For $k =1,2,3, \dots,$ let $X_k$ be a binomial distribution with parameters $k$ and $0.5.$ Fix $0<a<1$ and assume that $a\neq 0.5.$ $\left\lfloor \cdot\right\rfloor$ denotes the greatest integer function. Is the following statement true ?

\begin{equation}\label{eq:eqn1} \text{lim}_{k \rightarrow \infty}k\cdot\Pr\left(X_{k-1}=\left\lfloor ka\right\rfloor = 0\right) \end{equation}

I think it is true and I have the following rough proof of the above statement. The following is an attempt for a proof.

From Stirling's approximation for large values of $n$ and $x,$ we have (here $X$ is a binomial distribution with parameters $n$ and $p$) \begin{align*} P(X=x) &= \binom{n}{x}p^x(1-p)^{n-x}\\ &\approx \frac{1}{\sqrt{2\pi np(1-p)}}\exp\left[-\frac{1}{2}\frac{(x-np)^2}{np(1-p)} \right] \end{align*} Assume that $a > 0.5.$ This implies that there exists a $\eta >0$ such that $a=0.5+\eta.$ For sufficiently large $k,$ we have: \begin{align*} k\cdot\Pr\left(X_{k-1}=\left\lfloor ka\right\rfloor \right) &\approx \frac{k}{\sqrt{2\pi (k-1)(0.5)^2}}\exp\left[-\frac{1}{2}\frac{(\left\lfloor k(0.5+\eta)\right\rfloor-(k-1)0.5)^2}{(k-1)(0.5)^2} \right]\\ &\approx \sqrt{\frac{2}{\pi}}\frac{k}{\sqrt{k-1}}\exp\left[-\frac{1}{2}\frac{(k(0.5+\eta)-(k-1)0.5)^2}{(k-1)(0.5)^2} \right]\\ &= \sqrt{\frac{2}{\pi}}\left(\frac{k/\sqrt{k-1}}{\exp\left[\frac{2}{k-1}(0.5+k\eta)^2\right]}\right) \end{align*}

***** I am a little unsure in one of the above steps. For large $k,$ I think we can use $\left\lfloor ka\right\rfloor \approx ka.$ But am not completely sure about this. The remaining steps seem to be fine (but I could be wrong)*****

Clearly, the last expression in the above series of equations goes to zero as $k \rightarrow \infty.$ The proof for the case when $a<0.5$ is similar as above.

"I would be very grateful if someone can let me know, if the statement is true and the proof I have is correct (conditional on the statement being true)."

  • $\begingroup$ I think the idea is fine and works. The $\approx$ signs look not rigorous but you could probably replace them with $\le$ signs, e.g. using $ka-1<\lfloor ka\rfloor\le ka$. Also using WLLN might give a more direct proof. $\endgroup$
    – nejimban
    Commented Jan 1, 2022 at 11:53


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