Brownian increments
Brownian motion describes a complete random path \(W(t)\). A Brownian increment describes how much that path changes over one particular time interval.
What exactly is an increment?
Take two times, \(t\) and \(t+\Delta t\). The change in Brownian motion between them is
\[\boxed{\Delta W=W(t+\Delta t)-W(t)}.\]This has exactly the same structure as an ordinary change:
\[\Delta x=x_{\mathrm{new}}-x_{\mathrm{old}}.\]The difference is that \(\Delta W\) is random.
The distribution of an increment
For standard Brownian motion,
\[\boxed{\Delta W\sim N(0,\Delta t)}.\]Here \(N(0,\Delta t)\) means a normal distribution with
\[\text{mean}=0,\qquad\text{variance}=\Delta t.\]Therefore
\[\boxed{\operatorname{SD}(\Delta W)=\sqrt{\Delta t}}.\]| Property | Meaning |
|---|---|
| mean \(0\) | no preferred positive or negative direction |
| variance \(\Delta t\) | uncertainty grows with the length of the interval |
| standard deviation \(\sqrt{\Delta t}\) | sets the typical scale of the random change |
Why \(\sqrt{\Delta t}\), not \(\Delta t\)?
Start with a standard normal random variable
\[Z\sim N(0,1).\]If a random variable is multiplied by a constant \(a\), its variance is multiplied by \(a^2\):
\[\operatorname{Var}(aZ)=a^2\operatorname{Var}(Z)=a^2.\]We need the variance to equal \(\Delta t\), so
\[a^2=\Delta t,\qquad a=\sqrt{\Delta t}.\]Hence
\[\boxed{\Delta W=\sqrt{\Delta t}\,Z,\qquad Z\sim N(0,1)}.\]A step-by-step numerical example
Suppose the simulation step is
\[\Delta t=0.01.\]Then
\[\sqrt{\Delta t}=0.1.\]Generate one standard normal random number. Suppose
\[Z=-1.4.\]Then
\[\Delta W=0.1(-1.4)=-0.14.\]If the current Brownian value is \(W(t)=0.35\), the next value is
\[W(t+\Delta t)=0.35-0.14=0.21.\]A new independent normal number is generated for the next interval.
How interval length changes the distribution
For example,
| \(\Delta t\) | distribution | standard deviation |
|---|---|---|
| 1 | \(N(0,1)\) | 1 |
| 0.25 | \(N(0,0.25)\) | 0.5 |
| 0.01 | \(N(0,0.01)\) | 0.1 |
Independent increments
Brownian changes over non-overlapping intervals are independent. For example,
\[W(1)-W(0)\quad\text{and}\quad W(2)-W(1)\]are independent random variables.
A positive increment in the first interval does not make a positive or negative increment more likely in the second interval.
Same interval length, same distribution
The increment distribution depends only on the length of the interval:
\[W(t+h)-W(t)\sim N(0,h).\]Thus
\[W(2)-W(1)\sim N(0,1)\]and
\[W(101)-W(100)\sim N(0,1).\]Their locations in time differ, but both intervals have length 1.
Adding increments builds Brownian motion
Suppose the time grid is
\[t_0,t_1,t_2,\ldots\]and we generate independent increments
\[\Delta W_0,\Delta W_1,\Delta W_2,\ldots\]Then
\[W(t_{n+1})=W(t_n)+\Delta W_n.\]Starting from \(W(0)=0\),
\[\boxed{W(t_n)=\sum_{k=0}^{n-1}\Delta W_k}.\]So a simulated Brownian path is constructed by accumulating Brownian increments.
Why many small increments give the correct larger increment
If \(m\) independent increments each cover time \(\Delta t\), their sum covers total time \(m\Delta t\). Variances of independent increments add:
\[\operatorname{Var}\left(\sum_{k=1}^{m}\Delta W_k\right)=m\Delta t.\]Therefore the accumulated change has distribution
\[N(0,m\Delta t),\]exactly matching the Brownian rule for an interval of total length \(m\Delta t\).
\(\Delta W\) versus \(dW_t\)
In a numerical simulation we work with a finite time step and write
\[\Delta W=\sqrt{\Delta t}\,Z.\]In an SDE we write the infinitesimal notation
\[dW_t.\]The symbol \(dW_t\) represents the Brownian stochastic increment appearing inside stochastic calculus. It should not be treated as an ordinary differentiable quantity.
Why Brownian increments matter in an SDE
Consider
\[dX_t=f(X_t,t)\,dt+g(X_t,t)\,dW_t.\]Over a small numerical interval, this becomes approximately
\[\Delta X\approx f(X_t,t)\Delta t+g(X_t,t)\Delta W.\]Substituting the Brownian increment gives
\[\boxed{\Delta X\approx f(X_t,t)\Delta t+g(X_t,t)\sqrt{\Delta t}\,Z}.\]This is the central random step used by the Euler–Maruyama method.
Deterministic and stochastic increments scale differently
The deterministic part has size proportional to
\[\Delta t,\]whereas the Brownian part has characteristic size proportional to
\[\sqrt{\Delta t}.\]For very small \(\Delta t\), \(\sqrt{\Delta t}\) is much larger than \(\Delta t\). This difference in scaling is fundamental to stochastic calculus.
The connection with \((dW_t)^2=dt\)
Because a Brownian increment is of order \(\sqrt{\Delta t}\), its square is of order
\[(\Delta W)^2\sim\Delta t.\]In the stochastic-calculus limit, this leads to the Itô rule
\[\boxed{(dW_t)^2=dt}.\]The precise mathematical idea behind this is quadratic variation, discussed later.
A common misunderstanding
The statement
\[\Delta W\sim N(0,\Delta t)\]does not mean that \(\Delta W=0\) because its mean is zero. Individual increments are usually non-zero. Zero is the average over many possible increments.