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I'm trying to give a bound on the following formula: $$\frac1{2^i}\sum_{j=0}^i|2j-i| \binom{i}{j} \leq d\sqrt{i}$$ for some constant $d$.

This is how far I get: $$=\dfrac{\sum^{i/2}_{j=0} \ \binom{i}{j} (i - 2j) + \sum^{i}_{j=(i/2)+1} \ \binom{i}{j} (2j - i)}{2^i}$$ $$\leq \dfrac{2\sum^{i/2}_{j=0} \ \binom{i}{j} (i - 2j)}{2^i}$$ If I take the sum from j=0 to i instead of i/2, then the sum would be 0. So, I should use another way of upper bounding this formula.

Did
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YOUSEFY
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    The LHS is $E(|X_1+\cdots+X_i|)$ where $(X_k)$ is i.i.d. uniform on ${-1,+1}$. By Cauchy-Schwarz inequality, this is less than the square root of $E((X_1+\cdots+X_i)^2)=E(X_1^2)+\cdots+E(X_i^2)=i$. – Did Oct 21 '17 at 09:41
  • @Did Thank you! How did you figure out that this is probability! I did not mention that! you're really smart guy! it is amazing short answer! Best! – YOUSEFY Oct 21 '17 at 10:24

1 Answers1

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I try to go on with your approach. \begin{align} \sum_{j=0}^i|2j-i| \binom{i}{j} &=2\sum_{j=0}^{\lfloor i/2\rfloor}(i-2j) \binom{i}{j}\\ &=2i\sum_{j=0}^{\lfloor i/2\rfloor}\binom{i}{j}-4\sum_{j=0}^{\lfloor i/2\rfloor}j\binom{i}{j}\\ &=2i\sum_{j=0}^{\lfloor i/2\rfloor}\binom{i}{j}-4i\sum_{j=1}^{\lfloor i/2\rfloor}\binom{i-1}{j-1}\\ &=2i\sum_{j=0}^{\lfloor i/2\rfloor}\binom{i}{j}-4i\sum_{j=0}^{\lfloor i/2\rfloor-1}\binom{i-1}{j}\\ &=2i\binom{i}{\lfloor i/2\rfloor}+2i\sum_{j=0}^{\lfloor i/2\rfloor-1}\underbrace{\left(\binom{i-1}{j-1}-\binom{i-1}{j}\right)}_{\leq 0}\\ &\leq 2i\binom{i}{\lfloor i/2\rfloor}. \end{align} Now show that the inequality holds for $d=2$ (I am aware that the one-line proof in Did's comment show that $d=1$ suffices).

It remains to show that for $i\geq 1$, $$2i\binom{i}{\lfloor i/2\rfloor}\leq 2 \cdot 2^i\sqrt{i}\quad\text{or}\quad \binom{i}{\lfloor i/2\rfloor}\leq \frac{2^i}{\sqrt{i}}.$$

P.S. Take a look here: Elementary central binomial coefficient estimates

Robert Z
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