Extensions of Tower Property for Conditional Expectation Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space.
Let random variable $X \in \mathcal{F}$ and two sigma fields $\mathcal{F}_1, \mathcal{F}_2 \in \mathcal{F}$.
We consider the general case where neither one of $\mathcal{F}_1, \mathcal{F}_2$ is contained in the other.
Do we have $\mathbb{E}[\mathbb{E}[X | \mathcal{F}_1] | \mathcal{F}_2] = \mathbb{E}[X | \mathcal{F}_1 \cap \mathcal{F}_2]$?
If not, what is a counterexample?
On a related topic, let $X, Y, Z$ be pairwise independent random variables and $\mathcal{F}_1 = \sigma(X, Y), \mathcal{F}_2 = \sigma(X, Z)$.
If $X, Y, Z$ are not mutually independent, then $\mathbb{E}[\mathbb{E}[f(X, Y, Z) | \sigma(X, Y)] | \sigma(X, Z)] = \mathbb{E}[f(X, Y, Z) | \sigma(X)]$ doesn't hold (e.g. let $X, Y \stackrel{iid}{\sim} 2 * Bernoulli(0.5) - 1, Z = XY$ and $f(X, Y, Z) = Z$).
What if $X, Y, Z$ are mutually independent - does $\mathbb{E}[\mathbb{E}[f(X, Y, Z) | \sigma(X, Y)] | \sigma(X, Z)] = \mathbb{E}[f(X, Y, Z) | \sigma(X)]$ hold then?
 A: No to the first question.  Let $\Omega = \{1,2,3\}$, with all singletons having probability $1/3$.  Let $\mathcal F_1$ be the sigma algebra generated by $\{1,2\}$ and $\mathcal F_2$ be the sigma algebra generated by $\{1,3\}$.
Then $\mathbb E[\mathbb E[\chi_{\{1\}} | \mathcal F_1 ] | \mathcal F_2]$ maps $1$, $2$ and $3$ to $1/4$, $1/2$, and $1/4$ respectively, whereas $\mathbb E[\chi_{\{1\}} | \mathcal F_1 \cap \mathcal F_2]$ is the constant function taking the value $1/3$.
A: Yes to the second question. There is a bit more general result: if $\zeta$ is independent of $(\xi,\eta)$, then
$$
\mathbb{E}[\xi \mid \eta, \zeta] = \mathbb{E}[\xi\mid \eta].
$$
Indeed, for any bounded measurable $g(y,z)$,
$$
\mathbb{E}[\xi\, g(\eta,\zeta)] = \mathbb{E}[\mathbb{E}[\xi \,g(\eta,z)]|_{z=\zeta}] = \mathbb{E}[\mathbb{E}[\mathbb{E}[\xi\mid\eta] g(\eta,z)]|_{z=\zeta}] = \mathbb{E}[\mathbb{E}[\xi\mid\eta] g(\eta,\zeta)], 
$$
and $\mathbb{E}[\xi\mid\eta]$ is clearly $\sigma(\eta, \zeta)$-measurable, so $\mathbb{E}[\xi\mid\eta] = \mathbb{E}[\xi\mid \eta, \zeta ]$.
Now apply this to $\xi = \mathbb{E}[f(X,Y,Z)\mid X,Y]$,  $\eta = X$, $\zeta = Z$.
