When does a pure birth process explode?

Question

I'm trying to understand the criterion for a pure birth process to explode in finite time. Unfortunately, the script I'm studying only gives the result and not the proof. And I wasn't able to quickly solve it myself or find a suitable reference either.

Let $(X_t, t\geq 0)$ be a pure birth process on $\mathbb N_0$, starting in $0$, with rates $\lambda_i$. Then it should be true that $(X_t)$ explodes in finite time if and only if $\sum_i 1/\lambda_i < \infty$.

Now, this statement is as intuitive as it gets, but still I'd like to be able to see formally why this is true.

We can characterize the time of explosion as follows:

• $T_k = \inf\{t>0 : X_t = k \}, \qquad \forall k \in N$,
• $T_\infty = \sup_k T_k$.

Then I want to show:

1. $$\sum_{i\in \mathbb N} \frac{1}{\lambda_i} < \infty \implies \mathbb P_0(T_\infty < \infty) = 1.$$

2. $$\sum_{i\in \mathbb N} \frac{1}{\lambda_i} = \infty \implies \mathbb P_0(T_\infty < \infty) = 0.$$

How do I go about that?

Answer 1

$E(T_{\infty})=E(\sup_k T_k)=\sup_kE(T_k)=\sup_k\sum_{i=1}^k\frac{1}{\lambda_i}=\sum_{i=1}^{\infty}\frac{1}{\lambda_i}$.

If this sum is finite then $E(T_{\infty})$ is also finite and so $P(T_{\infty}<\infty)=1$.

The method for dealing with the case when the sum is infinite is taken from Kirkwood's "Markov Processes". Firstly we will use the fact that $$\prod_{n=1}^{\infty}\frac{1}{1+a_n}=0\quad\text{if and only if}\quad\sum_{n=1}^{\infty}a_n=\infty.$$ Since $T_k-T_{k-1}$ has an $\text{Exp}(\lambda_{k})$ distribution, $E(\exp(-(T_k-T_{k-1})))=1/(1+\lambda_k^{-1})$. We now have

$$ \begin{align*} E(\exp(-T_{\infty}))&=E(\exp(-\sum_{n=1}^{\infty}(T_n-T_{n-1}))) \\ &=\prod_{n=1}^{\infty}E(\exp(-(T_n-T_{n-1}))) \\ &=\prod_{n=1}^{\infty}\frac{1}{1+\frac{1}{\lambda_n}} \\ &=0. \end{align*} $$

As $\exp(-T_{\infty})$ is a non-negative random variable we must have $P(\exp(-T_{\infty})=0)=1$ and so $P(T_{\infty}=\infty)=1$.