Simulating Brownian Maxima Beyond the Time Grid

Why Grid-Point Maxima Miss Brownian Peaks

Simulating Brownian motion on a time grid gives its values at selected times, but the path between successive grid points is not linear. It can rise above both sampled endpoints and return before the next observation. Consequently, the largest simulated grid value is no larger than the true maximum over the time interval. The Brownian bridge gives an exact way to quantify and simulate these hidden peaks, conditional on the grid values already obtained.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space with filtration \((\mathcal{F}_t)_{t\ge0}\), and let \((B_t)_{t\ge0}\) be standard Brownian motion relative to this filtration: \(B_0=0\), its paths are continuous, and its increments over disjoint time intervals are independent centered normal variables with variance equal to the interval length. On an interval of length \(h>0\), conditional on its two endpoint values \(B_0=x\) and \(B_h=y\), the path has the Brownian bridge law from \(x\) to \(y\). Conditional statements about fixed endpoint values are understood through the conditional distribution given the endpoint, even though an individual endpoint value has probability zero.

The Brownian Bridge Crossing Probability

Suppose \(a>\max(x,y)\). The bridge starts and ends below \(a\), so a crossing can only occur between the endpoints. The exact conditional crossing probability is exponential in the product of the two endpoint distances from the level.

If \(a\le\max(x,y)\), then the crossing probability is \(1\), since an endpoint already reaches or exceeds \(a\). The strict condition \(a>\max(x,y)\) is the case in which the displayed exponential formula is used.

Proof. First consider Brownian motion started at \(x<a\), without conditioning on its endpoint. Its time-\(h\) endpoint has density \(p_h(y-x)\), where \(p_h(u)=(2\pi h)^{-1/2}\exp(-u^2/(2h))\). For \(y<a\), the reflection principle gives the joint endpoint density for paths that reach \(a\) and end at \(y\) as \(p_h(2a-y-x)\). To see the reflection argument, reflect a path after its first hitting time of \(a\). This maps paths that hit \(a\) and end at \(y\) to paths ending at \(2a-y\); symmetry and independent increments after the hitting time preserve their probabilities. Dividing the joint density of hitting and ending at \(y\) by the endpoint density gives the conditional probability given endpoint \(y\):

Conditioning Brownian motion started at \(x\) on its endpoint \(y\) gives the Brownian bridge from \(x\) to \(y\), so this ratio is the required bridge crossing probability. If \(a\le\max(x,y)\), an endpoint itself is at or above \(a\), which proves the remaining case. \(\square\)

Sampling an Interval Maximum

The crossing formula determines the entire conditional distribution of the maximum. For \(m\ge\max(x,y)\), the event \(M^{x,y}\le m\) is the complement of reaching or exceeding \(m\). Thus the conditional cumulative distribution function is

For \(m<\max(x,y)\), this cumulative distribution function is zero. At \(m=\max(x,y)\), the formula also gives zero. There is no point mass at the larger endpoint: for a bridge of positive duration, the maximum exceeds that endpoint with probability one.

Let \(U\) be uniform on \((0,1)\). Solving \(F_{x,y,h}(m)=U\) gives an exact way to sample the conditional maximum. Taking logarithms yields \((m-x)(m-y)=-\tfrac{h}{2}\log(1-U)\); solving this quadratic and selecting the root above both endpoints gives

The uniform variable is taken in the open interval \((0,1)\), so the logarithm is finite and the sampled maximum is strictly above both endpoints. The formula is an inverse-transform method: the cumulative distribution function is continuous and increasing from zero to one over its support, and substituting the displayed \(m\) into it returns \(U\).

Correcting a Maximum on a Full Time Grid

Let \(0=t_0<t_1<\cdots<t_n=T\) be a finite time grid, with interval lengths \(h_k=t_{k+1}-t_k>0\). The grid-only maximum is \(G=\max_{0\le k\le n}B_{t_k}\); the continuous maximum is \(M_T=\sup_{0\le t\le T}B_t\). Pathwise, \(G\le M_T\). Given all grid values, the bridge segments on the separate intervals are conditionally independent. Each segment’s conditional law depends on its two endpoints, and its maximum can be sampled using the inverse formula above.

Proof. Brownian independent increments imply the Markov property: given the value at a time, the future increments are independent of the past. Conditioning additionally on the endpoint of an interval gives the Brownian bridge law on that interval. For a finite collection of intervals, the conditional density of their path segments given all grid values factors into the product of their bridge densities: the unconditioned increment densities factor across intervals, and conditioning fixes the endpoint increment on each interval separately. Hence the bridge segments, and therefore their maxima, are conditionally independent. On every path, the maximum over \([0,T]\) is the largest of the interval maxima, because these intervals cover \([0,T]\). Drawing independent maxima from their conditional bridge laws therefore gives the stated conditional distribution of \(M_T\). \(\square\)

In practice, first simulate the Brownian values on the grid from Gaussian increments, as in Brownian motion on a time grid. Then, for each interval, draw an independent uniform variable and apply the inverse formula. Taking the maximum over all these draws gives a corrected continuous-time maximum. This generates the maximum, not the full path between grid points; additional bridge refinement is needed if values at intermediate times are also required.

Quantifying the Bias from Grid-Point Maxima

Because \(G\le M_T\), grid-point maxima systematically underestimate the continuous maximum. The expected bias is \(\mathbb{E}[M_T-G]\), when these expectations are finite. For a general grid, an exact representation follows by integrating tail probabilities. Since \(B_0=0\), both maxima are nonnegative. Using the conditional bridge distribution gives, for \(z\ge0\),

The formula applies on the event \(G\le z\), where each endpoint is at or below the level. If \(G>z\), the conditional probability is zero. Subtracting this conditional probability from \(\mathbf{1}_{\{G\le z\}}\) gives the conditional probability that the grid maximum is at most \(z\) but the continuous maximum exceeds it. Integrating over \(z\ge0\) yields an exact expression for the expected bias. For arbitrary grids, evaluating this expectation generally requires averaging over the random grid values; bridge correction provides a direct simulation method for doing so.

Common Pitfall: Interpolating Instead of Correcting

A straight line between two sampled values is not a Brownian bridge sample. For example, if both endpoints are zero over an interval of length one, linear interpolation has maximum zero. But the conditional probability that the true bridge reaches level one is \(e^{-2}\), not zero. Treating the linearly interpolated maximum as exact therefore discards genuine between-grid crossings. The correction is not to raise every grid maximum by a fixed amount; the bridge crossing probability depends on the interval length and on both endpoint distances from the level.

For a threshold above every sampled value, one can test whether the true path crossed it by applying the bridge crossing probability separately on each interval. To simulate the maximum itself, use the inverse-transform draw for each interval and then maximize those draws. The interval maxima must be independent conditional on the grid values, but they are not independent unconditionally because they share random endpoints.