I am trying to prove this form for pythagoras theorem:
Let $G \subset H \subset A$ and $X \in L^2(\Omega, A, P)$, then:
$E[(X-E[X/G])^2]=E[(X-E[X/H])^2]+E[(E[X/H]X-E[X/G])^2]$
My Idea is rewrite: $X-E[X/G] = X-E[X/H] + E[X/H]-E[X/G]$
But I am not sure about my solution. Could someone give me a clear and reliable solution? Thanks so much.
If you realise your idea, you will get two second moments of summands plus the following extra term: $$ 2\mathop{\mathbb E}\bigl[\bigl(X-\mathop{\mathbb E}[X\mid H]\bigr)\cdot\bigl(\mathop{\mathbb E}[X\mid H]-\mathop{\mathbb E}[X\mid G]\bigr)\bigr]. $$ Prove this is $0$. The second term $Z=\mathop{\mathbb E}[X\mid H]-\mathop{\mathbb E}[X\mid G]$ is measurable with respect to $H$. The definition of conditional expectation implies that for any $H$-measurable random variable $Z$
$$\tag{1}\label{1}\mathop{\mathbb E}[X\cdot Z] = \mathop{\mathbb E}[\mathop{\mathbb E}[X|H]\cdot Z]$$ Therefore $$\mathop{\mathbb E}\bigl[\bigl(X-\mathop{\mathbb E}[X\mid H]\bigr)\cdot Z\bigr]=0.$$
If you are not aware of the property (\ref{1}) of conditional expectations, lets go the other way. Find conditional expectation of the constant (npn-random) value of $\mathop{\mathbb E}\bigl[\bigl(X-\mathop{\mathbb E}[X\mid H]\bigr)\cdot Z\bigr]$ with respect to $H$: $$ \mathop{\mathbb E}\bigl[\bigl(X-\mathop{\mathbb E}[X\mid H]\bigr)\cdot Z\bigr] =\mathop{\mathbb E}\bigl[\mathop{\mathbb E}\bigl[\bigl(X-\mathop{\mathbb E}[X\mid H]\bigr)\cdot Z\bigr] \bigm| H\bigr] $$ $$=Z\cdot\underbrace{\bigl(\mathop{\mathbb E}[X\mid H] - \mathop{\mathbb E}\left[\mathop{\mathbb E}[X\mid H]\mid H\right]\bigr)}_{0}=0. $$