Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When a random variable $X$ has only one possible outcome $x_0$, the probability density function at $X=x_0$ is infinite, while the probability density at other locations is zero. Then the p.d.f is exactly a delta function $\Pr(X=x) = \delta(x=x_0)$.

However, when I tried to calculate the entropy of the random variable, the problem arises. How can I calculate the integral $\int_{-\infty}^{+\infty}{\delta(x-x_0)\log\delta(x-x_0) \, dx}$?

share|cite|improve this question
Mathematicians prefer to use measures rather than "densities" for probability distributions that are not absolutely continuous. $Pr(X=x) = 1 \ne \infty = \delta(x-x_0)$ – GEdgar Sep 4 '12 at 15:26
@Strin : I'd write "probability density function" in the title. "Distribution function" is usually construed as "cumulative distribution function". – Michael Hardy Sep 4 '12 at 17:48

Consider an absolutely continuous distribution with location parameter $x_0$ and scale parameter $\sigma$. We consider two such distributions: normal $\mathcal{D}_1 = \mathcal{N}\left(x_0, \sigma\right)$, and continuous uniform $\mathcal{D}_2 = \mathcal{U}\left(x_0-\sqrt{3} \sigma, x_0+\sqrt{3} \sigma\right)$. Distributions $\mathcal{D}_1$ and $\mathcal{D}_2$ have equal means and variances.

Carrying out the computation for the Shannon entropy: $H(\mathcal{D}) = \mathbb{E}\left(-\ln(f_\mathcal{D})\right)$ yields the following results: $$ \begin{eqnarray} H\left(\mathcal{D}_1\right) &=& \ln(\sigma)+\frac{\ln(2 \pi)}{2} + \frac{1}{2 \sigma^2} \mathbb{Var}(\mathcal{D}_1) = \ln(\sigma) + \frac{1}{2}\left(1 + \ln(2\pi)\right) \\ H\left(\mathcal{D}_2\right) &=& \mathbb{E}\left(\ln(2 \sqrt{3} \sigma)\right) = \ln(\sigma) + \ln\left(2 \sqrt{3}\right) \end{eqnarray} $$ Additionally, consider a Cauchy distribution $\mathcal{D}_3 = \mathrm{Cauchy}\left(x_0, \sigma\right)$: $$ H\left(\mathcal{D}_3\right) = \ln(\sigma) + \ln(\pi) + \frac{1}{\pi} \int_{-\infty}^\infty \frac{\ln\left(1+x^2\right)}{1+x^2}\mathrm{d}x = \ln(\sigma) + \ln(4 \pi) $$ Notice that in the limit $\sigma \to 0^+$, each distribution converges to a degenerate distribution, localized at $x_0$. For each distribution the entropy diverges to $-\infty$ as $\ln(\sigma)$.

Added Per OP's request, one can repeat the above argument for an arbitrary continuous random variable.

Let $f_X(x)$ be the pdf of a standardized random variables $X$ with zero mean, and unit variance. Consider $Y = \sigma X$ with pdf $f_Y(y) = f_X(\sigma x) \frac{1}{\sigma}$. Clearly the Shannon entropy of $Y$ is $H_Y = \log(\sigma) + H_X$ and $H_X$ is a independent of $\sigma$. As $\sigma \downarrow 0$, the distribution of $Y$ converges to a degenerate distribution, and Shannon entropy $H_Y$ tends to $-\infty$ as $\ln(\sigma)$.

share|cite|improve this answer
You showed this fact by computing several distributions. Is there any general theorem saying the entrophy of the delta function is $-\infty$? – Strin Sep 20 '12 at 15:15

Try to make a limit of random variables equidistributed in $[x_0-\varepsilon, x_0+\varepsilon]$. What is the entropy of these? What happens with $\varepsilon\to0$?

share|cite|improve this answer

My intuition says that the entropy of a delta function should be zero.

  1. Since it counts a system's micro states, entropy is always positive.
  2. Delta distribution function implies certainty of the value $X_0$ and thus $\ln(1)$ is zero.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.