expectation of products... I'm trying to build $m$ real random variables $X_1,\dots,X_m$ such that
$$\mathbb{E}[\prod_{i\in\alpha}X_i]=0\quad\forall\alpha\subsetneq\{1,\dots,m\}\,,$$
$$\mathbb{E}[\prod_{i=1}^mX_i]\neq 0\,.$$
I belive that such a choice should be possile, because in general the expectation of a product is not the product of expectations. Clearly $X_1,\dots,X_m$ will not be independent.
Does anybody have an idea?
 A: Hint: Presumably you do not want to allow $\alpha$ to be the empty set. Let $X_1,\dots,X_{m-1}$ be independent, taking values $1$ and $-1$ each with probability $\frac{1}{2}$. Let $X_m=1$ if the product of the $X_i$ with $i\ne m$ is $1$, and $-1$ otherwise.  
A: Let $S=\{-1,1\}^m$ and $S^+\subset S$ defined by $S^+=\{x\in S\mid \prod\limits_{i=1}^mx_i=1\}$ and $X=(X_i)_{1\leqslant i\leqslant m}$ uniformly distributed on $S^+$. Then $\prod\limits_{i=1}^mX_i=1$ almost surely hence $E\left[\prod\limits_{i=1}^mX_i\right]=1\ne0$. 
On the other hand, let $\alpha\subset\{1,\ldots,m\}$ be nonempty and not $\{1,\ldots,m\}$. Let $X_\alpha=\prod\limits_{i\in\alpha}X_i$. Choose $a$ in $\alpha$ and $b$ in $\{1,\ldots,m\}\setminus\alpha$. The set $S^+$ is invariant by the involution $s_{a,b}:S\to S$, $x\mapsto y$, defined by $y_i=x_i$ for every $i$ not in $\{a,b\}$, and by $y_a=-x_a$ and $y_b=-x_b$. Thus $Y=s_{a,b}(X)$ is distributed as $X$ and $Y_\alpha=-X_\alpha$ hence $E[X_\alpha]=E[Y_\alpha]=-E[X_\alpha]$, that is, $E[X_\alpha]=0$.
