# Binomial probabilities and limits

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

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

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)."

• 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. Commented Jan 1, 2022 at 11:53