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Let $p(x)$ be a continuous probablity density function on the interval $x\in[0,1]$, subject to the constraint $p(x)=0$ whenever $x\le0$ or $x\ge1$. Assume that the expected value of $p(x)$ is known and fixed:

$$\int_0^1 p(x)x\mathrm{d}x=\mu$$

What's the continuous probability density function $p(x)$ of maximum entropy satisfying these constrainsts? Note that I require that $p(x)$ is continuous on the interval $[0,1]$, so the answer cannot be a truncated exponential distribution, as Boltzmann's theorem would seem to suggest, because a truncated exponential distribution is discontinuous at the endpoints $x=0$ and $x=1$.

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This is answered at our sister site at https://math.stackexchange.com/questions/3123162/maximum-entropy-distribution-given-constrained-minimum-maximum-and-mean and the solution given is $f_{\mu}(x) = \frac{c}{e^c - 1} e^{cx} \mathbf{1}_{[0,1]}(x)$ where $c$ solves $\mu = 1 - \frac{1}{c} + \frac{1}{e^c - 1}$

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  • $\begingroup$ This is the solution given by Boltzmann theorem, which is not continuous. $\endgroup$
    – a06e
    Commented Dec 1, 2023 at 7:15
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    $\begingroup$ I think the conclusion is that there is no continuous local maxima of the entropy in this setting. $\endgroup$
    – a06e
    Commented Dec 1, 2023 at 15:50

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