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Could you prompt me, please, is it true?

Expected value of $XYZ$, $E(XYZ)$, is not always $E(X)E(Y)E(Z)$, even if $X$, $Y$, $Z$ are not correlated in pairs, because if $X$, $Y$, $Z$ are not correlated in pairs it doesn't entail that they are uncorrelated in aggregate (it is my idea)?

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    This has been asked before, I think. Searching...2011-12-17
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    Of course, independent implies uncorrelated. Hence: do you know an example where (X,Y) is independent, (Y,Z) is independent, (Z,X) is independent but (X,Y,Z) is not independent?2011-12-17
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    http://math.stackexchange.com/questions/90482/expectation-product-of-pairwise-uncorrelated-variables2011-12-17
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    David Mitra, thank you! I didn't this topic when I was searching for similar topics here. Didier Piau, thank you, I have understood it through the example below.2011-12-17

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Suppose $$ (X,Y,Z) = \begin{cases} (1,1,0) & \text{with probability }1/4 \\ (1,0,1) & \text{with probability }1/4 \\ (0,1,1) & \text{with probability }1/4 \\ (0,0,0) & \text{with probability }1/4 \end{cases} $$ Then $X,Y,Z$ are pairwise independent, and $E(X)E(Y)E(Z)=1/8\ne 0 = E(XYZ)$.

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    Thank you for this smart example!2011-12-17
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    This turns out to be a good counterexample for lots of questions relating independence and pairwise independence. I like to think of it as flipping two coins: $X$ is the outcome of the first coin, $Y$ is the outcome of the second coin, and $Z$ denotes whether the two coins came up the same or different.2011-12-17
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Let $\Omega=\{\omega_1,\omega_2, \omega_3, \omega_4\}$, $\mathcal{F}=2^{\Omega}$ and $\mathbb{P}(\{\omega_i\})=1/4$ for all $i$. It is easy to check that desired random variables are $$ X=1_{\{\omega_1,\omega_2\}},\quad Y=1_{\{\omega_1,\omega_3\}},\quad Z=1_{\{\omega_1,\omega_4\}} $$

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    Looking at this closely, it's not hard to see that it's really the same as the example I posted. But for me, at least, the notation-intensive style of writing doesn't help in understanding that.2011-12-17
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    Not at all, it was well known example.2011-12-17