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, $$ \lim_{n\to\infty}\max_{k:|k-np|<c_n n^{2/3}}|\delta_n(k)|=0 $$


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.

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

  • $\begingroup$ If $M>0$ is fixed, then isn't the conclusion that the ratio is $O(1)$ instead of the desired $o(1)$? $\endgroup$ – stats134711 Jan 18 '19 at 15:19
  • $\begingroup$ Fixed $M$ proves that the lower limit is not less than $1-1/M$. Since $M$ is arbitrary, the lower limit equals 1. $\endgroup$ – Fedor Petrov Jan 18 '19 at 17:21
  • $\begingroup$ 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? $\endgroup$ – stats134711 Jan 18 '19 at 18:58
  • $\begingroup$ 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. $\endgroup$ – Fedor Petrov Jan 18 '19 at 19:02
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    $\begingroup$ @stats134711 looks ok, yes $\endgroup$ – Fedor Petrov Jan 21 '19 at 19:59

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*}


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