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Given that the sum of two martingales, adapted to the same filtration, is also a martingale (see this ME stack answer), I would posit that the quadratic variation of martingale $M$ is also a martingale, since

$$ \langle M \rangle_t = - (M_t - \langle M \rangle_t) + M_t$$

Am I correct?

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  • $\begingroup$ \langle and \rangle for QV $\endgroup$ Apr 2 at 19:08

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The quadratic variation process is non-decreasing, so it is not a martingale (unless it is constant at $0$).

The flaw in your argument is that $M_t - \langle M\rangle_t$ is not a martingale, rather, $M_t^2 - \langle M\rangle_t$ is. Since $M_t^2$ is not a martingale (again, unless $M_t$ is constant), we can't use the result that the sum of martingales is a martingale in the equation $$ \langle M \rangle_t = -(M_t^2 - \langle M\rangle_t) + M_t^2.$$

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