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A Brownian bridge is a continuous-time B( t) whose probability distribution is the conditional probability distribution of a standard W( t) (a mathematical model of ) subject to the condition (when standardized) that W( T) = 0, so that the process is pinned to the same value at both t = 0 and t =  T. More precisely:

B_t := (W_t\mid W_T=0),\;t \in 0,T

The expected value of the bridge at any t in the interval 0,T is zero, with variance \frac{t(T-t)}{T}, implying that the most uncertainty is in the middle of the bridge, with zero uncertainty at the nodes. The of B( s) and B( t) is \min(s,t)-\frac{s\,t}{T}, or \frac{s(T-t)}{T} if s < t. The increments in a Brownian bridge are not independent.


Relation to other stochastic processes
If W(t) is a standard Wiener process (i.e., for t \geq 0, W(t) is normally distributed with expected value 0 and variance t, and the increments are stationary and independent), then

B(t) = W(t) - \frac{t}{T} W(T)\,

is a Brownian bridge for t \in 0,. It is independent of W(T) Aspects of Brownian motion, Springer, 2008, R. Mansuy, M. Yor page 2

Conversely, if B(t) is a Brownian bridge for t \in 0, and Z is a standard normal random variable independent of B, then the process

W(t) = B(t) + tZ\,

is a Wiener process for t \in 0,. More generally, a Wiener process W(t) for t \in 0, can be decomposed into

W(t) = \sqrt{T}B\left(\frac{t}{T}\right) + \frac{t}{\sqrt{T}} Z.

Another representation of the Brownian bridge based on the Brownian motion is, for t \in 0,

B(t) = \frac{T-t}{\sqrt T} W\left(\frac{t}{T-t}\right).

Conversely, for t \in 0,

W(t) = \frac{T+t}{T} B\left(\frac{Tt}{T+t}\right).

The Brownian bridge may also be represented as a Fourier series with stochastic coefficients, as

B_t = \sum_{k=1}^\infty Z_k \frac{\sqrt{2 T} \sin(k \pi t / T)}{k \pi}

where Z_1, Z_2, \ldots are independent identically distributed standard normal random variables (see the Karhunen–Loève theorem).

A Brownian bridge is the result of Donsker's theorem in the area of empirical processes. It is also used in the Kolmogorov–Smirnov test in the area of statistical inference.

Let K=\sup_{t\in0,1}|B(t)|, for a Brownian bridge with T = 1; then the cumulative distribution function of K is given by \operatorname{Pr}(K\leq x)=1-2\sum_{k=1}^\infty (-1)^{k-1} e^{-2k^2 x^2}=\frac{\sqrt{2\pi}}{x}\sum_{k=1}^\infty e^{-(2k-1)^2\pi^2/(8x^2)}.


Decomposition by zero-crossings
The Brownian bridge can be "split" by finding the last zero \tau_{-} before the midpoint, and the first zero \tau_{+} after, forming a (scaled) bridge over 0,\tau_{-}, an excursion over \tau_{-},\tau_{+}, and another bridge over \tau_{+},1. The joint pdf of \tau_{-},\tau_{+} is given by

\rho \left( \tau_{-},\tau_{+} \right) = \frac{1}{2 \pi \sqrt{\tau_{-} (1-\tau_{+}) (\tau_{+}-\tau_{-})^3 }}

which can be conditionally sampled as

\tau_{+} = \frac{1}{1+\sin^2 \left( \frac{\pi}{2} U_1 \right)} \in \left( \frac{1}{2}, 1\right)
\tau_{-} = \frac{U_2^2 \tau_{+}}{2\tau_{+} + U_2^2 -1} \in \left( 0,\frac{1}{2} \right)

where U_1,U_2 are uniformly distributed random variables over (0,1).


Intuitive remarks
A standard Wiener process satisfies W(0) = 0 and is therefore "tied down" to the origin, but other points are not restricted. In a Brownian bridge process on the other hand, not only is B(0) = 0 but we also require that B( T) = 0, that is the process is "tied down" at t = T as well. Just as a literal bridge is supported by pylons at both ends, a Brownian Bridge is required to satisfy conditions at both ends of the interval 0, T. (In a slight generalization, one sometimes requires B( t1) =  a and B( t2) =  b where t1, t2, a and b are known constants.)

Suppose we have generated a number of points W(0), W(1), W(2), W(3), etc. of a Wiener process path by computer simulation. It is now desired to fill in additional points in the interval 0, T, that is to interpolate between the already generated points. The solution is to use a collection of T Brownian bridges, the first of which is required to go through the values W(0) and W(1), the second through W(1) and W(2) and so on until the Tth goes through W( T-1) and W( T).


General case
For the general case when W( t1) = a and W( t2) = b, the distribution of B at time t ∈ ( t1t2) is normal, with

a + \frac{t-t_1}{t_2-t_1}(b-a)

and

\frac{(t_2-t)(t-t_1)}{t_2-t_1},

and the between B( s) and B( t), with s <  t is

\frac{(t_2-t)(s-t_1)}{t_2-t_1}.

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