Let $(W_t)_{t\ge0}$ a Wiener process. We want to find $p:=\mathbb{P}(W_{t} \text{ has no zeroes on $[a,b]$})$.

I've considered

$$p = \mathbb{P}(W_{t} > 0, t \in [a,b]) + \mathbb{P}(W_{t} < 0, t \in [a,b]) = 2\mathbb{P}(W_{t} < 0) $$

How it's better to step then? It's better to use Bachelier theorem? It would be great to some hint , to start with.

  • 1
    $\begingroup$ I don't see how you got a $\max$ in there after your final equal sign. I would say $\mathbb{P}(W_{t} > 0) + \mathbb{P}(W_{t} < 0) = \mathbb{P}(W_t \ne 0) = 1$. $\endgroup$ – GEdgar Mar 12 at 13:49
  • $\begingroup$ @GEdgar added explanation $\endgroup$ – openspace Mar 12 at 13:54

Approach I: By the continuity of the sample paths of Brownian motion, we have

$$\mathbb{P}(W_t \neq 0 \, \, \text{for all $t \in [a,b]$}) = \mathbb{P}(W_t < 0 \, \, \text{for all $t \in [a,b]$}) + \mathbb{P}(W_t > 0 \, \, \text{for all $t \in [a,b]$}).$$

By the symmetry of Brownian motion, this implies

$$\mathbb{P}(W_t \neq 0 \, \, \text{for all $t \in [a,b]$}) = 2\mathbb{P}(W_t < 0 \, \, \text{for all $t \in [a,b]$}). \tag{1}$$

Following the reasoning in this answer (with $[1,2]$ replaced by $[a,b]$) we obtain that

$$\mathbb{P}(W_t < 0 \, \, \text{for all $t \in [a,b]$}) = \mathbb{E} \left( \left[ 1-2\Phi\left(\frac{W_a}{\sqrt{b-a}} \right) \right] 1_{\{W_a<0\}} \right)$$ where $\Phi$ is the cdf of the centered Gaussian distribution with density $\varphi$. Hence,

$$\mathbb{P}(W_t < 0 \, \, \text{for all $t \in [a,b]$}) = \frac{1}{2} - 2 \int_{-\infty}^0 \Phi \left( \frac{\sqrt{a}}{\sqrt{b-a}} x \right) \varphi(x) \, dx.$$

Computing the latter integral (e.g. as in this answer) we conclude that

$$\mathbb{P}(W_t < 0 \, \, \text{for all $t \in [a,b]$}) = \frac{1}{\pi} \arctan \left( \frac{\sqrt{a}}{\sqrt{b-a}} \right).$$

Finally, by $(1)$,

$$\mathbb{P}(W_t \neq 0 \, \, \text{for all $t \in [a,b]$}) = \frac{2}{\pi} \arctan \left( \frac{\sqrt{a}}{\sqrt{b-a}} \right).$$

Approach II: (Following the proof in the monograph Brownian motion - An introduction to stochastic processes by Schilling & Partzsch, Lemma 11.23, 2nd edition)


$$\mathbb{P}(W_t \neq 0 \, \, \text{for all $t \in [a,b]$})= 1- \mathbb{P}(\exists t \in [a,b]: W_t = 0), \tag{2}$$

it clearly suffices to calculate tha latter probability. By the Markov property of Brownian motion, we have

$$ \mathbb{P}(\exists t \in [a,b]: W_t = 0) = \mathbb{E} \left( \mathbb{P}^{W_a}(\tau_0<b-a)\right)$$


$$\tau_0 := \inf\{t \geq 0; B_t = 0\}$$

is the hitting time of a Brownian motion $(B_t)_{t \geq 0}$ started at $B_0 = x$ under $\mathbb{P}^x$. Applying the reflection principle, we get

$$ \mathbb{P}(\exists t \in [a,b]: W_t = 0) = \mathbb{E} \left( \int_0^{b-a} \frac{|W_a|}{\sqrt{2\pi s^3}} \exp \left( - \frac{W_a^2}{2s} \right) \, ds \right).$$

Using $W_a \sim N(0,a)$ and applying Tonelli's theorem, we obtain that

\begin{align*} \mathbb{P}(\exists t \in [a,b]: W_t = 0) &= \frac{1}{2\pi} \int_0^{b-a} \int_{\mathbb{R}} \frac{|y|}{\sqrt{a s^3}} \exp \left( - \frac{1}{2} \left( \frac{1}{s} + \frac{1}{a} \right) y^2 \right) \, dy \, ds \\ &= \frac{1}{\pi} \int_0^{b-a} \frac{\sqrt{a}}{\sqrt{s} (s+a)} \, ds. \end{align*}

By a change of variables ($s=au^2$) this gives

$$\begin{align*} \mathbb{P}(\exists t \in [a,b]: W_t = 0) &= \frac{2}{\pi} \int_0^{\sqrt{(b-a)/a}} \frac{du}{1+u^2} \\ &= \frac{2}{\pi} \arctan \sqrt{\frac{b-a}{a}}. \end{align*}$$

As $$\arctan x = \frac{\pi}{2} - \arctan \frac{1}{x}, \qquad x>0,$$

we conclude from $(2)$ that

$$\mathbb{P}(W_t \neq 0 \, \, \text{for all $t \in [a,b]$}) = \frac{2}{\pi} \arctan \left( \frac{\sqrt{a}}{\sqrt{b-a}} \right).$$

  • $\begingroup$ Factor $\sqrt{\frac{a}{b - a}}$ is normalizing factor for dispersion ? $\endgroup$ – openspace Mar 12 at 14:45
  • $\begingroup$ @openspace What do you mean by dispersion in this context? If $a$ is small, then $W_a$ is (with high probability) still close to $0$ and hence the probability that there are no zeros in $[a,b]$ is relatively small ... this is encoded in the factor $\sqrt{a}$. The factor $\sqrt{b-a}$ is obviously a measure for the length of the time interval... the larger the time interval, the smaller the probability that there are no zeros. (The square root is coming from the scaling property, i.e. $W_t = \sqrt{t} W_1$ in distribution.) $\endgroup$ – saz Mar 12 at 14:53
  • $\begingroup$ I mean variance $\endgroup$ – openspace Mar 12 at 14:56
  • $\begingroup$ @openspace variance of what....? $\endgroup$ – saz Mar 12 at 14:58
  • $\begingroup$ Okey, this factor arises because of $Ф$ is normal cdf? $\endgroup$ – openspace Mar 12 at 15:01

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