Conditional Expectations and Chains of $\sigma$-algebras. Suppose $X$ is a random variable in $(\Omega, \mathcal{F}, P)$ and $G_1 \subset G_2 \subset \mathcal{F}$, with $EX^2<\infty$. Is is true that $E(X|G_2) \leq E(X |G_1)$? I'm trying to prove that under those conditions $E( [X-E(X|G_2)]^2) \leq E( [X-E(X|G_1)]^2)$. Since $EX^2 < \infty$, we can expand the expectations using linearity, look at the difference between the terms in the inequality and cancel out $EX^2$. I still get terms with $E(XE(X|G_i))$ and $E(E(X|G_i)^2)$.
Could someone provide a hint, please?
 A: 
Is is true that $E(X|G_2) \leq E(X |G_1)$?

Let $X_2=E(X|G_2)$ and $X_1=E(X |G_1)$, then $E(X_1)=E(X)=E(X_2)$ hence, if $X_2\leqslant X_1$ almost surely, then $X_1=X_2$ almost surely.
When $G_1\subset G_2$, this happens if and only if $X$ is $G_1$-measurable (in which case $X_1=X_2=X$ almost surely).
Unfortunately, we do not have enough information to see what is wrong with your idea of conditional expectations, which leads you to imagine that the properties that $G_2\subset G_1$ and that $E(X|G_2) \leqslant E(X |G_1)$ could be even related--but surely there is a serious misconception behind all this.

...under those conditions $E( [X-E(X|G_2)]^2) \leq E( [X-E(X|G_1)]^2)$.

This is true. An easy proof uses the fact that, for every $G\subset F$, the random variable $E(X\mid G)$ can be characterized as the minimizer in $L^2(G)$ of the functional $Y\mapsto E([X-Y]^2)$, that is, $E(X\mid G)$ is in $L^2(G)$ and, for every $Y$ in $L^2(G)$, 
$$
E([X-E(X\mid G)]^2)\leqslant E([X-Y]^2).
$$
Apply this to $G=G_2$ and $Y=E(X\mid G_1)$. This yields the result since, by definition, $E(X\mid G_1)$ is in $L^2(G_1)\subset L^2(G_2)$.
The "minimizing" property, if you ask, follows from the fact that $X-E(X\mid G)$ is orthogonal to every random variable in $L^2(G)$. Thus, if $Y$ is in $L^2(G)$, the decomposition $X-Y=X-E(X\mid G)+(E(X\mid G)-Y)$ where $E(X\mid G)-Y$ is in $L^2(G)$, and Pythagoras' theorem, yield
$$
E([X-Y]^2)=E([X-E(X\mid G)]^2)+E([E(X\mid G)-Y]^2),
$$
hence
$$
E([X-Y]^2)\geqslant E([X-E(X\mid G)]^2).
$$
A: Since $\mathbb EX^2$ is finite, we have 
$$\mathbb E(X-\mathbb E(X\mid\mathcal G_1))^2=\inf_{Y\in L^2(\mathcal G_1)}\mathbb E(X-Y)^2).$$
Since $\mathbb L^2(\mathcal G_1)\subset\mathbb  L^2(\mathcal G_2)$,  we are done. 
Notice that the relationship $\mathbb E(X\mid\mathcal G_2)\leqslant \mathbb E(X\mid\mathcal G_1)$ cannot hold almost surely: when $\mathcal G_1=\{\emptyset,\Omega\}$ and $\mathcal G_2=\mathcal F$, the LHS is $\mathbb EX$while the RHS is $X$. 
