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Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $\mathcal F$, $\mathcal G$ and $\mathcal H$ be $\sigma$-algebras on $\Omega$ with $\mathcal F\subseteq\mathcal G\subseteq\mathcal H\subseteq\mathcal A$
  • $(E,\mathcal E)$ be a measurable space
  • $X,Y,Z:\Omega\to E$ be $(\mathcal F,\mathcal E)$-, $(\mathcal G,\mathcal E)$- and $(\mathcal H,\mathcal E)$-measurable, respectively

Assume $$\operatorname P\left[Y\in B\mid\mathcal F\right]=\operatorname P\left[Y\in B\mid X\right]\;\;\;\text{for all }B\in\mathcal E\tag1.$$ How can we conclude that $$\operatorname E\left[1_{\left\{\:Y\:\in\:B\:\right\}}\operatorname E\left[1_{\left\{\:Z\:\in\:C\:\right\}}\mid Y\right]\mid\mathcal F\right]=\operatorname E\left[1_{\left\{\:Y\:\in\:B\:\right\}}\operatorname E\left[1_{\left\{\:Z\:\in\:C\right\}}\mid Y\right]\mid X\right]\tag2$$ for all $B,C\in\mathcal E$?

Seems to be easy, but I can't figure out how to do it.

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If we can prove that

$$\mathbb{E}(U \mid \mathcal{F}) = \mathbb{E}(U \mid X) \tag{3}$$

for any bounded function $U:\Omega \to \mathbb{R}$ which is measurable with respect to $\sigma(Y)$, then this gives $(2)$; simply choose $U := 1_{\{Y \in B\}} \mathbb{E}(1_{\{Z \in C\}} \mid Y)$ which is clearly bounded and $\sigma(Y)$-measurable.

The proof of $(3)$ is a standard monotone class argument:

  • It follows from $(1)$ that $(3)$ holds for functions $U$ of the form $U=1_{\{Y \in B\}}$.
  • Because of the linearity of the conditional expectation, this implies that $(3)$ holds for bounded step functions which are $\sigma(Y)$-measurable.

  • If $U$ is bounded and $\sigma(Y)$-measurable, there exists a sequence of $\sigma(Y)$-measurable step functions $(U_j)_{j \in \mathbb{N}}$ such that $U_j \to U$ and $|U_j| \leq |U|$. Since we already know that $(3)$ holds for each $j$, we can use the dominated convergence theorem to conclude that $$\mathbb{E}(U \mid \mathcal{F}) = \lim_{j \to \infty} \mathbb{E}(U_j \mid \mathcal{F}) = \lim_{j \to \infty} \mathbb{E}(U_j \mid X) = \mathbb{E}(U \mid X).$$

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  • $\begingroup$ You're right. We should even be able to generalize it as follows: Let $\mathcal G\subseteq\mathcal F\subseteq\mathcal A$ be $\sigma$-algebras on $\Omega$ and $X:\Omega\to E$ be $(\mathcal A,\mathcal E)$-measurable with $\operatorname P\left[X\in B\mid\mathcal F\right]=\operatorname P\left[X\in B\mid\mathcal G\right]$ for all $B\in\mathcal E$, then $\operatorname E\left[Y\mid\mathcal F\right]=\operatorname E\left[Y\mid\mathcal G\right]$ for all bounded $\sigma(X)$-measurable $Y:\Omega\to\mathbb R$; if I'm not missing something. $\endgroup$ – 0xbadf00d Oct 14 '18 at 15:09
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    $\begingroup$ @0xbadf00d Yeah sure. $\endgroup$ – saz Oct 14 '18 at 15:10
  • $\begingroup$ My question was motivated by a different problem. I've asked a separate question for that one. Maybe you can help there too. $\endgroup$ – 0xbadf00d Oct 14 '18 at 16:05

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