Minimum and maximum bound on mean of product of three pairwise uncorrelated random variables 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 $E(XYZ)$?
I have thougth to play with covariance but I stuck at repeating similiar actions with no effort and have no idea what to do.
 A: There is no bound, and the expectation of the product can be infinite in either direction.
As an initial example of a calculation rather than the bound, let $A,B,C,D$ be random variables taking the values $\pm \sigma$ with joint probabilities:


*

*$P(A=+\sigma, B=+\sigma, C=+\sigma, D=-\sigma)=\frac14$

*$P(A=+\sigma, B=-\sigma, C=-\sigma, D=+\sigma)=\frac14$

*$P(A=-\sigma, B=+\sigma, C=-\sigma, D=+\sigma)=\frac14$

*$P(A=-\sigma, B=-\sigma, C=+\sigma, D=-\sigma)=\frac14$


Then


*

*$E[A]=E[B]=E[C]=E[D]=0$

*$E[A^2]=E[B^2]=E[C^2]=E[D^2]=\sigma^2$ 

*$E[AB]=E[AC]=E[AD]=E[BC]=E[BD]=0$ 

*but $ABC=+\sigma^3$ so $E[ABC]=+\sigma^3$ 

*and $ABD=-\sigma^3$ so $E[ABD]=-\sigma^3$


For a second example, let $W$ be a random variable with a probability density function $f(w)=\dfrac{1}{\sqrt{3}x^4}$ for $w \gt \frac1{\sqrt{3}}$ independent of $A,B,C,D$.  This is a Pareto distribution with $w_{min}= \frac1{\sqrt{3}}$ and $\alpha=3$, so it has $E[W]=\frac{\sqrt{3}}{2}$, $\text{Var}(W)=\frac14$, $E[W^2]=1$ and $E[W^3]=+\infty$ 
Now let $X=AW$, $Y=BW$, $Z_1=CW$, $Z_2=DW$. Then 


*

*$E[X]=E[Y]=E[Z_1]=E[Z_2]=0$

*$E[X^2]=E[Y^2]=E[Z_1^2]=E[Z_2^2]=\sigma^2 E[W^2]=\sigma^2$ 

*$E[XY]=E[XZ_1]=E[XZ_2]=E[YZ_1]=E[YZ_2]=0$ 

*but $XYZ_1=ABCW^3=+\sigma^3 W^3$ so $E[XYZ_1]=+\sigma^3 E[W^3]=+\infty$ 

*and $XYZ_2=ABDW^3=-\sigma^3 W^3$ so $E[XYZ_2]=-\sigma^3 E[W^3]=-\infty$ 


With other distributions for $W$ with second moment $E[W^2]=1$ and a finite third moment $E[W^3]$, you can achieve any value for $E[XYZ]$
