Barrier Crossing Probabilities for Brownian Bridges

From a Path Event to an Endpoint Density

A Brownian bridge is constrained at both ends of its time interval, but its intermediate path remains random. To find the probability that it reaches a constant upper barrier, it is useful to compare two endpoint densities: the density of all paths with the prescribed endpoint, and the density of paths that hit the barrier and then end at that endpoint. Reflection identifies the second density explicitly.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Fix \(T>0\) and real numbers \(x,y\). A Brownian bridge from \(x\) to \(y\) over \([0,T]\) is the continuous Gaussian process with the law of Brownian motion started at \(x\), conditioned to have value \(y\) at time \(T\). Since the event \(x+B_T=y\) has probability zero, this means the regular conditional law given the endpoint, not ordinary conditioning on a positive-probability event. One representation of this bridge is

The centered Gaussian term \(B_t-\frac{t}{T}B_T\) has covariance \(\min(s,t)-st/T\) and is independent of \(B_T\), which gives the stated conditional bridge law. We use no stochastic integrals in this tutorial, so no integrand class is required. Write \(M=\sup_{0\le t\le T}X_t\). We use “reaches the barrier \(a\)” to mean \(M\ge a\).

Reflection and the Bridge Crossing Formula

Let \(a>\max(x,y)\), so the bridge starts and ends strictly below the barrier. Consider Brownian motion started at \(x\), denoted \(Y_t=x+B_t\), and its first hitting time \(\tau_a=\inf\{t\ge0:Y_t=a\}\). If a path hits \(a\) and ends at \(y<a\), reflect the part of the path after its first hit across the level \(a\). Its reflected endpoint is \(2a-y>a\). Brownian symmetry after the hitting time gives a one-to-one, probability-preserving correspondence between paths that hit \(a\) and end near \(y\), and paths that end near \(2a-y\). The endpoint transformation has absolute derivative \(1\), so there is no extra density factor.

Proof. First suppose \(a>\max(x,y)\). By reflection at \(\tau_a\), the density at \(y\) of Brownian paths started at \(x\) that hit \(a\) by time \(T\) is \(p_T(2a-y-x)\). More explicitly, for an endpoint interval \(I\) lying below \(a\), reflection identifies the probability of \(\{\tau_a\le T,\ Y_T\in I\}\) with the probability that \(Y_T\) lies in the reflected interval \(2a-I\), which is above \(a\). Every continuous path ending above \(a\) has hit \(a\). Taking endpoint densities gives the claimed density. The unrestricted endpoint density at \(y\) is \(p_T(y-x)\). Disintegrating with respect to the endpoint, the conditional bridge probability of hitting \(a\) is therefore the ratio of these densities. The regular conditional bridge law has the continuous Gaussian representation given above, so this density ratio gives its bridge crossing probability. Hence

The last equality follows by expanding the difference of squares: \((2a-y-x)^2-(y-x)^2=4(a-x)(a-y)\). If \(a\le\max(x,y)\), at least one endpoint is at or above \(a\), so the supremum over the closed interval \([0,T]\) is at least \(a\) for every bridge path. This proves both cases. \(\square\)

When \(a>\max(x,y)\), the probability of strictly exceeding \(a\) is the same as the probability of reaching it. Indeed, for each \(\varepsilon>0\), the event \(\{M>a\}\) contains \(\{M\ge a+\varepsilon\}\), and \(\{M>a\}\) is contained in \(\{M\ge a\}\). Letting \(\varepsilon\) decrease to zero in the formula shows that the two probabilities agree. Thus the maximum has no atom at any level strictly above both endpoints.

The Distribution of the Bridge Maximum

Proof. Since \(X_0=x\) and \(X_T=y\), every path has \(M\ge m_0\), so its distribution function is zero below \(m_0\). For \(m>m_0\), the event \(M\le m\) is the complement of \(M>m\). The crossing formula and the equality of strict and non-strict crossing probabilities above the endpoints give \(\mathbb{P}(M\le m)=1-\exp(-2(m-x)(m-y)/T)\). At \(m=m_0\), the right-hand limit is zero because at least one of \(m_0-x\) and \(m_0-y\) is zero. Hence the formula holds there too, and there is no atom at \(m_0\). Differentiating the distribution function for \(m>m_0\), using \(\frac{d}{dm}[-2(m-x)(m-y)/T]=-2(2m-x-y)/T\), gives the stated density. It is nonnegative because \(m>m_0\) implies \(2m-x-y>0\). Finally, the distribution function tends to \(1\) as \(m\) tends to infinity, so this density accounts for all probability above \(m_0\). \(\square\)

Endpoint Cases and a Common Error

The assumption \(a>\max(x,y)\) matters: it ensures the bridge begins and ends below the barrier, so reflection compares a hit-and-return path with a path whose reflected endpoint lies above the barrier. If \(a\le\max(x,y)\), reaching the barrier is automatic because an endpoint is already at or above it. The exponential formula should not be used to replace this endpoint argument outside its stated range.

Common mistake: reflecting the endpoint to \(a\) instead of \(2a-y\). For a bridge from \(0\) to \(0\) over \([0,1]\) with barrier \(1\), this error would use \(p_1(1)/p_1(0)=\exp(-1/2)\). Reflection is across the line at height \(1\), so the endpoint \(0\) becomes \(2\), not \(1\). The correct ratio is \(p_1(2)/p_1(0)=\exp(-2)\). The error substantially overestimates the crossing probability.

A second caution is that the bridge probability is not obtained by dividing by the probability of ending at exactly \(y\): that event has probability zero. The correct operation is to compare endpoint densities, or equivalently to use the regular conditional bridge law. In this setting reflection supplies the numerator density and the ordinary Brownian transition density supplies the denominator.