Maximum entropy distribution of a proportion with known mean and variance? Is it a beta? Given a proportion and its standard error, what distributional assumption minimizes assumptions/maximizes entropy?  Is it the beta (and can I use the method of moments to estimate its parameters)?  Or something else?
 A: It's a truncated Normal distribution.  This is a consequence of Boltzmann's Theorem.

The following analysis provides the details needed to implement a practical solution.
A Normal$(\mu,\sigma)$ distribution $F$ truncated to the interval $[0,1]$ arises by taking a standard Normal variable $X$ with probability distribution $\Phi$, scaling it by $\sigma$, shifting it to $\mu$, and truncating it to $[0,1]$.  Equivalently--working backwards--the original variable $X$ must have been truncated to the interval $[-\mu/\sigma, (1-\mu)/\sigma]$ where it had a total probability of
$$C = \Phi\left(\frac{1-\mu}{\sigma}\right) - \Phi\left(\frac{-\mu}{\sigma}\right),\tag{1}$$
expectation
$$\mu_1=\frac{1}{C\sqrt{2\pi}}\int_\frac{-\mu}{\sigma}^\frac{1-\mu}{\sigma} x\exp\left(\frac{-x^2}{2}\right)\mathrm{d}x,$$
and second (raw) moment
$$\mu_2 = \frac{1}{C\sqrt{2\pi}}\int_\frac{-\mu}{\sigma}^\frac{1-\mu}{\sigma} x^2\exp\left(\frac{-x^2}{2}\right)\mathrm{d}x.$$
Presumably your "standard error" is either $\sqrt{\mu_2-\mu_1^2}$ or some constant multiple of it.
These integrals can be computed in terms of
$$\mu_1(z) = \frac{1}{C\sqrt{2\pi}}\int_{-\infty}^z x\exp\left(\frac{-x^2}{2}\right)\mathrm{d}x = -\frac{1}{C\sqrt{2\pi}}\exp\left(-\frac{z^2}{2}\right)\tag{2}$$
and, integrating by parts,
$$\eqalign{
\mu_2(z) &= \frac{1}{C\sqrt{2\pi}}\int_{-\infty}^z (x)\left(x\exp\left(\frac{-x^2}{2}\right)\right)\mathrm{d}x \\
&= \frac{1}{C\sqrt{2\pi}}\left(x\left(-\exp\left(-\frac{x^2}{2}\right)\right)\mid_{-\infty}^z - \int_{-\infty}^z -\exp\left(-\frac{x^2}{2}\right)\mathrm{d}x \right)\\
&=-\frac{1}{C\sqrt{2\pi}}z\exp\left(-\frac{z^2}{2}\right) + \frac{1}{C}\Phi(z)\tag{3}.
}$$
Thus
$$\mu_1 = \mu_1\left(\frac{1-\mu}{\sigma}\right) - \mu_1\left(\frac{-\mu}{\sigma}\right)$$
and
$$\mu_2 = \mu_2\left(\frac{1-\mu}{\sigma}\right) - \mu_2\left(\frac{-\mu}{\sigma}\right).$$
These calculations $(1)$, $(2)$, and $(3)$ can be implemented in any software where exponentials, square roots,  and $\Phi$ are available.  This permits application in any fitting procedure, such as method of moments or maximum likelihood.  Either would require numerical solutions.
