X is a simple birth death process with birth rate $\lambda n$ and death rate $\mu n$
Embedded within a simple birth death process is a simple random walk.
Let $Y_n $ be the value of X at the time of the nth event.
$Y_0=X_0=1 $
Then
$Y_{n+1} = Y_n + 1 $ with probability $\lambda / (\lambda + \mu)$
And
$Y_{n+1} = Y_n-1 $ with probability $\mu / (\lambda + \mu)$
Provided $Y_n > 0$
Since X doesn't explode, extinction occurs in finite time iff $Y_n =0, n<\infty$
So $P[Y_n =0, n<\infty]. = 1$ if $\mu\geq\lambda$
Or $(\mu/\lambda)^X_0 $ if $\mu<\lambda$
where we use our analysis of discrete-time simple random walks,
How is the analysis of discrete simple random walks used here, I don't understand where these probabilities come from, how are they worked out?
Thanks
Best Answer
I would use branching process (which is discrete-time random walk) to solve this problem.
According to your description, an individual "gives birth to" $2$ individuals with the probability $\frac{\lambda}{\lambda + \mu}$; an individual "gives birth to" $0$ individuals with the probability $\frac{\mu}{\lambda + \mu}$.
This gives the generating function $$G(s) = \frac{\mu}{\lambda + \mu} + \frac{\lambda}{\lambda + \mu}s^2$$.
Extinction probability is the smaller root of the function
$$s = \frac{\mu}{\lambda + \mu} + \frac{\lambda}{\lambda + \mu}s^2$$.
After calculation, the equation has two roots: $1$, $\frac{\mu}{\lambda} $.
Indeed, if $\frac{\mu}{\lambda} \geq 1$, i.e. $\mu \geq \lambda$, the population will go extinct within a finite time(𝑃 [ 𝑌𝑛=$0$ , 𝑛<∞].=$1$ )