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?

  • 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, 2014 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, 2014 at 21:13

1 Answer 1


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 $$ Z_{t}=W_{t}+a-\frac{t}{T}\left(W_{T}-b+a\right) $$


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