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I am interested in the distribution of the time that the standard Brownian $W_t$ motion on $[0,1]$ satisfies the following inequality: $$W_t \ge stW(1)$$ For different values of $s$. I conjecture that the distribution is always a Beta distribution with both parameters equal (if they are Beta distributed they have to be equal because by symmetry the expected value of this time should be equal to $\frac{1}{2}$.
There are two special cases in which I can tell that the above is true: if $s=0$ then we have the usual question about the distribution of time BM spends above the $x$-axis in which case the answer is the $B(\frac{1}{2},\frac{1}{2})$ distribution. If $s=1$ the inequality can be transformed into: $$B_t=W_t-tW(1) \ge 0$$ Which asks about the distribution of the time that the Brownian bridge spends above the $x$-axis in which case the answer is $B(1,1)$ - the uniform distribution.
From the simulation I have conducted it seems that the result is true in general with $1$ being the highest parameter value. However, I don't have any proof nor any clue how to proceed.

Bartek
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  • This paper might be related: https://www.cambridge.org/core/journals/journal-of-the-london-mathematical-society/article/beta-variables-as-times-spent-in-0-by-certain-perturbed-brownian-motions/7F9593617DAF563E298A138A77E0FC6D# – Angela Pretorius Jul 07 '19 at 19:07
  • In my opinion, this is a good question for MathOverflow – Christophe Leuridan Sep 09 '22 at 19:12
  • The Brownian bridge defined by $B_t=W_t-tW_1$ is independent of $W_1$. Hence, if for each real number $\lambda$, we know the distribution of $|{t \in [0,1] : B_t> \lambda t}|$, we only need to can integrate it with regard to $P[(s-1)W_1 \in d\lambda]$. But I have no idea of how to find the distribution of $|{t \in [0,1] : B_t> \lambda t}|$. – Christophe Leuridan Sep 14 '22 at 18:17

1 Answers1

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Perhaps it may help to connect this process to a time-changed Wiener process so that you can more easily use the Wiener arcsine laws. Let

$$X_t = W_t - stW_1.$$

This is a Gaussian process with covariance function

$$c\left(u,t\right) = \min\left(u,t\right) - ut s\left(2-s\right).$$

Then the process

$$B_t = \left(1+ts\left(2-s\right)\right)X_{\frac{t}{1+ts\left(2-s\right)}}$$

is a Wiener process. If you know how much time $B$ spends above zero in various time intervals, can that help you say something about $X$?

sven svenson
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