Let $(W_t)_{t \geq 0}$ be a Brownian motion and let $\tau=$inf{$t\ge0:W_t=a+bt$} with $a<0$.
I want to compute $E[\tau]$ (and I already know that $E(\tau)<\infty$).
I want to say that I already read Expectation of the first passage time of $T_{a,b}$ [duplicate]
and Distribution of hitting time of line by Brownian motion
The reason I am not satisfied with the other posts is, that computing the distribution seems like an overkill to me (the exercise is from an exam and thus I think computing the distribution first is too time consuming). Furthermore was the solution only valid for $a>0$, which does not hold for my case. Additionally does the suggested solution from the first post: $E[\tau]=\frac{a}{|b|}e^{-ab}$ not hold for $a<0$ since this would imply a negative expectation for a strictly positive random variable.
If $b \leq 0$ then $$\tau \geq \inf\{t \geq 0; W_t = a\} =: T_a$$ implying $\mathbb{E}(\tau) \geq \mathbb{E}(T_a)=\infty$. Therefore we can assume from now on that $b>0$. If we set
$$X_t := W_t-bt$$
then
$$\tau = \inf\{t \geq 0; X_t=a\}.$$
Using that
$$M_t := \exp \left( - \alpha X_t - \alpha bt - \frac{1}{2} \alpha^2 t \right)$$
is a martingale for any $\alpha>0$ we find from the optional stopping theorem that
$$\mathbb{E}(M_{t \wedge \tau}) = \mathbb{E}(M_0)=1.$$
As $|M_{t \wedge \tau}| \leq e^{\alpha |a|}$ it follows from the dominated convergence theorem that $\mathbb{E}(M_{\tau})=1$, and so
$$\mathbb{E} \exp \left( - \left[ \alpha b + \frac{1}{2} \alpha^2 \right] \tau \right) = e^{\alpha a}.$$
Thus,
$$f(\lambda) := \mathbb{E}e^{-\lambda \tau} = \exp \left(- a \left[ b- \sqrt{b^2+2\lambda} \right] \right), \qquad \lambda>0.$$
Differentiating the expression with respect to $\lambda$ we find
$$\mathbb{E}(\tau) = - \frac{d}{d\lambda} f(\lambda) \bigg|_{\lambda =0} = \frac{-a}{b}.$$