I would like to proof the following having $\xi \in L^{2}(\Omega, \mathbb{A}, \mathbb{P})$ and the subspace $L^{2}(\Omega, \mathbb{B}, \mathbb{P})$:
$\mathbb{E}[(\xi-\mathbb{E}[\xi|\mathbb{B}])\cdot\eta] = 0$, $\forall \eta \in L^{2}(\Omega, \mathbb{B}, \mathbb{P})$
For me, the only reason why this is correct is that $(\xi-\mathbb{E}[\xi|\mathbb{B}])$ is the orthogonal projection from $L^{2}(\Omega, \mathbb{A}, \mathbb{P})$ to the subspace $L^{2}(\Omega, \mathbb{B}, \mathbb{P})$. Thus, calculating the inner product with an element of this subspace will result in 0.
However, in my setting I do not know what a projection is. Is there any other approach to proof this equation?
Thanks in advance,
MS
If you rearrange that equation, you have $E[\xi \eta] = E[E[\xi \mid \mathcal{B}] \eta]$ for all $\eta \in L^2(\Omega, \mathcal{B}, P)$ which is a known property of the conditional expectation. In fact, when $\eta$ is restricted to indicator functions of sets in $\mathcal{B}$, this is precisely the definition of the conditional expectation $E[\xi \mid \mathcal{B}]$. But it can also be extended to $\eta$ that are $\mathcal{B}$-measurable in the usual way (approximate $\eta$ by simple functions, etc.).