Prove that $P[B_{\tau_2 } > B_{\tau_1 } | B_{\tau_1 } ] = \frac{B_{\tau_1 } - f_2(B_{\tau_1 } , -1 )}{f_2(B_{\tau_1 } ,1 ) - f_2(B_{\tau_1 } , -1 )}$

Question

In what follows $B_t $ denotes a Brownian Motion.

My first question concerns the boject

$$P[B_{\tau_2 } > B_{\tau_1 } | B_{\tau_1 } ]$$

What probabilistic object is this referring to? Some regular conditional distribution?

For the main question the context is that we have a a stopping time $\tau_1 := \inf \{t>0 : \ B_t \in \{a, b\} \} $ and a measurable function $f_2: \ \mathbb R \times \{-1 , 1\} \to \mathbb R$ and have defined [it is assumed that $f_2( \bullet , -1 ) < f_2( \bullet, 1 ) $]

$$\tau_2 := \inf \{t>0 : \ B_t \in \{f_2(B_{\tau_1 }, -1), f_2(B_{\tau_1 } , 1) \} \} $$

Then it is claimed - referring to the Strong Markov property - that

$$P[B_{\tau_2 } > B_{\tau_1 } | B_{\tau_1 } ] = \frac{B_{\tau_1 } - f_2(B_{\tau_1 } , -1 )}{f_2(B_{\tau_1 } ,1 ) - f_2(B_{\tau_1 } , -1 )}$$

I think with some reference to the (known) fact that for $a < 0 < b $ and $\tau_{a, b } := \inf \{t>0 : B_t \in \{a, b \} \} $

$$P[B_{\tau_{a, b } } = b ]= \frac{a}{b-a}$$

How can I prove this?

Any help would be much appreciated!

Answer 1

What probabilistic object is this referring to? Some regular conditional distribution?

"Regularity" corresponds to the property of $\mathrm P(A\mid \mathcal C)$ as a function of $A$ (it should be a measure). Here, you have a lone event. So I the term "regular conditional distribution" is not relevant.

Per the question, thanks to the strong Markov property of $B$, $\{B_{t+\tau_1} - B_{\tau_1},t\ge 0\}$ is independent of $B_{\tau_1}$ and has the same distribution as $B$. Now we use the following property of conditional expectation (I'm not aware of any name for it):

If $X$ and $Y$ are independent, then, for any nice (say, bounded) jointly measurable function $f(x,y)$, $$ \mathrm E[f(X,Y) \mid Y] = \mathrm E[f(X,y)]\big|_{y=Y}. $$

Define $$ A(y) = \bigl\{x\in C([0,\infty)) \mid \exists t>0:\\ x(t) = f_2 (y,1) - y, \forall s\in [0,t], x(s) > f_2 (y,-1) - y\bigr\} $$ Then, $$ \mathrm P(B_{\tau_2} - B_{\tau_1} > 0 \mid B_{\tau_1} ) = \mathrm P( B_{\cdot +\tau_1} - B_{\tau_1} \in A(B_{\tau_1}) \mid B_{\tau_1} ). $$ Using the independence and the above property, $$ \mathrm P(B_{\tau_2} - B_{\tau_1} > 0 \mid B_{\tau_1} ) = \mathrm P( B_{\cdot +\tau_1} - B_{\cdot} \in A(y) )\big|_{y =B_{\tau_1}} = \mathrm P( B \in A(y) )\big|_{y =B_{\tau_1}}, \tag{1} $$ where the last equality holds thanks to the strong Markov property.

Now, $$ \mathrm P( B \in A(y) ) = \mathrm{P} (B_{\tau_{a,b}} = b) = \frac{-a}{b-a} $$ with $a = f_2(y,-1)-y$, $b = f_2(y,1)-y$ (you have a mistake here, as the probability can't be negative). Combining this with (1), $$ \mathrm P(B_{\tau_2} - B_{\tau_1} > 0 \mid B_{\tau_1} ) = \frac{B_{\tau_1} - f_2(B_{\tau_1},-1)}{f_2(B_{\tau_1},1) - f_2(B_{\tau_1},-1)}, $$ as required.