It is well-known fact, that if $X,Y$ are independent, integrable random variables then $E[Y|X]=E[Y]$. Next assume that $Y$ is centered and $E[Y|X]=0$. What reasonable conclusions can be made about the distributions of $Y$ and $X$? Particularly, I am interested in the case when $Y=X+Z$ and we know something about distribution of $X,Y,Z$ . Thanks.
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Recall that $\mathrm E(Y\mid X)=0$ is equivalent to the condition that $\mathrm E(Yu(X))=0$ for every measurable function $u$. If $u\equiv1$, one gets $\mathrm E(Y)=0$, as was to be expected, hence it is unnecessary to assume that $Y$ is centered. Likewise, $\mathrm E(YX)=0$ and $\mathrm E(YX^n)=0$ for every integer $n$. And $\mathrm E(Y;X\leqslant x)=0$ for every $x$. A canonical example such that $\mathrm E(Y\mid X)=0$ but $X$ and $Y$ are far from being independent is when $Y=\varepsilon v(X)$ with $v$ a measurable function and $\varepsilon=\pm1$ a centered Bernoulli random variable independent on $X$. |
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If $E[Y|X=x]=0$ for all $x$ in the support of $X$ then you know $E[Y]=0$ (though you knew that already). You also know that $E[XY] = 0$ , i.e. that $X$ and $Y$ are uncorrelated. You may not know more than that; for example you might have something such as $Y$ having a conditional normal distribution with mean $0$ and variance $\exp(\sin(X))$. If $Y=X+Z$ and $E[Y|X=x]=0$ for all $x$ in the support of $X$ then $E[Y|X=x]=E[Y|X=x]+E[Z|X=x]$, i.e. $E[Z|X=x]=-x$. So you have $E[Z]=-E[X]$ if that exists, though that is also obvious since $E[Y]=0$. So the covariance of $X$ and $Z$ is $-\text{Var}[X]$. |
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