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 $$\mathbb P(\forall t\in (a,b), B_t\neq 0\mid B_a=x)=\mathbb P(\forall t\in (0,b-a), B_t\neq -x\mid B_0=0) \ \ ?$$

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

2 Answers 2


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 $a<b$. 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

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


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

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

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


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

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.

  • $\begingroup$ 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 ? $\endgroup$
    – John
    Commented Nov 24, 2019 at 10:17
  • $\begingroup$ 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. $\endgroup$
    – John
    Commented Nov 24, 2019 at 10:20
  • $\begingroup$ @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. $\endgroup$
    – saz
    Commented Nov 24, 2019 at 13:54
  • $\begingroup$ Is it $\tau_x := \inf \{t>0; W_t=x\}$? $\endgroup$ Commented Nov 26, 2019 at 14:38
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    $\begingroup$ @dafinguzman Ah, now I see what you mean :-). Thank you! $\endgroup$
    – saz
    Commented Nov 26, 2019 at 18:37

I can answer your second question:

$$\mathbb \{\forall t\in (a,b), B_t\neq 0\mid B_a=x\} = \bigcap_{t \in (a, b)}\{B_t\neq 0\mid B_a=x\} = \bigcap_{t \in (a, b)}\{B_t \in \{0\}^c\mid B_a=x\}.$$ 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

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

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    $\begingroup$ Could you justify the 3rd equality ? They are indeed equal in law, but I would be surprised if the equality hold as sets. $\endgroup$
    – Surb
    Commented Nov 21, 2019 at 17:42
  • $\begingroup$ You are absolutely right, my bad $\endgroup$ Commented Nov 21, 2019 at 17:46
  • $\begingroup$ Why $\mathbb P(B_{t}\neq 0\mid B_a=x)=\mathbb P(B_{t-a}\neq 0\mid B_0=x)$ implies that $\mathbb P(\forall t\in (a,b), B_t\neq 0\mid B_a=x)=\mathbb P(\forall t\in (0,b-a), B_t\neq 0\mid B_0=x)$ ? $\endgroup$
    – John
    Commented Nov 21, 2019 at 17:59
  • $\begingroup$ @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 $B_{t-a} = Bs$ $\endgroup$ Commented Nov 21, 2019 at 18:11
  • $\begingroup$ 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 $\mathbb P(\bigcap_{t\in (a,b)}\{B_t\neq 0\mid B_a=x\})=\mathbb P(\bigcap_{t\in (0,b-a)}\{B_t\neq 0\mid B_0=x\})$. $\endgroup$
    – John
    Commented Nov 21, 2019 at 18:21

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