# Average of random variables is “more log-concave”

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 $$w_i$$.

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