Brownian Motion on a Time Grid from Gaussian Increments

Brownian Motion at Discrete Times

To simulate Brownian motion on a computer, one usually chooses finitely many times and generates the process values at those times. The key is to generate the increments correctly: over an interval of length $h$, a Brownian increment has mean $0$ and variance $h$. Thus it is a standard normal variable multiplied by $\sqrt{h}$, not by $h$. Adding these increments produces the values on the grid.

Here the goal is to construct and check the grid values, not to specify the path between grid points. Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space carrying independent standard normal random variables $Z_1,Z_2,\ldots$. For a fixed step size $h>0$, let $t_k=kh$, where $k=0,1,2,\ldots$. The simulated process $(W_k^h)_{k\ge0}$ is a discrete-time process with state space $S=\mathbb{R}$. Its value $W_k^h$ represents the Brownian value at time $t_k$. Use the filtration $\mathcal{F}_k=\sigma(Z_1,\ldots,Z_k)$, with $\mathcal{F}_0=\{\varnothing,\Omega\}$.

Definition: The Brownian grid construction with step size $h$ is $W_0^h=0$ and $W_k^h=W_{k-1}^h+\sqrt{h}\,Z_k$ for $k\ge1$. Equivalently, $W_k^h=\sum_{r=1}^k\sqrt{h}\,Z_r$. The increment from time $t_{k-1}$ to time $t_k$ is therefore $W_k^h-W_{k-1}^h=\sqrt{h}\,Z_k$.

A standard Brownian motion starts at $0$, has independent increments, and for $0\le s<t$ its increment over $[s,t]$ has a normal distribution with mean $0$ and variance $t-s$. The grid construction reproduces these properties at the grid times. Its covariance is the minimum of the two times, which is a useful check both in theory and in simulation.

Why the Square Root of the Step Size Appears

For each grid interval, $Z_k$ has mean $0$ and variance $1$. Multiplying by $\sqrt{h}$ gives an increment with mean $\sqrt{h}\,\mathbb{E}[Z_k]=0$ and variance $(\sqrt{h})^2\operatorname{Var}(Z_k)=h$. Since the increments are independent, their variances add. After $k$ steps, the elapsed time is $kh$, and the variance of the accumulated value is $kh$ as required.

Theorem (Grid distribution and covariance): For $h>0$, the construction above has $W_0^h=0$, independent increments, and $W_k^h-W_\ell^h\sim N(0,(k-\ell)h)$ whenever $0\le\ell<k$. For any nonnegative integers $j,k$, the pair $(W_j^h,W_k^h)$ is jointly Gaussian and $$ \mathbb{E}[W_k^h]=0,\qquad \operatorname{Cov}(W_j^h,W_k^h)=h\min(j,k)=\min(t_j,t_k). $$ Consequently, the vector of grid values has the same joint distribution as Brownian motion sampled at those grid times.

Proof. Each increment $W_r^h-W_{r-1}^h=\sqrt{h}\,Z_r$ is normal with mean $0$ and variance $h$. Distinct increments are independent because they are functions of distinct independent variables $Z_r$. For $\ell<k$, $W_k^h-W_\ell^h$ is the sum of $k-\ell$ independent normal variables, each with mean $0$ and variance $h$. The sum is normal, its mean is $0$, and its variance is $(k-\ell)h$.

Each $W_k^h$ is a linear combination of the independent normal variables $Z_1,\ldots,Z_k$. Any finite vector of such linear combinations is jointly Gaussian. Also, $W_k^h=\sqrt{h}\sum_{r=1}^k Z_r$, so its mean is $0$. In the covariance of $W_j^h$ and $W_k^h$, terms with different indices have covariance $0$ by independence and zero means. The shared terms have variance $h$ each. There are exactly $\min(j,k)$ shared indices, giving $\operatorname{Cov}(W_j^h,W_k^h)=h\min(j,k)$. Since $t_j=jh$ and $t_k=kh$, this equals $\min(t_j,t_k)$.

Brownian motion has jointly Gaussian values, mean zero, and covariance $\min(s,t)$ at times $s,t$. The constructed vector has those same means and covariances. A jointly Gaussian vector is determined by its mean vector and covariance matrix, so the grid values have the stated Brownian joint distribution. This establishes agreement at the selected times; it does not define the values between them. $\square$

The construction is also a direct simulation algorithm. For each step, generate a fresh independent standard normal value, multiply it by $\sqrt{h}$, and add the result to the current value. The use of a fresh independent draw at every step is essential: reusing a draw would not produce independent increments.

Worked Example: Four Equal Steps on the Unit Interval

Take $h=\tfrac14$, so the grid times are $0,\tfrac14,\tfrac12,\tfrac34,1$. The increment at each step is $\tfrac12 Z_k$, because $\sqrt{\tfrac14}=\tfrac12$. To show how a realization is accumulated, suppose the standardized normal outputs for this illustration are $1,-2,\tfrac12,-1$. These are specified draw values, not a claim that a continuous random variable takes them with positive probability as exact outcomes.

The four increments are, in order, $\tfrac12(1)=\tfrac12$, $\tfrac12(-2)=-1$, $\tfrac12(\tfrac12)=\tfrac14$, and $\tfrac12(-1)=-\tfrac12$. Starting from $W_0^h=0$, the cumulative values are $$ W_1^h=\tfrac12,\qquad W_2^h=\tfrac12-1=-\tfrac12,\qquad W_3^h=-\tfrac12+\tfrac14=-\tfrac14,\qquad W_4^h=-\tfrac14-\tfrac12=-\tfrac34. $$

Check the increments: subtracting consecutive values gives $\tfrac12,-1,\tfrac14,-\tfrac12$, exactly the increments calculated from the four draws. Each increment has variance $(\tfrac12)^2=\tfrac14$, the length of its grid interval. For example, the theoretical covariance of the values at times $\tfrac12$ and $\tfrac34$ is $\min(\tfrac12,\tfrac34)=\tfrac12$. In general, the covariance matrix for the values at $\tfrac14,\tfrac12,\tfrac34,1$ is $$ \begin{bmatrix} \tfrac14&\tfrac14&\tfrac14&\tfrac14\\ \tfrac14&\tfrac12&\tfrac12&\tfrac12\\ \tfrac14&\tfrac12&\tfrac34&\tfrac34\\ \tfrac14&\tfrac12&\tfrac34&1 \end{bmatrix}. $$ Each entry is the minimum of the corresponding pair of grid times, as required.

Checking a Grid with Unequal Time Steps

Equal spacing makes the implementation particularly simple, but the principle does not depend on equal steps. For deterministic times $0=t_0<t_1<\cdots<t_m$, define the interval lengths $\Delta_k=t_k-t_{k-1}$, and use independent standard normal variables to form increments $\sqrt{\Delta_k}Z_k$. The same calculation gives independent normal increments with variance equal to the corresponding interval length. The following example checks this calculation on a nonuniform grid.

Worked Example: Three Unequal Intervals

Consider the times $0,\tfrac14,1,2$. The successive interval lengths are $\tfrac14,\tfrac34,1$. The increments are therefore $\tfrac12Z_1,\tfrac{\sqrt{3}}2Z_2,Z_3$, respectively. For an illustrative set of standardized outputs $Z_1=2$, $Z_2=-1$, and $Z_3=\tfrac12$, the increments are $1,-\tfrac{\sqrt{3}}2,\tfrac12$.

Accumulating from zero gives $W(\tfrac14)=1$, $W(1)=1-\tfrac{\sqrt{3}}2$, and $W(2)=1-\tfrac{\sqrt{3}}2+\tfrac12=\tfrac32-\tfrac{\sqrt{3}}2$. In this notation, $W(t_k)$ denotes the value at the listed grid time. The covariance matrix at $\tfrac14,1,2$, obtained from the time-minimum formula, is $$ \begin{bmatrix} \tfrac14&\tfrac14&\tfrac14\\ \tfrac14&1&1\\ \tfrac14&1&2 \end{bmatrix}. $$

Check: the variance at time $1$ is the sum of the first two increment variances, $\tfrac14+\tfrac34=1$. The variance at time $2$ is $\tfrac14+\tfrac34+1=2$. The covariance between the values at times $1$ and $2$ is the variance of their shared first two increments, $\tfrac14+\tfrac34=1$; the last increment contributes to $W(2)$ but not to $W(1)$. These values match the displayed matrix.

Refining a Grid Without Changing Its Coarse-Time Law

A useful check is that subdividing a time interval does not alter the distribution of the total increment across that interval. For example, a single increment over length $h$ should have the same distribution as the sum of two independent increments over lengths $h/2$ and $h/2$. This is a statement about the values at the coarse times; it does not specify the intermediate path.

Proposition (Consistency under subdivision): Suppose a grid interval of length $h>0$ is divided into $r$ equal subintervals, where $r$ is a positive integer. The sum of the $r$ independent simulated subinterval increments has distribution $N(0,h)$, the same as a single increment over length $h$. Sums over distinct coarse intervals are independent.

Proof. Each subinterval increment has the form $\sqrt{h/r}\,Z_a$, with the $r$ standard normal variables independent. The sum of the $r$ increments is normal because it is a sum of independent normal variables. Its mean is $0$, and its variance is the sum of the $r$ variances: $$ \sum_{a=1}^r \frac{h}{r}=h. $$ It therefore has distribution $N(0,h)$. If two coarse intervals are distinct, their sums use disjoint collections of the independent normal variables, so the sums are independent. $\square$

Worked Example: Subdividing One Interval

Compare an increment over an interval of length $1$ with the sum of four increments over subintervals of length $\tfrac14$. In the subdivided construction, each increment is $\tfrac12 Z_a$, for $a=1,2,3,4$. The sum has mean $0$ and variance $$ 4\left(\tfrac12\right)^2=4\cdot\tfrac14=1. $$ Because it is a sum of independent normal variables, its distribution is $N(0,1)$, matching a single increment $\sqrt{1}Z\sim N(0,1)$ over the full interval.

For a direct illustrative calculation, take standardized outputs $1,-1,2,0$ on the four subintervals. The four increments are $\tfrac12,-\tfrac12,1,0$, whose sum is $1$. The total is the same as the cumulative sum of the four increments, and the variance calculation—not this one set of draw values—establishes equality in distribution.

Check: summing the displayed increments gives $\tfrac12-\tfrac12+1+0=1$. Their variances add to $\tfrac14+\tfrac14+\tfrac14+\tfrac14=1$, as required for a length-one interval.

Common Scaling Error and Scope of the Construction

A frequent mistake is to multiply each standard normal draw by $h$ rather than $\sqrt{h}$. For $h=\tfrac14$, that incorrect rule gives increments $\tfrac14 Z_k$, each with variance $(\tfrac14)^2=\tfrac1{16}$. Four such increments cover elapsed time $1$, but their total variance is $4\cdot\tfrac1{16}=\tfrac14$, not $1$. The simulated values would therefore be too concentrated. The correction is to use $\sqrt{h}=\tfrac12$, giving four increment variances of $\tfrac14$ and total variance $1$.

Another useful distinction is between a grid-valued simulation and a continuous-time Brownian path. The construction gives exactly the Brownian joint distribution at the selected grid times. It does not, by itself, say what the process does between successive times. Drawing straight line segments between the computed values is an additional interpolation rule; it is not a consequence of the Gaussian-increment construction. When assessing a grid simulation, first check that each increment has mean $0$, variance equal to its time-step length, and independence from the other increments. Then check that the resulting covariance at grid times is $\min(s,t)$.

Check Your Understanding

Use the increment formula and the covariance calculation to answer these questions.

For a step size $h=\tfrac19$, what multiple of a standard normal variable gives one increment, and what is its variance?
On the grid $0,\tfrac12,1$, find the covariance between the values at times $\tfrac12$ and $1$.
Explain why increments over two disjoint collections of grid intervals are independent in this construction.
If an interval of length $2$ is divided into four equal pieces, what is the variance of each increment and of their sum?
Why does knowing the simulated values on a grid not, on its own, determine the path between grid points?