Pitman probability poisson approximation

Question

On pg118 of Pitman's Probability, a derivation of the Poisson approximation to $P(k)$ (the binomial probability of getting $k$ successes over $n$ trials, where the success probability $p$ is nearly 0 and $n$ is large) is presented in terms of the consecutive odds ratios $R(k) = P(k)/P(k-1)$. The idea is to calculate $P(k) = P(0)R(1) \cdots R(k-1)R(k)$. One thing in this presentation that I don't really follow is his approximation for $R(k)$. He calculates $$R(k) = P(k)/P(k-1) = \frac{n - k + 1}{k}\frac{p}{1-p} = \frac{np}{k}\frac{(1-(k-1)/n)}{1-p} \approx \frac{\mu}{k}.$$ Implicit in this calculation is the approximation $$\frac{(1-(k-1)/n)}{1-p} \approx 1,$$ which I don't understand. The reason I don't understand how this can always be so is that, in Pitman's discussion, there are no conditions on $k$ other than the obvious $0 \leq k \leq n$. So if we take $k$ to be nearly $n$, then the numerator in the fraction directly above goes to zero while its denominator goes to 1. Another, more understandable, derivation was presented in class, but I want to know what the reasoning is behind $\frac{np}{k}\frac{(1-(k-1)/n)}{1-p} \approx \frac{\mu}{k}$ in this particular explanation, because I'm not seeing it.

Answer 1

The "truth" about this approximation is to actually consider sequences $p_n$ of probabilities such that

• $np_n \stackrel{n \to \infty}{\longrightarrow}0$

From this it follows that

• $p_n = o\left(\frac{1}{n}\right)$ ($p_n$ goes to $0$ even if you multiply it by $n$)

Then, for any fixed $k$ you have

$$\frac{1-\frac{k-1}{n}}{1-p_{\color{blue}{n}}} = \frac{1-\frac{k-1}{n}}{1-o\left(\frac{1}{n}\right) } \stackrel{n \to \infty}{\longrightarrow}\frac{1-0}{1-0} = 1$$