1 What is a Brownian Bridge?
A Brownian bridge is a continuous-time gaussian process \((X_{t})\) whose probability distribution is the conditional probability distribution of a standard Brownian Motion \(B_t\) 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.
[!definition] Brownian Bridge The Brownian Bridge is an SDE \((X_{t})\) characterized by the dynamics \[\begin{cases}dX_{t}=\frac{b-X_{t}}{1-t}dt+dB_{t}\\ X_{0}=a,\end{cases}\] for \(a,b \in \mathbb{R}\) and \(0 \leq t < 1\).
This is the Brownian bridge starting from \(a\) and ending at \(b\) that has the same finite-dimensional distribution as the BM conditioning on starting from \(a\) at time \(0\) and ending at \(b\) at time \(1\). The Brownian bridge starting from \(0\) at time \(0\) and ending at \(0\) at time \(1\) has the simple form as \(B_t−tB_1\).
We can verify that the solution is given by \[ X_{t}=a(1-t)+bt+(1-t)\int_{0}^t \frac{1}{1-s}dB_{s}. \] First note that from standard results and the Itô Formula we have that \[ G:=\int_{0}^t \frac{1}{1-s}sB_{s} \sim\mathcal{N}\left( 0, \frac{1}{1-t} \right), \] and so the characteristic function of \((1-t)G\) is given by \[ \phi(p)=\phi_{G}(p(1-t))=e^{\frac{1}{2(1-t)}p^2(1-t)^2}=e^{\frac{1}{2}p^2(1-t)}\to 1~~\text{as}~~t \to 1. \] Hence \((1-t)G\stackrel{\mathcal{L}}{\to} 0\) as \(t \to 1\) and since the limit is a constant \((1-t)G\stackrel{\mathbb{P}}{\to} 0\) as \(t \to 1\). Finally, we notice that this stochastic integral is the sum of countably many independent random variables, hence from [[durrett-2-5-10]] we have \((1-t)G\stackrel{\text{a.s.}}{\to} 0\) as \(t \to 1\) and \(Y_{t}\stackrel{\text{a.s.}}{\to}b\).
2 0⃣ Standard Brownian Bridge (zero-to-zero)
A special case is the Brownian Bridge from 0 to 0 which has dynamics given by \[ \begin{cases} X_{t}=W_{t}-tW_{1} \\ X_{0}=0. \end{cases} \] The Brownian bridge is sometimes called the tied-down Brownian Motion (or tied-down Wiener process). It is useful for modeling a system that starts at some given level and is expected to return to that level at some specified future time. We note that \[ \begin{align} X_{0} & =W_{0}-0W_{1} \\ X_{1} & = W_{1}-1W_{1}. \end{align} \] The mean is clearly \(\mathbb{E}X_{t}=\mathbb{E}B_{t}-t\mathbb{E}B_{1}=0\) and the covariance is computed as \[ \begin{align} \text{Cov}(X_{s}, X_{t}) &=\mathbb{E}[X_{s}X_{t}]-\underbrace{\mathbb{E}[X_{s}]\cdot \mathbb{E}[X_{t}]}_{=0} = \mathbb{E}[(B_{s}-sB_{1})(B_{t}-tB_{1})] \\ & = \mathbb{E}[B_{s}B_{t}]-s\mathbb{E}[B_{1}B_{t}]-t\mathbb{E}[B_{s}B_{1}]+st\mathbb{E}[B_{1}^2] \\ & = \sigma^2s-s\sigma^2t-t\sigma^2s+st \sigma^2=\sigma^2s(1-t). \end{align} \] where we have used the fact that \(\mathbb{E}[B_{s}B_{t}]=\sigma^2(s\wedge t)\).
Thus, the Brownian bridge is not a wide-sense stationary process because the covariance is not a function of only the difference between s and t.
3 Transition Probability of a Brownian Bridge
The distribution of \(X_{t}\) is the same as the conditional distribution of BM conditioned on \(B_{1}=0\), which is a Gaussian process. To see this, consider the process \(Y_t = B_t | B_1 = 0\) and let’s derive the conditional distribution of \((B_s, B_t)\) where \(0 \leq s \leq t \leq 1\).
Consider the transformation from \((B_t, B_1 - B_t)\) to \((B_s, B_t)\). The joint density is: \[ \begin{align} f_{B_{s}, B_{1}-B_{t}}(x, y) & = f_{B_{s}}(x) f_{B_{1}-B_{t}}(y) \\ & = \frac{1}{\sqrt{2\pi s}} e^{-x^{2}/2s} \cdot \frac{1}{\sqrt{2\pi(1-t)}} e^{-y^{2}/2(1-t)} \end{align} \] since \(B_t\) and \(B_1 - B_t\) are independent. The conditional density with respect to \(B_1\) is: \[ f_{B_{s}, B_{t} | B_{1}}(x, y) = \frac{f_{B_{s}, B_{1}}(x, y)}{\sqrt{2\pi(1-s)} \cdot e^{-(x^{2}-1)y^{2}/2}} . \]
For the mean, we have: \[ \mathbb{E}Y_t = \mathbb{E}[B_t | B_1 = 0] = 0. \]
For the covariance kernel, for \(s < t\): \[ \begin{align} \text{Cov}(Y_s, Y_t) & = \text{Cov}(B_s, B_t | B_1 = 0) \\ & = \mathbb{E}[B_s B_t | B_1 = 0] \\ & = \mathbb{E}[\mathbb{E}[B_s B_t | B_1] | B_1 = 0] \\ & = \mathbb{E}[B_s \mathbb{E}[B_t | B_1] | B_1 = 0] \\ & = \mathbb{E}[B_s B_t | B_1 = 0] \\ & = s(1-t). \end{align} \] By the Markov property of BM, we have proved that the finite-dimensional distributions of these two processes are the same.