Ill give some background first before asking questions.(the text below is straight out of the book)

Each individual in the population is assumed give birth at an exponential rate of $\lambda$ in addition ,there is a an exponential rate of increase $\theta$ due to external source of immigration. Hence the total birth rate where there are $n$ persons in the system is $n\lambda + \theta$ . Deaths are assume to occur at an exponential rate $\mu$ for each member of the population, so $\mu_n = n\mu$.

Let $X(t)$ denote the population size at time $t$. Suppose $X(0)= i$ and let $M(t) = E[X(t)]$ . So they will determine $M(t)$ by deriving and then solving a differential equation that is satisfies.

we start by deriving an equation for $M(t+h)$ by conditioning on $X(t)$ this yields:

$M(t+ h) = E[X(t+h)] = E[E[X(t+h)\vert X(t)]]$

Now,given the size of the population at time $t$ then, ignoring events whose probability is $o(h)$, the population at time $t+h$ will either increase in size by 1 if a birth or immigration occurs in $ (t,t+h)$ , or, decrease by 1 if a death occurs in this interval, or remain the same if neither of these two possibilities occurs that is given $X(t)$

$X(t+h)$= \begin{cases} X(t) + 1 , with- probability & [\theta + X(t)\lambda]h + o(h) \\ X(t) - 1, with-probability & X(t)\mu h + o(h)\\ X(t), with-probability & 1-[\theta + X(t)\lambda + X(t)\mu]h +o(h) \end{cases}

therefore, $E[X(t+h) \vert X(t)] = X(t) + [\theta + X(t)\lambda - X(t)\mu]h + o(h)$

..... ..... .....(text continues)


$\textbf{(1)}$, i understand the first two cases , but the last case i don' t quiet get: $X(t)$, with-probability $1-[\theta + X(t)\lambda + X(t)\mu]h +o(h)$. can someone explain this?

$\textbf{(2)}$How do i interpret this statement: $E[X(t+h) \vert X(t)] = X(t) + [\theta + X(t)\lambda - X(t)\mu]h + o(h)$

these are my two questions.


(1) the probabilities have to add up to 1, that determines (within $o(h)$) the value of the probability measure for the last case...

(2) The meaning of the left-hand side is, given that I know what $X_t$ is, what do I expect $X_{t+h}$ will be? And the right hand side gives you roughly what it should be. Not surprisingly, it directly depends on $X_t$, slightly altering it by some amount...

  • $\begingroup$ sorry there was a typo i corrected it : (minus sign) $E[X(t+h) \vert X(t)] = X(t) + [\theta + X(t)\lambda - X(t)\mu]h + o(h)$ $\endgroup$ – Danny Apr 28 '14 at 15:35
  • $\begingroup$ about the first question, i was thinking i need to add up case 1 and 2 and then subtract 1 from it but then the $o(h)$ should have a negative sign? @gt6989b $\endgroup$ – Danny Apr 28 '14 at 15:39
  • $\begingroup$ @Danny $o(h)$ and $-o(h)$ is the same thing. $\endgroup$ – gt6989b Apr 28 '14 at 15:55
  • $\begingroup$ is it that simple to say that when u add up case 1 and case 2 then u get $2\times o(h) = o(h)$ and then subtract the third case from case 1 and 2 then u get $-o(h) = o(h) $ since any linear combinations of $o(h)$ is $o(h)$ @gt6989b $\endgroup$ – Danny Apr 28 '14 at 16:00
  • $\begingroup$ @Danny yes, $o(h)$ includes any constant factor... $\endgroup$ – gt6989b Apr 28 '14 at 17:28

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