Quadratic variation of double Itô/deterministic integral

Question

Let $f$ be a deterministic bounded function, $W_t$ be a standard Brownian motion and define
$$
M_t = \int_0^t \int_s^t f(r) \, d W_r \, ds.
$$
This quantity is a martingale, as, considering the filtration $\mathcal F_t$ induced by the BM and $t' < t$
\begin{align}
\mathbb E(M_t \mid \mathcal F_{t'}) &= \int_0^{t'} \mathbb E\left(\int_s^t f(r) \, d W_r \mid \mathcal F_{t'}\right) ds \\
&\quad + \int_{t'}^t \mathbb E\left(\int_s^t f(r) \, d W_r \mid \mathcal F_{t'}\right) ds \\
&= \int_0^{t'} \mathbb \int_s^{t'} f(r) \, d W_r \, ds \\
&\quad + \int_{t'}^t \mathbb E\left(\int_s^t f(r) \, d W_r \right) ds \\
&= M_{t'} + 0 = M_{t'}.
\end{align}
My question is, is it possible to determine the quadratic variation $\langle M \rangle_t$ of $M_t$?

Answer 1

We have $\forall t \geq 0$, \begin{align} M_t &= \int_0^t \int_s^t f(u) dW_u ds \\ &=\int_0^t \left(\int_s^t f(u)dW_u+\int_0^s f(u)dW_u-\int_0^s f(u)dW_u\right) ds \\ &=\int_0^t \left(\int_0^t f(u)dW_u\right)ds-\int_0^t\left(\int_0^s f(u)dW_u\right)ds \tag{By linearity} \\ \end{align} Denote here $I_t= \int_0^tf(u)dW_u$, we have then \begin{align} M_t &=\int_0^t I_tds -\int_0^tI_sds \\ &=tI_t -\left(tI_t - \int_0^t sdI_s\right) \tag{Integration by parts}\\ &= \int_0^tsdI_s \end{align} Now $\langle{M}\rangle_t=\int_0^ts^2f(s)^2ds$. (Note that we can also use stochastic fubini).