I have some difficultes with an exercise that I must do. I try to post it there, maiby someone can help me to resolve it.
Let ${X(t)}_{t>0}$ on $\{0,1,2,3\}$ a birth end death process, with $\lambda(s)=(3-s)^2$ and $\mu(s)=s^2+s$.Furthermore we know that $P(X(0)=3)=1$. Determine:
1) $E[X(t)]$.
2) $Var[X(t)]$.
I tried to do the part one, but I'm stopped and I don't know how I can go one.
Let $M(t)=E[X(t)]$, I determine $M(t)$.
$M(t+h)=E[X(t+h)]=E[E[X(t+h)|X(t)]]$
$X(t+h)=\begin{cases}X(t)+1, \textrm{with probability } (3-X(t))^2h+o(h)\\X(t)-1,\textrm{with probability } (X(t)^2+X(t))h+o(h)\\X(t), \textrm{with probability } 1-[(3-X(t))^2+X(t)^2+X(t)]h+o(h)\end{cases}$
So $E[X(t+h)|X(t)]=(X(t)+1)P(X(t)+1)+(X(t)-1)P(X(t)-1)+X(t)P(X(t))=X(t)+[9-7X(t)]h+o(h)$
$M(t+h)=M(t)+[9-7M(t)]h+o(h)$
$\frac{M(t+h)-M(t)}{h}=9-7M(t)+\frac{o(h)}{h}$
I do this limit for $h\rightarrow0$ and i obtain this differential equation: $M'(t)=9-7M(t)$ (*)
Let $h(t)=9-7M(t)$,
so $h'(t)=-7M'(t)$
So I can rewrite the differential equation (*) in this way: $-\frac{h'(t)}{7}=h(t)$ or $\frac{h'(t)}{h(t)}=-7$
Integration yields: $\log[h(t)]=-7t+c$ or $h(t)=K\exp{-7t}$.
So $9-7M(t)=K\exp{-7t}$.
Now to detrmine K I think that I must use the condition that $P(X(0)=3)=1$, but I don't know how I use it. However I'm not sure that my reasoning is corect. I hope that someone can explain it to me. Thank you for your help.