Show that $M$ is a martingale

Question

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.

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Answer 1

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.)