Brown Bridge

Question:

Let \(B_t\) be standard Brownian motion, find the conditional distribution \(B_s|B_t\) for \(0<s<t\) .

If \(B_s\) is known, \(B_t \sim N(?,?)\),consider the increment(Note they are independent):

\[ \begin{align*} B_t &= B_s + (B_t - B_s) \\ E(B_t) &= E(B_s) + E(B_t - B_s | B_s) = B_s \\ Var(B_t) &= E(B^2_t) - (E(B_t))^2 \\ &= E\left(B^2_s + (B_t - B_s)^2 + 2B_s(B_t - B_s)\right) - B_s^2 \\ &= B_s^2 + E((B_t - B_s)^2) + 0 - B_s^2 \\ &= B_s^2 + E((B_t - B_s)^2) \\ &= B_s^2 + (t - s) = t-s \end{align*} \]

Whereas Brown Bridge is when \(B_t\) known，assume it:

\[ \begin{align} B_s = B_s-\lambda B_t+ \lambda B_t \\ Cov(B_t-\lambda B_t,B_t) = 0 = s-\lambda t,\lambda = \frac{s}{t} \\ B_s = B_s- \frac{sB_t}{t} + \frac{sB_t}{t} \end{align} \]

which shows that \(B_s - \frac{sB_t}{t}\)is independent with \(B_t\)

\[ E(B_s|B_t) = E(B_s-\frac{sB_t}{t}|B_t)+ \frac{sB_t}{t}=\frac{s}{t} B_t \]

By law of total expectation.

Compute its variance:

\[ Var(B_s|B_t) = Var( B_s-\frac{s}{t}B_t |B(t)) = Var(B_s-\frac{s}{t}B_t), \text{as independence} \]

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