A linear growth model with immigration

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)

$\textbf{questions:}$

$\textbf{(1)}$, 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...
• sorry there was a typo i corrected it : (minus sign) $E[X(t+h) \vert X(t)] = X(t) + [\theta + X(t)\lambda - X(t)\mu]h + o(h)$ – Danny Apr 28 '14 at 15:35
• 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 – Danny Apr 28 '14 at 15:39
• @Danny $o(h)$ and $-o(h)$ is the same thing. – gt6989b Apr 28 '14 at 15:55
• 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 – Danny Apr 28 '14 at 16:00
• @Danny yes, $o(h)$ includes any constant factor... – gt6989b Apr 28 '14 at 17:28