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.
Best Answer
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:
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 $$\mathbb{E}(U \mid \mathcal{F}) = \lim_{j \to \infty} \mathbb{E}(U_j \mid \mathcal{F}) = \lim_{j \to \infty} \mathbb{E}(U_j \mid X) = \mathbb{E}(U \mid X).$$