Let $X_1$ and $X_2$ be independent random variables. Let $Y$ be a random variable such that $E(Y\mid X_1)=0$ and $E(Y\mid X_2)=0$. Under what conditions is it true that $E(Y\mid X_1,X_2)=0$?


The condition is on the joint distribution of $\left(X_{1},X_{2}\right)$. Let $f$ be the probability density function.

Let $h\left(x_{1},x_{2}\right)$ be a function. Assume for $i=1$ or $i=2$:

$\int_{x_{i}} h\left(x_{1},x_{2}\right) f\left(x_{1},x_{2}\right) dx_{i} = 0 \implies h\left(x_{1},x_{2}\right) = 0$.

(The assumption is called "completeness" in the nonparametric instrumental variables literature in econometrics.)

Let $x_{-i}$ be the $x$ that is not $x_{i}$.

Then, because $\mathbb{E}\left[Y | X_{i}\right] = \mathbb{E}\left[\mathbb{E}\left[Y | X_{1},X_{2}\right] | X_{i}\right] = 0$, we have:

$\int_{x_{i}} \mathbb{E}\left[Y | X_{1}=x_{1},X_{2}=x_{2}\right] \frac{f\left(x_{1},x_{2}\right)}{f_{i}\left(x_{i}\right)} dx_{-i} = 0 \iff \int_{x_{i}} \mathbb{E}\left[Y | X_{1}=x_{1},X_{2}=x_{2}\right] f\left(x_{1},x_{2}\right) dx_{-i} = 0$.

Which, by assumption, implies $\mathbb{E}\left[Y|X_{1},X_{2}\right] = 0$.


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