Let $W_t$ be a standard Wiener process, and $0\leq a < b$. Let $\hat{W}_t:=W_{a+t}-W_a$. Then $\hat{W}_t$ is also a standard Wiener process. I think that the following should be true:

$$\mathbb P\left(\forall s\in[a,b], W_s\neq 0\right)=\mathbb P\left(\sup_{ 0\leq v\leq b-a} \hat {W}_v < |W_a|\right)=\mathbb P\left(\sqrt{b-a}|Y|\leq \sqrt{a}|X|\right),$$

where $X$ and $Y$ are independent standard normal random variables. I have tried several approaches, but I think I am missing the right one.


1 Answer 1


The event $\{\forall s\in[a,b], W_s\not=0\}$ is the disjoint union of $\{\forall s\in[a,b], W_s>0\}$ and $\{\forall s\in[a,b], W_s<0\}$, which have the same probability, by symmetry. The latter event is equal to as $$ \{\sup_{0\le t\le b-a}\hat W_t<-W_a\}, $$ and (by design) the supremum is independent of $W_a$ and has the law of the absolute value of a Brownian motion at time $b-a$; namely the normal distribution with mean $0$ and variance $b-a$. Therefore the probability of the displayed event is $$ P[\sqrt{b-a}|Y|<-\sqrt{a}X\,], $$ where $X$ and $Y$ are independent standard normals. The required probability is twice this last displayed probability, and this in turn is equal to $$ P[\sqrt{b-a}|Y|<\sqrt{a}|X|\,], $$ again using symmetry.


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