If $T_{U,V}=\inf\{t:B_t \notin (U,V)\}$ does $\int E[B_{T_{U,V}}|U,V]dP=\int E[B_{T_{u,v}}]dP_{U,V}$

Question

I have the following question:

if $T_{U,V}=\inf\{t:B_t \notin (U,V)\}$ does $$\int E[B_{T_{U,V}}|U,V]dP=\int E[B_{T_{u,v}}]dP_{U,V}$$ - or if it makes the difference does $\int E[1_{\{-\infty ,x]} B_{T_{U,V}}|U,V]dP= \int E[1_{\{-\infty ,x]}B_{T_{u,v}}]dP_{U,V}$

The context is the following: $B_t $ is the one-dimensional Brownian motion and I want to calculate its distribution $P[B_{T_{U,V}} \le x]$ for a stopping time $T_{U,V}=\inf\{t:B_t \notin (U,V)\}$. To do se we condition on the random interval $(U,V)$ to be able to use the known idstribution of $(U,V) $ via the result cited below. $P_{U,V } $ denotes the distribution of the random vector $(U,V)$ for which we may assume that $U<0<V $.

I know the following result. Given two independent random variables $X,Y$ tanking values in the measurable spaces $(D,\mathcal E)$ and $(E,\mathcal E)$ respectively and a real valued, bounded $\mathcal D \otimes\mathcal E $-measurable function $\phi $ then, $$\int E[\phi(X,Y)|X]dP=\int E[\phi(x,Y)]d P_X(dx)$$

To apply this to the case above we would ned to write $\omega \mapsto B_{T_{U(\omega),V(\omega)} } (\omega) $ as a map of the form $\omega \mapsto \phi(B_t(\omega),T_{U(\omega),V(\omega) })$ where $\phi (f,t)=f(t) $ maps a function $f $ to its evaluation att the point $t $. [or should it be $\phi $ maps $(f,(u,v))$ to $f(\inf \{t:f(t)\notin (u,v) \})$?]

I have the following questions: (1) how should the map $\phi $ be formulated? and (2) how do we check the conditions it neeeds to satisfy?

Thanks in advance!

Answer 1

Since stopping times typically depend on a path of a process (and not only the value at some particular time), your formula for the conditional expectation is not too useful. However, there is a " functional" version of this formula which comes in handy; you can find it for instance in the book Brownian motion - an introduction to stochastic processes by Schilling & Partzsch (Lemma A.3).

Let $(\Omega,\mathcal{A})$ and $(S,\mathcal{S})$ be measurable spaces. Let $\mathcal{F},\mathcal{H}$ be sub-$\sigma$-algebras of $\mathcal{A}$ and let $X: \Omega \to S$, $\Psi: S \times \Omega \to \mathbb{R}$ be random variables with the following properties:

• $\mathcal{F}$ and $\mathcal{H}$ are independent,
• $X: (\Omega,\mathcal{F}) \to (S,\mathcal{S})$ is measurable,
• $\Psi: (S \times \Omega, \mathcal{S} \otimes \mathcal{H}) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ is measurable and bounded.

Then $$\mathbb{E}(\Psi(X(\cdot),\cdot) \mid \mathcal{F}) = h(X)$$ where $$h(x) := \mathbb{E}(\psi(x,\cdot)), \qquad x \in S.$$

Let's check that we can apply this result in your framework. Set

• $S := \mathbb{R}^2$ and $\mathcal{S} := \mathcal{B}(\mathbb{R}^2)$
• $X(\omega) := (U(\omega),V(\omega))$,
• $\Psi((u,v),\omega) := B_{T_{u,v}}(\omega)$,
• $\mathcal{F} := \sigma(X)$,
• $\mathcal{H}:=\sigma(B_t; t \geq 0)$.

By assumption, $\mathcal{F}$ and $\mathcal{H}$ are independent. Moreover, it is trivial that $X$ is $\sigma(X)$-measurable. Let's assume for the moment that we already know that $\Psi$ is measurable (as specified above) and that $\Psi$ is bounded which is e.g. satisfied if $U$ and $V$ are both bounded. Applying the above result we then get

$$\mathbb{E}(B_{T_{U,V}} \mid U,V) = \mathbb{E}(B_{T_{u,v}}) \mid_{(u,v) = (U,V)}$$

which implies in particular that

$$\int \mathbb{E}(B_{T_{U,V}} \mid U,V) \, d\mathbb{P} = \int \mathbb{E}(B_{T_{u,v}}) \, dP_{U,V}(u,v).$$

The assumption on the boundedness of $U$ and $V$ is quite natural since it ensures that all appearing integrals are well-defined.

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Proof of the measurability of $\Psi$: For $x \neq 0$ set

$$\tau_x := \inf\{t >0; B_t > \text{sgn}(x) \cdot x\}.$$ Let's first consider $x \geq 0$. Since $$\tau_x = \inf\{t>0, \sup_{s \leq t} B_s> x\}$$ we see that $x \mapsto \tau_x(\omega)$ is the generalized inverse of the increasing and right-continuous mapping $t \mapsto \sup_{s \leq t} B_s(\omega),$ and therefore $x \mapsto \tau_x$ is also right-continuous. As $\omega \mapsto \tau_x(\omega)$ is $\mathcal{H}$-measurable for each fixed $x \geq 0$, this implies that $$([0,\infty) \times \Omega, \mathcal{B}[0,\infty) \otimes \mathcal{H}) \ni (x,\omega) \mapsto \tau_x(\omega) \in ([0,\infty),\mathcal{B}[0,\infty))$$ is measurable (it's an approximation argument, e.g. as in the proof of joint measurability of Brownian motion). A similar reasoning applies for $x \leq 0$, and so we get that $$(\mathbb{R} \times \Omega, \mathcal{B}(\mathbb{R}) \otimes \mathcal{H}) \ni (x,\omega) \mapsto \tau_x(\omega) \in ([0,\infty),\mathcal{B}[0,\infty))$$is measurable. This entails that $$(\mathbb{R}^2 \times \Omega, \mathcal{B}(\mathbb{R}^2) \otimes \mathcal{H}) \ni ((u,v),\omega) \mapsto \tau_u(\omega) \in ([0,\infty),\mathcal{B}[0,\infty))$$ is measurable which, in turn, gives that

$$(\mathbb{R}^2 \times \Omega, \mathcal{B}(\mathbb{R}^2) \otimes \mathcal{H}) \ni ((u,v),\omega) \mapsto T_{u,v}(\omega)=\min\{\tau_u(\omega),\tau_v(\omega)\} \in ([0,\infty),\mathcal{B}[0,\infty)) \tag{1}$$ is measurable. On the other hand, also the mapping $$(\mathbb{R}^2 \times \Omega, \mathcal{B}(\mathbb{R}^2) \otimes \mathcal{H}) \ni ((u,v),\omega)\mapsto \omega \in (\Omega,\mathcal{H}) \tag{2}$$ is measurable. By basic properties of the product-$\sigma$-algebra we conclude that $$(\mathbb{R}^2 \times \Omega, \mathcal{B}(\mathbb{R}^2) \otimes \mathcal{H}) \ni ((u,v),\omega) \mapsto (T_{u,v}(\omega),\omega) \in ([0,\infty) \times \Omega,\mathcal{B}[0,\infty) \otimes \mathcal{H}) \tag{3}$$ is measurable. Finally, we recall that the Brownian motion is progressively measurable, i.e. $$([0,\infty) \times \Omega,\mathcal{B}[0,\infty) \otimes \mathcal{H}) \ni (t,\omega) \mapsto B_t(\omega) \in (\mathbb{R},\mathcal{B}(\mathbb{R})) \tag{4}$$ is measurable. Hence, the composition of the mappings in $(3)$ and in $(4)$ is measurable, i.e. $$(\mathbb{R}^2 \times \Omega, \mathcal{B}(\mathbb{R}^2) \otimes \mathcal{H}) \ni ((u,v),\omega) \mapsto B_{T_{u,v}}(\omega)$$ is measurable.