Let $m$ a probability measure, $f$ a positive measurable function (one can assume it is bounded, the existence of the moments is not a problem here).

Is $m(f^3) \le m(f^2) m(f)$?


No. Consider $([0,1],\mathcal{B}([0,1]),\lambda|_{[0,1]})$ and $f(x) := 1+x$. Then $$\int_0^1 (1+x)^n \, dx = \frac{1}{n+1} (1+x)^{n+1} \bigg|_{x=0}^1 = \frac{2^{n+1}-1}{n+1}$$

for any $n \in \mathbb{N}$. Hence,

$$\frac{15}{4} = \int_0^1 (1+x)^3 \, dx > \left( \int_0^1 (1+x)^2 \, dx \right) \cdot \left( \int_0^1 (1+x) \, dx \right) = \frac{7}{3} \cdot \frac{3}{2} = \frac{7}{2}.$$


No. Actually, for every probability measure $m$ and nonnegative function $f$, $$m(f^3)\geqslant m(f^2)\cdot m(f),$$ with equality if and only if $f$ is ($m$-almost surely) constant.

Hence, checking any example would have shown that the conjecture is wrong.

  • $\begingroup$ and why is there such an inequality? $\endgroup$ – mookid Apr 13 '14 at 16:40
  • $\begingroup$ Because $f^2$ and $f$ are increasing functions of $f$ hence they are positively correlated. I might have already explained this several times on the site. $\endgroup$ – Did Apr 13 '14 at 18:42

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