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For a continuous local martingale $X$ and stopping time $T$, show that the quadratic variation $\langle X\rangle^T=\langle X^T\rangle$. Here $X^T$ is the stopped process, denoted by $X^T=X_{t\land T}$.


It is enough to show that $$ \langle X\rangle_{T\land t}=\langle X^T\rangle_t $$

I try to use the fact that $X_{t\land T}^2-\langle X\rangle_{t\land T}=(X^T)^2-\langle X\rangle_{t\land T}=(X^2)^T-\langle X\rangle_{t\land T}$ is a continuous local martingale. But how to go the next step? Thanks.

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1 Answer 1

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$(X^T_t)^2-\langle X\rangle^T_t = X^2_{T\wedge t}-\langle X\rangle _{T\wedge t}$ is a local martingale, being the local martingale $X^2_t-\langle X\rangle_t$ stopped at time $T$. But $\langle X^T\rangle$ is the unique continuous increasing process that compensates $(X^T)^2$; namely, which when subtracted from $(X^T)^2$ yields a local martingale. This uniqueness forces $\langle X^T\rangle = \langle X\rangle^T$.

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