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Let $I_t=\int_0^t f_tdB_t,$ where $(f_t,t\ge 0)$ is a bounded process, $|f_t|\leq M$ almost surely for all $t \ge 0.$ Show that $$\mathcal{P}\left[\sup_{0\leq t\leq T}|I_t|>\lambda\right]\leq \exp\left(-\frac{\lambda^2}{2M^2T}\right).$$

Hint: Define $Y_t^{\alpha}=\exp\left(\alpha I_t-\frac{1}{2}\int_{0}^t f^{2}(s)ds\right)$, where $\alpha\in \mathbb{R}.$ First use $Y^{\alpha}$ to get an upper bound, then optimize over choice of $\alpha$ to get the smallest bound.

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@Byron Schmuland I fixed the typo –  user9641 Apr 15 '11 at 23:47
    
@user9641: Isn't this an application of the Martingale PDE condition? –  Jonas Teuwen Apr 16 '11 at 23:04
    
Use Doob's inequality for martingales. –  Ilya Apr 17 '11 at 19:35
    
@Gortaur can you write in more details? –  user9641 Apr 17 '11 at 21:16
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Maybe it will be useful for you to apply the following Doob's inequality. If $Z_t$ is a martingale then for all $p\geq 1$ $$ \mathsf{P}(\sup\limits_{t\leq T}Z_t\geq \lambda) \leq \frac{1}{\lambda^p}\mathsf{E}|Z_T^p|. $$

Your process $Y^\alpha_t$ is a

1) martingale if $\alpha^2 = 1$

2) supermartingale if $\alpha^2<1$

3) submartingale if $\alpha^2>1$.

For $Z_t$ which is a sub- or supermartingale you can use $$ \mathsf{P}(\sup\limits_{t\leq T}Z_t\geq \lambda) \leq \frac{1}{\lambda}\sup\limits_{t\leq T}\mathsf{E}|Z_t|. $$

As I understand you should try all these bounds for different $\alpha$ and find the best one (which seems to be reached in the case $|\alpha|= 1$).

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