The <martingale product identity> says that $N_tK_t-[N,K]_t$ is a <martingale>. Passing to the terminal values of the square-integrable martingales and using $N_0=K_0=0$ gives
$$
\mathbb E[N_\infty K_\infty]=\mathbb E[N,K]_\infty.
$$
The <quadratic covariation> identity for a <stochastic integral> is
$$
[H\mathbin\cdot M,K]_t=\int_0^tH_s\,d[M,K]_s.
$$
Applying the same product identity to $H\mathbin\cdot M$ and $K$ therefore gives
$$
\mathbb E[(H\mathbin\cdot M)_\infty K_\infty]=\mathbb E[H\mathbin\cdot M,K]_\infty.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 202 / 2 / c / i / Solution

For continuous local martingales $M$ and $N$, the <Itô product rule> says that
$$
M_tN_t-M_0N_0-[M,N]_t
$$
is a <local martingale>. If the martingales are square-integrable and converge in $L^2$, then
$$
\mathbb E[M_\infty N_\infty]=\mathbb E[M_0N_0]+\mathbb E[M,N]_\infty.
$$

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