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This is a homework question and I need suggestion how to approach it. We have given the transitions

  1. $\ i\rightarrow i+1$ with rate $\lambda(i)$ where $\ i \ge 1$
  2. $\ i\rightarrow i-1$ with rate $\mu(i)(i-1)$ where $\ i \ge 2$

I am starting the forward equation like this:

$$\ p_j(t) = [1-(\lambda_jh+\mu_jh +o(h))p(t)]+ \lambda(j-1)p(j-1)(t)h + ...$$

and I cannot really continue from here. I need to get to the point where I can show that

$$\ G(z, t) = \sum P(N(t)= j|N(0) = a) z^j $$

satisfies

$$\frac{\partial G}{\partial t} = z \left( z-1\right) \left(\lambda\frac{\partial G}{\partial z}- \mu\frac{\partial^2 G}{\partial z^2}\right)$$

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  • $\begingroup$ Would stats.stackexchange.com/questions/46389/… be of any help? $\endgroup$
    – whuber ♦
    Commented Jan 29, 2014 at 16:24
  • $\begingroup$ $p_j(t)=[1-\lambda h+\mu(j+1)h +o(h)]jp_j(t) +\lambda(j-1)hp_j-_1(t)+\mu j(j+1)hp_j+_1(t)$ $\endgroup$
    – Aknur
    Commented Jan 29, 2014 at 20:54
  • $\begingroup$ Is transition 2 right? $\endgroup$
    – Glen_b
    Commented Jan 29, 2014 at 23:25
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    $\begingroup$ The left hand side should be $p_j(t+h)$ then expand the RHS and take $p_j(t)$ to the other side as a negative. Then divide both sides by h, take the limit as h -> infinity so that the LHS becomes $\frac{d}{dt}{p_j(t)}$. After that multiply both sides by $z^j$ and sum both sides over j=0 to infinity. $\endgroup$
    – Clair Crossupton
    Commented Jan 30, 2014 at 21:19

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