On the volcanic island of Montserrat the number of species increases(by immigration from neighbouring islands) at rate α. However, at rate η the volcano explodes, and all life is wiped out, although the island immediately becomes hospitable again. A.Find the equilibrium proportion of time on which there there is no life on the island.
B.If currently there is a single species on the island, what is the expected maximum number of species before the first volcanic eruption?
For the part A, I have written
$q(n,n) = -(\eta + \alpha)$ n=1,2,... = $-\alpha$
$q(n,0) = \eta $. n=1,2...
$q(n,n+1) = \alpha$
Then I solve $\pi Q = 0 $
To get $\pi_k = \alpha^k.k/(\alpha + \eta)^{(k+1)}$
So does this mean the answer to A is $\pi_0 $?
I am totally clueless about B
Thanks
The answer to A can be obtained intuitively. Let's say you start at time $0$ and wait for the first individual to arrive (ignore the explosions before the first arrival because the island becomes hospitable immediately after). After the first arrival (at whatever time), the expected time for which there is life on the island is $1/\eta$ since that is the expected time after which the volcano will explode (exponential distribution). Now just after a wiping out event, the expected time delay for the next arrival of individuals will be after $1/\alpha$. This is a renewal process, so the average fraction of time that there will be no life on the planet is $\eta/(\eta + \alpha)$. You can check this answer by taking the limit of large $\eta$ (the explosions are occurring with smaller and smaller delay) and see that the average proportion of time when there is no life approaches 1.
The answer to B can be obtained by integrating the poisson distribution (with arrival rate $\alpha$) against the exponential distribution of the time at which the next explosion occurs. The probability that there were $k$ individuals at the time of explosion given that there was 1 at the beginning is...
$P(S = k) =\int_0^\infty e^{-\alpha t}\dfrac{(\alpha t)^{k-1}}{(k-1)!} \eta e^{-\eta t} \,dt = \dfrac{\eta \alpha^{k-1}}{(\alpha + \eta)^k}$
The mean of this distribution is $1 + \alpha/\eta$
The problem can be interpreted from the framework of queueing theory as well.