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TheMoney Loop
The math of randomness

Brownian motion explained

From a 50/50 random walk to continuous Brownian motion, and why uncertainty grows with the square root of time.

Play with it

Random walk to Brownian motion. Shrink a coin-flip walk until it becomes continuous, and watch uncertainty grow with √t.

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4

Each step moves ±0.500, so more steps means smaller moves.

4

The red line on the chart.

Spread at t = 41.99theory: √4 = 2.00
Average position-0.03theory: 0
Dark line: one walkDashed: ±1σ = ±√tDotted: ±2σ

Where 3,000 walks are at time t = 4

Bars: simulated positions. Dashed curve: normal distribution with mean 0 and variance 4.

At t = 4, the walks spread out by 1.99, close to √4 = 2.00. Double the time to t = 8 and the spread grows to √8 = 2.83, only √2 ≈ 1.41 times wider, not twice. With so few steps the walk is jagged and lands only on a grid of values.

Try this

  • Start at 4 steps per unit and slide up to 1000. Watch the jagged walk turn into a continuous path.
  • Set t to 4, then 8. Does the spread double, or grow by about 41%?
  • Flip new coins a few times. Each walk is different, but the cloud keeps the same shape.
  • At low step counts the histogram is spiky. Raise the steps and it settles onto the bell curve.

The math

Wt∼N(0, t)σ(Wt)=tW_t \sim \mathcal{N}(0,\, t) \qquad \sigma(W_t) = \sqrt{t}
WtW_t
position of the walk at time t
N(0,t)\mathcal{N}(0, t)
normal distribution, mean 0, variance t
t\sqrt{t}
spread grows with the square root of time

Quick check

Two questions to see if it clicked. Answers explain themselves.

If you double the time horizon, what happens to the uncertainty of Brownian motion?
At a fixed time t, Brownian motion's position follows which distribution?