Weird probability on Yule-Furry pure birth process

Question

Essentially, I'm trying to calculate $\Pr(\text{1 birth in } [t,t+\Delta t) \mid N(t)=n)$, given that $\Pr(\text{1 birth in } [t,t+\Delta t) \mid N(t)=1)=\beta\Delta t+o(\Delta t)$. This is related to a pure birth Yule-Furry process (the pure death case should be equivalent).

I'm reading Mark Kot's Elements of Mathematical Ecology, and the way the author does it is as such:

$$\Pr(\text{1 birth in } [t,t+\Delta t) \mid N(t)=n) = n(\beta\Delta t+o(\Delta t))(1-\beta\Delta t + o(\Delta t))^{n-1}$$

which makes sense. But the simplification in the following line is not as straightforward:

$$n(\beta\Delta t+o(\Delta t))(1-\beta\Delta t + o(\Delta t))^{n-1} = n\beta\Delta t+o(\Delta t)$$

This is actually pretty easy to prove by induction but I can't prove directly how the author goes from the LHS to the RHS. I tried using the binomial theorem, but the calculations get pretty heavy and I'll eventually get lost with all the symbols involved.

Answer 1

I think your problem might be with understanding the $o(\Delta t)$ notation. Here is the Wikipedia article for that

Let's use the Multinomial theorem and open the brackets for $(1-\beta\Delta t + o(\Delta t))^{n-1}$. Then each summand will be of the form $${n-1\choose {i\ \ j\ \ k}}1^i(-\beta\Delta t)^j(o(\Delta t))^k$$ where $i,j,k\geq 0$ and $i+j+k=n-1$. Out of all combinations for $i,j,k$, the ones where $i=n-1,j=k=0$ and $i=n-2,j=1,k=0$ give the only terms whose ratio to $\Delta t$ doesn't vanish as $\Delta t\rightarrow 0$ (which is basically what being $o(\Delta t)$ is). But these terms are $1$ and $-C\beta\Delta t$! (here $C$ is just some coefficient). Then the LHS is basically $$n(\beta\Delta t+o(\Delta t))(1-C\beta\Delta t + o(\Delta t)) = n\beta\Delta t+o(\Delta t)$$

Check this by plugging any other $i,j,k$ combination to see for yourself.

Hope this helps.

Answer 2

the key step is $(1-x)^{-1} = 1 + O(x)$ for $|x|\to 0$. This is nothing but Mean Value Theorem, or if you wish, Taylor's Theorem(and if you wish, by uniqueness of Taylor series, identifying a geometric series). If you plug $\beta \Delta t+o(\Delta t)$ into this, the quantity to estimate becomes

\begin{align} (n\beta \Delta t+ o(\Delta t))(1-\beta \Delta t+o(\Delta t))^{-1} &= (n\beta \Delta t+ o(\Delta t))\big(\color{red}1+ \color{blue}{ \underbrace{O(\beta \Delta t+o(\Delta t))}_{=O(\Delta t)}}\big)\\ &= \color{red}{n\beta \Delta t+ o(\Delta t)} +\underbrace{\color{blue}{(n\beta \Delta t+ o(\Delta t)) O( \Delta t)}}_{=O(\Delta t^2)=o(\Delta t)} \\&= n\beta \Delta t+ o(\Delta t) \end{align}

as $\Delta t \to 0$.