6
$\begingroup$

I know from Karatzas & Shreve (1991) that a Brownian Bridge $B(t)$ from $a$ to $b$ on time interval $[0,T]$ satisfies:

$$B(t)=a(1-t/T) + b*t/T + [W(t) - W(T)*t/T]$$

where $W(t)$ is a standard one-dimensional Brownian motion.

By the above equation we can get its distribution.

My question is what's the distribution of the Brownian Bridge $B(t)$ from $a$ to $b$ on time interval $[T_1, T_2]$?

Any idea or reference?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ The starting point doesn't matter. You're conditioning on the Brownian Motion being $a$ at time $T_1$, so there's no variance there. Just do the calculation on $[0, T_2 - T_1]$. $\endgroup$ – quasi Apr 22 '14 at 23:14
  • $\begingroup$ Thanks for your reply. But what about t? t should be on time interval [T1, T2]. So I think do the calculation with t -> t-T1 and T -> T2-T1. $\endgroup$ – Jack Apr 30 '14 at 21:13
2
$\begingroup$

A Brownian bridge can be built simply from a forced process. For example, if we define the process $Z$ by

$$ Z_{t}=\left(\dfrac{T-t}{T-t_0}\right)\left(a-W_{t_0}\right)+\left(\dfrac{t-t_0}{T-t_0}\right)\left(b-W_{T}\right)+W_{t}\;\;\;; t\in[t_0,T] $$

Where $W$ is a standard Brownian motion.

The process $Z$ is a Brownian bridge satisfying the following conditions: $Z_{t_0}=a$ and $Z_T=b$, in particular, if we put $t_0=0$, Then the process $Z$ becomes $$ Z_{t}=W_{t}+a-\frac{t}{T}\left(W_{T}-b+a\right) $$

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.