The Brownian Bridge

You're simulating a price path from today to option expiry and you already know one specific thing: it must end exactly at the strike, because you're checking how much time the path spends near that level along the way. Simulating ordinary Brownian motion and then discarding every path that doesn't happen to land there would be absurdly wasteful — almost none would land exactly on target. The fix is to directly simulate a process that's forced to end where you want, keeping all the random wiggle in between: the Brownian bridge.

The analogy: a rope pinned at both ends, shaken in the middle

Picture Brownian motion as a rope, free at the far end, being shaken randomly — it can end up anywhere. A Brownian bridge is the same rope, but now pinned down at both ends at fixed heights, and still being shaken randomly in the middle. The randomness in the interior is exactly the same kind of noise as ordinary Brownian motion — jagged, unpredictable — but its magnitude tapers off as you approach either pinned end, since the rope has less and less freedom to wander the closer it gets to a point that's already fixed.

Writing it down

Let $W_t$ be standard Brownian motion on $[0,T]$. The Brownian bridge from 0 to $b$ is Brownian motion conditioned on $W_T = b$:

$$X_t = W_t - \frac{t}{T}\big(W_T - b\big), \qquad 0 \le t \le T.$$

In words: take an ordinary Brownian path, then subtract off a straight-line correction that grows linearly from 0 (at $t=0$, no correction needed) to exactly cancel whatever $W_T$ actually was (at $t=T$, the full correction), forcing the endpoint to land at $b$ instead of wherever the original unconstrained path happened to end. The result has mean $\frac{t}{T}b$ (a straight-line interpolation between the two pinned points) and variance

$$\text{Var}(X_t) = \frac{t(T-t)}{T}.$$

In words: the bridge's uncertainty is zero at both endpoints (it's pinned, so there's no randomness left there) and largest exactly in the middle, at $t=T/2$ — the point furthest from either anchor, where the process has had the most "room" to wander before being reined back in.

Sample a few plain Brownian paths above; each ends wherever it happens to land. A Brownian bridge is what you'd get by taking many such paths that all happen to pass through the same two endpoints and looking only at their shape in between — same jaggedness throughout, but the spread of possible values visibly narrows to zero as each path approaches either pinned end.

Worked example 1: variance at the midpoint, numerically

Take $T=4$ (say, four days to expiry) and look at the midpoint $t=2$. Variance is $\frac{2(4-2)}{4} = \frac{4}{4}=1$, so the standard deviation of the bridge's position at the midpoint is exactly 1 — the same as an unconditioned Brownian motion would have at $t=2$ if it started from 0 with no endpoint constraint ($\text{Var}(W_2)=2$, actually — so the bridge's midpoint variance of 1 is exactly half of the unconstrained value). Now check $t=1$: $\text{Var}(X_1) = \frac{1\times3}{4} = 0.75$. And $t=3$ (by symmetry, since $3 = T-1$): also $0.75$. The variance curve is symmetric and bulges in the middle — confirmed by three numbers computed directly from the formula.

Worked example 2: simulating a bridge from an ordinary path

Suppose an ordinary simulated Brownian path over $[0,4]$ happens to produce $W_1=0.8$, $W_2=1.5$, $W_3=1.1$, $W_4=2.0$, and you want the bridge pinned to end at $b=0$ instead of wherever this path landed ($W_4=2.0$). The correction at each time is $\frac{t}{4}(W_4 - 0) = \frac{t}{4}\times 2.0 = 0.5t$. So $X_1 = 0.8 - 0.5(1) = 0.3$; $X_2 = 1.5 - 0.5(2) = 0.5$; $X_3 = 1.1 - 0.5(3) = -0.4$; $X_4 = 2.0 - 0.5(4) = 0$, exactly landing at the target. Notice the correction grows linearly (0.5, 1.0, 1.5, 2.0) while the underlying jaggedness of the original path is fully preserved — the bridge is the original noise, minus a straight ramp, nothing more exotic than that.

What this means in practice

• Path-dependent option pricing (barrier options, Asian options) that needs to know how much time a path spends above or below a level, given it starts and ends at known points, is exactly the Brownian bridge's setting — and the reflection principle (see The Reflection Principle) gives closed forms for barrier-crossing probabilities inside a bridge.
• Variance reduction in Monte Carlo: simulating bridges directly, rather than simulating full unconstrained paths and discarding the ones that don't hit a target endpoint, is enormously more efficient — this is the core idea behind Brownian-bridge path construction used in quasi-Monte Carlo pricing.
• Missing-data interpolation: if you know a price at the start and end of a gap (say, around an exchange halt) but not in between, a Brownian bridge is the standard, statistically honest way to simulate plausible interior paths consistent with both known endpoints.