I have a linear birth death process with birth rates $\lambda n$ and death rates $\mu n$ .
Let r(t) be the probability of extinction by time t.
If there is 1 individual alive at time 0 explain why
r(t) = $\int^t_0 e^{-(\mu+\lambda)s}(\mu+\lambda r(t-s)^2)ds$
With r(0)=0
I have done this but the next bit is to derive a differential equation for r and solve it (at least in the case where $\mu\neq\lambda)$ and show that if $\mu>\lambda $ then r(t)–> 1 whereas if $\mu<\lambda $ then r(t) –> $\mu/\lambda$
I wanted to proceed as we do similar things in lectures.
So step 1 : multiply by $e^{(\lambda +\mu)t}$
This gives
$e^{(\lambda +\mu)t}r(t)$ = $\int^t_0 e^{(\mu+\lambda)(t-s)}(\mu+\lambda r(t-s)^2)ds$
Then I let t-s = k to get
$e^{(\lambda +\mu)t}r(t)$ = $\int^t_0 e^{(\mu+\lambda)(k)}(\mu+\lambda r(k)^2)dk$
I then differentiate both sides
$e^{(\lambda +\mu)t}r'(t)+ (\lambda +\mu)r(t)e^{(\lambda +\mu)t}$ = $e^{(\mu+\lambda)(t)}(\mu+\lambda r(t)^2)$
Divide all terms by $ e^{(\mu+\lambda)(t)}$
Rearrange to get
$dr/(r(t)(r(t)\lambda -\lambda -\mu)) = dt\mu$
Then separate into partial fractions
This gives $[-1/(\lambda+\mu r(t))] + [\lambda/((\lambda+\mu)(r(t)\lambda-\lambda-\mu))] = dt\mu$
Integrate both sides between t and 0
$[-1/(\lambda+\mu).lnr(t) + 1/(\lambda+\mu).ln(r(t)\lambda-\lambda-\mu)]^{r(t)}_{r(0)}= \mu t$
When I evaluate this at the limits and rearrange I get
$[\lambda r(t)-\lambda-\mu]/r(t) = -(\mu + \lambda)e^{\mu t(\lambda + \mu)} = D$
Finally I rearrange to get $r(t) = (\lambda+\mu)/(\lambda – D)$ which does not have the qualities required, what have a done wrong?
Best Answer
The trouble is that you
rearrangewrong... In fact, $$ r'(t)+(\lambda+\mu)r(t)=\mu+\lambda r(t)^2, $$ hence $$ \frac{r'(t)}{\lambda r(t)^2-(\lambda+\mu)r(t)+\mu}=1. $$ The decomposition into simple fractions reads $$ \frac1{\lambda r^2-(\lambda+\mu)r+\mu}=\frac1{\lambda -\mu}\left(\frac1{r-1}-\frac{\lambda}{\lambda r-\mu}\right). $$ Can you finish?