# Brownian Motion Hitting Time of a line with a negative axis intercept

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]

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.

• Using martingale techniques it is not too difficult to compute $$f(\lambda) := \mathbb{E}e^{-\lambda \tau}, \qquad \lambda>0$$ see this question: math.stackexchange.com/q/2107249/36150. Once you have computed $f$, you can use that $$\mathbb{E}(\tau) = - \frac{d}{d\lambda} f(\lambda) \big|_{\lambda=0}.$$ – saz Jan 31 '18 at 11:37
• When I am using that approach I am getting a negative expectation: $E[e^{-\lambda \tau}]=e^{-a(b+\sqrt{b^2+2\lambda})}$ with $\lambda=-a\alpha+1/2\alpha^2>0.$ Thus $E[\tau]=\frac{a}{\sqrt{b^2}}e^{-ab}<0.$ – Andrew Jan 31 '18 at 12:20
• Or is it than we choose $\sqrt{b^2}=-b$ (if b is positive) such that the expectation is bigger than 0. Than the sign of $a$ still has a large impact on how $e^{-ab}$ explodes which seems a bit wierd to me but I guess it is fine. – Andrew Jan 31 '18 at 12:27
• What do you mean by "choose "$\sqrt{b^2} = -b$ (if b is positive)"... Obviously, $\sqrt{b^2} = -b$ is plainly wrong if $b>0$. If you follow the approach, you get the correct result, see the answer below. – saz Jan 31 '18 at 20:17

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

$$\mathbb{E}(\tau) = - \frac{d}{d\lambda} f(\lambda) \bigg|_{\lambda =0} = \frac{-a}{b}.$$

• I think what confused me is that you need two different approaches depending if $a>0$ or $a<0$, in particular for $a>0$ you use a "classic" GBM and for $a<0$ you use that $-W_t$ is a Brownian Motion as well. I understand how you derive the solution, but I still have two problems: 1) Why is your solution wrong for a>0? (In particular I don't see where we are using that a<0) 2) Consider the two stopping times $\tau_1=$inf{$t\ge0:W_t=t-1$}, $\tau_2=$inf{$t\ge0:W_t=10t-1$}. It holds that $\tau_2\le \tau_1.$ But from your result we have that $E[\tau_1]=e<E[\tau_2]=\frac{1}{10}e^{10}$ – Andrew Feb 1 '18 at 9:46
• I got it. I think your solution is not correct in the last step. After differentiating and setting $\lambda=0$, we get $e^{-ab+a\sqrt{b^2}}=1$ , in particular $E[\tau]=-\frac{a}{b},$ if $a<0$. – Andrew Feb 1 '18 at 9:53
• @Andrew 1) We are using $a<0$ to conclude that $X_{t \wedge \tau} \geq a$, and so $|M_{t \wedge \tau}| \leq e^{\alpha |a|}$. 2) Yeah, you are right about that, the $e^{-ab}$ cancels... – saz Feb 1 '18 at 10:57