Comparing Squared Difference of a Random Variable and it's mean, or it's mean conditioned on a $\sigma$-field. I have $X \in L^2$, and I want to show $E[(X-E[X|G])^2] \leq E[(X-E[X])^2]$
All I've tried is expanding and simplifying, which gives me wanting to show:
$E[E[X|G]^2]-2E[XE[X|G]] \leq E[E[X]^2] - 2E[XE[X]]$
I'm not sure where to go from here. 
Another thing is I thought $E[X]$ is supposed to minimize $E[(X-\alpha)^2]$, but it seems here we're disproving that by showing $E[X|G]$ does (maybe?) better than $E[X]$ at minimizing. 
Is there a way to give an intuitive reason for the thing being proven?
 A: The high road to this result, which explains at the same time your second question, is to note that $$E[(X-E[X])^2]=\inf\{E[(X-a)^2]\,\mid\,a\in\mathbb R\},$$ while
$$E[(X-E[X|G])^2]=\inf\{E[(X-Y)^2]\,\mid\,Y\in L^2(G)\},$$
hence the latter, being an infimum on at least as many random variables, is not larger than the former.
Edit: Pushing a little the idea explained by @saz, one sees that Pythagoras theorem implies $$E[(X-E[X])^2]=E[(X-E[X|G])^2]+E[(E[X|G]-E[X])^2].$$
A: @Did already explained that the claim follows from the very definition of the conditional expectation. Alternatively, we can obtain the result performing some straightforward calculations:
Obviously,
$$\mathbb{E}[(X-\mathbb{E}(X \mid G))^2] = \mathbb{E}(X^2)-2 \mathbb{E}(X \cdot \mathbb{E}(X \mid G)) + \mathbb{E}( \mathbb{E}(X \mid G)^2).$$
It follows from the tower property that
$$\begin{align*} \mathbb{E}(X \cdot \mathbb{E}(X \mid G)) &= \mathbb{E}[\mathbb{E}\big(X \cdot \underbrace{\mathbb{E}(X \mid G)}_{G-\text{measurable}} \mid G \big)] = \mathbb{E}\big( \mathbb{E}(X \mid G)^2 \big). \end{align*}$$
Hence,
$$\mathbb{E}[(X-\mathbb{E}(X \mid G))^2] = \mathbb{E}(X^2)-\mathbb{E}( \mathbb{E}(X \mid G)^2). \tag{1}$$
Finally, we observe that by Jensen's inequality
$$\mathbb{E}( \mathbb{E}(X \mid G)^2) \geq \left[\mathbb{E}\big( \mathbb{E}(X \mid G) \big) \right]^2 = (\mathbb{E}(X))^2. \tag{2}$$
Combining $(1)$ and $(2)$ finishes the proof.
