There are three pairwise uncorrelated random variables $X, Y, Z$
$$E(X) = E(Y) = E(Z) = 0$$
$$E(X^2) = E(Y^2) = E(Z^2) = \sigma^2$$
How we could find minimum and maximum bound on $E(XYZ)$?
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$\begingroup$ Holder's Inequality yields $|E(XYZ)|\le (E(|X|^3)E(|Y|^3)E(|Z|^3))^{1/3}$ and example in the comments shows that the bound is tight. $\endgroup$– A.S.Dec 21, 2015 at 20:48
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I think one cannot find any useful bounds with these assumptions. For example, consider a positive r.v. $\xi$ such that $E\xi^2<\infty$, but $E\xi^3=\infty$. Then, let $\eta_1,\eta_2,\eta_3$ be i.i.d.r.v. (also independent of $\xi$), $P[\eta_k=\pm 1]=1/2$. Set $X=\eta_1\xi$, $Y=\eta_2\xi$, $Z=\eta_3\xi$; then we have $E(X)=E(Y)=E(Z)=E(XY)=E(XZ)=E(YX)=0$, so they are centered and uncorrelated. But $E(XYZ)$ does not exist.
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$\begingroup$ Ok, and if we also assume that $E(XYZ)$ exist? $\endgroup$– AcapelloDec 20, 2015 at 16:42
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$\begingroup$ Then I don't know. Maybe, play with inequalities like $\frac{1}{3}(|XY|+|YZ|+|XZ|)\geq |XYZ|^{2/3}$?.. $\endgroup$ Dec 20, 2015 at 16:49
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3$\begingroup$ Modified version of this: Let $\eta_1$ and $\eta_2$ be independent, and set $\eta_3=\eta_1 \eta_2$. The variables are still uncorrelated, but now $E(XYZ)=E(\xi^3)$. Even if $E(\xi^3)$ is finite, it could be anything. $\endgroup$ Dec 20, 2015 at 17:13
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$\begingroup$ @Kevin Not anything per se. This shows that Holder's bound $|E(XYZ)|\le (E(|X|^3)E(|Y|^3)E(|Z|^3))^{1/3}$ is tight. $\endgroup$– A.S.Dec 21, 2015 at 20:47