Is conditional expectation with respect to two sigma algebra exchangeable?

Question

$(\Omega, \mathcal{F}, P)$ is a probability space. $X$ is a r.v. defined on it, and $\mathcal{G}_1, \mathcal{G}_2$ are two $\sigma$-algebra, can we claim the following: $$ \mathbb{E}\{\mathbb{E}[X|\mathcal{G}_1]|\mathcal{G}_2\}=\mathbb{E}\{\mathbb{E}[X|\mathcal{G}_2]|\mathcal{G}_1\}=\mathbb{E}[X|\mathcal{G}_1\cap\mathcal{G}_2]. $$ If this is not true, why the following is true: $X_1,X_2,\ldots,X_n$ are independent r.v. on the same probability space, $Z=f(X_1,X_2,\ldots,X_n)$ is another r.v. And we have: $$ \mathbb{E}\{\mathbb{E}[Z|X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_n]|X_1,\ldots,X_i\}=\mathbb{E}[Z|X_1,\ldots,X_{i-1}]. $$

Answer 1

A fair die is rolled with outcome $X$. The sample space is $\Omega=\{\omega_1,\omega_2,\omega_3,\omega_4,\omega_5,\omega_6\}$, where all outcomes are equally likely and $X(\omega_i)=i$ for $1\leq i\leq 6$.

Let $\cal P$ be the $\sigma$-algebra generated by $\{\omega_1,\omega_2,\omega_3\}$ and $\{\omega_4,\omega_5,\omega_6\}$. Let $\cal H$ be the $\sigma$-algebra generated by $\{\omega_1,\omega_2\}$, $\{\omega_3,\omega_4\}$, and $\{\omega_5,\omega_6\}$.

Then \begin{eqnarray*}\mathbb{E}(\mathbb{E}(X\,|\,{\cal P})\,|\,{\cal H})(\omega_i)&=&\cases{2& for $i=1,2$\cr 7/2& for $i=3,4$\cr 5& for $i=5,6$.}\\[10pt] \mathbb{E}(\mathbb{E}(X\,|\,{\cal H})\,|\,{\cal P})(\omega_i)&=&\cases{13/6 & for $i=1,2,3$\cr 29/6& for $i=4,5,6$.}\end{eqnarray*}