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I am trying to prove the following:

$${}^nC_k \geq \frac{1}{n+1}2^{nH(x)}$$

for $x = \frac{k}{n}$ and $ 0 \leq k \leq n$. The entropy function is the binary entropy function with $H(0) = H(1) = 0$.

I have already proved that $ {}^nC_k \leq 2^{nH(x)}$

I tried to use this along with the equation $\sum_{i=0}^{n} {}^nC_i x^i (1-x)^{n-i} = 1$

So this becomes $\sum_{i=0}^{n} {}^nC_i x^i (1-x)^{n-i} = 1 \le \sum_{i=0}^{n} 2^{nH(x)} x^i (1-x)^{n-i}$

So, $$\sum_{i=0}^{n}\frac{{}^nC_i}{ 2^{nH(\frac{m}{n})}} = 1 \le n+1$$

And then each individual term must be $\le n+1$, but this does not help. I am not sure how to approach this. I found a similar question, but wasn't able to understand the answer.

learning
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1 Answers1

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To expand on the answer linked:

Note that $$\int_0^1 x^k (1 - x)^{n-k} \,dx = \frac{1}{(n+1)\binom{n}{k}},$$

either by Wolfram Alpha or induction via integration by parts. If we note that the function $f(x) = x^k ( 1- x)^{n-k}$ achieves a maximum on $[0,1]$ at $x = \frac{k}{n}$, then we get the upper bound of $$\int_0^1 f(x)\,dx \leq 1 \cdot f(k/n).$$

Noting that $$f(k/n) = \left(\frac{k}{n}\right)^k \left(1 - \frac{k}{n}\right)^{n-k} = 2^{-n H_2(k/n)}$$ gives $$\frac{1}{(n+1)\binom{n}{k}} \leq 2^{-n H_2(k/n)}.$$

Rearranging terms completes it.

Marcus M
  • 11,229
  • Thanks! Is there any way to do it with that summation in my question? I'm curious because I ran into this during self-study. – learning Nov 15 '17 at 21:16