A problem on conditional expectation and inner product in a Hilbert space

Question

Let $X$ and $Y$ be two random elements in a Hilbert space. Then I have seen in a paper to use the following $$E[\langle X,Y \rangle] = E[\langle E[X|Y], Y\rangle].$$ Here $E[X|Y]$ is also an element of the Hilbert space.

Answer 1

In a Hilbert space $H$, the expectation of a random element $X$ is defined to be the element $z$ such that for every $x \in H$,
$$E[\langle x, X \rangle] = \langle x, \mu\rangle.$$ and we denote $\mu$ by $E[X]$. Therefore, for every $y$ in $H$, \begin{align} & E[\langle E(X|Y), y\rangle] \\ = & \langle E[E(X|Y)], y\rangle \\ = & \langle E(X), y \rangle. \tag{$*$} \end{align}

Now replace $y$ in $(*)$ by $Y$ to get $$E[\langle E(X|Y), Y\rangle] = \langle E(X), Y\rangle \tag{$**$}$$

Now take expectations on both sides of $(**)$ and notice that the left side of $(**)$ is already a nonrandom constant, therefore, $$E[\langle E(X|Y), Y\rangle] = E[\langle E(X), Y\rangle] = \langle E(X), E(Y)\rangle.$$

In the last equality we used the definition for $E(Y)$.

To show that $E[\langle X, Y \rangle] = \langle E(X), E(Y) \rangle$. Again start with for any $y \in H$, we have $$E[\langle X, y\rangle] = \langle E(X), y\rangle.$$ Similar replacement, then taking expectations on both sides give that $$E[\langle X, Y \rangle] = E[\langle E(X), Y\rangle] = \langle E(X), E(Y) \rangle.$$

Answer 2

Yes. Yes it is.

$$\begin{align}\mathsf E\big(\langle X , Y\rangle\big) & = \mathsf E\Big(\mathsf E\big(\langle X , Y\rangle\mid \color{blue}{Y}\big)\Big) \\[1ex] & = \mathsf E\Big(\big\langle \mathsf E(X\mid \color{blue}{Y}) , Y\big\rangle\Big) \end{align}$$

So what's the problem?