how to show that if $E(Yh(X))=0$ for any (measurable) function $h(X)$, then $E(Y|X)=0$?

Question

in a lecture note I read about the statement: $E(Y|X)=0$ if and only if $E(Yh(X))=0$ for any measurable function $h(X)$ . The proof of the "only if" part is trivial by law of iterated expectations. I'm wondering how to prove the "if" part, that is, how to arrive at $E(Y|X)=0$ from $E(Yh(X))=0$ for any measurable function $h(X)$? Thanks!

Answer 1

Let $A = \{E[Y \mid X] \ge 0\}.$ As $E[Y \mid X]$ is $\sigma(X)$-measurable, we have $A \in \sigma(X)$, so $A = X^{-1}(B)$ for some $B \in \mathcal B$, and the function $1_A = 1_B(X)$ is a measurable function of $X$. We then have $$E[\max(0, E[Y \mid X])] = E[E[Y\mid X]1_A] = E[E[Y1_A \mid X]] = E[Y1_A] = 0,$$ so $\max(0, E[Y \mid X])$ is a non-negative function integrating to zero, and hence zero almost everywhere. The same follows for $\min(0, E[Y \mid X])$, and their sum $E[Y \mid X]$.

Answer 2

$\newcommand{\d}{\,\mathrm{d}}\newcommand{\pr}{\mathrm{Pr}}$$E(Y|X)$ is defined / can be defined as "the" $\sigma(X)$-measurable function $f:\Omega\to\overline{\Bbb R}$ satisfying: $$\int_Af\d\pr=\int_AY\d\pr$$

For all $A\in\sigma(X)$. Suppose that for all measurable functions $h:\Bbb R\to\overline{\Bbb R}$ we have $E(Yh(X))=0$ i.e. $$\int_{\Omega}Y(\omega)\cdot h(X(\omega))\d\pr=0$$

Take any Borel set $B\subseteq\overline{\Bbb R}$ and put $A:=X^{-1}(B)\in\sigma(X)$. Let $h:=\chi_B$.

Then: $$0=\int_{\Omega}Y(\omega)\cdot\chi_{B}(X(\omega))\d\pr=\int_AY\d\pr$$

Since sets of the form $A=X^{-1}(B)$ generate $\sigma(X)$, and we have seen that $E(Y|X)$ must integrate to zero on all the generators, it follows from the basic limit theorems and the fact $Y$ is obviously integrable that $E(Y|X)$ must integrate to zero on every element of $\sigma(X)$ and so is, up to $\sigma(X)$-almost sure equivalence, the constant zero function.

Answer 3

Taking $h(X)=E[Y|X]$ we would have: $$E(YE[Y|X])=E(E[Y|X]^2)=0$$ and therefore: $$E[Y|X]=0$$