In short, the function I am interested in is the following: $$I_n(p) = \sum_{k=0}^n \binom{n}{k} p^k (1-p)^{n-k} \left[h(p) - h\left(\tfrac{k}{n}\right)\right],$$ where $h(x) \triangleq -x \log x - (1-x) \log (1-x)$ is the standard binary entropy function. I am able to prove that for large $n$ this function attains a local maximum around $p \approx \frac{1.338}{n}$ at which $I_n(p) \approx \frac{0.84}{n}$ in case binary logarithms are used. By assuming that $p = \frac{\alpha}{n}$ for $\alpha = O(1)$ and using a Poisson approximation of the binomial distribution, the value $\alpha \approx 1.338$ is a (numerically found) maximizing value of $$\max_{\alpha > 0} \left\{\sum_{z=1}^{\infty} \frac{\alpha^z}{z!} e^{-\alpha} (z \log z) - \alpha \log \alpha\right\}.$$ Note that the latter expression does not contain $n$ anymore; I am interested in large-$n$ asymptotics of $I_n(p)$ which leads to the above, $n$-independent expression.

What I would like to show is that this point $p = \frac{\alpha}{n}$ is actually not just a local, but a global maximum, i.e., there are no values $p = \omega(\frac{1}{n})$ with even higher asymptotic values of $I_n(p)$ for large $n$. So my question can be formulated as:

Can we prove that for large $n$, $\max\limits_{p \in [0,1]} I_n(p) \to I_n(\frac{\alpha}{n})$ with $\alpha \approx 1.338184$?

Numerical inspection of $I_n(p)$ for say $n = 100\,000$ shows that $I_n(p) \propto \frac{1}{n}$ is nice and smooth, has two global maxima at $\frac{\alpha}{n}$ and $1 - \frac{\alpha}{n}$ with value $\frac{0.84}{n}$ and a local minimum at $p = \frac{1}{2}$ with value $\frac{0.72}{n}$ (and two trivial minima at $0$ and $1$ with value $0$). The figure below sketches what $I_n(p)$ (scaled with a factor $n$) looks like for small $n$. As $n$ increases, the function might be best described as a flat line at height $0.72$ with two peaks close to $0$ and $1$ of height $0.84$.

enter image description here

I've tried various approaches using e.g. Taylor expansions of the entropy, but I am not able to prove that "order terms" are actually small for arbitrary $p$. Also, a bound given here using Jensen's inequality is not sharp enough; it's roughly a factor $2$ too loose to prove the above statement.

Any help would be appreciated!

Attempt 1

For large $n$, the dominating terms in the summation in $I_n$ come from values $k \approx np$. Expanding $h(\frac{k}{n})$ around $k = np$ leads to $$h\left(\frac{k}{n}\right) = h(p) + \left(\frac{k}{n} - p\right) h'(p) + \frac{1}{2}\left(\frac{k}{n} - p\right)^2 h''(p) + \frac{1}{6}\left(\frac{k}{n} - p\right)^3 h^{(3)}(p) + \dots.$$ Substituting this back into the definition of $I_n(p)$, this leads to $$I = \sum_{k=0}^n \binom{n}{k} p^k (1-p)^{n-k} \left[-\left(\tfrac{k}{n} - p\right) h'(p) - \tfrac{1}{2}\left(\tfrac{k}{n} - p\right)^2 h''(p) - \tfrac{1}{6}\left(\tfrac{k}{n} - p\right)^3 h^{(3)}(p) - \dots\right].$$ Now the term $\frac{k}{n}$ leads to a factor $\frac{np}{n} = p$ when pulled out of the summation, so the first term disappears. As the second term is simply the variance (scaled by a factor $1/n^2$), this term results in $\frac{1}{2n} p(1-p)h''(p)$. As $h'(p) = \log_2(\frac{1-p}{p})$ and $h''(p) = -1/[p(1-p)\ln 2]$, the second term contributes $1/(2n \ln 2) \approx \frac{0.72}{n}$ as expected. So we have: $$I_n(p) = \frac{1}{2n \ln 2} + \sum_{k=0}^n \binom{n}{k} p^k (1-p)^{n-k} \left[- \tfrac{1}{6}\left(\tfrac{k}{n} - p\right)^3 h^{(3)}(p) - \dots\right].$$ At this point, what remains is to show that all remaining order terms are small, if $np(1-p) = \omega(1)$.

Attempt 2

Using the approach described here, consider the sum $$S = \sum_{k=0}^n \binom{n}{k} p^k (1-p)^{n-k} \frac{k}{n} \log_2 \frac{k}{n}.$$ First, we can cancel the $\frac{k}{n}$ with factors in the binomial coefficient, and pull out a factor $p$ to obtain $$S = p \sum_{k=1}^n \binom{n-1}{k-1} p^{k-1} (1-p)^{n-k} \log_2 \frac{k}{n} = p \sum_{k=0}^{n-1} \binom{n-1}{k} p^{k} (1-p)^{n-k-1} \log_2 \frac{k+1}{n}.$$ Applying Jensen's inequality, we obtain $$S \leq p \log_2\left(\sum_{k=0}^{n-1} \binom{n-1}{k} p^{k} (1-p)^{n-k-1} \frac{k+1}{n}\right) = p \log_2 \frac{(n-1)p + 1}{n}.$$ Pulling out the term $\log_2 p$, we obtain $$S \leq p \log_2 p + p \log_2\left(1 + \frac{1 - p}{np}\right).$$ Applying the same procedure to the other term in $I_n(p)$, this leads to $$I_n(p) \leq h(p) + \sum_{k=0}^n \binom{n}{k} p^k (1-p)^{n-k} \left[\frac{k}{n} \log_2 \frac{k}{n} + \frac{n-k}{n} \log_2 \frac{n-k}{n}\right] \\ = h(p) + \left[p \log_2 p + p \log_2\left(1 + \frac{1 - p}{np}\right)\right] + \left[(1-p) \log_2 (1-p) + (1-p) \log_2\left(1 + \frac{p}{n(1-p)}\right)\right] \\ = p \log_2\left(1 + \frac{1 - p}{np}\right) + (1-p) \log_2\left(1 + \frac{p}{n(1-p)}\right)$$ Using $\ln(1 + x) \leq x$ for all $x$, this leads to $$I_n(p) \leq \frac{1 - p + p}{n \ln 2}.$$ Unfortunately this is a factor $2$ off, as $I_n(p) \sim \frac{1}{2n \ln 2}$ for almost all $p$.

  • 1
    $\begingroup$ Do you mean $h(x) = - x \log x - (1-x) \log(1-x)$? $\endgroup$ Mar 16, 2015 at 17:39
  • $\begingroup$ @Robert: Yes, I'm sorry. I'll update it when I'm at my PC. (And the result is independent of the base of the $\log$s.) $\endgroup$
    – TMM
    Mar 16, 2015 at 18:31
  • 1
    $\begingroup$ For larger $p$, restricting the sum to $\log n$ standard deviations from the mean is enough ($h(x)$ being bounded), then within those limits a few terms of the Euler-Maclaurin expansion will get it to the accuracy you need. Unfortunately the integral doesn't seem to have a closed form. $\endgroup$ Mar 17, 2015 at 2:12
  • 1
    $\begingroup$ @Pietro: It's perhaps best described in terms of $J_n(p) \triangleq n \sup\limits_{p \in [0,1]} I_n(p)$ as $\lim\limits_{n \to \infty} J_n(p) = 0.84\dots$ where $0.84\dots$ is some numerical constant. So take the maximum over $p$ for fixed $n$, and then consider large-$n$ asymptotics of the resulting expression. $\endgroup$
    – TMM
    Mar 18, 2015 at 0:39
  • 1
    $\begingroup$ OK, so you want an estimate on the uniform convergence of the Bernstein polynomials of $h(x)$, since $\sup_{p\in[0,1]}I_n(p )=\|h−B_n h\|_\infty$. $\endgroup$ Mar 18, 2015 at 11:22

2 Answers 2


By symmetry I can assume $p\le\frac12$. I will use natural logs.

Put $f(x)=h(p)-h(x)$ and $b_k = \binom{n}{k} p^k(1-p)^{n-k}$. Define $k_0=\lceil pn/2\rceil$ and $k_1=n-k_0$.

The plan is: find polynomials $f_0(x),f_1(x)$ such that $f_0(x)\le f(x)\le f_1(x)$ for $k_0/n\le x\le k_1/n$. Then make bounds on the various parts of $$\sum_{k=0}^{k_0-1} b_k(f(k/n)-f_0(k/n)) + \sum_{k=0}^n b_k f_0(k/n) + \sum_{k=k_1+1}^n b_k (f(k/n)-f_0(k/n)) \le \sum_{k=0}^n b_k f(k/n) \le \sum_{k=0}^{k_0-1} b_k(f(k/n)-f_1(k/n)) + \sum_{k=0}^n b_k f_1(k/n) + \sum_{k=k_1+1}^n b_k (f(k/n)-f_1(k/n)) .$$ Since $f^{(iv)}(x) = 2/x^3+2/(1-x)^3$, we can use Taylor's theorem with remainder to get $$f_j(x) = (\ln p -\ln(1-p))(x-p) + \frac{(x-p)^2}{2p(1-p)} - \frac{(1-2p)(x-p)^3}{6p^2(1-p)^2} + \frac{j(8-8p+2p^2+p^3)(x-p)^4}{3p^3(2-p)^2}$$ for $j=0,1$. Maple now tells us $$\sum_{k=0}^n b_k f_j(k/n)=\frac{1}{2n}-\frac{(1-2p)^2}{6p(1-p)n^2} + \frac{Bj}{n^2},$$ where $$B = \frac{(1-p)^2(8-8p+2p^2+p^3)}{p(2-p)^2} + \frac{(1-p)(8-8p+2p^2+p^3)(1-6p+6p^2)}{3p^2(2-p)^2n}.$$ This much is $1/(2n)+O(1/(pn^2))$.

In the range $0\le k\le k_0-1$, $b_k$ is dominated by a geometric series with ratio $\frac12$ and $f(x)-f_j(x)=O(p\ln p)$. Using the Stirling approximation for $b_k$, we find that $$\sum_{k=0}^{k_0-1} b_k(f(k/n)-f_j(k/n))=o(1/n)$$ for $p\ge (2+\epsilon)\ln\ln n/n$.

In the range $k_1+1 \le k\le n$, $b_k\le 2^{-3n/4}$ (using the assumption $p\le \frac12$) and $f(x)-f_j(x)=O(p^{-3})$, so this part is negligible compared to the previous part.

In summary, $$\sum_{k=0}^n b_k f(k/n) = \frac{1+o(1)}{2n}$$ for $(2+\epsilon)\ln\ln n/n\le p\le 1-(2+\epsilon)\ln\ln n/n$, any $\epsilon\gt 0$.

For the case $p=a/n$ with $a=O(\ln\ln n)$, use $\binom{n}{k}=\frac{n^k}{k!}\left(1-\binom{k}{2}/n -O(k^3/n^2)\right)$ and simple bounds on the tail to find $$\sum_{k=0}^n b_k f(k/n) = \sum_{k=0}^{(\ln n)^2} \frac{a^k}{k!}\left(1-\frac{\binom{k}{2}}{n}-\frac{ak}{n}\right)f(k/n) + O((\ln n)^{O(1)}/n^2).$$

  • $\begingroup$ Thanks for the answer. How exactly did you get the expressions for $f_j(x)$? What bound on the remainder term did you use there? $\endgroup$
    – TMM
    Mar 19, 2015 at 12:18
  • $\begingroup$ Since I was assuming $p\le\frac12$, the largest value of the fourth derivative of $f$ in the interval $[\frac p2,1-\frac p2]$ occurs at $x=\frac p2$. And the fourth derivative is positive in the interval. I used that in the standard formula for the remainder term. $\endgroup$ Mar 19, 2015 at 23:55

EDIT: For reasons explained by TMM in the comment below, these bounds aren't in and of themselves good enough. I'm leaving them here because the comment is valuable, but this doesn't work as an answer.

Assuming that your Taylor expansion/Poisson arguments are enough to handle the case $np$ bounded, I believe the case $np \rightarrow \infty$ can be taken care of using the law of large numbers/Chebyshev. We have $$I_n(p) = E( H(p) - H(X))$$ where $X$ is $\frac{1}{n}Bi(n,p)$. Assume WLOG $p \leq \frac{1}{2}$, and take arbitrary $t<p$. We can bound $H(p)-H(X)$ above by $H(p)-H(t)$ for $t \leq X \leq 1-t$, and above by $1$ in general. This means $$I_n(p) \leq [H(p)-H(t)] + P(X<t) + P(X>1-t)$$

If, say, $t=p-(np)^{-1/3}$ and $np \rightarrow \infty$, then by Chebyshev's inequality the latter two terms are both at most $(np)^{-1/3}$, while the first term goes to $0$ because $H$ is continuous.

  • 2
    $\begingroup$ Maybe I should have added this, but $I_n(p)$ scales as $1/n$ on all of $[0,1]$. Using binary logarithms, for $p \approx 1.34/n$ we have $I_n(p) \approx 0.84/n$ and for other values of $p$ (say $p = 1/2$) we have $I_n(p) \approx 0.72/n$. So finding an upper bound with terms that only weakly converge to $0$ is not enough. (And since the values of $I_n(p)$ for $0 \ll p \ll 1$ are only $16\%$ smaller than the value at the maximum, you'll need to use very tight bounds to prove the result.) $\endgroup$
    – TMM
    Mar 16, 2015 at 20:34

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