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Let $B$ be typical Brownian motion with $\mu >0$ and $x \in \mathbb{R}$. $X(t):=x+B(t)+\mu t$, for each $t\geqslant 0$, Brownian motion with velocity $\mu$ that starts at $x$. For $r \in \mathbb{R}$, $T(r):=\inf\{s\geqslant 0:X(s)=r\}$ and $\phi(r):=\exp(-2\mu r)$. Show that $M(t):=\phi(X(t))$ for $t\geqslant 0$ is martingale.

I hope you can help me....! Thanks in advance!!!

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What is the role of $T$ in this problem? – Kuba Helsztyński May 4 '13 at 0:14
    
I don't know.... Maybe we have to show that M(t) is martingale adapted to T(r). Or what do you mean??? :/ – evinda May 4 '13 at 12:39
up vote 1 down vote accepted

Hint By definition, we have

$$\phi(X_t) = \exp(-2\mu \cdot (x + B_t + \mu \cdot t))$$

Hence

$$\mathbb{E}(\phi(X_t) \mid \mathcal{F}_s) = \exp(-2\mu \cdot (x +\mu \cdot s +B_s)) \cdot \mathbb{E} \left( \exp( -2 \mu \cdot (B_t-B_s)) \mid \mathcal{F}_s \right)$$

for all $t \geq s$. Use the independence of the increments and the knowledge about exponential moments of Gaussian random variables to calculcate the remaining conditional expectation.

(As Kuba Helsztyński wrote in his comment, the role of the stopping time $T$ is not clear.)

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Thank you very much!!! :) – evinda May 4 '13 at 23:36
    
@evinda You are welcome. By the way, you can mark answers as the accepted answer by clicking on the check box outline to the left of the answer. – saz May 5 '13 at 9:53

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