The Expectation of the Product of two Stochastic Integrals Suppose $ M, N\in \mathcal{H}^{2,c} $, $K \in L^2(M)$ and $ F \in L^2(N)$. Show that for each $ t \in [0,\infty]$ we have
$$  \mathop{\mathbb{E}}\left[ \left(  \int_{0}^{t}K_{s}dM_{s}\right ) \left( \int_{0}^{t}F_{s}dN_{s}\right ) \right] = \mathop{\mathbb{E}} \left[   \int_{0}^{t}K_{s}F_{s}d\langle M,N \rangle_{s} \right] $$
I have struggled with this for a while! Many thanks!
 A: $$\left(  \int_{0}^{t}K_{s}dM_{s}\right ) \left( \int_{0}^{t}F_{s}dN_{s}\right ) - \bigg \langle  \int_{0}^{\cdot}K_{s}dM_{s}, \int_{0}^{\cdot}F_{s}dN_{s}\bigg \rangle_t$$
is a martingale and hence 
$$\mathbb{E}\bigg[\left(  \int_{0}^{t}K_{s}dM_{s}\right ) \left( \int_{0}^{t}F_{s}dN_{s}\right ) \bigg] = \mathbb{E}\bigg[\bigg \langle  \int_{0}^{\cdot}K_{s}dM_{s}, \int_{0}^{\cdot}F_{s}dN_{s}\bigg \rangle_t \bigg] = \mathbb{E} \bigg[\int_0^t K_s F_s d \langle M,N \rangle_s \bigg]$$
with the last inequality following by the defining property of the Ito integral with respect to the quadratic variation.

Seeing that the first expression is indeed a martingale is a standard exercise (and a good one to make sure you understand well so I'll try to be detailed here). 
Note first that an Ito integral against an element of $\mathcal{H}^{2,c}$ is again an element of $\mathcal{H}^{2,c}$ so that 
$$\bigg\{\int_0^t K_s dM_s, \int_0^t F_s dN_s \bigg\} \subset\mathcal{H}^{2,c}$$
It follows from Doob's Maximal inequality that
\begin{align}
\mathbb{E}\bigg[ \sup_{t \geq 0} \bigg|\int_0^t K_s dM_s \int_0^t F_s dN_s \bigg| \bigg] \leq& \mathbb{E}\bigg[ \sup_{t \geq 0} \bigg(\int_0^t K_s dM_s \bigg)^2 \bigg]^{\frac12} \mathbb{E}\bigg[ \sup_{t \geq 0} \bigg(\int_0^t F_s dN_s \bigg)^2 \bigg]^{\frac12}
\\ \leq& 4 \sup_{t \geq 0} \mathbb{E}\bigg[ \bigg(\int_0^t K_s dM_s \bigg)^2 \bigg]^{\frac12} \sup_{t \geq 0}\mathbb{E}\bigg[ \bigg(\int_0^t F_s dN_s \bigg)^2 \bigg]^{\frac12}
\\ <& \infty
\end{align}
and so $\sup_{t \geq 0} \bigg|\int_0^t K_s dM_s \int_0^t F_s dN_s \bigg| \in L^1(\mathbb{P})$. Also $\bigg \langle \int_0^\cdot K_s dM_s, \int_0^\cdot F_s dN_s \bigg \rangle_\infty \in L^1(\mathbb{P})$ and so the continuous local martingale 
$$L_t = \left(  \int_{0}^{t}K_{s}dM_{s}\right ) \left( \int_{0}^{t}F_{s}dN_{s}\right ) - \bigg \langle  \int_{0}^{\cdot}K_{s}dM_{s}, \int_{0}^{\cdot}F_{s}dN_{s}\bigg \rangle_t$$
satisfies $|L_t| \leq \sup_{t \geq 0} \bigg|\int_0^t K_s dM_s \int_0^t F_s dN_s \bigg| + \bigg \langle \int_0^\cdot K_s dM_s, \int_0^\cdot F_s dN_s \bigg \rangle_\infty \in L^1(\mathbb{P})$ for all times $t$ (and hence also at stopping times) since the quadratic variation is increasing. This implies that $L_t$ is a true martingale, since if $L_t^{\tau_n}$ is a martingale for each $n$ and $\tau_n \to \infty$ as $n \to \infty$ then we can now use the D.C.T. to pass to the limit $n \to \infty$ in 
$$\mathbb{E}[L_t^{\tau_n} \mid \mathcal{F}_s] = L_s^{\tau_n}$$ to conclude.
This is an example of a general fact; if $M,N \in \mathcal{H}^2$ then $MN - \langle M,N \rangle$ is a uniformly integrable martingale. The proof of this goes exactly as the special case above.
