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Theorem 7.37: Let $M (t)$ be a continuous martingale, null at zero, such that $[M, M ](t)$ is non-decreasing to $∞$, and $τ_t:=\inf\{s : [M, M ](s) > t\}$. Then the process $B(t) = M (τ_t )$ is a Brownian motion with respect to the filtration $\mathcal{F}_{τ_t}$. Moreover, $[M, M ](t)$ is a stopping time with respect to this filtration, and the martingale $M$ can be obtained from the Brownian motion $B$ by the change of time $M (t) = B([M, M ](t))$

Proof. Let $M (t)$ be a local martingale. $τ_t$ defined as in the statement are finite stopping times since $[M, M ](t) → ∞$. Thus $\mathcal{F}_{τ_t}$ are well defined. Note that $\{[M, M ](s) ≤ t\} = \{τ_t ≥ s\}$. This implies that $[M, M ](s)$ are stopping times for $\mathcal{F}_{τ_t}$ . Since $[M, M ](s)$ is continuous $[M, M ](τ_t ) = t. Let X(t) = M(τ_t )$. Then it is a continuous local martingale since $M$ and $[M, M]$ have the same intervals of constancy (see the comment following Theorem 7.28). Since $M^2(t)-[M,M](t)$ is a martingale, we obtain $EX^2 (t) = E[X, X](t) = E[M, M ](τ_t ) = t$. Thus $X$ is a Brownian motion by Levy’s characterization. The second part is proven as follows. Recall that $M$ and $[M, M ]$ have the same intervals of constancy. Thus $X([M, M ](t)) = M (τ_{[M,M ](t)} ) = M (t)$.

There are two things I don't understand in this proof. First isn't $M(τ_{[M,M ](t)} ) = M (t)$ a consequence of the fact that $[M,M]$ is non-decreasing rather than the fact that "$M$ and $[M, M ]$ have the same intervals of constancy". Second I am not sure why $[X, X](t) = [M, M ](τ_t)$. Indeed, if $\delta_n$ denotes the size of a partition, $$[X, X](t)=\lim\limits_{\delta_n\to0}\sum_{i=1}^n (X(t_{i+1})-X(t_i))^2=\lim\limits_{\delta_n\to0}\sum_{i=1}^n (M(\tau_{t_{i+1}})-X(\tau_{t_i}))^2$$ however it is not clear to me that this is the same thing as $[M,M](\tau_t)$ which I understand as the quadratic variation on a partition of the random interval $[0,\tau_t]$.

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I think I understand now why $[X,X](t)=[M,M](\tau_t)$. One has to consider the characterization of the quadratic variation as the fact $[M,M]$ is the unique process such that $M^2(t)-[M,M](t)$ is a martingale. Then evaluating this at the stopping time $\tau_t$ by the stopping time theorem, $M^2(\tau_t)-[M,M](\tau_t)$ is also a martingale meaning exactly that $X^2(t)-[M,M](\tau_t)$ is a martingale but then by the uniqueness of the Doob-Meyer decomposition we have $[X,X](t)=[M,M](\tau_t)$

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