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The following is an exercise from a book on Continuous martingales and Brownian motion by Revuz and Yor.

Let $B$ be the standard linear Brownian motion.

1) For any probability measure $\mu$ on $\mathbb R$ prove that there is a $\mathcal{F}^B_{\frac{1}{2}} $ measurable random variable $Z$ such that the distribution of $Z$ equals $\mu$, that is $P \circ Z = \mu$

2) Define a $\mathcal{F}^B_t$-stopping time $T$ by

$$T = \inf \{t \ge 1 : B_t =Z \} $$

Prove that the distribution of $B_T$ is $\mu$ and that $E[T]= \infty$

For 1) one of course thinks about the result that for any probability measure $\mu$ on $\mathbb R$ there exists a random variable with distribution $\mu$. We need to 1) make sure that we may take this random variable to be map from the same probability space as $B$ maps from, and 2) make sure that it is $\mathcal{F}^B_{\frac{1}{2}} $ measurable.

The construction for the above result is to let $Z: \mathbb R \to \mathbb R, \ Z(x)=x$, which then certainly is Borel measurable. Does there exist a construction of Brownian motion as a stochastic process on $\mathbb R$? Or rather we should take some construction of $B$ on the trace sigma algebra on $C(\mathbb R ) \cap \mathbb R^{[0,1) } $ and then composite $Z$ with some projection $\pi_t$ from $\mathbb R^{[0,1) } $ to $\mathbb R$? How does $\mathcal{F}^B_{\frac{1}{2}} $ come in to this?

For 2) one would think that the method is to prove that for any $a < b$, $\{B_T \in (a,b) \} = \{Z \in (a,b) \} $ and then the claim about the distributions being equal will follow. How do we show that $E[T]=\infty$?

Most grateful for any help provided!

MrFranzén
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  • Even if this might answer your question, for your information, this is the celebrated "Skorokhod embedding problem", you can have a look at it on the web and read the work of Obloj on the multiple solutions and extensions to this subject : https://arxiv.org/pdf/math/0401114.pdf – TheBridge Oct 21 '19 at 15:38
  • Actually I know of the Skorokhod embedding problem and that and any form of solution would be much harder to prove as an exercise. But here we have slighty weaker conditions. No assumption is made that $\mu $ is the distribution of a centered random variable. – MrFranzén Oct 21 '19 at 15:46
  • Instead we get a result that $E[T]= \infty $ from which we eg couldn't prove Donsker's Theorem. And I think to show this would be one point of the exercise. – MrFranzén Oct 21 '19 at 15:47

1 Answers1

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The stopping time $T_a = \inf\{t\ge0:B_t=a\}$ has infinite expectation for any $a\ne 0$. Moreover, the probability that $T_a\le \tau$ is decreasing as a function of $a$ for any constant $\tau$.

So to prove that $\mathbb{E}[T]=\infty$, note that $\mathbb{P}[|Z-B_1|>1]\ge\mathbb{P}[|B_1|>1]$ since $Z$ is independent of $B_1$ and the distribution of $B_1$ is same as the distribution of $-B_1$ and therefore $Z$ cannot on average Vd closer to $B_1$ than zero is to $B_1$.

Write $W_t=B_{t+1}-B_1$. Then $T-1=\inf\{t\ge0:W_t=Z-B_1\}$.

Then $\mathbb{E}[T-1]\ge \mathbb{E}[\inf\{t\ge0:W_t=1\}]\mathbb{P}[Z-B_1>1]+\mathbb{E}[\inf\{t\ge0:W_t=-1\}]\mathbb{P}[Z-B_1<-1]=\mathbb{E}[\inf\{t\ge0:W_t=1\}]\mathbb{P}[|Z-B_1|>1]$ which is the product of an infinite term with a non-zero term.

Angela Pretorius
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  • Thank you! You get the bounty, but I'm having a realy hard time following your arswer! First I cannot understand your motivation for why $P[|Z-B_1|>1] \ge P[|B_1|>1] $, is it possible to phrase this in some other way? Secondly we have that $B_{t+1 } -B_1 \sim B_t $ so wouldn't we have to put $\inf {t \ge 0 : W_t = Z + B_1 } $ for it to have the same distribution as $T $? – MrFranzén Oct 31 '19 at 13:09
  • For your first question, I'm not sure how to explain this but the value of $x$ which minimises $\mathbb{P}[|x-B_1|>1]$ is $x=0$ because the distribution of $B_1$ is centered around zero. For your second question, I realise that i have made an error and I have edited my answer. Thanks for pointing that out. – Angela Pretorius Oct 31 '19 at 16:36
  • Reading this again today, I still cannot get my head around how $\inf {t \ge 0: B_t = Z } = \inf {t \ge 0: W_t = Z - B_1 } $. Could you please tell how you reason here? – MrFranzén Apr 10 '20 at 05:34
  • The two stopping times are not actually equal, but they have the same distribution. I'll edit to make this a bit clearer. – Angela Pretorius Apr 10 '20 at 06:38
  • You claim $\inf {t \ge 0: B_t= Z} \overset{D}{=} \inf {t \ge 0: B_{t+1 } - B_1 = Z - B_1}$. There might be some steps in between but this seams to me to claim that $B_{t+1 } \overset {D } {= } B_{t } $, which isn't true. Or how do you derive this? – MrFranzén Apr 10 '20 at 07:37
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    I see what you mean. I think what I really meant to say was that $T-1=\inf{t\ge 1 : B_t=Z}-1=\inf{t\ge 0: B_{t+1}=Z}$. Let me know if you spot any more errors. – Angela Pretorius Apr 10 '20 at 17:53