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Question:
Let $X(t)$ be a continuous-time Markov chain on all non-negative integers with generator matrix $Q$ having for all $i\geq 0:$ $$ q_{i,i}=-i(\lambda +\mu ) \qquad q_{i,i+1}=\lambda i \qquad q_{i,i-1}=\mu i $$ (this last rate does not have any meaning for $i=0)$ and all other $q_{i,j}=0.$ Here $\lambda >0$ and $\mu >0.$ We write $$ h(t)=\Bbb{P}(X(t)=0|X(0)=1). $$ Show that $$ h(t)=\int _{0}^t e^{-(\lambda +\mu )u}\Big\{\mu +\lambda (h(t-u))^2\Big\}\ du $$

Context:
I'm trying some exercises to prepare for my exam. This is an old exam question.

Effort:
I've done quite some work, not sure if it the quickest route, but according to my calculation it suffices to show that $$P(X(t)=0 | X(0)=2)= P(X(t)=0 | X(0)=1)^{2} \tag{*}.$$

But I don't know how to show that. I think my calculations are correct, not sure if you guys want to see it, will make this post a bit chaotic. As I think those calculations are correct I'm most interested in showing that $(*)$ is true. Any ideas ?

Kasper
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2 Answers2

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Are we allowed to note that this is the dynamics of a population such that every living individual gives birth to one new individual at rate $\lambda$ and dies at rate $\mu$, independently of the other living individuals? Because this is what rates proportional to the population are modelling, actually...

Then one sees that a population of $i\geqslant1$ individuals at time $0$ is extinct at time $t$ if and only if the $i$ independent subpopulations issued from each of these individuals living at time $0$ is extinct at time $t$. By independence, this yields $$P(X(t)=0\mid X(0)=i)=P(X(t)=0\mid X(0)=1)^i.$$

Did
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Okay, here are my calcuations. I hope it is clear how to justify the (backwards step), it becomes such a hairy chaos, if I write everything out explicitly, but I think I've made no mistakes.

\begin{gather*} \Bbb{P}(X(t)=0|X(0)=1)=\int _{0}^t e^{-(\lambda +\mu )u}\Big(\mu +\lambda (h(t-u))^2\Big)\ du \\ \Uparrow \\ \int _{0}^t f_{J_{1}}(u) \Bbb{P}(X(t)=0 | X(0)=1, J_{1}=u) du = \int _{0}^t e^{-(\lambda +\mu )u}\Big(\mu +\lambda (h(t-u))^2\Big)\ du \\ \Uparrow \\ f_{J_{1}}(u) \Bbb{P}(X(t)=0 | X(0)=1, J_{1}=u) = e^{-(\lambda +\mu )u}\Big(\mu +\lambda (h(t-u))^2\Big) \\ \Uparrow \\ \Bbb{P}(X(t)=0 | X(0)=1, J_{1}=u) = \frac{\mu +\lambda (h(t-u))^2}{\lambda +\mu } \\ \Uparrow \\ \Bbb{P}(X(t)=0 \mid \substack{X_{0}=1\\ J_{1}=u \\ X(J_{1})=0}) \cdot \Bbb{P}(X(J_{1})=0| \substack{X_{0}=1 \\J_{1}=u}) + \Bbb{P}(X(t)=0 \mid \substack{X_{0}=1\\ J_{1}=u \\ X(J_{1})=2}) \cdot \Bbb{P}(X(J_{1})=2| \substack{X_{0}=1 \\J_{1}=u}) =\frac{\mu }{\lambda +\mu }+\frac{\lambda (h(t-u))^2}{\lambda +\mu } \\ \Uparrow \\ \Bbb{P}(X(t)=0 \mid \substack{X_{0}=1\\ J_{1}=u \\ X(J_{1})=0}) \cdot \Bbb{P}(X(J_{1})=0| \substack{X_{0}=1 \\J_{1}=u}) = \frac{\mu }{\lambda +\mu } \qquad\text{and}\qquad \Bbb{P}(X(t)=0 \mid \substack{X_{0}=1\\ J_{1}=u \\ X(J_{1})=2}) \cdot \Bbb{P}(X(J_{1})=2| \substack{X_{0}=1 \\J_{1}=u})=\frac{\lambda (h(t-u))^2}{\lambda +\mu } \\ \Uparrow \\ \Bbb{P}(X(t)=0 \mid \substack{X_{0}=1\\ J_{1}=u \\ X(J_{1})=0}) =1 \qquad\text{and}\qquad \Bbb{P}(X(t)=0 \mid \substack{X_{0}=1\\ J_{1}=u \\ X(J_{1})=2})=h(t-u))^{2} \\ \Uparrow \\ \Bbb{P}(X(t-u)=0 \mid \substack{X(0)=2})=h(t-u))^{2} \\ \Uparrow \\ P(X(t)=0 | X_{0}=2)= P(X(t)=0 | X_{0}=1)^{2} \\ \end{gather*}

Kasper
  • 13,528