Suppose I have three uncorrelated random variables $X, Y$ and $Z$ (discrete or continuous) such that
$$\newcommand{\Cov}{\mathrm{Cov}}\Cov(X,Y)=0;\quad \Cov(Y,Z)=0;\quad \Cov(X,Z)=0 \tag{$\ast$}$$
I am to prove or disprove the fact that
$$E(XYZ) = E(X)\cdot E(Y)\cdot E(Z)$$
It is evening here already and my head is somewhat dumb; now the obvious step I took was that
$$\Cov(XY, Z) = E(XYZ) - E(XY)\cdot E(Z)$$
Now that $X$ and $Y$ are uncorrelated, it implies that
$$\Cov(XY, Z) = E(XYZ) - E(X)\cdot E(Y)\cdot E(Z)$$
If I could prove or disprove that from $(\ast)$ implies $\Cov(XY,Z)=0$, then I would succeed.
Can you please help me with that?
Thank you in advance!
Suppose $X$ and $Y$ are independent and identically distributed and $$ \begin{cases} X = 1 & \text{with probability }1/2, \\ X = 0 & \text{with probability }1/2, \end{cases} $$ and let $Z$ be the mod-$2$ sum of $X$ and $Y$. Find $E(X)$, $E(Y)$, $E(Z)$ and all three covariances. Then find $E(XYZ)$.
(This is the same example I put into this initial edit of a Wikipedia article in 2004, and it's still prominent in the current version. But in that article it served a slightly different purpose.)