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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.

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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.

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  • $\begingroup$ I think you misread the sign . In order to apply the polarization identity we need a negative sign . I am writing from the phone so I am not writing precisely but I hope you got my point $\endgroup$ Commented May 26, 2018 at 14:53
  • $\begingroup$ You're right, I had misread the sign. But with the sign as it is, why isn't this trivial since $\frac12 (\langle X+Y \rangle + \langle X-Y \rangle) = \langle X \rangle + \langle Y \rangle$? $\endgroup$ Commented May 26, 2018 at 16:10
  • $\begingroup$ The identity says $\frac{1}{4}(\langle X+Y \rangle_t -\langle X-Y \rangle_t) =\langle X,Y \rangle_t$ $\endgroup$ Commented May 26, 2018 at 16:20
  • $\begingroup$ Yes, I know what the polarisation identity says. That is not the identity I stated in the previous comment. $\endgroup$ Commented May 26, 2018 at 16:21
  • $\begingroup$ I am sorry I didint read the previous comment. It didnt update on my computer.I am rereading now again the proof $\endgroup$ Commented May 26, 2018 at 16:23

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