# Why $\mathbb P(\forall s\in [a,b], B_s\neq 0\mid B_a=x)=\mathbb P(\forall s\in [0,b-a], B_s\neq -x\mid B_0=0)$?

I want to compute $$\mathbb P(\forall s\in (a,b), B_s\neq 0),$$ but I already don't see why $$\{\forall s\in (a,b), B_s\neq 0 \}$$ is $$\mathcal F_b$$ measurable. Also, I know that if $$t>s$$ then $$\mathbb P(B_t\in A\mid B_s=x)=\mathbb P(B_{t-s}\in A\mid B_0=x).$$ By why from this we can get $$\mathbb P(\forall t\in (a,b), B_t\neq 0\mid B_a=x)=\mathbb P(\forall t\in (0,b-a), B_t\neq -x\mid B_0=0) \ \ ?$$

• For your first question, using continuity of the BM yields $$\{\forall t\in [a,b], B_t\neq 0\}=\bigcap_{q\in \mathbb Q}\{B_q\neq 0\}.$$
– Surb
Commented Nov 21, 2019 at 17:41

First of all, mind that we are not dealing with classical conditional probabilities, that is, $$\mathbb{P}(A \mid B_s=x) \neq \frac{\mathbb{P}(A \cap \{B_s=x\})}{\mathbb{P}(B_s=x)}.\tag{1}$$

This is pretty obvious from the fact that $$\mathbb{P}(B_s=x)=0$$ for all $$x \in \mathbb{R}$$, $$s>0$$. The expression on the left-hand side of $$(1)$$ is defined as follows: It follows from properties of the conditional expectation that there exists a measurable mapping $$g:\mathbb{R} \to \mathbb{R}$$ such that $$\mathbb{P}(A \mid B_s) := \mathbb{P}(A \mid \sigma(B_s))=g(B_s) \quad \text{a.s.}$$ Now one defines $$\mathbb{P}(A \mid B_s = x) := g(x).$$ (This definition is not specific for Brownian motion; more generally, this is the way to define $$\mathbb{P}(A \mid X)$$ for an event $$A$$ and some random variable $$X$$).

Let's return to your original problem. Fix some Borel set $$A$$ and some $$a. Recall that the restarted process

$$W_t := B_{t+a}-B_a, \qquad t \geq 0,$$

is a Brownian motion which is independent of $$(B_t)_{t \leq a}$$. Clearly,

$$B_t(\omega) \neq 0 \, \, \text{for all t \in (a,b)} \iff W_t(\omega) \neq -B_a(\omega) \, \, \text{for all t \in (0,b-a)}.$$

If we set $$\tau_x := \inf\{t>0; W_t=x\}$$, we can write this equivalently as

$$B_t(\omega) \neq 0 \, \, \text{for all t \in (a,b)} \iff \tau_{-B_a(\omega)}(\omega)\geq b-a.$$

Hence,

$$\mathbb{P}\left(B_t \neq 0 \, \, \text{for all t \in (a,b)} \mid B_a \right) = \mathbb{P}(\tau_{-B_a} \geq b-a \mid B_a).$$

Since $$\omega \mapsto \tau_{-x}(\omega)$$ is independent of $$\mathcal{F}_a^B:=\sigma(B_s; s \leq a)$$ for all $$x \in \mathbb{R}$$ and $$B_a$$ is $$\mathcal{F}_a^B$$-measurable, it follows that

$$\mathbb{P}(\tau_{B_a} \geq b-a \mid B_a) = g(B_a)$$

where

$$g(x) = \mathbb{P}(\tau_{-x} \geq b-a); \tag{2}$$

see e.g. the result which I stated in the first part of this answer. Combining the previous two identites, we get $$\mathbb{P}\left(B_t \neq 0 \, \, \text{for all t \in (a,b)} \mid B_a \right)=g(B_a)$$

for $$g$$ defined in $$(2)$$. Hence,

$$\mathbb{P}\left(B_t \neq 0 \, \, \text{for all t \in (a,b)} \mid B_a=x \right)=g(x), \tag{3}$$

as I explained earlier. By definition,

\begin{align*} g(x) = \mathbb{P}(\tau_{-x} \geq b-a) &= \mathbb{P} \left(W_t \neq -x \, \, \text{for all t \in (0,b-a)}\right) \end{align*}

Using the fact taht both $$(B_t)_{t \geq 0}$$ and $$(W_t)_{t \geq 0}$$ are Brownian motions, which means in particular that they are continuous with probability $$1$$ and have the same finite dimensional distributions, we conclude that

\begin{align*} g(x) = \mathbb{P} \left( \bigcap_{q \in \mathbb{Q} \cap (a,b)} \{W_q\neq-x\} \right) &= \mathbb{P} \left( \bigcap_{q \in \mathbb{Q} \cap (a,b)} \{B_q\neq-x\} \right) \\ &= \mathbb{P} \left(B_t \neq -x \, \, \text{for all t \in (0,b-a)}\right)\end{align*}

Plugging this into $$(3)$$ proves the assertion.

• Thank you for your answer. Three questions : 1) the link you put under the formula (2) doesn't work. I guess, it's a proof of $\mathbb E[X\mid \mathcal G]=\mathbb E[X]$ whenever $X$ is independent of $\mathcal G$, no ? 2) When you write $\mathbb P(\tau_{-x}\geq b-a)$ or $\mathbb P(B_t\neq -x \text{ for all$t$})$, do you mean $\mathbb P(\tau_{-x}\geq b-a\mid B_0=0)$ and $\mathbb P(B_t\neq -x\text{ for all$t$}\mid B_0=0)$ ? 3) Where did you use Markov property ?
– John
Commented Nov 24, 2019 at 10:17
• Notice that the proof where this is used is in the proof of arc-sine law in the book Brownian motion of Schilling and Partzsch.
– John
Commented Nov 24, 2019 at 10:20
• @John 1) I fixed the broken link. 2) Note that the Brownian motion starts at $B_0=0$ with probability $1$, and so $\mathbb{P}(B_0=0)=1$. For this reason, we have $\mathbb{P}(A) = \mathbb{P}(A \mid B_0=0)$ for any event $A$, i.e. it doesn't matter whether we condition on $B_0=0$ or not. 3) The proof uses that the restarted Brownian motion $(W_t)_{t \geq 0}$ is a Brownian motion and independent of $(B_t)_{t \leq a}$. That's where the Markov property pops up.
– saz
Commented Nov 24, 2019 at 13:54
• Is it $\tau_x := \inf \{t>0; W_t=x\}$? Commented Nov 26, 2019 at 14:38
• @dafinguzman Ah, now I see what you mean :-). Thank you!
– saz
Commented Nov 26, 2019 at 18:37

$$\mathbb \{\forall t\in (a,b), B_t\neq 0\mid B_a=x\} = \bigcap_{t \in (a, b)}\{B_t\neq 0\mid B_a=x\} = \bigcap_{t \in (a, b)}\{B_t \in \{0\}^c\mid B_a=x\}.$$ Now, by translation invariance of $$B_t$$ we get that the law of $$B_t$$ started at $$B_a=x$$ is equal to the law of $$B_{t-s}$$ started at $$B_0=x$$, so $$\mathbf P \left( \cap_{t \in (a, b)}\{B_t \in \{0\}^c\}\mid B_a=x\right) = \mathbf P \left(\cap_{t \in (a, b)}\{B_{t-a} \in \{0\}^c\}\mid B_0=x\right)$$ This last set is equal to $$\bigcap_{t \in (0, b-a)}\{B_t \in \{0\}^c\mid B_0=x\} \\= \{\forall t\in (0,b-a), B_t\neq 0\mid B_0=x\} \\= \{\forall t\in (0,b-a), B_t - x\neq -x\mid B_0-x=0\}$$

Now, the process $$B_t - x$$ with $$B_t$$ started at $$B_0 = x$$ has the same distribution as another brownian motion $$W_t$$ started at $$W_0 = 0$$, so the set above has a probability equal to

$$\mathbf P(\forall t\in (0,b-a), W_t \neq -x\mid W_0 =0).$$

• Could you justify the 3rd equality ? They are indeed equal in law, but I would be surprised if the equality hold as sets.
– Surb
Commented Nov 21, 2019 at 17:42
• You are absolutely right, my bad Commented Nov 21, 2019 at 17:46
• Why $\mathbb P(B_{t}\neq 0\mid B_a=x)=\mathbb P(B_{t-a}\neq 0\mid B_0=x)$ implies that $\mathbb P(\forall t\in (a,b), B_t\neq 0\mid B_a=x)=\mathbb P(\forall t\in (0,b-a), B_t\neq 0\mid B_0=x)$ ?
– John
Commented Nov 21, 2019 at 17:59
• @John if it's clear for you that $\mathbf P \left(\bigcap_{t \in (a, b)}\{B_t \in \{0\}^c\mid B_a=x\}\right) = \mathbf P \left(\bigcap_{t \in (a, b)}\{B_{t-a} \in \{0\}^c\mid B_0=x\}\right)$ then it amounts to changing the variable in the intersection, say put $s = t-a$, then $s\in(0, b-a)$ and $B_{t-a} = Bs$ Commented Nov 21, 2019 at 18:11
• No, this is not clear. Why this hold from $\mathbb P(B_t\neq 0\mid B_a=x)=\mathbb P(B_{t-a}\neq 0\mid B_0=x)$ ? Even, when $\mathbb P(A_1)=\mathbb P(A_2)$ and $\mathbb P(B_1)=\mathbb P(B_2)$ it's not true that $\mathbb P(A_1\cap A_2)=\mathbb P(B_1\cap B_2)$. So there is no reason for me that $\mathbb P(\bigcap_{t\in (a,b)}\{B_t\neq 0\mid B_a=x\})=\mathbb P(\bigcap_{t\in (0,b-a)}\{B_t\neq 0\mid B_0=x\})$.
– John
Commented Nov 21, 2019 at 18:21