# For a continuous local martingale $X$ and stopping time $T$, show that the quadratic variation $\langle X\rangle^T=\langle X^T\rangle$.

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.

$$(X^T_t)^2-\langle X\rangle^T_t = X^2_{T\wedge t}-\langle X\rangle _{T\wedge t}$$ 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 $$\langle X^T\rangle = \langle X\rangle^T$$.