# A question on relationship between quadraticc variation of right continuous martingales?

How can I show that the following relation hold for square-integrable right continuous martingales $X,Y$ starting at zero $$\frac{1}{2}(\langle X+Y \rangle_t(\omega)-\langle X+Y \rangle_s(\omega))+\frac{1}{2}(\langle X-Y \rangle_t(\omega)-\langle X-Y \rangle_s(\omega)) \le\big(\langle X \rangle_t(\omega)-\langle X \rangle_s(\omega)+\langle Y \rangle_t(\omega)-\langle Y \rangle_s(\omega)\big) \text{ }P \text{ a.s.}$$ where $(\langle X\rangle_t)_{t \ge 0}$ is the unique upto indistinguishability, non decreasing process which makes $X^2-\langle X\rangle$ a martingale (where X is a square integrable right continuous martingale starting at $0$ almost surely)i.e the increasing part in the Doob-Meyer decomposition of $X^2$. I am trying to solve exercise 1.5.7(iv) in Karatzas and shreve and if I can show this I can complete the proof.

We even have equality. It holds that $$\frac12 (\langle X+Y \rangle + \langle X-Y \rangle) = \langle X \rangle + \langle Y \rangle$$ by symmetry and bilinearity of the covariation bracket. The desired equality immediately follows.
• You're right, I had misread the sign. But with the sign as it is, why isn't this trivial since $\frac12 (\langle X+Y \rangle + \langle X-Y \rangle) = \langle X \rangle + \langle Y \rangle$? Commented May 26, 2018 at 16:10
• The identity says $\frac{1}{4}(\langle X+Y \rangle_t -\langle X-Y \rangle_t) =\langle X,Y \rangle_t$ Commented May 26, 2018 at 16:20