Yule process intuitive question

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!

Answer 1

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 $\mathbb{E}(T_n)=\frac1{\lambda}\left(1+\frac12+\frac13+\dots+\frac1{n-1}\right)\to\infty$.

Answer 2

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. $\square$

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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 $n = 10$.

Answer 3

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 $\mathbb{E}(X_t)\leq\sum_{k=0}^{\infty}\mathbb{E}(Y_k)<\infty$.