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Given that $\mathcal{G}$ is a sub-sigma field. $Z=\mathbb E(X|\mathcal{G})$, how can we show that $X$ is independent of $\mathcal{G}$ given $Z$?

I am struggling about the interpretation of this result.

By definition, we only need to show that given $A\in \sigma(X)$, $B\in \mathcal{G}$ that $\mathbb P(AB|\sigma(Z))=\mathbb P(A|\sigma(Z))\mathbb P(B|\sigma(Z))$ , but I don't know how to deal with it then..

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This is true when $Z$ is the conditional law of $X$ given ${\cal G}$, not the conditional expectation. – Stéphane Laurent Sep 4 '12 at 8:54
@StéphaneLaurent : do you mean for any given set $A\in \mathcal{F}$ that $P(X\in A|\mathcal{G})$? – Julie Sep 5 '12 at 14:54
I have just posted the answer – Stéphane Laurent Sep 5 '12 at 20:43
up vote 6 down vote accepted

The result does not hold in general. If it did, $\mathrm E(X\mid\mathcal G)=0$ would imply that $X$ is independent of $\mathcal G$. But this is wrong. For a counterexample, consider $X=UV$ where $V\geqslant0$ almost surely and $V$ is nondegenerate, $U$ is a centered Bernoulli random variable independent of $V$, and $\mathcal G=\sigma(V)$. Then $\mathrm E(X\mid\mathcal G)=\mathrm E(U)V=0$ but $X$ is not independent of $\mathcal G$ since $\mathrm E(X^2V)=\mathrm E(V^3)$ is not $\mathrm E(X^2)\mathrm E(V)=\mathrm E(V^2)\mathrm E(V)$.

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As shown by did, the result is not true for $Z=E(X\mid {\cal G})$. But the result holds true for $Z={\cal L}(X \mid {\cal G})$.


  • For any Polish space $E$, the space $P(E)$ of probability on $E$ is Polish when equipped with the usual topology of convergence in distribution.

  • When $X$ is an $E$-valued random variable, then for any $\sigma$-field ${\cal G}$ the conditional law ${\cal L}(X \mid {\cal G})$ always exists and it is a $P(E)$-valued random variable.

  • A random variable $X$ is conditionally independent of a $\sigma$-fielsd ${\cal B}$ given a $\sigma$-field ${\cal G} \supset {\cal B}$ if and only if ${\cal L}(X \mid {\cal G}) = {\cal L}(X \mid {\cal B})$. This equality is equivalent to $\Pr(X \in A \mid {\cal G}) = \Pr(X \in A \mid {\cal B})$ for any Borel set $A$ (and it is more appealing to say that $X$ depends on ${\cal G}$ only through ${\cal B}$), and then the equivalence to the conditional independence follows from a well-known characterization of conditional independence (see e.g. Kallenberg's book).

  • This fact holds when ${\cal B}=\sigma({\cal L}(X \mid {\cal G}))$ is the $\sigma$-field generated by ${\cal L}(X \mid {\cal G})$. In fact ${\cal B}= \sigma\big(E(f(X)\mid {\cal G}) \mid f(X) \in L^1 \big)$ (hence the $\sigma$-field ${\cal B}$ is considerably bigger than the $\sigma$-field generated by the conditional expectation $E(X\mid {\cal G})$), therefore the equality $\Pr(X \in A \mid {\cal G}) = \Pr(X \in A \mid {\cal B})$ holds because $\Pr(X \in A \mid {\cal G})$ is ${\cal B}$-measurable.

In addition, the $\sigma$-field generated by ${\cal L}(X \mid {\cal G})$ actually is the smallest $\sigma$-field ${\cal B} \subset {\cal G}$ such that $X$ is conditionally independent of ${\cal B}$ given ${\cal G}$.

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