Show that the quadratic variation $\left[W\right]$ of a $Q$-Wiener process $W$ is given by $\left[W\right]_t=tQ$ for all $t\ge0$

Question

Let

• $U$ be a separable $\mathbb R$-Hilbert space
• $Q$ be a bounded, linear, nonnegative and self-adjoint operator on $U$
• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(\mathcal F_t)_{t\ge0}$ be a filtration of $\mathcal A$

Now, let $(W_t)_{t\ge0}$ be a $Q$-Wiener process on $(\Omega,\mathcal A,\operatorname P)$ with respect to $\mathcal F$, i.e.

1. $W$ is an $\mathcal F$-adapted $U$-valued stochastic process on $(\Omega,\mathcal A,\operatorname P)$
2. $W_t-W_s$ is independent of $\mathcal F_s$ for all $t\ge s\ge0$
3. $W_t\sim\mathcal N(0,tQ)$ for all $t\ge0$

You don't need to be familiar with the Gaussian law in 3.: The crucial properties are:

4. $\operatorname E\left[\langle W_t,u\rangle_U\right]=0$
5. $\operatorname E\left[\langle W_t,u\rangle_U\langle W_t,v\rangle_U\right]=t\langle Qu,v\rangle_U$

for all $u,v\in U$ and $t\in[0,T]$.

I want to show the following: Let $u,v\in U$ $\Rightarrow$ $$X_t:=\langle W_t,u\rangle_U\langle W_t,v\rangle_U-t\langle Qu,v\rangle_U\;\;\;\text{for }t\ge0$$ is an $\mathcal F$-martingale.

How can we show that?

Clearly, we need to use that $W$ is an $\mathcal F$-martingale. Letting $t\ge s\ge0$, I've tried to write $$\operatorname E\left[\langle W_t\color{blue}{-W_s},u\rangle_U\langle W_t,v\rangle_U\color{blue}{+\langle W_s,u\rangle_U\langle W_t,v\rangle_U}\mid\mathcal F_s\right]=\operatorname E\left[\langle W_t-W_s,u\rangle_U\langle W_t,v\rangle_U\mid\mathcal F_s\right]+\langle W_s,u\rangle_U\operatorname E\left[\langle W_t,v\rangle_U\mid\mathcal F_s\right]\tag1\;,$$ but I don't know what I need to do with the first term on the right-hand side.

Answer 1

Observe that $$X_t=\langle W_t-W_s,u\rangle_U\langle W_t-W_s,v\rangle_U+\langle W_t-W_s,u\rangle_U\langle W_s,v\rangle_U\\+\langle W_s,u\rangle_U\langle W_t-W_s,v\rangle_U+\langle W_s,u\rangle_U\langle W_s,u\rangle_U-t\langle Qu,v\rangle_U.$$ Using 1., 2. and 4., it follows that $$\mathbb E\left[X_t\mid\mathcal F_s\right]=\mathbb E\left[\langle W_t-W_s,u\rangle_U\langle W_t-W_s,v\rangle_U\right]+\langle W_s,u\rangle_U\langle W_s,u\rangle_U-t\langle Qu,v\rangle_U.$$ It remains to show that $$\mathbb E\left[\langle W_t-W_s,u\rangle_U\langle W_t-W_s,v\rangle_U\right]=(t-s)\langle Qu,v\rangle_U,$$ which follows from 2. and 5.