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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 $t>0$: $$P(T_{a,b}<t)=P(\exists s \le t: B(s)=as-b)\le P(\exists s \le t: B(s)<at-b) = 2\cdot P(B(t)<at-b)$$ 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?

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    $\begingroup$ Why are you allowed to send $b \to -\infty$? Aren't $a$ and $b$ fixed? $\endgroup$
    – paulinho
    May 25, 2020 at 14:29
  • $\begingroup$ $a,b$ 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) $\endgroup$ May 25, 2020 at 14:34

1 Answer 1

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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 $\lambda \to 0$.

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