$B$ denotes Brownian motion and the hitting time I am interested in is $$\tau = \inf\{t \geq 0: B_t = b\sqrt{a+t}\}$$ where $a,b >0$. I first want to show that $\tau < \infty$ almost surely. I am going to use Khintchine's law of the iterated logarithm (LIL). Define

$$\Omega := \left\{\omega: \limsup_{t\to\infty } \frac{B_t(\omega)}{\sqrt{2t\log\log t}} = 1 \right\}$$ The LIL states that $P(\Omega) = 1$. I will show that $\tau(\omega) < \infty$ for every $\omega \in \Omega$. Fix $\omega \in \Omega$ and some small $\varepsilon > 0$. Then there exists $T(\omega,\varepsilon) > 0$ such that $$\left\lvert\sup_{t > T } \frac{B_t(\omega)}{\sqrt{2t\log\log t}} -1\right\rvert < \varepsilon$$ This in turn implies that $$\sup_{t > T } \frac{B_t(\omega)}{\sqrt{2t\log\log t}} > (1-\varepsilon)$$ Then there must exist some $u > T$ such that $$B_u(\omega)> (1-\varepsilon)\sqrt{2u\log\log u}$$

Since $\sqrt{2t\log\log t}$ is greater than $b\sqrt{a+t}$ for $t$ large enough, I can choose $\varepsilon$ small enough to make sure $\sqrt{2u\log\log u} > b\sqrt{a+u}$ so that $\tau(\omega) < u$. I got a bit sloppy towards the end but I hope the idea is correct. I just wanted to have this verified. My main concern is in the part where I dropped the supremum.

My second question is about $E[\tau]$. It is easy to show that $E[\tau] = \infty$ whenever $b \geq 1$ by Wald's identity. But I am trying to show that $E[\tau] < \infty$ whenever $b < 1$. The hint in the book is that $E[\tau\wedge n] \leq \frac{ab^2}{1-b^2}$ for $n \geq 1$. Obviously, if I can show this inequality then I would be done immediately by monotone convergence but I haven't been able to prove that this inequality holds. I made some efforts below but I find it very hard to believe that $E[\tau] < \infty$ whenever $b < 1$. For a hitting time against a constant boundary, i.e. $\inf\{t\geq 0: B_t = x\}$, the expected value is not finite. How can it possibly be finite for a growing boundary? Anyway, here is what I tried so far.

I know that $B_{\tau \wedge n}$ is lower and upper bounded by the running minimum and running maximum Brownian motion at time $n$. So $E[B_{\tau \wedge n}^2] < \infty$. Therefore, I can use Wald's identity to write

$$E[\tau \wedge n] = E[B_{\tau \wedge n}^2]$$ and focus on $E[B_{\tau \wedge n}^2]$. However, I don't see how I can make use of the fact that $b < 1$. By time scaling property of Brownian motion I have $$\tau \stackrel{\text{d}}{=} b^{2n} \inf\left\{t \geq 0: B_t = b\sqrt{\frac{a}{b^{2n}}+t}\right\}$$ If I denote $f(a,b) := E[\tau]$, I can write $$f(a,b) = b^{2n}f\left(\frac{a}{b^{2n}},b\right)$$ Letting $n \to \infty$ does not help because I don't know what happens to $f\left(\frac{a}{b^{2n}},b\right)$ in the limit (it is increasing for sure but beyond that I don't have much else to say on it).

Applying the time scaling property in a slightly different way I get

$$\tau \stackrel{\text{d}}{=} ab^2 \inf\{t \geq 0: B_t = \sqrt{1+b^2t}\}$$ The stopping time on the RHS looks easier to study but I don't see how this brings me closer to the answer.

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    $\begingroup$ I highly doubt that $\mathbf{E}[\tau] < \infty$. If $T=\inf\{t\geq0:B_t=b\sqrt{a}\}$, then we know that $T \leq \tau$ but $T$ has infinite expectation for the Brownian motion started at $0$. I suspect that the definition of $\tau$ should be $$\tau = \inf\{t\geq 0: \lvert B_t \rvert=b\sqrt{a+t}\},$$ and then the the computation in the answer below makes sense. $\endgroup$ – Sangchul Lee May 2 '18 at 15:19
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    $\begingroup$ @SangchulLee That was my suspicion as well. I mentioned this in the question. Maybe I send the authors of the book an email about this. Hopefully, it is just a typo. $\endgroup$ – Calculon May 2 '18 at 15:21
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    $\begingroup$ @SangchulLee: I agree. Your $T$ explains the trouble. $\endgroup$ – hypernova May 2 '18 at 15:24

Edit: This is an invalid proof

(I recommend @SangchulLee's comment below the original post)

For the $\mathbb{E}\tau<\infty$ part of your question, perhaps you could go for the proof as follows.

Upon your result $\mathbb{E}\tau\wedge n=\mathbb{E}B_{\tau\wedge n}^2$, it follows that $$ \mathbb{E}\tau\wedge n=\mathbb{E}B_{\tau\wedge n}^2\le\mathbb{E}B_{\tau}^2=\mathbb{E}\left(b\sqrt{a+\tau}\right)^2=ab^2+b^2\mathbb{E}\tau. $$ Note that the RHS of the above inequality is a fixed value, whether or not $\mathbb{E}\tau$ is finite. Thus $$ \mathbb{E}\tau=\lim_{n\to\infty}\mathbb{E}\tau\wedge n\le ab^2+b^2\mathbb{E}\tau. $$ This, together with $0<b<1$, eventually leads to $$ \mathbb{E}\tau\le\frac{ab^2}{1-b^2}<\infty. $$ By the way, since $\mathbb{E}\tau<\infty$, optional stopping theorem applies again and finally results in $$ \mathbb{E}\tau=\frac{ab^2}{1-b^2}. $$

  • $\begingroup$ Thanks for your answer. How did you justify $E[B_{\tau \wedge n}^2] \leq E[B_{\tau}^2]$? $\endgroup$ – Calculon May 2 '18 at 15:01
  • $\begingroup$ @Calculon: Oops, that was a typo (should be deleted). $\tau\wedge n\le\tau$ implies $B_{\tau\wedge n}\le b\sqrt{a+\tau\wedge n}\le b\sqrt{a+\tau}$. $\endgroup$ – hypernova May 2 '18 at 15:15
  • $\begingroup$ What if $B_{\tau \wedge n}$ is negative? You cannot square both sides of an inequality and expect the direction to stay the same. $\endgroup$ – Calculon May 2 '18 at 15:17
  • $\begingroup$ @Calculon: You are right. Let me reconsider this. $\endgroup$ – hypernova May 2 '18 at 15:19

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