# Interpolation of Brownian motion

Suppose you have obtained a sample path of Brownian motion using discrete step size $$\Delta t$$. Yet what happens at a finer scale? While in the experimental setup it would be impossible to say how the Brownian particle actually moved in between $$t$$ and $$t + \Delta t$$, one could numerically generate probable paths particle took in that time period. We do this by relying on the fractal nature of the Brownian motion.

## Interpolation method

If we know $$W(0)$$ and $$W(1)$$, we could try obtaining $$W(0.5)$$ by straightforward linear interpolation:

$$W(0.5) = \frac{W(0) + W(1)}{2} .$$

Yet this would have an obvious drawback - $$W(t)$$ for any $$t \in \left( 0, 1 \right)$$ would very much predictable.

Any one familiar with Euler method for the stochastic differential equations could come up with another approach. If

$$W(1) = W(0) + \xi ,$$

with $$\xi$$ being sample from a standard normal distribution, then the following should hold, too:

$$W(0.5) = W(0) + \sqrt{0.5} \cdot \xi .$$

Notice that motion increments between $$t=0$$ and $$t=0.5$$ would have a correct distribution (normal distribution with zero mean and $$0.5$$ variance). But motion increments between $$t=0.5$$ and $$t=1$$ would have higher variance ($$1.5$$).

To get the correct formula we need to combined both approaches. Let:

$$W(0.5) = \frac{W(0) + W(1)}{2} + \sqrt{0.25} \cdot \xi . \label{eq:correct}$$

## Interactive app

The app below uses \eqref{eq:correct} to generate a sample Brownian motion trajectory between two points $$x(0)$$ and $$x(256)$$, which you select. First the app generates the mid-point between $$x(0)$$ and $$x(256)$$, which is $$x(128)$$. Then it generates $$x(64)$$ and $$x(192)$$. And so on, it will continuously add the missing mid-points until the trajectory is known at all integer times.

The app won't generate the trajectory instantaneously on purpose. After filling in all current mid-points it will pause for $$1$$ second. After that second it will fill in all the new mid-points. It will continue doing so until sample trajectory is known for unit step size $$\Delta t = 1$$.