Show that if $(M_t)$ is a continuous local martingale, then $M^2$ has quadratic variation $$\langle M^2\rangle_t=4\int_0^tM_s^2\mathrm{d}\langle M\rangle_s$$

It seems to me that $M^2$ is a semi-martingale with decomposition $M_t^2 = (M_t^2-\langle M\rangle_t)+\langle M\rangle_t$, so I have $\langle M^2\rangle_t=\langle (M^2-\langle M\rangle)\rangle_t$. But I don't know how to proceed.

Any help is appreciated!!


1 Answer 1


By Ito's formula, $$M_t^2 = 2\int_0^t M_s dM_s + \int_0^t d\langle M,M \rangle_s$$ so \begin{align*} \langle M^2,M^2\rangle_t &= \left\langle 2\int_0^t M_s dM_s ,2\int_0^t M_s dM_s \right\rangle \\ &= 4\left\langle \int_0^t M_s dM_s ,\int_0^t M_s dM_s \right\rangle \\ &= 4\int_0^t M_s^2 d\langle M,M\rangle_s. \end{align*}


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