Define the Brownian bridge on [0,1] as W conditioned on W_1 = 0. Give Cov(B_s, B_t), the variance at time 1/2, and the stochastic differential equation the bridge satisfies. Explain why the drift blows up at t = 1.

Define the Brownian bridge on [0,1] as W conditioned on W_1 = 0. Give Cov(B_s, B_t), the variance at time 1/2, and the stochastic differential equation the bridge satisfies. Explain why the drift blows up at t = 1.

Approach: Condition a bivariate normal to get the covariance, then write the bridge as an integral against dW and differentiate to recover the drift.

Cov(B_s, B_t) = min(s,t) - st, Var(B_{1/2}) = 1/4, and the bridge solves dB_t = -B_t/(1-t) dt + dW_t with B_0 = 0. The conditional distribution of the Gaussian vector (W_s, W_t) given W_1 = 0 shifts the mean to zero and subtracts the outer product of the covariances with W_1 divided by Var(W_1), giving min(s,t) - st directly. At t = 1/2 this is 1/2 - 1/4 = 1/4, half the unconditioned variance. The representation B_t = W_t - t W_1 has the same covariance, and applying Ito to it produces the linear drift -B_t/(1-t) dt. The coefficient 1/(1-t) diverges at the right endpoint because the process must be pulled to a fixed value with no time remaining, and the divergence is integrable in the sense that ∫_0^1 |B_t|/(1-t) dt is finite almost surely, since |B_t| itself is of order sqrt(1-t) near the endpoint.

Follow-up: What is the distribution of the maximum of a Brownian bridge on [0,1], and how is it used to correct a discretely monitored barrier option?

Key concepts: Brownian bridge, conditional distribution, covariance, linear drift.