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I have some question related to expectation and Brownian motion.

For $a,b > 0$ we define stopping time $T_{a,b}= \inf \{t \geq 0: B_t < bt-a \}. $

I have to show that for $b=1$: $E[e^{1/2 T_{a,1}}] = e^a.$

I am a little bit confused how to start this, because I usually had problems where $T$ is hitting time, but here $B_t < bt-a$, so I am not sure how to do this. Thanks for any help.

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    $\begingroup$ Hint: $e^{B_t -t/2}$ is a martingale. $\endgroup$ – zhoraster Mar 18 '17 at 14:50
  • $\begingroup$ @zhoraster Clearly, $\mathbb{E}\exp(B_{t \wedge T}-t \wedge T/2)=1$ ... however, I don't see why we can let $t \to \infty$ (as far as I can see neither dominated convergence nor monotone convergence is applicable). What am I missing...? $\endgroup$ – saz Mar 18 '17 at 21:16
  • $\begingroup$ The Brownian motion with drift $-1$, namely $B_t-t$ hits $-a$ with probability $1$. This fact in combination with Girsanov's theorem will give you the formula you seek. $\endgroup$ – John Dawkins Mar 19 '17 at 1:27
  • $\begingroup$ @saz , zhoraster I also tried to start with the fact that $e^{B_t - t/2}$ is martingale, but I don't know how would I come to wanted result of $e^a$. Jown Dawnkins This first part it is clear to me. Could you explain a little bit more how would I use Girsanov's theorem on this in particular? Thanks. $\endgroup$ – User1999 Mar 19 '17 at 8:21
  • $\begingroup$ @saz I don't know if I understood your question okay, but I don't think you would need to send $t \to \infty $. You can only use the fact that $e^{B_t - t/2}$ is martingale and than use optional stopping theorem which says that $E[e^{B_T -T/2}] = E[e^{B_0 - 0}]=1.$ But now that I know this, I still don't know how would that get me to the result. $\endgroup$ – User1999 Mar 19 '17 at 9:04
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Elaboration: Abbreviate $T=T_{a,1}$ and $Z_t=\exp(B_t-t/2)$. There is no doubt that $P[T<\infty]=1$. Because $Z=(Z_t)$ is a martingale, so too is $Z$ stopped at time $T$; because $Z_0=1$, $$ 1=E[Z_{T\wedge t}]=E[e^{T/2-a};T\le t]+E[e^{B-t-t/2}; t<T]. $$ The first term on the right side of this display converges to $E[e^{T/2-a}]$ as $t$ increases to $+\infty$ by monotone convergence. It remains to show that the second term converges to $0$. By Girsanov's theorem, because $Z_t1_{\{t<T\}}$ is $\mathcal F_t$-measurable, $$ E[e^{B_t-t/2};t<T]=E[Z_t; t<T]=Q[t<T], $$ where $Q$ is the law of the Brownian motion with drift +1 (that is, the law of $B_t+t$). Thus $$ \lim_{t\to+\infty}E[e^{B_t-t/2};t<T]= Q[T=+\infty]. $$ Finally, $Q[T=+\infty]=0$, because the hitting timeof the boundary $t-a$ by a Brownian motion with unit positive drift has the same distribution as the hitting time of $-a$ by a standard Brownian motion, and the latter hitting time is finite, a.s.

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