# Concentration of measure for local martingale

Can someone tell me, where I can find a proof of the following fact:

Let $M$ be a continuous local martingale with $M_0=0$. Then we have $$P \left(\max_{s \leq t} \; M_s \geq y, \ [M]_t \leq C \right) \leq \exp \left(-\frac{y^2}{2C} \right)$$ for every $t, y, C> 0$.

Here I denote by $[M]_t$ the quadratic variation of $M$.

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## 1 Answer

By a time change argument, we can express $M_t=B([M]_t)$ where $(B(w))_{w\geq 0}$ is a Brownian motion. The result now follows from the maximal inequality for Brownian motion.

\begin{eqnarray} P \left(\max_{s \leq t} \; M_s \geq y, \left[M\right]_t \leq C \right) &=& P \left(\max_{s \leq t} \; B([M]_s) \geq y, \ [M]_t \leq C \right) \cr &\leq& P \left(\max_{w \leq C} \; B(w) \geq y\right) \cr &\leq& \exp \left(\frac{-y^2}{2C} \right) \end{eqnarray}

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Great 3 line argument !!! –  TheBridge Dec 5 '11 at 8:04