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I have a sequence of random variables $Y_t$ with the same distribution. Is there a way to show that they are independent (i.e., are an iid sequence), if we know that:

$\forall i \in \{1,\ldots,n\}: E[Y_i \mid Y_1,\ldots,Y_{i-1}]=E[Y_i]$

?

Thanks for your answers.

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I think if you know that $E[g(Y_i)\mid Y_1,\ldots,Y_{i-1}] = E[g(Y_i)]$ for every bounded continuous function $g$, that's enough, but I don't think your weaker condition is enough. –  Michael Hardy Jul 23 '12 at 1:17

1 Answer 1

No. Let $(X,Y)$ be uniformly distributed on the unit disk. Then $X$ and $Y$ have the same distribution, but are not independent. For every $-1<y<1$ we have $\mathbb{E}(X\,|\,Y=y)=0$. The random variable $\mathbb{E}(X\,|\,Y)$ is identically equal to zero, and hence $\mathbb{E}(X\,|\,Y)=\mathbb{E}(X)$.

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