Proving a statement in quadratic variation that ${\langle X \rangle}^{\tau} = \langle X^{\tau} \rangle$

Question

Let $\tau$ be a stopping time and $X$ be a continuous local martingale. Let $\langle \cdot \rangle$ denote the quadratic variation. We want to show that $${\langle X \rangle}^{\tau} = \langle X^{\tau} \rangle.$$

Does anyone know how to justify this statement, as it is not obvious at all from the definition of quadratic variation?

Answer 1

Suppose that $(X_t)_{t \geq 0}$ is a continuous martingale. We know that the compensator $(\langle X \rangle_t)_{t \geq 0}$ is the unique increasing predictable process such that

$$X_t^2 - \langle X \rangle_t$$

is a martingale. It follows from the optional stopping theorem that

$$X_{t \wedge \tau}^2 - \langle X \rangle_{t \wedge \tau} = (X^{\tau}_t)^2 - \langle X \rangle^{\tau}_t$$

is a martingale. As the compensator is unique, we conclude

$$\langle X \rangle^{\tau}_t = \langle X^{\tau} \rangle_t$$

for all $t \geq 0$.