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If $W_t$ is a standard Wiener process, Levy's Arcsine Law gives us the CDF of the random variable

$$ \int_0^1 \delta[W_t \ge 0] dt $$

(where $\delta[\text{event}]$ is $1$ if the event happens and $0$ otherwise) as $\frac{2}{\pi}\arcsin(\sqrt{x}), x\in[0,1]$. I am aware there is a generalization to $W(t)\ge\alpha$ instead of just $W(t)\ge 0$, and also a generalization to Brownian Motion with drift (i.e. $W_t+\mu t$ where $\mu$ is some constant).

Is there a similar formula for Brownian Bridges? I.E. if $B_t, t\in[0,1]$ is a Brownian Bridge from $B_0=0$ to $B_1=b$ (the simplest case would be $b=0$, but hopefully we can generalize to non-horizontal bridges), can we say anything about the distribution of the following random variable?

$$ \int_0^1 \delta[B_t \ge \alpha] dt $$

I know if $b=0,\alpha=0$ then the distribution is uniform. There might be a way to transform the drifted Brownian motion result into a Brownian Bridge result, but I couldn't think of such an idea. (In particular, the Bridge distribution for $b=0, \alpha=0$ does not coincide with the Motion distribution for $\mu=0, \alpha=0$).

Juan Carlos Ortiz
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  • I also tried using the standard proof for Levy's Law using the Feynman-Kac formula and applying the same idea to this problem, but I wasn't successful. – Juan Carlos Ortiz Feb 16 '24 at 17:51
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    I saved this to try to find answer. Some useful resources seem to be: 1)last few pages in "Arcsin Laws and Feynman-Kac " https://sites.stat.washington.edu/jaw/COURSES/520s/523/HO.523.20/Feynman-Kac-Arcsin-v2.pdf .They mention arcsin distributions for Bridges.

    Another is "The distribution of local times of a brownian bridge" by J.Pitman. He goes over the explicit law for local time.

     – Thomas Kojar Feb 22 '24 at 22:18

1 Answers1

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Potential approach: We study the occupation time $T_{r}:=\int_{0}^{1}1_{B_{t}> r}dt$ following the notes Arcsin Laws and Feynman-Kac for Brownian bridge. As mentioned in Transition density of Brownian bridge using generators, the generator is

$$A(f)=\frac{-x}{1-t}f'+\frac{1}{2}f''$$

and the transition density satisfies

$$-\frac{x}{1-t}\frac{\partial }{\partial x}p(t,x)+\frac{1}{2}\frac{\partial^{2} }{\partial^{2} x}p(t,x)=\frac{\partial }{\partial t}p(t,x).$$

So following Arcsin Laws and Feynman-Kac, we have the analogous equation

$$u_{t}=\left\{\begin{matrix} \frac{1}{2}u_{xx}-\frac{x}{1-t}u_{x}-\lambda u&x> r\\\frac{1}{2}u_{xx} -\frac{x}{1-t}u_{x}&x\leq r \end{matrix}\right.$$

Using again the Laplace transform $\hat{u}(\alpha,x)$, we get a PDE system: for $x> r$

$$-1+\alpha\hat{u}(\alpha,x)+\frac{d\hat{u}(\alpha,x)}{d\alpha}= \frac{1}{2}(\hat{u}_{xx}(\alpha,x)+\frac{d\hat{u}_{xx}(\alpha,x)}{d\alpha})-x\hat{u_{x}}(\alpha,x)-\lambda (\hat{u}(\alpha,x)+\frac{d\hat{u}(\alpha,x)}{d\alpha}) ,$$

and $x\leq r$ $$-1+\alpha\hat{u}(\alpha,x)+\frac{d\hat{u}(\alpha,x)}{d\alpha}= \frac{1}{2}(\hat{u}_{xx}(\alpha,x)+\frac{d\hat{u}_{xx}(\alpha,x)}{d\alpha})-x\hat{u_{x}}(\alpha,x).$$

Since this is again a PDE of similar form, it is unclear if this can be solved explicitly. If anybody has some suggestions, I will be happy to try them out.

Thomas Kojar
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  • Hi Thomas, thanks for your insight. Sorry if I'm being silly here, but where exactly are we using a Brownian Bridge as opposed to just Brownian Motion? I understand you're changing the "cutoff" from $0$ to $r$ in your equations, which is where we get $A$ and $B$ from. But where exactly are we conditioning on $W_1=0$ (or something else to get a bridge, like using $W_t-tW_1$)? – Juan Carlos Ortiz Feb 27 '24 at 18:38
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    @JuanCarlosOrtiz, you are right, I forgot to add the term involving the generator for Brownian bridge. I will try a bit but might delete. – Thomas Kojar Feb 27 '24 at 21:12
  • No worries, thanks for the clarification. By the "generator", do you mean the $-tW_1$ term, or something more ellaborate? – Juan Carlos Ortiz Feb 27 '24 at 21:27
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    @JuanCarlosOrtiz see updated answer – Thomas Kojar Feb 27 '24 at 21:28
  • Thanks, that's really helpful! Since the process is no longer time homogeneous like Brownian Motion, shouldn't the transition probabilities have two time arguments, not just one? And I guess two $x$-arguments as well? Or are you fixing the "left" time to be $0$ and the "left" $x$ to also be $0$? Thus $p(t,x)$ is the probability density of going from $t=0, B_t=0$ to $t=t, B_t=x$? – Juan Carlos Ortiz Feb 28 '24 at 15:40
  • It seems the main complication arising here, that doesn't arise in the standard (Motion) case, is solving a PDE (using $f:=\hat{u}$ for simplicity) that involves the terms $f, f_{xx}, f_{xx\alpha}, f_x, f_\alpha, 1$. Most of the coefficients are constants except for $xf_x$. This seems like a very difficult PDE to find a closed-form solution for, and even if we did, we'd have to "glue it" together at the boundary $x=0$ to the solution to $-1+\alpha f=\frac12 f_{xx}$ (which does have a closed-form solution). Maybe this problem is too hard to solve – Juan Carlos Ortiz Feb 28 '24 at 16:30
  • Specially because we now need partial derivatives of $f$ both in terms of $x$ and $\alpha$, so general ODE theory won't help, and because there's even the cubic term $f_{xx\alpha}$. As far as I know the general literature only offers numerical methods to solve this kinds of PDEs – Juan Carlos Ortiz Feb 28 '24 at 16:31
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    @JuanCarlosOrtiz yes that generator is for the excursion from zero to zero. – Thomas Kojar Feb 28 '24 at 19:43
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    It depends on your problem. This is a nice linear pde to simulate numerically. – Thomas Kojar Feb 28 '24 at 19:44
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    I would also look into the literature Feynmac Kac for Brownian bridge. To see how they study those pdes. – Thomas Kojar Feb 28 '24 at 19:45