# Entropy of a binomial distribution

How do we get the functional form for the entropy of a binomial distribution? Do we use Stirling's approximation? According to Wikipedia, the entropy is $\frac1 2 \log_2 \big( 2\pi e\, np(1-p) \big) + O \left( \frac{1}{n} \right)$.

As of now, my every attempt has been futile so I would be extremely appreciative if someone could guide me or provide some hints for the computation.

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A comment: the entropy of the normal distribution with variance $\sigma^2$ is ${1 \over 2} \log (2\pi e \sigma^2)$, which can be computed by a fairly straightforward integration. Perhaps using Stirling's approximation you can reduce the computation of the entropy of the binomial to this same integral plus some error terms. (I haven't actually tried to do this.) –  Michael Lugo Nov 25 '12 at 19:51
We are interested in the sum $$H = -\sum_{k=0}^n {n\choose k}p^k(1-p)^{n-k} \log_2\left[{n\choose k}p^k(1-p)^{n-k} \right].$$ For $n$ large we can use the de-Moivre-Laplace theorem, $$H \simeq -\int_{-\infty}^\infty dx \, \frac{1}{\sqrt{2\pi}\sigma} \exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right] \log_2\left\{\frac{1}{\sqrt{2\pi}\sigma} \exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right] \right\},$$ where $\mu = n p$ and $\sigma^2 = n p(1-p)$. Thus, $$\begin{eqnarray*} H &\simeq& \int_{-\infty}^\infty dx \, \frac{1}{\sqrt{2\pi}\sigma} \exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right] \left[\log_2(\sqrt{2\pi}\sigma) + \frac{(x-\mu)^2}{2\sigma^2} \log_2 e \right] \\ &=& \log_2(\sqrt{2\pi}\sigma) + \frac{\sigma^2}{2\sigma^2} \log_2 e \\ &=& \frac{1}{2} \log_2 (2\pi e\sigma^2) \end{eqnarray*}$$ and so $$H \simeq \frac{1}{2} \log_2 \left[2\pi e n p(1-p)\right].$$ Higher order terms can be found, essentially by deriving a more careful (and less simple) version of de-Moivre-Laplace.