1,2,\ldots\}$, with success probability $e^{-\lambda t}$.

Proof. Set $q=e^{-\lambda t}$, so $0<q<1$ when $t>0$. For $n=1$, solving $p_1'=-\lambda p_1$ with $p_1(0)=1$ gives $p_1(t)=e^{-\lambda t}$, as claimed. Now suppose the formula holds for $n-1$. The integral recurrence gives

\begin{aligned} p_n(t) &=\int_0^t (n-1)\lambda e^{-\lambda s} (1-e^{-\lambda s})^{n-2}e^{-n\lambda(t-s)}\,ds\\ &=(n-1)\lambda e^{-n\lambda t} \int_0^t e^{(n-1)\lambda s}(1-e^{-\lambda s})^{n-2}\,ds. \end{aligned}

With $u=e^{\lambda s}-1$, we have $du=\lambda e^{\lambda s}ds$, $e^{(n-1)\lambda s}(1-e^{-\lambda s})^{n-2}ds=u^{n-2}du/\lambda$, and the endpoints are $u=0$ and $u=e^{\lambda t}-1$. Hence

\begin{aligned} p_n(t) &=(n-1)e^{-n\lambda t}\int_0^{e^{\lambda t}-1}u^{n-2}\,du\\ &=e^{-n\lambda t}(e^{\lambda t}-1)^{n-1}\\ &=e^{-\lambda t}(1-e^{-\lambda t})^{n-1}. \end{aligned}

This proves the formula for every $n$ by induction. It is a probability distribution because, writing $r=1-e^{-\lambda t}$, the geometric series gives $\sum_{n\ge1}e^{-\lambda t}r^{n-1}=e^{-\lambda t}/(1-r)=1$. At $t=0$, the initial distribution is $p_1(0)=1$ and $p_n(0)=0$ for $n\ge2$, separately. Finally, the recurrence determines each $p_n$ uniquely: given $p_{n-1}$, the equation for $p_n$ is a first-order linear differential equation with specified initial value, whose solution is the integral above. Thus the computed probabilities are the solution of the forward recurrences. $\square$

The recurrence also provides a direct check of the proposed formula. For $t>0$, put $q=e^{-\lambda t}$, so $q'=-\lambda q$. Differentiating $p_n=q(1-q)^{n-1}$ gives

p_n'(t)=\lambda q(1-q)^{n-2}(nq-1),\qquad n\ge2.

On the other hand, substituting $p_{n-1}=q(1-q)^{n-2}$ and $p_n=q(1-q)^{n-1}$ into the right side of the recurrence gives $\lambda q(1-q)^{n-2}[(n-1)-n(1-q)]$, which is the same expression. For $n=1$, differentiation gives $p_1'=-\lambda p_1$. The initial values also follow from the formula and its limit as $t$ decreases to zero.

Using the Geometric Law

For $t>0$, put $q=e^{-\lambda t}$. Summing the geometric series yields, for an integer $m\ge1$,

\mathbb{P}(X_t\le m)=1-(1-q)^m,\qquad \mathbb{P}(X_t\ge m)=(1-q)^{m-1}.

These formulas count the states inclusively: $X_t\le m$ means the states $1,\ldots,m$, and $X_t\ge m$ includes state $m$. They are consequences of the one-time distribution, not claims that the process is stationary. The probabilities change with $t$, and at $t=0$ the population is exactly one.

Worked Example: Probabilities When the Success Parameter Is One-Half

Take $\lambda=2$ and $t=\tfrac{\ln 2}{2}$. Then $\lambda t=\ln2$, so $q=e^{-\lambda t}=e^{-\ln2}=\tfrac12$, and $1-q=\tfrac12$. Therefore

p_1(t)=\tfrac12,\qquad p_2(t)=\tfrac12\cdot\tfrac12=\tfrac14,\qquad p_3(t)=\tfrac12\cdot\left(\tfrac12\right)^2=\tfrac18.

The probability of a population of at most three is $\tfrac12+\tfrac14+\tfrac18=\tfrac78$. The probability of at least four individuals is $(1-q)^3=(\tfrac12)^3=\tfrac18$. Check: $\mathbb{P}(X_t\le3)+\mathbb{P}(X_t\ge4)=\tfrac78+\tfrac18=1$; the two events partition the positive integer states.

Worked Example: A Tail Probability at a Different Time

Take $\lambda=1$ and $t=\ln3$. Then $q=e^{-\ln3}=\tfrac13$ and $1-q=\tfrac23$. The probabilities of populations one, two, and three are

p_1(t)=\tfrac13,\qquad p_2(t)=\tfrac13\cdot\tfrac23=\tfrac29,\qquad p_3(t)=\tfrac13\cdot\left(\tfrac23\right)^2=\tfrac4{27}.

Adding them gives $\mathbb{P}(X_t\le3)=\tfrac9{27}+\tfrac6{27}+\tfrac4{27}=\tfrac{19}{27}$. The probability of at least four individuals is $(\tfrac23)^3=\tfrac8{27}$. Check: $\tfrac{19}{27}+\tfrac8{27}=1$, and the tail formula uses the inclusive event $X_t\ge4$.

The geometric form also makes conditional probabilities straightforward. Conditioning on $X_t\ge m$ is meaningful here because $(1-q)^{m-1}>0$ for $t>0$ and finite $m$. For any integer $r\ge1$,

\mathbb{P}(X_t=m+r-1\mid X_t\ge m) =\frac{q(1-q)^{m+r-2}}{(1-q)^{m-1}} =q(1-q)^{r-1}.

Thus, given that the population is at least $m$, the amount by which it exceeds $m-1$ has the same geometric probability formula as the original population. This is a property of the geometric distribution at a fixed time; it does not say that the process after time $t$ is independent of its population at time $t$.

Worked Example: Conditioning on a Population Threshold

Use $\lambda=2$ and $t=\tfrac{\ln2}{2}$ again, so $q=\tfrac12$. Find the probability that $X_t=5$, given that $X_t\ge3$. The numerator is

\mathbb{P}(X_t=5)=\tfrac12\left(\tfrac12\right)^4=\tfrac1{32}, \qquad \mathbb{P}(X_t\ge3)=\left(\tfrac12\right)^2=\tfrac14.

Since the conditioning event has positive probability, division is valid, and $\mathbb{P}(X_t=5\mid X_t\ge3)=\tfrac{1}{32}/\tfrac14=\tfrac18$. Check: the conditional formula has $m=3$, $r=3$, so it gives $q(1-q)^{r-1}=\tfrac12(\tfrac12)^2=\tfrac18$, agreeing with the direct calculation.

Common Pitfalls and a Differential Check

A common error is to use the rate out of state $n$ as the rate into $n$. The rate out is $n\lambda$, but the only incoming transition comes from $n-1$, at rate $(n-1)\lambda$. Consequently, the correct recurrence is $p_n'=(n-1)\lambda p_{n-1}-n\lambda p_n$, not $n\lambda p_{n-1}-n\lambda p_n$.

Another useful safeguard is to check derivatives at a selected time by both sides of the recurrence. For $\lambda=1$, $p_2(t)=e^{-t}(1-e^{-t})$, and differentiation gives $p_2'(t)=-e^{-t}+2e^{-2t}$. At $t=\ln2$, this is $-\tfrac12+2(\tfrac14)=0$. The recurrence's right side is $p_1-2p_2=\tfrac12-2(\tfrac14)=0$, as required. In particular, the derivative at that time is zero, not $-\tfrac14$.

The assumption $\lambda>0$ gives the geometric formula for positive times. If $\lambda=0$, there are no births and $X_t=1$ with probability one for every $t$; this is a separate degenerate case. For $\lambda>0$, the formula approaches the initial population law as $t$ decreases to zero. The population distribution derived here concerns a single time $t$; long-run growth and moments are separate questions.

Check Your Understanding

Use the forward recurrence and the geometric formula to answer the following questions. In each probability statement, attend to whether the endpoint state is included.

For $n=4$, write the forward recurrence, identifying separately the incoming and outgoing rates.
With $\lambda=1$ and $t=\ln2$, compute $p_1(t),p_2(t),p_3(t)$, and $\mathbb{P}(X_t\ge4)$.
For general $t>0$, derive $\mathbb{P}(X_t\le m)$ by summing the probabilities for states $1$ through $m$.
For $\lambda=3$, find a positive time $t$ for which $e^{-\lambda t}=\tfrac14$, and compute $\mathbb{P}(X_t=2)$.
Explain why the recurrence uses $(n-1)\lambda$ for incoming probability flow but $n\lambda$ for outgoing probability flow.

The Yule Process Population Distribution

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