If we take $X,Y,Z$ to be square integrable martingales starting at zero, we want to show that for any $\alpha\in\mathbb{R}$ we have
- $\langle X + Y , Z \rangle = \langle X,Z\rangle + \langle Y, Z \rangle$;
- $\langle X ,Y+ Z \rangle = \langle X,Y\rangle + \langle X, Z \rangle$;
- $\langle \alpha X,Z\rangle = \langle X, \alpha Z \rangle= \alpha\langle X , Z \rangle$.
We only need to show (1.) and (3.) as the symmetry of $\langle X,Y\rangle$ is clear from the definition \begin{align*}\langle X,Y \rangle:= \tfrac14 \left(\langle X + Y \rangle + \langle X -Y\rangle\right) = \tfrac12 \left(\langle X+Y \rangle - \langle X \rangle - \langle Y \rangle\right). \end{align*}
I tried to derive (1.) and (3.) by using that the Doob—Meyer decomposition $$ Y = A + M $$ gives an unique natural increasing process $A=:\langle Y \rangle$ that makes $Y-A$ a martingale. This, unfortunately, did not yield any viable results to prove (1.) or (3.).
Does anyone know the proof or have a reference to a book where it is proved? I believe the proof is straightforward, so I guess I'll understand how this is done by only reading a proof.
Many thanks in advance!
$\langle Z \rangle$ is the unique natural increasing process such that
$$Z^2 - \langle Z \rangle$$
is a martingale. Applying this for $Z = X \pm Y$, we find that
$$X \cdot Y - \langle X,Y \rangle = \frac{1}{4} ((X+Y)^2- \langle X+Y \rangle) + \frac{1}{4} ((X-Y)^2 - \langle X-Y \rangle)$$
is a martingale. Consequently, if $A$ is of bounded variation and natural, then $$A = \langle X,Y \rangle \iff X \cdot Y - A \qquad \text{is a martingale.} \tag{1}$$
To prove the first statement, it suffices by $(1)$ to show that
$$(X+Y) \cdot Z - (\langle X,Z \rangle + \langle Y,Z \rangle)$$
is a martingale. But this follows directly from the fact that both
$$X Z - \langle X,Z \rangle \qquad \text{and} \qquad Y Z - \langle Y,Z \rangle$$
are martingales. The proof of 3. is very similar.
Reference: Jean Jacod, Albert Shiryaev: Limit Theorems for Stochastic Processes, Theorem I.4.2