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Problem (1)

Suppose $\phi_i\in [0,\pi/2]$ are drawn uniformly for $1\le i\le n$, and $\sum_{i=1}^n w_i=1$, $w_i\ge 0$. Show that the pdf $p_1$ of the random variable $$\phi = \sin^{-1}\left(\sqrt{\sum_{i=1}^n w_i \sin^2 \phi_i}\right)$$ satisfies $-(\ln p_1)''\ge 0$. (This is implied by $p_1$ attaining maximum at $\phi=\pi/4$.)

Alternate formulation (2)

Let $X_1,\ldots, X_n$ be random variables drawn from the distribution Beta(1/2,1/2) (which has pdf $p(x)=\frac1{\sqrt{x(1-x)}}$), and suppose $\sum_{i=1}^n w_i=1$, $w_i\ge 0$. Let the pdf of the random variable $\sum_{i=1}^n w_i X_i$ be $q(x)$.

Show that $-(\ln q)''(1/2) \ge -(\ln p)''(1/2)=-4$.

Discussion

To see that (1) and (2) are equivalent, note that by change of variable the pdf in (1) is $ p_1(\phi) = q(\sin^2\phi)\sin\phi\cos\phi$, and $-(\ln p_1)''(\pi/4) = -(\ln q)''(1/2) + 4$.

Say that a pdf $p$ is $\alpha$-log-concave if $- (\ln p)''(x)\ge \alpha$. One expects that the pdf of an average of random variables to be more concentrated and more log-concave than the individual pdfs.

For example, Theorem 3.7 here says that if $X,Y$ have pdf's that are $a$-log-concave, then $w_1X+w_2Y$ has a pdf that is $\frac{1}{\frac{w_1^2}{a}+\frac{w_2^2}{a}}=\frac{a}{w_1^2+w_2^2}$-log-concave. The difficulty here is that the distribution Beta(1/2,1/2) is more log-convex away from the point of interest 1/2, and is in fact log-convex everywhere.

We also know that in the limit as $\max w_i\to 0$, by the central limit theorem, the distribution (suitably scaled) approaches a normal distribution. Here, I'm not requiring that the distribution becomes significantly more concentrated, but I do need something that works for any weights $w_i$.

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  • $\begingroup$ Are the $w_i$'s positive? $\endgroup$ – Iosif Pinelis Jun 5 '20 at 20:59
  • $\begingroup$ @IosifPinelis Yes, I added this to the problem statement. $\endgroup$ – Holden Lee Jun 5 '20 at 22:37
  • $\begingroup$ Also the log-concavity is equivalent to $p_1(\sqrt{xy})\ge \sqrt{p_1(x)p_1(y)}$, yes? $\endgroup$ – Matt F. Jun 6 '20 at 2:14
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    $\begingroup$ @MattF. The left hand side of your inequality should be $p_1((x+y)/2)$. $\endgroup$ – Holden Lee Jun 7 '20 at 12:28

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