Random processes in Continuous time. Given that $\beta = \frac{4}{5}*\mu$ I have calculated that the birth rate $= 0.4$ and the death rate $= 0.5$.
If the initial population $X(0)=6$, how many events (births and deaths) are there until the population dies out?
This is an embedded process $\{X_n;\:n=0,1,...\}$.
Is this possible to work out with so little information? It's all I have.
Let $N_n$ be the number of events until extinction given that $X(0)=n$. Then, $N_0=0$ and conditioning on the next event, we have for $n\geq 1$,
\begin{align} & N_n = 1 + \dfrac{4}{9}N_{n+1} + \dfrac{5}{9}N_{n-1} \\ \text{or, re-arranging, }\qquad & 4N_{n+1} - 9N_n + 5N_{n-1} = -9 \qquad\qquad(1). \end{align}
Note also that to reach $0$ from state $n$ requires going in turn to states $n-1,n-2,\ldots,0,$ and the number of events in each of these sub-paths has the same distribution as $N_1$, independently of the number of events in the other sub-paths. This means that,
$$N_n = nN_1.$$
Substituting this into $(1)$ gives:
\begin{align} -9 &= 4(n+1)N_1 - 9nN_1 + 5(n-1)N_1 \\ &= -N_1 \\ \therefore\quad N_1 &= 9. \end{align}
So we have the solution $N_n = 9n$ and in particular our required value is $N_6 = 54$.