Conditional probability for zeros of Brownian motion

Question

I wonder to know how to proof the following statement related with zeros of brownian movement in a given interval.

Let $a, b \in \mathbb{R}$, both having the same sign. Then prove that:
$$
\mathbb{P} [X_{s} \ne 0, \forall s ∈ (0,t) \mid X_0 = a, X_t = b ]= 1−e^{−2ab/t}
$$

That is, the probability that Brownian motion does not attain the value of zero in a given interval, where the motion is non null and has the same sign at the initial and final point of the interval.

I have tried to apply the reflection principle and the distribution of the minimum
$$
\mathbb{P}[M(t)\geq c]
= 2P[B(t)\geq c]
= 2 \frac{1}{\sqrt{2\pi t}} \int_c^{\infty} \exp \left(\frac{-x^2}{2t}\right)dx
$$

Answer 1

Let $(B_t)_{t \geq 0}$ be a Brownian motion (started at $B_0=0$) and denote by $M_t=\sup_{s \leq t} B_s$ and $m_t = \inf_{s \leq t} B_s$ the running maximum and minimum, respectively. The joint law of $(M_t,m_t,B_t)$ has a density with respect to Lebesgue measure, and this density can be calculated explicitly; this is known as 'Lévy's triple law'; you can find the result for instance in Chapter 6 in the Brownian motion book by Schilling & Partzsch. As a consequence of this, we can compute the density of the joint law of $(m_t,B_t)$:

Theorem: For $t>0$ the distribution of $(m_t,B_t)$ has a density with respect to Lebesgue measure; it is given by $$p_{m_t,B_t}(x,y) = - 2 \frac{2x-y}{\sqrt{2\pi t^3}} \exp \left( - \frac{(2x-y)^2}{2t} \right) 1_{\{x \leq 0\}} 1_{\{x \leq y\}} .$$

Now fix $a<0$ and $b>a$. It follows from the (a.s.) continuity of the sample paths of Brownian motion that

\begin{align*} \mathbb{P} \left( \forall s <t: B_s \neq a \mid B_t = b\right) &= \mathbb{P}(m_t>a \mid B_t = b). \end{align*}

Recalling that $$\mathbb{E}(f(X,Y) \mid Y=y) = \int f(x,y) p_{X \mid Y}(x|y) \, dy$$ where $p_{X \mid Y}(x|y):= \frac{p_{X,Y}(x,y)}{p_Y(y)}$ is the conditional density of $X$ given $Y$, it follows that

$$ \mathbb{P} \left( \forall s <t: B_s \neq a \mid B_t = b\right) = \int_a^0 q(x \mid b) \,dx$$where $$q(x \mid b) := \frac{p_{m_t,B_t}(x,b)}{p_{B_t}(b)}$$ is the conditional density of $m_t$ given $B_t$. Because of the above theorem, we have

$$q(x \mid b) = - 2 \frac{2x-b}{t} \exp \left( - \frac{2x^2-2bx}{t} \right) 1_{\{x \leq 0\}} 1_{\{x \leq b\}}.$$

Hence,

\begin{align*}\mathbb{P} \left( \forall s <t: B_s \neq a \mid B_t = b\right) &=-\frac{2}{t} \int_a^{\min\{0,b\}} (2x-b) \exp \left(-\frac{2x^2-2xb}{t} \right) \, dx \\ \end{align*}

If we set $y := 2x^2-2bx$, i.e. $dy/dx=4x-2b=2(2x-b)$, then we get

\begin{align*}\mathbb{P} \left( \forall s <t: B_s \neq a \mid B_t = b\right) &= - \frac{1}{t} \int_{2a^2-2ab}^{2(b \wedge 0)^2-2b (b \wedge 0)} \exp \left(-\frac{y}{t} \right) \, dy \\ &= 1-\exp \left( \frac{-2a^2+2ab}{t} \right). \tag{2}\end{align*}

Now take, say, $c,d>0$. By shifting the path of Brownian motion by $c$, we get

$$\mathbb{P}^c \left( \forall s \in (0,t): B_s \neq 0 \mid B_t = d\right) = \mathbb{P}^0(\forall s \in (0,t): B_s \neq -c \mid B_t = d-c).$$

Using $(2)$, we get

$$\mathbb{P}^c \left( \forall s \in (0,t): B_s \neq 0 \mid B_t = d\right) = 1-\exp \left( \frac{- 2 c^2 +2 (-c)(d-c)}{t} \right) = 1-\exp \left(-\frac{2cd}{t} \right).$$

This proves the assertion for positive numbers. If $c,d$ are both negative, then we can use the symmetry of Brownian motion, i.e. the fact that $W_t :=-B_t$ is also a Brownian motion, and this gives immediately the desired result.