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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    $\begingroup$ 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.) $\endgroup$ Nov 25, 2012 at 19:51

1 Answer 1


This answer follows roughly the suggestion of @MichaelLugo in the comments.

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.

  • $\begingroup$ How is the integration done in the second to last step? $\endgroup$ Jul 15, 2014 at 18:55
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    $\begingroup$ @user26872 is there any analogue for multinomial distribution? Thanks in advance! $\endgroup$ Jan 14, 2016 at 13:28
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    $\begingroup$ @EgorovaLena: There is a generalized de-Moivre-Laplace theorem that should be useful in this regard. See these notes, for example. $\endgroup$
    – user26872
    Jan 14, 2016 at 19:58
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    $\begingroup$ Why is the "dx" before the term being integrated. I know that it is a multiplication, but doesn't convention put the increment of integration at the far right side of the integral? $\int{sin \left( x \right) \cdot dx}$ vs. $\int{dx \cdot sin \left( x \right) }$. Is there a reason for the difference? $\endgroup$ Mar 9, 2016 at 15:54
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    $\begingroup$ @EngrStudent: I tend to think about integration as an operation that acts to the right like differentiation, rather than from outside to inside as it is often written: $\int_{y_1}^{y_2} dy\int_{x_1}^{x_2} dx(\ldots)$ rather than $\int_{y_1}^{y_2}\int_{x_1}^{x_2}(\ldots)dx\,dy$. This is probably due to my background in physics, where this notation is common. $\endgroup$
    – user26872
    Mar 11, 2016 at 0:11

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