Interpolating asymptotic expression for logarithm of middle binomial sums

Question

Define $S(k,2n)=\sum_{i=-k}^k\binom{2n}{n+i}$ at every $k\in\{0,\dots,n\}$.

We know $$\ln(S(\gamma\mbox{ } n^\gamma,2n))\asymp(2n\ln2-\frac12\ln\pi-\frac12\ln n)$$ at $\gamma\rightarrow0$ and $$\ln(S(\gamma\mbox{ } n^\gamma,2n))=2n\ln 2$$ at $\gamma\rightarrow1$.

Is there a good approximating expression that interpolates for $$\ln(S(\gamma\mbox{ } n^\gamma,2n))$$ at general $\gamma\in(0,1)$ between $\gamma=0$ and $\gamma=1$ here in the sense that gives information on at what $\gamma$ we may expect $2n\ln2 - f(n)$ for a given $f(n)<\frac12\ln\pi+\frac12\ln n$?

Denote $\gamma n^{\gamma}=m^{\beta}$ and $2n=m^{\alpha}$. What is the right description when $\beta<\frac\alpha2$ and $\frac\alpha2 - \frac1{c\log m}<\beta$ holds at a fixed $c>0$? Does $$m^\alpha\ln2 - (\frac\alpha2-\beta)\ln m -\frac12\ln\frac\pi2$$ seem an accurate estimate?

Answer 1

I think the best intuition is to translate it to a question about random walk. Let $S_{n}$ be a simple random walk with $n$ steps. Then $$S(k,2n) = 4^{n} \mathbb{P}(|S(2n)| \leq k).$$

Thus, for $\gamma > 1/2$, we have $S(\gamma n^\gamma,2n) \sim 4^n$ and for $\gamma < 1/2$ we have $$S(\gamma n^{\gamma},2n) \sim \frac{\gamma n^{\gamma} 4^{n}}{\sqrt{ \pi n}}.$$

At $\gamma = 1/2$, the central limit theorem takes over and so we calculate $$S(n^{1/2}/2,2n) = 4^{n} \mathbb{P}(|S(2n)| \leq n^{1/2}/2) = 4^n \mathbb{P}(\frac{|S(2n)|}{\sqrt{2n}} \leq 2^{-3/2}) \sim 4^n \mathbb{P}(|Z| \leq 2^{-3/2})$$

where $Z$ is a standard normal variable.

Answer 2

An approximation that works well is $$ \frac{S(k,2n)}{\binom{2n}{n}} \sim \sqrt{\pi \, n}\, \text{erf }(k/\sqrt{n})+\exp{(-k^2/n)}.$$ For instance, letting $n=100$ and letting $k$ run from 0 to $n,$ the maximum error is about 0.12%. For $n=10^3,$ max err is 0.014% and for $n=10^4,$ max err is 0.0016%.

I derived it from the gaussian approximation of the central binomial, and Euler-McLaurin, that is:

$$ (*) \quad \quad \frac{\binom{2n}{n+m}}{\binom{2n}{n}} \sim \exp{(-m^2/n)} $$ $$ \sum_{m=-k}^{k}\frac{\binom{2n}{n+m}}{\binom{2n}{n}} \sim \int_{-k}^{k}\exp{(-m^2/n)} \,dm + \exp{(-k^2/n)} + ... $$ where the (...) indicate the terms in sum involving the Bernoulli numbers and derivatives of the gaussian. This expression is probably better than it deserves, as I've only included the first asymptotic term in (*) and neglected the (...) terms. This analysis is not rigorous, but might give you an avenue to explore.