# How to interpret the little-$o$ notation in this definition of the Poisson process?

In the book Probability and Random Processes by Grimmett and Stirzaker, a Poisson process is defined to be any process $(N(t))_{t\in[0,\infty)}$ with values in $\mathbb{N}_0$ satisfying following three properties:

$\hspace{20pt}$(a) $N(0) = 0$ and $s<t\Rightarrow N(s)\leq N(t)$

$\hspace{20pt}$(b) $\mathbb{P}\left(N(t+h)=n+m|\hbox{ }N(t) =n\right) =\begin{cases} o(h) & \text{if } m>1,\\ \lambda h + o(h) &\text{if } m=1,\\ 1 - \lambda h + o(h) & \text{if }m=0\\ \end{cases}$

$\hspace{20pt}$(c) if $s<t$ then $N(t)-N(s)$ is independent of the times of emissions on $[0,s]$.

I am having problems with understanding the meaning $o(h)$ in (b), so my question is:

How do we interpret $o(h)$ in this definition?

I know $o(h)$ in the classical sense is supposed to be just a function such that $\lim_{h\to 0}\frac{o(h)}h=0$. But here, $o$ is used three times and I suspect it has a different meaning each time. (Since otherwise, we would necessarily have $o\equiv0$.)

So, I did some thinking and reinterpreted (b) to read:

$\hspace{20pt}$(b') There exist functions $o_1, o_2, o_3$, such that for $i=1,2,3$:$$\lim_{h\to 0}\frac{o_i(h)}h=0$$ $\hspace{38pt}$and $$\mathbb{P}\left(N(t+h)=n+m|\hbox{ }N(t) =n\right) =\begin{cases} o_1(h) & \text{if } m>1,\\ \lambda h + o_2(h) &\text{if } m=1,\\ 1 - \lambda h + o_3(h) & \text{if }m=0.\\ \end{cases}$$

But I am completely confused about the order of quantifiers in this statement (which is why I left this rewording a bit vague). Should the functions $o_1,o_2,o_3$ be the same for all $t,n$ and $m>1$? (That is: should we take a separate function $o_m$ for each m? Or even a separate function $o_{t,n,m}$ for each triple $(t,n,m)$?)

Does this even matter or are these definitions miraculously equivalent?

I would really like to know what exactly it is this definition is trying to define. Thanks in advance for any helpful suggestions.

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Yes, the convention for using little-o notation is that every $o(\cdots)$ in an equation can stand for a different function that is eventually smaller than the identity.

Also, usually the precise function is allowed to depend in arbitrary ways on anything that is not mentioned in the "\cdots".

Consider, for example the not unusual definition of the derivative of $f(x)$ at $x_0$ being the (unique) $\lambda$ such that $f(x_0+h)=f(x_0)+\lambda h + o(h)$. Here it is implicit that the $o(\cdots)$ function can depend on $x_0$ but not on $h$.

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So, if I understand correctly, $o$ can be a different function $o_{t,m,n}$ for every triple $(t,m,n)$? Ok, this pretty much completely answers my question. Thanks. –  Dejan Govc Feb 12 '12 at 21:42
Yes, that's how I would interpret it (absent some very specific indications to the contrary). –  Henning Makholm Feb 12 '12 at 21:47

Once you've found exact formulas for the Poisson process, it could help your intuition to go back and write the functions explicitly. I will re-index the $o$ functions to match $m$ as follows: $$\mathbb{P}\left(N(t+h)=n+m|\hbox{ }N(t) =n\right) =\begin{cases} 1 - \lambda h + o_0(h) & \text{if }m=0\\ \lambda h + o_1(h) &\text{if } m=1,\\ o_m(h) & \text{if } m>1,\\ \end{cases}$$

From the explicit formulas, we have $$\begin{eqnarray*} o_0(h)&=&e^{-\lambda h}-1+\lambda h\\ o_1(h)&=&\lambda h(e^{-\lambda h}-1)\\ o_m(h)&=&(\lambda h)^m\, e^{-\lambda h}/m! \ \mbox{ for } m>1\\ \end{eqnarray*}$$

As you see, these functions depend on $m$ and $\lambda$, but not $n$ or $t$.

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However, the fact that they don't depend on $n$ or $t$ is an "accidental" consequence of all of the definitions together. It is not something one should have been able to read syntactically off part (b) of the definition. –  Henning Makholm Feb 12 '12 at 22:56
Also, oughtn't there be an $m$ somewhere in your expression for $o_1$? Surely the probability of the process going from $0$ to $m$ in time $h$ cannot be the same for every $m\ge 2$. –  Henning Makholm Feb 12 '12 at 22:59
@Henning Quite right. It is because the Poisson process is homogeneous in space and time. Things won't always work out so well. –  Byron Schmuland Feb 12 '12 at 22:59
@Hennig Thanks for the correction. –  Byron Schmuland Feb 12 '12 at 23:09
Thanks, this indeed helps my intuition. (And justifies my suspicion of things being miraculously equivalent. =)) –  Dejan Govc Feb 13 '12 at 0:09