I am trying to use De Moivre–Laplace theorem to approximate $$1 - \sum_{k=0}^{n} {n \choose k} p^{k}(1-p)^{n-k} \log\left(1+\left(\frac{p}{1-p}\right)^{n-2k}\right)$$

The idea of an approximation is that we don't have the sum term which is difficult to calculate if $n$ is high.

Using the De Moivre–Laplace theorem gets us that: $${n \choose k} p^{k}(1-p)^{n-k} \approx \frac{1}{\sqrt{2 \pi np(1-p)}}e^{-\frac{(k-np)^2}{2np(1-p)}}$$ Now we see that \begin{align} F &= 1 - \sum_{k=0}^{n} {n \choose k} p^{k}(1-p)^{n-k} \log\left(1+\left(\frac{p}{1-p}\right)^{n-2k}\right) \\&\approx 1 - \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi np(1-p)}}e^{-\frac{(x-np)^2}{2np(1-p)}}\log_2\left(1+\left(\frac{p}{1-p}\right)^{n-2x}\right) dx \end{align}

my calculation is inspired by Entropy of a binomial distribution

If one has an other suggestion to approximate $F$ or get a closed for i would like to hear those. So far i've tried approximating $F$ with a least squares method using a tanh function as the fit function.

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    $\begingroup$ Missed $dp$ in the integral $\endgroup$ – Yuri Negometyanov Aug 4 '18 at 20:22
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    $\begingroup$ I didnt get it. There is something seriously wrong here. The sum formula is eventually independent of $k$ but dependent on $p$ and the final integral is independent of $p$ but dependent on $k$. What does $k$ mean after the integration? sounds strange right? $\endgroup$ – Seyhmus Güngören Aug 4 '18 at 21:06
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    $\begingroup$ I guess $k$ is eventually $x$ and the integration over $p$ is probably a typo. I do not get the limit $-\infty..+\infty$ though when originally we have $0..n$. $\endgroup$ – Diger Aug 4 '18 at 21:12
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    $\begingroup$ Actually it doesnt matter. So for $p<0.5$ use the approximation $\log(1+y)\approx y$ and for large $y$, we have $\log(1+y)\approx \log(y)$. One more thing you have $y=a^{(f(x))}$ and you can write this as $\exp(a,f(x))$, then in the integral you have the multiplication of two exponentials. I think these forms are well studied. $\endgroup$ – Seyhmus Güngören Aug 4 '18 at 22:07
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    $\begingroup$ take $p$ small enough and $n$ large enough. I think it should work. Btw. It will be very good, if you could put the results that you already have to the question description. Plus: the answer of wolfram.alpha for the evaluation of this integral, just the link. $\endgroup$ – Seyhmus Güngören Aug 4 '18 at 22:28


Let WLOG the inequality $$q=\dfrac p{1-p}\in(0,1)\tag1$$ is valid. Otherwise, the corresponding opposite events can be reversed.

This allows to present the issue expression in the form of \begin{align} &S(n,p)=1 - (1-p)^n\sum_{k=0}^{n} {n \choose k} q^k\log\left(1+q^{n-2k}\right),\tag2\\[4pt] \end{align} or \begin{align} &=1 - (1-p)^n\sum_{k=0}^{n} {n \choose k}q^kq^{n-2k} - (1-p)^n\sum_{k=0}^{n} {n \choose k}q^k\left(\log\left(1+q^{n-2k}\right)-q^{n-2k}\right)\\[4pt] &=1 - (1-p)^n(1+q)^n - (1-p)^n\sum_{k=0}^{n} {n \choose k}q^k\left(\log\left(1+q^{n-2k}\right)-q^{n-2k}\right)\\[4pt] &S(n,p)= - (1-p)^n\sum_{k=0}^{n} {n \choose k}q^k\left(\log\left(1+q^{n-2k}\right)-q^{n-2k}\right).\tag3\\[4pt] \end{align} Formula $(3)$ can simplify the calculations, because it does not contain the difference of the closed values.

$$\color{brown}{\textbf{How to calculate this.}}$$

Note that the sum of $(3)$ contains both the positive and the negative degrees of $q.$ This means that in the case $n\to \infty$ the sum contains the terms of the different scale.

The calculations in the formula $(3)$ can be divided on the two parts.

$\color{green}{\textbf{The Maclaurin series.}}$

The Maclaurin series for the logarithmic part converges when the term $\mathbf{\color{blue}{q^{n-2k} < 1}}.$ This corresponds with the values $k<\frac n2$ in the case $\mathbf{q<1}$ and with the values $k>\frac n2$ in the case $\mathbf{q>1}.$ Then the Maclaurin series in the form of $$\log(1+q^{n-2k}) = \sum_{i=1}^\infty\frac{(-1)^{i+1}}{i}q^{(2n-k)i}\tag4$$ can be used.

If $\mathbf{\color{blue}{q^{n-2k} > 1}},$ then $$\log(1+q^{n-2k}) = \log(q^{2n-k}(1+q^{k-2n})) = (2n-k)\log q + \log(1+q^{k-2n}).\tag5$$

If $\mathbf{\color{blue}{q^{n-2k} = 1}},$ then $LHS(4) = \log2.$

If $\mathbf{\color{blue}{q^{n-2k} \lesssim 1}},$ then $$\log(1+q^{2n-k}) = \log\frac{1+r}{1-r} = 2r\sum_{i=0}^\infty\frac{(-1)^i}{2i+1}r^{2i},\quad \text{ where } r=\frac{q^{2n-k}}{2+q^{2n-k}}\approx\frac{q^{2n-k}}3,\tag6$$ and can be used some terms of the series.

$\color{green}{\textbf{The double summations.}}$

After the substitution of the $(4)$ or $(5)$ to $(3)$ the sums can be rearranged. For example, $$\sum_{k=0}^{L}{n \choose k}q^k\sum_{i=1}^\infty\frac{(-1)^{i+1}}{i}q^{(2n-k)i}= \sum_{i=1}^\infty\frac{(-1)^{i+1}}{i}\sum_{k=0}^{L}{n \choose k}q^kq^{(2n-k)i}$$ $$= q^{n+1}\sum_{i=1}^\infty\frac{(-1)^{i+1}}{i}\sum_{k=0}^{L}{n \choose k}\left(q^{i+1}\right)^{n-k},$$ wherein the order of the summation can be chosen, taking in account the given data.

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    $\begingroup$ hmmmm, i am quite curious now. Does the limit converge to 0 as $n \rightarrow \infty?$ $\endgroup$ – Kees Til Aug 11 '18 at 0:15
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    $\begingroup$ @KeesTil If $k<\frac n2,$ then we have suitable Maclaurin series. If $k>\frac n2,$ then the factor $(1-p)^n$ provides the convergence. $\endgroup$ – Yuri Negometyanov Aug 11 '18 at 0:37
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    $\begingroup$ @KeesTil Yes, that's valid. Fortunately, $p$ and $1-p$ are the possibilities of the opposite events, whose designations can be changed for this task. $\endgroup$ – Yuri Negometyanov Aug 11 '18 at 19:10
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    $\begingroup$ @KeesTil I see that $S(p)=S(1-p).$ Please check this. Thanks for the useful comments. $\endgroup$ – Yuri Negometyanov Aug 11 '18 at 19:29
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    $\begingroup$ I don't get the last sentence where you day that the inner sums can be calculated via the de Moivre Laplace theorem. How can we do this, $q+1 \neq 1$. I use this definition btw: en.wikipedia.org/wiki/De_Moivre%E2%80%93Laplace_theorem $\endgroup$ – Kees Til Aug 12 '18 at 10:33

The expression looks very much like the Bernstein approximation of a function ($1-f(x)$) on $[0,1]$. But the argument (in fact the degree $n-2k$) of $log$ function ruins everything.

Here is a quick idea. Denote $$ y(p)=\sum_{k=0}^{n} {n \choose k} p^{k}(1-p)^{n-k} \log\left(1+\left(\frac{p}{1-p}\right)^{n-2k}\right). $$

Let us assume that we can represent $y(p)$ in the form $y(p)=\sum_{m=0}^\infty y_m p^m$, where $y_m$ are constants not depending on $p$.

Note that $y(p)=y(1-p)$. Let us consider the equation $$ y(p)+y(1-p)=f(p). \tag{eq1}\label{eq1} $$ Although we can write out the expression for $f(p)$, let us think that we don't know how $f(p)$ looks like. But for sure, $f(p)$ must satisfy $f(p)=f(1-p)$. It is know (see for example http://eqworld.ipmnet.ru/en/solutions/fe/fe1116.pdf) that equations like \eqref{eq1} have a solution, for example, $$ \tag{eq2}\label{eq2} y(p)=f(p) \sin^2({\pi p \over 2}). $$ By expanding $\sin^2({\pi p \over 2})$ into the Maclaurin series we get $$ y(p)=f(p) \sum_{m=1}^\infty {(-1)^{m+1} 2^{2m-1} \over (2m)!} {p^{2m} \pi^{2m} \over 2^{2m}}. $$

Let us assume that $f(p)$ is an analytic function i.e. $f(p)=\sum_{m=0}^\infty {f^{(m)}(0)\over m!} p^m$. By writing \eqref{eq2} in the series form we have: $$ \sum_{m=0}^\infty y_m p^m = \left ( \sum_{m=0}^\infty {f^{(m)}(0)\over m!} p^m \right ) \left ( \sum_{m=1}^\infty {(-1)^{m+1} \over (2m)!} {p^{2m} \pi^{2m} \over 2} \right ). $$

From this relation it may be possible to find the expressions for $f^{(m)}(0)$ through $y_m$ by equating the coefficients at $p^m$. If this works out, we go back to the right part of \eqref{eq2} and try to find how many terms in the product $$ \left ( f^{(0)}(0) + f^{(1)}(0) p + f^{(2)}(0) {p \over 2} + \dots \right ) \left ( \sum_{m=1}^\infty {(-1)^{m+1} \over (2m)!} {p^{2m} \pi^{2m} \over 2} \right ). $$

yield the approximate value.

  • $\begingroup$ so i can't use this because of the "n-2k" term? $\endgroup$ – Kees Til Aug 7 '18 at 18:07
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    $\begingroup$ It is impossible (at least I don't see how) to use Bernstein approximation because of the (n-2k) degree. But the idea above is the one you can try. $\endgroup$ – rrv Aug 8 '18 at 5:21
  • $\begingroup$ I tried this method, however my terms were very complicated and i could not see something that was simplified. My Mathemathica code: A = FullSimplify[ Series[Sum[ Binomial[n, k]*p^k (1 - p)^(n - k) Log[2, 1 + (p/(1 - p))^(n - 2 k)], {k, 0, n}] , {p, 0, 2}]] B = FullSimplify[ Series[Sum[((-1)^(m + 1))/2 m!*(p^(2*m)*Pi^(2*m)/(2)), {m, 1, Infinity}], {p, 0, 2}]] $\endgroup$ – Kees Til Aug 8 '18 at 9:30

We can apply a method similar to this. Since the summand has a sharp peak around $k = n/2$, we can take an expansion valid for large $n$ and for $k$ close to $n/2$ and then, also due to the tails being small, extend the summation range indefinitely:

$$a_k = \binom n k p^k q^{n - k} \ln \left( 1 + \left( \frac p q \right)^{n - 2 k} \right), \quad q = 1 - p, \\ a_{n/2 + i} \sim \sqrt {\frac 2 {\pi n}} \left( 2 \sqrt {p q} \right)^n \left( \frac p q \right)^i \ln \left( 1 + \left( \frac q p \right) ^{2 i} \right), \\ \sum_{k = 0}^n a_k \sim \sqrt {\frac 2 {\pi n}} \left( 2 \sqrt {p q} \right)^n \sum_{i = -\infty}^\infty \left( \frac p q \right)^i \ln \left( 1 + \left( \frac q p \right) ^{2 i} \right), \\ n \to \infty, p \text{ fixed}, 0 < p < 1, p \neq \frac 1 2.$$

  • $\begingroup$ i don't really get it right now, but you now have a summation term from $-\infty$ to $\infty$, does that not make the computation expensive? $\endgroup$ – Kees Til Aug 11 '18 at 18:08
  • $\begingroup$ Depends on what your goal is. If $p$ is fixed, the sum over $i$ is a constant. $\endgroup$ – Maxim Aug 11 '18 at 18:17
  • $\begingroup$ can you write out the sum for fixed $p$, i don't see the solution directly $\endgroup$ – Kees Til Aug 11 '18 at 18:30
  • $\begingroup$ Do you mean a closed form? This infinite sum probably doesn't have one. $\endgroup$ – Maxim Aug 11 '18 at 18:45
  • $\begingroup$ too bad my p is not fixed so i dont think this will work for me :( $\endgroup$ – Kees Til Aug 11 '18 at 18:53

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