Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$ and $M,N\in\mathcal V$ (see definition below$^2$) be càdlàg $\mathcal F$-martingales on $(\Omega,\mathcal A,\operatorname P)$.
I would like to show $$\operatorname E\left[M_tN_t\right]=\operatorname E\left[\sum_{s\in(0,\:t]}\Delta M_s\Delta N_s\right]\tag4\;\;\;\text{for all }t\ge0.$$
Let $M^-$ and $N^-$ (see definition below$^1$) are clearly $\mathcal F$-adapted and left-continuous; hence $\mathcal F$-predictable. Moreover, since a regular function on a compact interval is bounded, they are pathwise locally bounded. So, they satisfy $(2)$ below$^3$.
We are currently missing a suitable assumption $(3)$, i.e. \begin{align}\operatorname E\left[\int_{[0,\:t]}|N^-|\:{\rm d}|M|\right]<\infty\tag5;\\\operatorname E\left[\int_{[0,\:t]}|N^-|\:{\rm d}|M|\right]<\infty\tag6\end{align} for all $t>0$. Assuming this, we obtain by the result below$^3$ that $M^-\cdot N$ and $N^-\cdot M$ are $\mathcal F$-martingales. Hence, $$\operatorname E\left[(M^-\cdot N)_t\right]=\operatorname E\left[(M^-\cdot N)_0\right]=\operatorname E\left[M_0N_0\right]\tag7$$ and hence $$\operatorname E\left[\int_{(0,\:t]}M^-\:{\rm d}N\right]=\operatorname E\left[(M^-\cdot N)_t\right]-\operatorname E\left[M_0N_0\right]=0\tag8.$$ Analogously, $$\operatorname E\left[\int_{(0,\:t]}N^-\:{\rm d}M\right]=0\tag9.$$
This yields $$\operatorname E\left[M_tN_t\right]=\operatorname E\left[M_0N_0\right]+\operatorname E\left[\sum_{s\in(0,\:t]}\Delta M_s\Delta N_s\right]\;\;\;\text{for all }t\ge0.\tag{10}$$ So,
- Do we need to assume $M_0=N_0=0$ in order to obtain $(4)$?
- Is there a handy assumption ensuring $(5)$ and $(6)$?
- Did I make any mistake in my argumentation?
$^1$ If $f,g:[0,\infty)\to\mathbb R$ are right-continuous and of locally bounded variation, then $$f(t)g(t)=f(0)g(0)+\int_{(0,\:t]}f(s-)\:g({\rm d}s)+\int_{(0,\:t]}g(s-)\:f({\rm d}s)+\sum_{s\in(0,\:t]}\Delta f(s)\Delta g(s)\tag1$$ As usual, $f(t-):=\lim_{s\to t-}f(s)$ for $t>0$, $f(0-):=f(0)$ and $\Delta f(t):=f(t)-f(t-)$ for $t\ge0$. We also write $$f^-(t):=f(t-)\;\;\;\text{for }t\ge0.$$
$^2$ Let $\mathcal V$ denote the set of $V:\Omega\times[0,\infty)\to\mathbb R$ which are $\mathcal A\otimes\mathcal B([0,\infty))$-measurable and for which $V(\omega)$ is right-continuous and of locally finite variation for all $\omega\in\Omega$. If $A\in\mathcal V$ and $X:\Omega\times[0,\infty)\to\mathbb R$ with $X(\omega)\in\mathcal L^1_{\text{loc}}(A)$ for all $\omega\in\Omega$, then $$(X\cdot A)_t:=\int_{[0,\:t]}X\:{\rm d}A\;\;\;\text{for }t\ge0.$$
$^3$ If $M\in\mathcal V$ is an $\mathcal F$-martingale, then $X\cdot M$ is an $\mathcal F$-martingale for all $\mathcal F$-predictable $X:\Omega\times[0,\infty)\to\mathbb R$ with \begin{align}(|X(\omega)\cdot|M|)_t<\infty&\;\;\;\text{for all }\omega\in\Omega\tag2\\\operatorname E\left[(|X|\cdot|M|)_t\right]<\infty\tag3\end{align} for all $t>0$.