Yule process intuitive question

I have a Yule process with $$n$$ individuals.

There is no death, so the death rate is $$\mu_n$$ $$=$$ $$0$$ for all $$n$$.

Each individual gives birth to a new individual independently after waiting for $$\text{Exponential}(\lambda)$$ amount of time.

So the birth rate is $$\lambda_n = n\lambda$$ for all $$n \ge 1$$.

I get this by considering the minimum waiting time, as there are $$n$$ waiting times here, each distributed as $$\text{Exponential}(\lambda)$$.

Now, my question is, in this process, does the population become infinite in a finite amount of time? Is that possible, and if so, how should I think about it?

Any hints or advice will be very helpful. Thank you so much!

Let $$\tau_k$$ denote the time between $$(k-1)$$th birth and $$k$$th birth. In light of the exponential race, $$\tau_k$$'s are independent and each $$\tau_k$$ is distributed as $$\text{Exponential}((n+k-1)\lambda)$$.

Now, let $$T_k = \sum_{j=1}^{k} \tau_j$$ denote the time to the $$k$$th birth, measured from the start. Then the population blows up to infinity in a finite time if and only if $$T_{\infty} := \lim_{k\to\infty} T_k$$ is finite.

In this regard, we claim that $$T_{\infty} = \infty$$ almost surely, hence the population remains finite at all time. To this end, we establish a much stronger result:

Theorem. For each given initial population $$n \geq 1$$, $$T_k - \mathbf{E}[T_k] \quad \text{converges to a finite random variable} \quad Y \tag{1}$$ as $$k\to\infty$$ almost surely.

Since $$\mathbf{E}[T_k] \sim \frac{\log k}{\lambda}$$ as $$k \to \infty$$ for each fixed $$n$$, this shows that $$T_k \to \infty$$ as $$k \to \infty$$.

Proof of Theorem. Let $$\delta_k = \tau_k - \mathbf{E}[\tau_k]$$ and note that

$$|\delta_k| \geq \frac{1}{\lambda} \qquad \iff \qquad \tau_k \geq \frac{1}{(n+k-1)\lambda} + \frac{1}{\lambda}$$

with probability one. (This is because $$\mathbf{E}[\tau_k] = \frac{1}{(n+k-1)\lambda} \leq \frac{1}{\lambda}$$.) Then a direct computation shows that

\begin{align*} \sum_{k=1}^{\infty} \mathbf{P}(|\delta_k| \geq \tfrac{1}{\lambda}) &= \sum_{k=1}^{\infty} \mathbf{P}(\tau_k \geq \tfrac{1}{(n+k-1)\lambda} + \tfrac{1}{\lambda}) = \sum_{k=1}^{\infty} e^{-n-k}, \\ \sum_{k=1}^{\infty} \mathbf{E}[\delta_k \mathbf{1}_{\{ |\delta_k| \leq \tfrac{1}{\lambda} \}} ] &= \sum_{k=1}^{\infty} \mathbf{E}[\tau_k \mathbf{1}_{\{ \tau_k \leq \tfrac{1}{(n+k-1)\lambda} + \tfrac{1}{\lambda} \}} ] = - \sum_{k=1}^{\infty} \frac{e^{-k-n} (k+n+1)}{(k+n-1) \lambda}, \\ \sum_{k=1}^{\infty} \mathbf{Var}[\delta_k \mathbf{1}_{\{ |\delta_k| \leq \tfrac{1}{\lambda} \}} ] &\leq \sum_{k=1}^{\infty} \mathbf{Var}[ \tau_k ] = \sum_{k=1}^{\infty} \frac{1}{\lambda^2 (k+n-1)^2}. \end{align*}

Since all of these three series converge, it follows that

\begin{align*} Y = \lim_{k\to\infty} T_k - \mathbf{E}[T_k] = \sum_{j=1}^{\infty} (\tau_j - \mathbf{E}[\tau_j]) \end{align*}

converges by Kolmogorov's three-series theorem. $$\square$$

Remark. The theorem tells that the number $$N(t)$$ of individuals at time $$t$$ is roughly proportional to $$e^{\lambda t}$$, which can also be verified by other means.

Below shows ten numerical simulations of the sample path $$k \mapsto T_k$$ together with the graph of $$\log(1+\frac{k}{n})$$ when $$\lambda = 1$$ and $$n = 10$$.

Yule process does not explode. The time it takes to go from $$n$$ individual to $$n+1$$ individuals, $$\tau_n$$, is exponential with parameter $$\lambda n$$. Hence $$\mathbb{E}\tau_n=\frac1{\lambda n}$$, $$\mathbb{E}\tau_n^2=\frac2{\lambda^2 n^2}$$, and by the Paley–Zygmund inequality with $$\theta=1/2$$ $$\mathbb{P}\left(\tau_n>\frac1{2\lambda n}\right)\ge \frac34 \times \frac{\left(\frac1{\lambda n}\right)^2}{\frac2{\lambda^2 n^2}}=\frac38.$$

Starting from one individual, time $$T_n$$ to reach $$n$$ individuals is thus $$T_n=\tau_1+\tau_2+\dots+\tau_{n-1}$$ (note that $$\tau_i$$'s are independent). Each $$\tau_i>\frac1{2\lambda i}$$ with probability at least $$3/8$$, hence for those integer $$i\in B_k:=((k-1)\sqrt{n},k\sqrt{n}]$$, $$k=1,2,\dots,\lfloor \sqrt{n}\rfloor$$, the probability that less than $$2/8$$ of them are less than $$\min_{i\in B_k}\frac1{2\lambda i}>\frac1{2\lambda k\sqrt{n}}$$ is exponentially small in $$\mathrm{card}(B_k)\sim\sqrt{n}$$ (i.e., like $$e^{-c\sqrt{n}}$$, $$c>0$$) e.g. by the large deviation theory. Since $$\sum_n \sqrt{n} e^{-c\sqrt{n}}<\infty$$ by the Borel-Cantelli lemma only finitely many such events occur and hence a.s. for all large $$n$$ we have $$T_n>\sum_{k=1}^{\lfloor \sqrt{n}\rfloor}\frac{\mathrm{card}(B_k)}{2\lambda k\sqrt{n}} =\sum_{k=1}^{\lfloor \sqrt{n}\rfloor}\frac{1}{2\lambda k} \sim \ln n\to\infty\quad \text{ as }n\to\infty\text{ almost surely.}$$

Note: a more intuitive (but not completely rigorous) way to see it, is to observe that $$\mathbb{E}(T_n)=\frac1{\lambda}\left(1+\frac12+\frac13+\dots+\frac1{n-1}\right)\to\infty$$.

• Your lower bound on $T_n$ seems not aligning with the fact $\mathbb{E}[T_n]\sim\frac{1}{\lambda}\log n$. In fact, we can show that $T_n\sim\frac{1}{\lambda}\log n$ almost surely as $n\to\infty$. Nov 29, 2023 at 10:31
• Indeed, there was a mistake (corrected now) Nov 29, 2023 at 12:26

Let $$X_t$$ be the expected number of individuals after time $$t$$, and suppose that $$\mathbb E(X_t)<\infty$$. Now we have $$\mathbb{E}(X_{2t}\mid X_t=x)=x\mathbb{E}(X_t)$$, since each of the $$x$$ individuals at time $$t$$ evolves as an independent Yule process of rate $$\lambda$$ for the next time $$t$$. It follows that $$\mathbb{E}(X_{2t})=\mathbb{E}(X_t)^2<\infty$$. Therefore if $$\mathbb{E}(X_t)<\infty$$ for some $$t>0$$, then in fact $$\mathbb{E}(X_t)<\infty$$ for all $$t>0$$, and so almost surely the process doesn't explode.

Now let's bound $$\mathbb{E}(X_t)$$ for some fixed small $$t$$. Rather than stopping at $$t$$, it's easier to work with the same process where each individual lives for exactly time $$t$$, and count the total number of individuals produced; this is an upper bound on $$X_t$$. In this process each individual independently has $$\mathrm{Po}(\lambda t)$$ offspring before dying. Write $$Y_k$$ for the number of individuals in the $$k$$th generation. By conditioning on $$Y_k$$, we have $$\mathbb{E}(Y_{k+1})=\lambda t\mathbb{E}(Y_k)$$, i.e. $$\mathbb{E}(Y_k)=(\lambda t)^k$$. Thus for $$t<1/\lambda$$ we have $$\mathbb{E}(X_t)\leq\sum_{k=0}^{\infty}\mathbb{E}(Y_k)<\infty$$.