• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $\mathcal F\subseteq\mathcal A$ be a $\sigma$-algebra on $\Omega$
  • $(E_i,\mathcal E_i)$ be a measurable space
  • $X_i$ be an $(E_i,\mathcal E_i)$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$
  • $f:E_1\times E_2\to E_3$ be $(\mathcal E_1\otimes\mathcal E_2,\mathcal E_3)$-measurable
  • $X_3:=f(X_1,X_2)$

Assuming $X_2$ is independent of $\mathcal F$, are we able to show that $X_3$ is conditionally independent of $\mathcal F$ given $X_1$, i.e. $$\operatorname P\left[X_3\in B_3,F\mid X_1\right]=\operatorname P\left[X_3\in B_3\mid X_1\right]\operatorname P\left[F\mid X_1\right]\;\;\;\text{almost surely}\tag1$$ for all $B_3\in\mathcal E_3$ and $F\in\mathcal F$?

Let $B_3\in\mathcal E_3$ and $F\in\mathcal F$. We need to prove that $$\operatorname P\left[X_1\in B_1,X_3\in B_3,F\right]=\operatorname E\left[1_{\{\:X_1\:\in\:A\:\}}\operatorname P\left[X_3\in B_3\mid X_1\right]\operatorname P\left[F\mid X_1\right]\right]\tag2.$$ What's the easiest way to show $(2)$? Maybe we are able to reduce the problem to the case $f^{-1}(B_3)=A_1\times A_2$ for some $A_i\in\mathcal E_i$, but I'm missing the right argument for that.

EDIT: If necessary, feel free to impose a stronger notion of measurability on $f$.


1 Answer 1


This is false. Indeed, consider Bernstein's example:

  • $\Omega = \{\omega = (\omega_1,\omega_2)\mid \omega_i \in \{0,1\}\}$,
  • $\mathrm P(\omega) = 1/4, \omega \in \Omega$,
  • $X_i(\omega) = \omega_i$, $i=1,2$;
  • $X_3 = (X_1+X_2) \mod 2$, $\mathcal F = \sigma(X_3)$.

Then, $X_1,X_2,X_3$ are pairwise independent, but not jointly independent. So for $B_3 = \{0\}$ and $F = \{X_3 = 0\}$ $$ \mathrm P \left[X_3=0,F\mid X_1\right] = \mathrm P [F] = 1/2\neq 1/4 = \mathrm P[F]^2 = \mathrm P \left[X_3=0\mid X_1\right] \cdot \mathrm P \left[F\mid X_1\right]. $$

Note that in this counterexample we even have that $X_1$ and $X_2$ are independent and $X_1$ and $\mathcal F$ are independent. So it is hard to imagine an additional assumption which would make this true (except that the pair $(X_1,X_2)$ is independent of $\mathcal F$, which makes this trivial).


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