# Expectation of Stopping Time for a Brownian Motion with a drift

Let $$a,b>0$$ and define the stopping time $$T_{a,b}$$ for Brownian Motion as $$T_{a,b}:=inf\{t>0:B(t)=at-b\}$$ Compute $$E[T_{a,b}]$$.

My idea:

I think $$E[T_{a,b}]=\infty$$.

If that was not the case, then by Wald´s Lemma $$E[T_{a,b}]=Var(B(T_{a,b}))=Var(aT_{a,b}-b)=a^2Var(T_{a,b})$$ for all $$b>0$$.

But by reflection principle, for any fixed $$t>0$$: $$P(T_{a,b} and that converges to $$0$$ as $$b \rightarrow -\infty$$.

Could that help me conclude $$E[T_{a,b}]=\infty$$ and is that assumption even correct in the first place?

• Why are you allowed to send $b \to -\infty$? Aren't $a$ and $b$ fixed? May 25, 2020 at 14:29
• $a,b$ are fixed but arbitrary. And I attempted to show that if indeed $E[T]<\infty$ then the Expectation is identical for all $b$ which is counterintuitive (and which I tried to disprove) May 25, 2020 at 14:34

For fixed $$a,b>0$$ set $$f(t) := at-b$$. Clearly, $$f(0)=-b<0 = B_0$$, and therefore it follows from the continuity of the sample paths of Brownian motion that

$$\mathbb{P}(T_{a,b} \geq t) \leq \mathbb{P}(B_t\geq f(t)).$$

For sufficiently large $$t$$, we have $$f(t)=at-b>0$$, and using $$B_t \sim N(0,t)$$ we find that

$$\mathbb{P}(T_{a,b} \geq t) \leq \frac{1}{\sqrt{2\pi t}} \frac{1}{(at-b)} \exp \left(- \frac{(at-b)^2}{2t} \right)$$

for large $$t$$. The right-hand side decays exponentially as $$t \to \infty$$, and therefore it follows that

$$\int_0^{\infty} t^k \mathbb{P}(T_{a,b} \geq t) \, dt < \infty$$

for all $$k \geq 1$$. This implies that $$\mathbb{E}(T_{a,b}^k)<\infty$$ for all $$k \geq 1$$, i.e. $$T_{a,b}$$ has finite moments of arbitrary order.

To compute the moments explicitly, you can use the Laplace transform $$\mathbb{E}e^{-\lambda \tau}$$ of $$\tau$$. First compute the Laplace transform (see this question) and then differentiate with respect to $$\lambda$$ and let $$\lambda \to 0$$.