| Brownian motion Y(t) is stationary: the
differences
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| Y(t + h) - Y(t)
|
| are independent of t.
|
| To illustrate this visually, we sample a Brownian
motion simulation and compute increments
|
| incr1 = Y(t1 + h) - Y(t1),
incr2 = Y(t2 + h) - Y(t2),
...,
incr1000 = Y(t1000 + h) - Y(t1000).
|
| We must take the ti so ti+1 ≥ ti + h;
if the sampling increments overlap, we should not expect independence.
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| Then we plot the points
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| (incr2, incr1), ...,
(incr1000, incr999).
|
| If the increments are
independent of one another, the points should lie in an approximately
circular cloud, denser near the center.
|
| The left picture shows such a plot
for h = 1, the right for h = 2.
|
|
