Given a probability distribution on the interval $[0,1]$, is there any relationship between the quantity $$\sup_{S}{\mathbb{E}(X|X\in S)^{2}\Pr(X\in S)}$$ over all measurable subsets $S$, and the quantity $\mathbb{E}X^2$? I ask because the case where $S=[0,1]$ means comparing $(\mathbb{E}X)^2$ and $\mathbb{E}X^2$, which is obviously a well-known thing, but it seems like the bad distributions all have concentrations in a subset, which might be alleviated by taking a sup over all $S$.



For a given value of $\text{Pr}(S)$, we maximize $\mathbb E[X \mid X \in S]$ if $S$ is an interval $(y, 1]$ or $[y,1]$. Thus if $\mu$ is the probability distribution of $X$, your quantity is $$ \sup_{0 \le y \le 1} \; \max\left(\frac{\left(\int_{(y,1]} x\; d\mu(x)\right)^2}{\int_{(y,1]}\; d\mu(x)}, \frac{\left(\int_{[y,1]} x\; d\mu(x)\right)^2}{\int_{[y,1]}\; d\mu(x)}\right) $$ At least for absolutely continuous distributions with strictly positive density, an extremum will occur at some $y$ such that $$ 2 y = \frac{\int_y^1 x \; d\mu(x)}{\int_y^1 \; d\mu(x)}$$

  • $\begingroup$ Well that's certainly helpful! Using this fact, I can conduct numerical simulations, which suggest that the ratio between these two quantities, $$\sup_{S}\frac{\left[\mathbb{E}(X|X\in S)\right]^{2}\Pr(x\in S)}{\mathbb{E}X^{2}}$$ approaches a constant in the limit $\mathbb{E}X^{2}\to 0$. Does that sound plausible? $\endgroup$ – Jennifer Lung Dec 24 '17 at 17:05
  • $\begingroup$ It looks as if the $x$ shouldn't be squared in your formulas? (whence neither the $y$ in the end) $\endgroup$ – H. H. Rugh Dec 25 '17 at 12:41
  • $\begingroup$ @H.H.Rugh Thanks: I don't know how those squares crept in. $\endgroup$ – Robert Israel Dec 25 '17 at 20:15

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