Brownian Bridge First Passage Time via a Time-Changed Wiener Process
Summary
The document sketches a way to derive the first time a Brownian bridge from zero to a terminal level reaches that level. It first rescales the horizon to one and represents the bridge as a standard zero-to-zero bridge plus a linear term. A standard bridge can in turn be expressed using a Wiener process with a time change, which transforms the bridge’s hitting time into a transformed Wiener hitting time.
The Wiener hitting time has a Lévy density; applying the time transformation and its density adjustment yields a proposed density for the bridge passage time. The answer notes that the resulting density has an unusual shape for several example levels, but offers no independent validation. A second response points to a thesis as further reading. The derivation is a useful stochastic-process method, though the proposed formula is presented as a sketch and should be checked before use.
Key ideas
- A Brownian bridge ending at a level can be written as a linear term plus a standard bridge.
- A standard bridge admits a representation through a time-changed Wiener process.
- The bridge passage time is obtained by transforming the Wiener process hitting time.
- The proposed density follows from the Lévy hitting-time density and a change of variables.
- The derivation is only a sketch, and the proposed density is not independently verified in the document.
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Full text
# Brownian Bridge's first passage time distribution
# Brownian Bridge's first passage time distribution
Let's say we have a Brownian Bridge $Y_{b,T}(t)$ such that $Y_{b,T}(0)=0$, $Y_{b,T}(T)=b$.
Let's say we are interested in the first passage time of $Y_{b,T}(t)$ at level $b$: $\tau_b = \{\min \tau; Y_{b,T}(\tau)=b\}$.
How could I calculate the distribution of $\tau_b$?
## Answer by Kiwiakos (score 3, accepted)
https://quant.stackexchange.com/a/14219
What about this sketch of an answer: Let's put $T=1$ in your formula to simplify the notation. Then $Y_b(t)$ is a Brownian bridge where $Y_b(0)=0$ and $Y_b(1)=b$.
This can be written as $Y_b(t) = b\ t + Y_0(t)$, that is to say the standard Brownian bridge (from zero to zero) with an added drift $b\ t$.
The standard Brownian bridge can be written in terms of a time changed Wiener process $W$, namely $$ Y_0(t) = (1-t)\ W\left(\frac{t}{1-t}\right)$$
The hitting time $\tau$ that you are interested in can be expressed as $$\tau_{Y_b}(b) = \inf \{t : Y_b(t) = b\} = \inf\{t : b\ t + (1-t)\ W\left(\frac{t}{1-t}\right) = b \} = \inf\{t : W\left(\frac{t}{1-t}\right) = b \} $$
Hence, the hitting time of the Brownian bridge is the hitting time of a time changed Wiener process. That is to say, if $$\tau_W(b) = \inf\{s : W(s) = b \}$$ then $$\frac{\tau_{Y_b}(b)}{1-\tau_{Y_b}(b)} = \tau_W(b) \Rightarrow \tau_{Y_b}(b) = \frac{\tau_W(b)}{1+\tau_W(b)} $$
For a standard Wiener process the hitting time $\tau_W(b)$ follows a Levy distribution with density $$ f_W(\tau; b) = \frac{b}{\sqrt{2\pi\tau^3}} \exp \left\{- \frac{b^2}{2\tau} \right\}$$ hence the density of the hitting time of the Brownian bridge will be $$ f_{Y_b}(\tau; b) = \frac{b}{\sqrt{2\pi\tau^3(1-\tau)}} \exp \left\{- \frac{b^2(1-\tau)}{2\tau} \right\} $$
Hope this is right.
Edit: The density (if correct) for $b=\{0.25, 0.5, 1, 2\}$ looks quite funky actually!
## Answer by Plamen (score 1)
https://quant.stackexchange.com/a/31117
It may be overkill, but you may find the following PhD thesis by Peter Hieber of some use.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.