I'm reading Sheldon Ross and he gives two proofs of the same result: that given some assumptions, $N(t)$ has a Poisson distribution with mean $\lambda t$. The first proof is in chapter 4 and introduces the pmf of the Poisson and the second proof comes many pages later in chapter 9 and talks about the Poisson Process.

My specific question

In the screenshot below, the blue box says $P(N(h) = 0) = 1 - \lambda h - o(h)$ whereas the red box says $P(N(h) = 0) = 1 - \lambda h + o(h)$. Can you help explain why are they different?

My attempt

In both proofs, the $o(h)$ part eventually goes to zero so the sign doesn't seem that important. But I am trying to build okay understanding.

I am just learning about little o notation, but I don't think $o(h) = - o(h)$ because all probabilities have to be between $0$ and $1$ and $P(N(h) \ge 2) = o(h)$. From reading related posts I know $o(h)$'s can be different functions and all be called $o(h)$ (which is confusing for a noob like me) but I don't think the signs can be flipped as arbitrarily.

From book

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$f(h)=o(h)$ is an abbreviation for $\frac {f(h)} h \to 0$. So there is absolutely no difference between $f(h)=o(h)$ and $f(h)=-o(h)$. The statement has nothing to do with the sign of the right side.

  • $\begingroup$ Thanks for your help. One of the assumptions is $P(N(h) \ge 2) = o(h)$. Since all probabilities have to be between zero and one doesn't it cause issues to say $P(E) = -o(h) = -f(h)/h$? $\endgroup$ – HJ_beginner Oct 19 '18 at 23:21
  • 1
    $\begingroup$ You are thinking that a notation like $-o(h)$ implies that we are dealing with negative numbers . That is not correct. For example $-h^{2}=o(h)$ because $\frac {-h^{2}} h=-h \to 0$. $\endgroup$ – Kavi Rama Murthy Oct 19 '18 at 23:26
  • $\begingroup$ Hmmm interesting... I will meditate on this answer. $\endgroup$ – HJ_beginner Oct 19 '18 at 23:28

If you say a function $f(h)$ is "little-o of $g(h)$" it means that $f$ goes to $0$ faster than $g$ when $h$ goes to $0$, that is (and just like @Kavi noted in his answer)

$$\lim_{h \to 0} \frac{f(h)}{g(h)} = 0$$

It is common to use $g(h) = h^p$ and thus often one sees "$f$ is $\mathcal{o}(h^p)$" and, in particular, "$f$ is $\mathcal{o}(h)$"

Of course that, if $f$ is $\mathcal{o}(h)$, then $-f$ is also $\mathcal{o}(h)$ because

$$\lim_{h \to 0} \frac{f(h)}{h} = - \lim_{h \to 0} \frac{-f(h)}{h} = 0$$

And writing $+\mathcal{o}(h)$ or $-\mathcal{o}(h)$ can be read as "add something that decays faster than $h$" or "subtract something that decays faster than $h$" but when we are thinking of the decay, the sign won't actually matter, because $0$ has no sign... So in writing $\pm \mathcal{o}(h)$ we aren't strictly compromising on the sign.

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    $\begingroup$ Thanks for your explanation. That definitely helps. $\endgroup$ – HJ_beginner Oct 19 '18 at 23:37

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