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Let $f$ be the density of a log-concave probability distribution on the interval $[0,1]$ (with respect to Lebesgue measure). To be concrete, suppose that $f(x) = \exp( - \varphi(x))$, for some convex function $\varphi: [0,1] \to \mathbb{R}$.

Is it true that the largest variance of a random variable with density $f$, is attained by the uniform density? (that is, the largest variance is $\frac{1}{12}$). If it is true, is there a simple proof?

If the density is only required to be unimodal, then the supremum over the variance is $\frac19$ (the proof here is wrong). In case the density itself is concave, there is a proof here.

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  • $\begingroup$ What is the mistake in the CrossValidated proof? $\endgroup$
    – Aryeh Kontorovich
    Commented Aug 23, 2016 at 7:20
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    $\begingroup$ The maximum of the expression $\frac13 \mu (2 - 3 \mu)$ over the interval $0 \le \mu \le \frac12$ is $\frac19$ and not $\frac{1}{12}$. $\endgroup$
    – sometempname
    Commented Aug 23, 2016 at 7:24

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