Diagnosing Brownian Simulations with Scaling and Covariance

What a Brownian Simulation Should Preserve

A simulated Brownian path may look irregular and plausible while having the wrong distribution. A reliable check therefore examines quantities with known targets: the means and variances of increments, dependence between disjoint increments, covariance between path values, and the way these quantities change when the time grid is refined. These checks can reveal a scaling error that a visual inspection misses.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space with a filtration \((\mathcal{F}_t)_{t\ge0}\). A standard Brownian motion \((B_t)_{t\ge0}\) starts at zero, has continuous paths, and has independent increments such that, for \(0\le s<t\), \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). On an equally spaced grid with step size \(h>0\), define the increments \(\Delta_k=B_{kh}-B_{(k-1)h}\). They are independent, each with mean \(0\) and variance \(h\).

Proof. The increment statement is part of the definition of Brownian motion. To obtain the covariance formula, suppose first that \(s\le t\). Write \(B_t=B_s+(B_t-B_s)\). The second term is independent of \(B_s\) and has mean zero, so \(\operatorname{Cov}(B_s,B_t)=\operatorname{Cov}(B_s,B_s)+\operatorname{Cov}(B_s,B_t-B_s)=\operatorname{Var}(B_s)+0=s\). If \(t\le s\), the same argument with \(s\) and \(t\) exchanged gives covariance \(t\). Thus the covariance is \(\min(s,t)\). For increments on disjoint intervals, independence follows from the independent-increments property; independent variables with finite means have covariance zero. \(\square\)

This theorem separates two checks that are often confused. Brownian levels at different times are generally dependent: for \(s\le t\), their covariance is \(s\). In contrast, increments on nonoverlapping intervals are independent, not merely uncorrelated. A diagnostic should therefore be applied to the increments when checking independent step noise, and to the levels when checking the covariance structure of the path.

Scaling Across Time Grids

The variance of an increment is its time-step length. Consequently, halving the step size halves the increment variance and multiplies its standard deviation by \(1/\sqrt{2}\). The increment should be generated as \(\sqrt{h}Z\), where \(Z\sim N(0,1)\), rather than as \(Z\) itself at every resolution.

Proof. Set \(Y_t=c^{-1/2}B_{ct}\). Then \(Y_0=0\), and its paths are continuous because those of \(B\) are continuous. For \(0\le s<t\), \(Y_t-Y_s=c^{-1/2}(B_{ct}-B_{cs})\). The increment inside the parentheses is normal with mean zero and variance \(c(t-s)\), so the scaled increment is normal with mean zero and variance \(t-s\). Increments of \(Y\) over disjoint intervals are independent, because the corresponding increments of \(B\) are over disjoint intervals and scaling does not change independence. Thus \(Y\) satisfies the defining finite-dimensional distribution properties of standard Brownian motion. \(\square\)

Scaling suggests a useful practical comparison: divide an increment by the square root of its step length. If the simulation is correct, the resulting standardized increments should have distribution \(N(0,1)\) at every grid resolution. For path values, another equivalent comparison is that \(B_{ct}/\sqrt{c}\) should have the same distribution as \(B_t\). Neither check says that two finite simulated datasets will look identical; they say what distribution each dataset should approximate.

Using Repeated Simulations to Separate Error from Variation

A single path supplies only a small number of increments at a given resolution. A clearer diagnostic is to simulate \(R\) independent paths and examine the increment at the same grid location across paths. Those \(R\) increments are independent replicates with mean \(0\) and variance \(h\). Their sample mean will usually not be exactly zero.

If \(Y_1,\ldots,Y_R\) are such replicates and \(\bar Y=R^{-1}\sum_{r=1}^R Y_r\), independence gives \(\operatorname{Var}(\bar Y)=h/R\): expanding the variance of the sum yields \(R\) variance terms, each \(h\), and zero cross-covariance terms. Hence the standard deviation of the sample mean is \(\sqrt{h/R}\). This is the Monte Carlo standard error for the mean under the stated independence and Brownian increment assumptions. It is a scale for expected sampling fluctuation, not a guarantee that every observed mean lies within any fixed multiple of it.

For variance checks, calculate the sample variance across independent paths and compare it with \(h\). For independence checks, examine pairs of increments from nonoverlapping intervals across paths, or assess their joint distribution rather than relying only on a sample correlation. A correlation near zero does not establish independence for arbitrary variables. Here, Brownian increments are jointly Gaussian, so zero covariance of disjoint increments is consistent with their independence, but a faulty simulator need not produce jointly Gaussian increments.

Diagnosing a Scaling Bug

The most consequential grid error is to add a standard normal variable at every step without multiplying by \(\sqrt{h}\). This produces increments with variance \(1\), regardless of the time step. The simulated process then accumulates variance according to the number of steps, rather than elapsed time. Checking the same terminal time on two grids makes the error easy to expose.

Covariance Checks and a Common Mistake

The formula \(\operatorname{Cov}(B_s,B_t)=\min(s,t)\) provides a direct check on path levels. It also explains why subtracting two correlated levels requires a covariance term. A common mistake is to treat levels at different times as independent and add their variances when computing the variance of their difference.

For example, at times \(\tfrac12\) and \(1\), Brownian motion has \(\operatorname{Var}(B_{1/2})=\tfrac12\), \(\operatorname{Var}(B_1)=1\), and \(\operatorname{Cov}(B_{1/2},B_1)=\tfrac12\). The incorrect independence calculation gives \(\operatorname{Var}(B_1-B_{1/2})=1+\tfrac12=\tfrac32\). The correct calculation is \(1+\tfrac12-2(\tfrac12)=\tfrac12\), which equals the length \(1-\tfrac12=\tfrac12\) of the increment interval. The missing covariance term is not a minor adjustment: omitting it changes the target variance.

When checking a simulation, use this sequence:

• Check that increments have empirical mean near \(0\) and variance near \(h\).
• Check that increments from disjoint intervals show no systematic dependence; use multiple independent paths when possible.
• Check level covariance against \(\min(s,t)\), not against zero at all distinct times.
• Repeat at a second grid resolution and verify that increment variance changes in proportion to \(h\), while terminal variance at a fixed time remains unchanged.

These are diagnostic comparisons, not proofs that a simulator is correct. Monte Carlo variation can produce visible discrepancies, especially with few paths. Conversely, matching a few means and variances does not rule out every defect: a simulator can have the right marginal distributions and still have incorrect dependence. Interpret several checks together, and investigate discrepancies that persist as the number of independent paths increases or that change systematically with grid resolution.