It occurs in Durrett's proof of Skorokhod embedding that he needs the following.

Suppose that we have a Brownian motion $B_t$ in $1$-d that starts at $0$. (To be clear, it is not necessarily constructed on the Wiener space, but I do require it to be everywhere continuous. With that assumption, then we have that $T_{b, a}=\inf\{t\geq 0 : B_t \notin (a, b)\}$ is a stopping time wrt the standard right continuous (RC) filtration.) If $U, V$ are RVs with $U\geq0$ and $V\leq0$ then supposedly the function $T_{U, V}$ is measurable. (with respect to the original sigma algebra for which $U, V$ are RVs.) Notice that this does not make it a stopping time.

I need help with that.

He also then uses the optional stopping theorem for continuous RC martingales with RC filtrations. This theorem implies that if $T$ is the stopping time and $X_t$ is the martingale, then $X_T$ is measurable wrt the sigma algebra of the stopping time $T$ and also for all $t\geq0$ we have $X_T=E(X_t\mid \mathfrak{F}_T)$ on $\{T<t\}$ a.s.

Durrett only says that by conditioning on the values of $(U, V)$, we can apply this to our situation above to conclude that (By now he has already started assuming the measurability of $T_{U, V}$.)

$$E(T_{U, V})=E\lbrace E(T_{U, V}\mid (U, V))\rbrace=E(-UV).$$

He has already shown that $E(T_{b,a})=-ab$. Where I object is it seems like his manipulation is "treating $U, V$ as constant" just because they are the subject of being "conditioned upon." I don't understand this, especially since $U, V$ are allowed to take values in continua.

(For those with the book, see p. 384 and the surrounding.)


1 Answer 1


Let $X\in\mathcal X$ and $Y\in\mathcal Y$ be independent random elements on some joint probability space.

You can rewrite Fubini's theorem as: for every measurable function $f:\mathcal X\times \mathcal Y\to\mathbb R$ $$\mathbb E(f(X,Y)|X=x) =\mathbb Ef(x,Y).$$

So for your case as $B_{T_{U,V}}$ is a function of the two random elements $B_t\in C^0(\mathbb R)$ and $(U,V)\in\mathbb R^2$ we have

$$\mathbb E\left(\left. B_{T_{U,V}} \right|U=u,V=v\right) = \mathbb E\left(B_{T_{u.v}} \right). $$

  • $\begingroup$ I'm a bit confused about how this applies here, but first of all does it imply measurability? $\endgroup$
    – Jeff
    May 30, 2013 at 17:03
  • $\begingroup$ @Jeff Measurability of what with respect to what? $\endgroup$
    – Tim
    May 30, 2013 at 17:40
  • $\begingroup$ $T_{U, V}$ with respect to the underlying sigma algebra. $\endgroup$
    – Jeff
    May 30, 2013 at 17:43
  • $\begingroup$ $[T_{U,V} <t] = \bigcup_{\alpha,\beta\in\mathbb Q} [\alpha < V < U < \beta]\cap[T_{\beta,\alpha}<t]$ $\endgroup$
    – Tim
    May 30, 2013 at 17:53
  • $\begingroup$ I don't think this is quite right. I'm toying with replacing your union with intersections, and strict with nonstrict inequalities. As it stands, the problem is that perhaps some path exist $(U, V)$ but not anything more. Don't forget that because this is a question of measurability, null sets matter. (Because there's not even a notion of null set until you put the probability measure in.) So BMs are just paths that start somewhere, and they can have local max/min. $\endgroup$
    – Jeff
    May 30, 2013 at 19:42

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