# Showing $o(1)$ convergence for ratio of successive binomial tail probabilities

For a Binomial$$(n,p)$$ random variable $$X$$, I'm interested in showing that $$\frac{P(X>c)}{P(X>c-1)}=1-o(1)$$ uniformly in $$c\in\mathcal{R}$$, where $$\mathcal{R}$$ is the range of interest (Note that $$c$$ will vary with $$n$$). The $$o(1)$$ rate is meant as $$n\to\infty$$.

Now I have the following results (Note that $$q=1-p$$):

Result 1 For $$0\leq k\leq n$$, set $$P(X=k)=\frac{1}{\sqrt{2\pi pq n}}\exp\left(-\frac{(k-np)^2}{2npq} \right)(1+\delta_n(k))$$ Then for every positive real sequence $$\{c_n\}$$ approaching zero, $$\lim_{n\to\infty}\max_{k:|k-np|

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Result 2 Suppose that $$\{a_n\}$$ is a sequence of real numbers such that $$\lim_{n\to\infty}a_n=+\infty$$ and $$\lim_{n\to\infty}a_n n^{-1/6}=0$$. Then $$P(X\geq np+a_n\sqrt{npq})\sim \frac{1}{a_n\sqrt{2\pi}}\exp(-a_n^2/2)$$ where "$$\sim$$" means asymptotic equivalence.

Now, \begin{align} \frac{P(X>c)}{P(X>c-1)}&=\frac{P(X>c-1)-P(X=c)}{P(X>c-1)}\\ &=1-\frac{P(X=c)}{P(X\geq c)} \end{align}

EDIT From the the above results, the range $$\mathcal{R}$$ can be at least $$np$$ and at most $$np+c_{n}\sqrt{npq}$$, for some $$c_n=o(n^{1/6})$$. I can certainly show that at the extremes of the range, the ratio is $$o(1)$$. However, I think I also need to show that either a) for some value $$\tilde{c}$$ in between the extremes, $$P(X=\tilde{c})/P(X\geq \tilde{c})=o(1)$$ or that b) the ratio itself is monotonic (based on some numerical experiments, I think it is increasing in $$c$$). I've tried to go through the route of b) and show that $$P(X=c)/P(X\geq c)\leq P(X=c+1)/P(X\geq c+1)$$, but can't seem to get the math to work out.

• When you write $o(1)$, you mean something goes to 0 as what happens? – Anthony Quas Jan 13 '19 at 21:48
• @AnthonyQuas, as $n\to\infty$. I've updated the post. – stats134711 Jan 13 '19 at 21:51

I think, if $$c=np+o(n)$$, you may simply use $$P(X=c+1)=P(X=c)(1+o(1))$$, and so $$P(X\in \{c,c+1,\dots,c+M-1\})=(M+o(1))P(X=c)$$ for any fixed $$M$$.
• If $M>0$ is fixed, then isn't the conclusion that the ratio is $O(1)$ instead of the desired $o(1)$? – stats134711 Jan 18 '19 at 15:19
• Fixed $M$ proves that the lower limit is not less than $1-1/M$. Since $M$ is arbitrary, the lower limit equals 1. – Fedor Petrov Jan 18 '19 at 17:21
• Would it be more rigorous to write, let $\varepsilon>0$, and choose $M>0$, such that $M>1/\varepsilon$? Then $(M+o(1))^{-1}<1/M<\varepsilon$. This would then result in $P(X=c)/P(X\geq c)<\varepsilon$. I guess I am hung up on the fact that say if $M$ is fixed at the beginning, then how can we make it arbitrary at the end? – stats134711 Jan 18 '19 at 18:58
• It is essentially the same argument. We fix $M$ and prove that $\liminf \frac{P(X>c)}{P(X>c-1)}\geqslant 1-\frac1M$. After proving this, we may remember that $M$ could be fixed arbitrary, thus $\liminf \frac{P(X>c)}{P(X>c-1)}=1$. You suggest to say it another way: denote $\liminf \frac{P(X>c)}{P(X>c-1)}=1-\varepsilon$, then choose $M>\varepsilon^{-1}$ and get a contradiction. – Fedor Petrov Jan 18 '19 at 19:02
Going off Fedor's answer, for $$c$$ at most $$np+o(n)$$, we may use the fact $$P(X=c+1\mid p)=P(X=c\mid p)(1+o(1))$$. It follows that for any $$M>0$$, $$P(c\leq X\leq c+M-1\mid p)=(M+o(1))P(X=c\mid p)$$. In other words, if we let $$\varepsilon_1>0$$, then there exists $$N_{\varepsilon_1}>0$$, such that when $$n\geq N_{\varepsilon_1}$$, $$\left\vert\frac{P(c\leq X\leq c+M-1\mid p)}{P(X=c\mid p)}-M\right\vert <\varepsilon_1$$ This implies $$P(c\leq X\leq c+M-1\mid p)\geq P(X=c\mid p)(M-\varepsilon_1)$$. Now let $$\varepsilon_2>0$$. Choosing $$M>\varepsilon_1+\varepsilon_2^{-1}$$, we have \begin{align*} \frac{P(X=c\mid p)}{P(X\geq c\mid p)}&\leq \frac{P(X=c\mid p)}{P(c\leq X\leq c+M-1\mid p)}\\ &\leq \frac{P(X=c\mid p)}{P(X=c\mid p)(M-\varepsilon_1)}\leq \varepsilon_2 \end{align*}