This is similar in spirit to Sum of squares of middle binomial sums or 'Truncated mean' of binomial coefficients under binomial distribution but gives some total estimates. Though the other one was amenable to computations this by nature looks formidable. I do not have much intuition on how fast $f(m,n)$ can grow?

$\mu=1+\epsilon$ where $\epsilon>0$ holds.

$n<m$ holds.

Is there a good bound for $$\log_2\Bigg({\sum_{i_1,\dots,i_{m/n}=-\sqrt{\mu n\ln n}}^{\sqrt{\mu n\ln n}}\binom{n}{\frac n2 +i_1}\dots\binom{n}{\frac n2 +i_{m/n}}}{\mathbb P(\frac n2+i_1)\dots\mathbb P(\frac n2+i_{m/n})}\Bigg)?$$

where $\mathbb P(\frac n2+i)$ is under bionmial distribution and thus is $\frac{\binom{n}{\frac n2 +i}}{2^n}$ and thus this expression is 'truncated joint moment $\binom{n}{\frac n2 +i_j}$ from $j\in\{1,\dots,{m/n}\}$'.

Clearly this is $m-f(m,n)$ (seen from similar nature of terms and bounds from Sum of squares of middle binomial sums or 'Truncated mean' of binomial coefficients under binomial distribution) for some function $f$. How fast can $f(m,n)$ grow?

Conjecture: $\exists c>1: f(m,n)>c\frac mn$.

Question: Can $c>c'\ln n$ hold in above at some $c'>1$?


The OP has changed the original question. This change invalidates my previous answer. The answer below is to the changed question.

For any natural $m$ and $n$ such that $m/n$ is also natural, the expression now to be bounded is \begin{equation} L:=\frac mn\,\log_2 T, \end{equation} where $T$ is just as in the question here. In the corresponding answer there, it was shown that $$\log_2T=n-\log_2\sqrt{(1+o(1))\pi n};$$ (the asymptotics everywhere here are as $n\to\infty$). Hence, $$L=m-\frac mn\,\log_2\sqrt{(1+o(1))\pi n}.$$

  • $\begingroup$ I was reading $\ln\sqrt n$ as $\sqrt{\ln n}$. $\endgroup$ – VS. Jan 19 '20 at 20:39

For any natural $m$ and $n$ such that $m/n$ is also natural, the expression to be bounded is \begin{equation} L:=\frac mn\,\log_2(U_n/2^n), \end{equation} where \begin{equation} U_n:=\sum_{k\colon\,|k-n/2|\le u}\binom nk^3 \end{equation} and \begin{equation} u:=\sqrt{\mu n\ln n}. \end{equation} Letting \begin{equation} h_k:=\frac{k-n/2}{n/2}\quad\text{and}\quad t_k:=h_k\sqrt{2n}=\frac{k-n/2}{\sqrt{n/8}}, \end{equation} by Stirling's formula we have \begin{align} \binom nk&\sim\frac{2^n}{\sqrt{\pi n/2}}\exp\{-n[(1+h_k)\ln(1+h_k)+(1-h_k)\ln(1-h_k)]\} \\ &\sim\frac{2^n}{\sqrt{\pi n/2}}\exp\{-n[h_k^2+O(h_k^3)]\} \\ & \sim\frac{2^n}{\sqrt{\pi n/2}}\exp\{-t_k^2/2\}; \end{align} everywhere here, the asymptotics are for $n\to\infty$ and $k$ such that $|k-n/2|\le u$.

Hence, \begin{align*} U_n&\sim\frac{2^{3n}}{\pi^{3/2}n}\;\sum_{k\colon\,|t_k|\le u/\sqrt{n/8}}e^{-3t_k^2/2}(t_{k+1}-t_k) \\ &\sim\frac{2^{3n}}{\pi^{3/2}n}\;\int_{|t|\le u/\sqrt{n/8}}e^{-3t^2/2}\,dt \\ &\sim\frac{2^{3n}}{\pi n}\sqrt{\frac23}. \end{align*} Here the transition from the sum to the integral is possible because $t_{k+1}^2-t_k^2=(t_{k+1}-t_k)(t_{k+1}+t_k)\le\frac1{\sqrt {n/8}}\,(2u/\sqrt{n/8}+\frac1{\sqrt{n/8}})=o(1)$.

So, \begin{align} L&=\frac mn\,\Big(2n+\log_2\Big(\frac1{(\pi+o(1)) n}\sqrt{\frac23}\,\Big)\Big) \\ &=2m-\frac mn\,\log_2\frac{(\pi+o(1)) n}{\sqrt{2/3}}. \end{align}

Thus, your conjecture will hold if (and only if) you replace $m-f(m,n)$ by the correct expression $2m-f(m,n)$, with the main term $2m$ rather than $m$.

  • $\begingroup$ I think the quantity inside the logarithm is at most $2^m$. $\endgroup$ – VS. Jan 19 '20 at 5:55
  • $\begingroup$ Sorry I made a mistake. If you look at the other problem I either had a square form or a moment form. I mixed the two and I removed the square now and so for this do we have $m-\Omega(\frac mn\ln n)$? $\endgroup$ – VS. Jan 19 '20 at 5:59
  • $\begingroup$ I see only $m\sqrt{\ln n}/n$ when combined with other problem. Perhaps $\ln n$ is not correct. $\endgroup$ – VS. Jan 19 '20 at 6:05
  • $\begingroup$ @VS. : You should not change the question so as to invalidate an answer, especially if the answer required a non-negligible effort. If you have any additional questions, you should ask them in separate posts. Anyway, I have now given an answer to the changed question as well. If any specific step in either answer is unclear, please let me know. $\endgroup$ – Iosif Pinelis Jan 19 '20 at 14:48
  • $\begingroup$ I like your answer. Only thing I made a typo mistake from copying in problem. Essentially what happens if there are no squares? $\endgroup$ – VS. Jan 19 '20 at 20:33

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