Definition of Cylindrical Brownian Motion and Spatial Correlation

Question

From Gawarecki and Mandrekar, Stochastic Differential Equations in Infinite Dimensions:

We call a family $\{ W_t \}_{t\geq 0}$ defined on a filtered probability space $(\Omega, \mathcal{F}, \{\mathcal{F}_t\}_{t \geq 0}, P)$ a cylindrical Wiener process in a Hilbert space $K$ if:

1. For an arbitrary $t\geq0$, the mapping $W_t:K \rightarrow L^2(\Omega, \mathcal{F}, P)$ is linear.
2. For an arbitrary $k \in K$, $W_t(k)$ is an $\mathcal{F}_t$-Brownian motion.
3. For arbitrary $k, k' \in K$ and $t \geq 0$, $E(W_t(k)W_t(k')) = t \langle k, k'\rangle_K$.

Exercise 2.2 Show that $E(W_t(k)W_s(k')) = (t \wedge s) \langle k, k' \rangle_K$ and conclude that $W_t(f_j), j=1, 2, ...$, where $\{f_j\}_{j=1}^\infty$ is an orthonormal basis in $K$, are independent Brownian motions.

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I am new to functional analysis and don't see how to start this exercise. I can see that the third point is useful, but how does one begin? My thought was the following (assuming $t \leq s$:

$$ \begin{align*} E(W_t(k)W_s(k')) &= E(E(W_t(k)W_s(k')|\mathcal{F}_t))\\ &= E(W_t(k)E(W_s(k')|\mathcal{F}))\\ &= E(W_t(k)W_t(k')\\ &= t\langle k, k' \rangle_K \end{align*} $$ by LIE, point two, and point three, respectively. The opposite holds for $s \leq t$. Is this the right approach? Then the independence follows from orthogonality.

Answer 1

For posterity's sake, here's my answer (mainly copied from above and using the comment from @zhoraster):

When $t\leq s$ $$ \begin{align*} E(W_t(k)W_s(k')) &= E(E(W_t(k)W_s(k')|\mathcal{F}_t))\\ &= E(W_t(k)E(W_s(k')|\mathcal{F}))\\ &= E(W_t(k)W_t(k')\\ &= t\langle k, k' \rangle_K \end{align*} $$ by LIE, point two, and point three, respectively. The opposite holds for $s \leq t$. Thus $$ \begin{align*} E(W_t(k)W_s(k')) &= (t\wedge s)\langle k, k' \rangle_K \end{align*} $$ where $\wedge$ is the $\min$ operator. Since $\{f_j\}_{j=1}^\infty$ is an orthonormal basis, $W(f_j)$ is an $\mathcal{F}_t$-Brownian motion for all $j$ by 2. Then, using the above result, $E(W_t(f_j)W_s(f_i)) = (t\wedge s)\langle f_j, f_i \rangle_K = 0$ for all $i \neq j$ by orthoganality. Thus, the Brownian motions are independent.