Here's what I have:

Let X be a simple birth-death process where individuals have independent $\text{Exp}(\mu)$ lifetimes and, during their lifetime give birth at rate $λ$ independently of other individuals.

Let $T = \inf\{t ≥ 0: X_{t} = 0\}$ be the extinction time for the population.

I have to find the density of T.

Earlier in the question I had to show that if $G(s, t) = E(s^{X_{t}})$, then $G$ satisfies :

$$ \frac{∂}{∂t}G(s, t) = (λs − \mu)(s − 1) \frac{∂}{∂s}G(s, t)$$

which I did.

The hint is that the solution to the above is given by:

$$\frac{\mu(s − 1) − (\lambda s − \mu)e^{-(\lambda-\mu)t}}{\lambda(s − 1) − (\lambda s − \mu)e^{-(\lambda-\mu)t}} $$ or

$$\frac{\lambda t(s-1)-s}{\lambda t(s-1)-1}$$

depending on whether $\mu \neq \lambda$ or $\mu = \lambda$.

Another observation is that $G(0, t) = P(X_t = 0)$ and that $$\inf\{t ≥ 0: X_{t} = 0\} \leq k \Leftrightarrow X_k=0$$

I just can add all those things up to a solution.

Any help is appreciated

  • Note that $G(s,t) = E[s^{X_t}]$ is the probability generating function of $X_t$. Therefore, $$P(X_t = k) = \frac{d^kG(s,t)}{ds^k}(0,t)$$ As a special case, $G(0,t)=P(X_t=0)$.

  • Next, the hint tells you that $P(T \leq t) = P(X_t=0)$. Can yo you use the previous item to determine $P(T \leq t)$?

  • Finally, use $$f_{T}(t)=\frac{dP(T \leq t)}{dt}$$

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  • $\begingroup$ $P(T \leq k) = G(0,k)$ but then if I've not made a mistake, the density comes out quite nasty $\endgroup$ – asdf Nov 13 '17 at 15:00
  • $\begingroup$ I don't understand your comment. $\endgroup$ – madprob Nov 13 '17 at 15:02
  • $\begingroup$ We have that $P(T \leq k)=G(0,k)$ which is given by the above solutions. But differentiating this doesn't give anything sensible. Or I've made a mistake somewhere $\endgroup$ – asdf Nov 13 '17 at 15:03
  • $\begingroup$ You didn't indicate anywhere in your question that you had tried this path already. How can I know? $\endgroup$ – madprob Nov 13 '17 at 15:04
  • $\begingroup$ No way you could, but I was just wondering whether we are expected to get something familiar or is this badly looking in general $\endgroup$ – asdf Nov 13 '17 at 15:05

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