# Does $\text E[\text E[X\mid\mathcal F]\mid\mathcal G]=\text E[\text E[X\mid\mathcal G]\mid\mathcal F]$ hold when $\mathcal G\not\subseteq\mathcal F$?

Let $X$ be a random variable on a measurable space $(\Omega,\mathcal A)$ and $\mathcal F$, $\mathcal G$ two sub-$\sigma$-algebras of $\mathcal A$. Assume that $$Y:=\operatorname E\left[X\mid\mathcal G\right]$$ is $\mathcal F$-measurable. I'm curious whether or not $(\ast)$ in $$Y=\operatorname E\left[Y\mid\mathcal F\right]\stackrel{(\ast)}=\operatorname E\left[\operatorname E\left[X\mid\mathcal F\right]\mid\mathcal G\right]\tag 1$$

holds. I know, that if $\mathcal G\subseteq\mathcal F$, then $(1)$ is just the tower property of conditional expectation.

I've tried the following, but I'm not sure if I've made a mistake: Let $Z:=\operatorname E\left[Y\mid\mathcal F\right]$. Then, $$Z=\operatorname E\left[Z\mid\mathcal A\right]\stackrel{\text{tower property}}=\operatorname E\left[\operatorname E\left[Z\mid\mathcal A\right]\mid\mathcal G\right]\stackrel{\text{tower property}}=\operatorname E\left[Z\mid\mathcal G\right]\;.\tag 2$$ Why do I think that this might be wrong? Well, I don't understand why we should otherwise force $\mathcal G\subseteq\mathcal F$ in the statement of the tower property, since we could always switch to $\mathcal A$ in an intermediate steplike in $(2)$.

Suppose $X$ can have two states, state $A$ and state $B$, with values 0 and 1 respectively. Consider $\mathcal F$ and $\mathcal G$ to be related to $A$ and $B$. Try to make it as simple as possible, so you don't have to take some complicated expectation.
• Do you think, that this is a counterexample? If so, could you say what in $(2)$ is wrong? – 0xbadf00d Sep 30 '15 at 17:33
• I think the second use of the tower property? The first equality holds because $\mathcal G \subseteq \mathcal A$; ignoring the first equality, for the second to hold I think you'd need $\mathcal A \subseteq \mathcal G$ in the same way. (Note that $\{\mathcal G = \mathcal A\} \iff \{\mathcal G \subseteq \mathcal A, \mathcal A \subseteq \mathcal G\}$, and of course $(2)$ ignoring the left-hand equality clearly holds if $\mathcal A = \mathcal G$. – Sam T Sep 30 '15 at 17:54
• The first equality is correct, since $Z$ is $\mathcal A$-measurable. – 0xbadf00d Sep 30 '15 at 17:58
• The second equality is wrong, since we would need $\mathcal A\subseteq\mathcal G$. So, you're right. – 0xbadf00d Sep 30 '15 at 18:00