In a course I am studying on Stochastic Processes, I encountered the following exercise:
Let $X_t = B_t + ct$ for some $c \in \mathbb{R}$ and where $B$ is a standard Brownian Motion. Now define $T_x$ to be the first hitting time of $x$. More formally: $$ T_x = \inf \{ t > 0 : X_t = x \}$$ Calculate $\mathbb{E}(\exp (- \lambda T_x ))$ for $\lambda > 0$ (ie. find the Laplace transform of $T_x$).
I have found that $M_t = \exp (\theta X_t - \lambda t)$ the choices of $\theta$ that ensure that $M$ is a martingale are $$\theta_1 = -c + \sqrt{c^2 + 2 \lambda} \space \text{ and } \space \theta_2 = -c - \sqrt{c^2 + 2 \lambda}$$
I know that the provided solution to problem is $$\mathbb{E}(\exp (- \lambda T_x )) = \exp (x(c+\sqrt{c^2 + 2 \lambda})) = \exp (-x \theta_2)$$
However, I am unsure of how to prove the result.
It feels as though I ought to begin by substituting the hitting time into the definition of $M$ to yield: $$M_{T_x} = \exp (\theta X_{T_x} - \lambda T_x)$$ and work on some simplification from here.
However, I am conscious of the fact that if I use the Martingale property, then I can no longer take $\theta = 0$ which would simplify $M_{T_x}$ to $\exp ( - \lambda T_x)$ (the desired expectation).
Is there an alternative approach that I am missing? I would be grateful for any help with this problem.