# 1st Yr Probability: Question about the Poisson Process in my book, what is $P(N(h) = 0)$

Background

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 ## 2 Answers

$$f(h)=o(h)$$ 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.

• 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 $P(E) = -o(h) = -f(h)/h$? – HJ_beginner Oct 19 '18 at 23:21
• 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 $\frac {-h^{2}} h=-h \to 0$. – Kavi Rama Murthy Oct 19 '18 at 23:26
• Hmmm interesting... I will meditate on this answer. – 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.

• Thanks for your explanation. That definitely helps. – HJ_beginner Oct 19 '18 at 23:37