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I am having a hard time with the intuition behind some of the representation theorems dealing with Brownian Motion. I think if someone can simply explain the intuition behind this theorem then everything will fall into place:

Theoreom:

For any $0\le a <b$ and any finite random variable $X\in\mathscr{F}_a$, there is a stopping time $\tau$ with $a\le\tau<b$ such that \begin{align*}X=\int_a^{\tau}\frac{1}{b-t}dB_t\end{align*}

That is the theorem. One thing I am having a massive problem with is understanding how this can be true since the right side of the above equality is a Gaussian process, but I don't see why $X$ needs to be a Gaussian Process. Can anyone explain this?

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2 Answers 2

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For any deterministic $T>0$ the random variable

$$Y_T := \int_a^T \frac{1}{b-t} \, dB_t$$

is Gaussian, but for a stopping time $\tau$ the integral

$$Y_{\tau} = \int_a^{\tau} \frac{1}{b-t} \, dB_t$$

does not need to be Gaussian. Just consider Brownian motion itself: For each fixed $T>0$ we know that $B_T$ is Gaussian, but this does not imply that $B_{\tau}$ is Gaussian for any stopping time $\tau$ (consider e.g. $\tau := \inf\{t>0; B_t = 1\}$, then $B_{\tau}=1$ almost surely).

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  • $\begingroup$ Ah ok, that makes a lot more sense actually. In the book I am reading the author says that $\tau =inf\{t>0: Y_t=X\}$ must be less than $b$ and the reasoning is that $limsup_{t\to\infty}Y_t=\infty$ and $liminf_{t\to\infty}Y_t=-\infty$. Why are these two facts enough to show that? $\endgroup$ Commented Jul 29, 2016 at 13:51
  • $\begingroup$ @MathStudent I'm confused with your notation. What is $Y_t$...? If you use the same notation as I do in my answer, then your definition of $\tau$ doesn't make sense (because it is self-redundant)... or is this $\tau$ different from the $\tau$ in the definition of $X$...? $\endgroup$
    – saz
    Commented Jul 29, 2016 at 14:04
  • $\begingroup$ Sorry, it's the same as the $Y_t$ you wrote for $t\ge a$ and 0 for $t<a$. The claim in the proof I am looking at says that the $\tau$ I wrote in the first comment is the $\tau$ we are looking for, and that as long as we can show that $P(\tau<b)=1$ then we are done. $\endgroup$ Commented Jul 29, 2016 at 14:07
  • $\begingroup$ @MathStudent What do you mean by "the $\tau$ we are looking for"...? In the definition of $\tau$ there is $X$ (which, I take it, is defined in your opening post)... however in the definition of $X$ there is also a $\tau$... which means that the definition is kind of circular. Do you see why I am confused? $\endgroup$
    – saz
    Commented Jul 29, 2016 at 14:21
  • $\begingroup$ $\tau$ is not defined in my post. The theorem is stating that for any random variable $X\in\mathscr{F}_a$ there is a $\tau$ such that the $X=\int_a^\tau\frac{1}{b-t}dB_t$. $X$ can be any random variable that is at least $\mathscr{F}_a$ measurable. $\endgroup$ Commented Jul 29, 2016 at 14:25
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Consider the process $M_t:=\int_a^{a+t}{1\over b-s}\,dB_s$, $0\le t<b-a$. This is a martingale (of the shifted filtration $(\mathscr F_{a+t})_{0\le t<b-a}$) with (deterministic) quadratic variation $q(t):=\int_a^{a+t}(b-s)^{-2}\,ds$, which increases to $+\infty$ as $t$ increases to $b-a$. As such, by the Dambis-Dubins-Schwarz theorem, the process $Z_s:=M_{q^{-1}(s)}$, $0\le s<\infty$, is a standard Brownian motion. Consequently, $\limsup_{s\to\infty}Z_s=+\infty$ and $\liminf_{s\to\infty}Z_s=-\infty$. And therefore $\limsup_{t\uparrow b-a}M_t=+\infty$ and $\liminf_{t\uparrow b-a}M_t=-\infty$. Finally, define $\tau:=\inf\{u>a: M_{u-a}=X\}$. In view of the limit behavior of $M_t$ as $t\uparrow b-a$ just noted, $\tau\in[a,b)$ a.s. And because $X$ is $\mathscr F_a$-measurable, $\tau$ is a stopping time of the filtration $(\mathscr F_t)$. Finally, it is clear that $\int_a^\tau(b-s)^{-1}\,dB_s=M_{\tau-a}=X$.

(This representation is the basic idea behind an interesting paper of R.M. Dudley "Wiener Functionals as Ito Integrals (Annals of Probability, vol. 5 (1977) pp. 140-141); http://projecteuclid.org/download/pdf_1/euclid.aop/1176995898 )

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