# Prove a simple property of conditional expectation

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.

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

• 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. – 0xbadf00d Oct 14 '18 at 15:09
• @0xbadf00d Yeah sure. – saz Oct 14 '18 at 15:10
• My question was motivated by a different problem. I've asked a separate question for that one. Maybe you can help there too. – 0xbadf00d Oct 14 '18 at 16:05