Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of $(-1,\infty)$-valued random variables. Is it possible to have $$ \forall \, n \colon \ \mathbb{E}[X_n \, \vert \, X_1, \ldots, X_{n-1}] = 0 $$ without the $(X_n)$ being independent?
I know that there are examples which work for two random variables and without the assumption on the range. Typically, one needs uncorrelated but dependent random variables. Do such examples also exist in this specific setting?
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I am afraid Henry already posted an answer. But to elaborate, take any sequence $X_{n}$ of random variables and let $Y$ be a random variable independent of $X_{n}$'s and with $E[Y]=0$
Then define $Z_{1}=X_{1}$ , $Z_{2}=X_{2}$ and $Z_{3}=Yf(X_{2},X_{1})$ where $f$ is an arbitrary measurable function from $\Bbb{R}^{2}\to \Bbb{R}$
Then $E(Z_{3}|Z_{1},Z_{2})=E(Yf(X_{2},X_{1})|Z_{1},Z_{2}))=f(X_{1},X_{2})E(Y|Z_{1},Z_{2})=f(X_{1},X_{2})E(Y)=0$ where the last equality is due to the tower law for conditional expectation.So you can take $f$ to be any function you like for example $f(x,y)=x^{2}+y^{2}$ or $f(x,y)=x^{\pi}y^{e}(x+y)^{12345}$ and you'll find that $Z_{3}$ is not independent of $X_{1},X_{2}$ but still the conditional expectation will be $0$.
Note that directly by definition of conditional expectation , if $X$ is $\mathcal{G}$ measurable , then $E(Y\cdot X|\mathcal{G})=XE(Y|\mathcal{G})$
Try something like $Y_n =\pm1$ with equal probability iid, with $X_1=Y_1$ and $$X_n=Y_n\left(X_{n-1}+1\right).$$
Clearly you could replace the bracketed term with most other functions of $X_1, \ldots, X_{n-1}$ and still have dependency with a conditional expectation of $0$ thanks to the $Y_n$.