# Need some clarification for the proof that a martingale is a Brownian motion with change of time.

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 $$[0,\tau_t]$$.

## 1 Answer

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 $$[X,X](t)=[M,M](\tau_t)$$