Expected value of a norm I am currently studying the Monte Carlo methods for solving PDEs with random coefficients. My problem here is basically just doing with some algebraic properties of the expected value function which I cannot work it out myself.
$V$ is a Hilbert space. Suppose $u(x,\omega)$ is a random field associated to the probability space $(\Omega,\Sigma,\mathbb{P})$ and we denote the mean field $\bar{u}$ as
\begin{equation*}
\bar{u}(x)=\int_{\Omega}u(x,\omega)d\mathbb{P}(\omega).
\end{equation*}
Also, let $\bar{u}_N = \frac{1}{N}\sum_{i=1}^{n}u_i$, where $u_i$'s are independent and identical to the law of $u$. It concludes that
\begin{equation*}
E(\|\bar{u}-\bar{u}_N\|^2_V) = \frac{1}{N}E(\|u-\bar{u}\|^2_V).
\end{equation*}
May I know how should I get to the conclusion there? My way of doing it is to expand the term on the left and make use of the independent relationship between $u_i$, but somehow, I cannot show the equality. This text comes from the paper titled, Convergence rates of best N-term Galerkin approximations for a class of elliptic sPDEs.
 A: I'm going to assume that $V$ is separable, as I'm not even sure how to define the necessary objects in the non-separable case (plus, if this is for sPDEs, everything is going to be separable).
In this case, we have
\begin{align*}
\mathbf{E} \lVert \bar{u} - \bar{u}_N\rVert^2
&= \mathbf{E} \Bigl\lVert \frac{1}{N}\sum_{i=1}^N(\bar{u} - u_i)\Bigr\rVert^2\\
&= \frac{1}{N^2}\sum_{i=1}^N \mathbf{E} \lVert \bar{u} - u_i\rVert^2 + \frac{2}{N^2}\sum_{i<j} \mathbf{E}\langle \bar{u} - u_i, \bar{u} - u_j\rangle \\
&= \frac{1}{N} \mathbf{E} \lVert u - \bar{u}\rVert^2 + \frac{2}{N^2}\sum_{i<j} \mathbf{E}\langle \bar{u} - u_i, \bar{u} - u_j\rangle.
\end{align*}
To show that the latter sum is $0$, let $\{e_i\}$ be a Schauder basis for $V$ so that we have
\begin{align*}
\mathbf{E}\langle\bar{u} - u_i, \bar{u} - u_j\rangle
&= \mathbf{E}\sum_{n=1}^\infty \langle\bar{u} - u_i, e_n\rangle \langle\bar{u} - u_j, e_n\rangle \\
&= \sum_{n=1}^\infty  \mathbf{E} \langle\bar{u} - u_i, e_n\rangle \mathbf{E}\langle\bar{u} - u_j, e_n\rangle \\
&= \sum_{n=1}^\infty   \langle\bar{u} - \mathbf{E} u_i, e_n\rangle \langle\bar{u} - \mathbf{E}u_j, e_n\rangle \\
&= 0.
\end{align*}
Here, we used the fact that $u$ is Bochner-integrable to exchange the sum with the expectation and the Riesz representation theorem to see that $\mathbf{E} \langle\bar{u} - u_i, e_n\rangle = \langle\bar{u} - \mathbf{E}u_i, e_n\rangle$.
