What is the maximum entropy probability distribution if the support is a fixed interval (e.g. $[-1,1]$) with an already known variance?

If we know the support is a fixed interval, then the maximum entropy prob. distribution is the uniform distribution. But if we also add the variance constraint, I'm not sure what the answer is.



1 Answer 1


Denote the support by $[a,b]$ and the variance by $\sigma^2$. Repeating the variational calculus argument that shows the normal distribution maximizes the differential entropy (see here for instance), probability distributions of variance $\sigma^2$ and supported in $[a,b]$ that attain the largest possible entropy should have a density function of the form

$$ p(x)=\frac{e^{-\lambda(x-\mu)^2}}{\int_a^be^{-\lambda(t-\mu)^2}{\rm{d}}t}\quad (a\leq x\leq b)\tag{$\star$} $$ where $\lambda$ and $\mu$ satisfy

$$ \begin{cases} \int_a^bte^{-\lambda(t-\mu)^2}{\rm{d}}t=\mu\int_a^be^{-\lambda(t-\mu)^2}{\rm{d}}t,\\ \int_a^b(t-\mu)^2e^{-\lambda(t-\mu)^2}{\rm{d}}t=\sigma^2\int_a^be^{-\lambda(t-\mu)^2}{\rm{d}}t. \end{cases} $$ The first equation holds automatically if $\mu$ is at the center of the interval $(a,b)$. Assuming that $a,b$ are finite and the variational problem has a unique solution, then $\mu$ must be $\frac{a+b}{2}$: If $p:[a,b]\rightarrow(0,\infty)$ is the density function for the maximizer, then $x\in [a,b]\mapsto p(a+b-x)$ defines another probability distribution with the same support, variance and differential entropy. Then, the first equation above becomes irrelevant and one only needs to solve $$ \int_a^b\left(t-\frac{a+b}{2}\right)^2e^{-\lambda\left(t-\frac{a+b}{2}\right)^2}{\rm{d}}t=\sigma^2\int_a^be^{-\lambda\left(t-\frac{a+b}{2}\right)^2}{\rm{d}}t \tag{$\star\star$} $$ for $\lambda$.

Update) Following the comment by KConrad, a more general system of equations should be considered. Let's write the density function as

$$ p(x)=\frac{e^{-(\lambda x^2+\beta x)}}{\int_a^be^{-(\lambda x^2+\beta x)}{\rm{d}}t}\quad (a\leq x\leq b), \tag{$\star\star\star$} $$ In $(\star)$, it was implicitly assumed that $-\frac{\beta}{2\lambda}$ coincides with the mean $\mu$. Given the form $(\star\star\star)$ for the density function, the system of equations becomes

$$ \begin{cases} \int_a^bte^{-(\lambda t^2+\beta t)}{\rm{d}}t= \mu\int_a^be^{-(\lambda t^2+\beta t)}{\rm{d}}t,\\ \int_a^b(t-\mu)^2e^{-(\lambda t^2+\beta t)}{\rm{d}}t= \sigma^2\int_a^be^{-(\lambda t^2+\beta t)}{\rm{d}}t. \end{cases} $$

As argued before, assuming that the solution is unique, $p(x)$ must coincide with $p(a+b-x)$. When $p(x)$ is of the form $(\star\star\star)$, by comparing the coefficients of $x$ in the exponent from the numerator, $p(a+b-x)\equiv p(x)$ implies $-\frac{\beta}{2\lambda}=\frac{a+b}{2}$. Moreover, $\int_a^btp(t){\rm{d}}t=\int_a^btp(a+b-t){\rm{d}}t$ implies that the mean $\mu=\int_a^btp(t){\rm{d}}t$ is equal to $\frac{a+b}{2}$. Just like before, when $\mu=-\frac{\beta}{2\lambda}=\frac{a+b}{2}$, the first equation holds automatically because $\int_a^b\left(t-\frac{a+b}{2}\right)e^{-\lambda\left(t-\frac{a+b}{2}\right)^2}{\rm{d}}t=0$. So it only remains the second equation which would be same as $(\star\star)$, and should be solved for $\lambda$ to derive a density function of the form $$ p(x)=\frac{e^{-\lambda\left(x-\frac{a+b}{2}\right)^2}}{\int_a^be^{-\lambda\left(t-\frac{a+b}{2}\right)^2}{\rm{d}}t}. $$

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    $\begingroup$ In your equation to determine the mean, you are using $\mu$ in two different ways: the $\mu$ in the exponent need not be the mean $\mu$ outside the integral on the right side. On the interval $[0,1]$ with $p(x) = e^{-(x-1/3)^2}/\int_0^1 e^{-(x-1/3)^2}\,dx$, the mean $\int_0^1 xp(x)\,dx$ is around $.47407$, not $1/3$. $\endgroup$
    – KConrad
    Apr 14 at 14:39
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    $\begingroup$ So just the normal distribution conditioned on being within this interval? $\endgroup$ Apr 14 at 15:49
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    $\begingroup$ @KConrad I edited my answer. $\endgroup$
    – KhashF
    Apr 14 at 15:52
  • $\begingroup$ This may not work if the known variance $\sigma^2 > \frac{(b-a)^2}{12}$ unless you allow $\lambda$ to go negative; if you do, this would not be a truncated normal distribution $\endgroup$
    – Henry
    Apr 27 at 9:59

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