As per Wikipedia:

The Cauchy distribution is the maximum entropy probability distribution for a random variate $X$ for which

$$ {\displaystyle \operatorname {E} [\log(1+(X-x_{0})^{2}/\gamma ^{2})]=\log 4} $$

or, alternatively, for a random variate $X$ for which

$$ {\displaystyle \operatorname {E} [\log(1+(X-x_{0})^{2})]=2\log(1+\gamma ).} $$

Is there an interpretation of this constraint? I understand how maximizing entropy with this constraint leads to the Cauchy Distribution, but what does this constraint mean? I suspect that it is in some way a constraint on the variance of the distribution, or even on the entropy itself, but I don't quite get it. Any insight would be greatly appreciated.


1 Answer 1


I would say the result has been found backward, namely that, using the general property that the maximum entropy distribution under the constraint $\mathbb{E}[f(X)]=\alpha$ is given by the density $$\pi(x)=C\,\exp\{\lambda f(x)\}$$ when $C$ and $\lambda$ are determined by the conditions $$\int_\mathcal{X} \exp\{\lambda f(x)\} \text{d}x = C^{-1}\quad \text{and}\quad \int_\mathcal{X} f(x) \exp\{\lambda f(x)\} \text{d}x = \alpha C^{-1}$$(where $\text{d}x$ is understood as the chosen dominating measure (e.g., the Lebesgue measure).

Hence, if one wants an arbitrary distribution with density proportional to $\exp\{\lambda f(x)\}$ to become a maximum entropy distribution, it is enough to choose the constraint as $\mathbb{E}[f(X)]=\alpha$, or conversely to define $$f(x)=\log(\pi(x)$$ and find the value of $$\mathbb{E}[\log\pi(X)]$$

  • $\begingroup$ Thanks, but that's not quite what I was asking. I understand that this constraint was probably reverse-engineered in order to make the Cauchy the maximum entropy distribution of something , but I was hoping to find out what the meaning of that something was (if it is meaningful). $\endgroup$ Jan 26, 2018 at 20:05
  • $\begingroup$ I do not think it has a useful interpretation, but I may be wrong! $\endgroup$
    – Xi'an
    Jan 26, 2018 at 20:48
  • $\begingroup$ I suspect you are right, but that's what I was afraid of! $\endgroup$ Jan 26, 2018 at 20:57

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