# Interpolation of Brownian motion (part 2)

Last time we have discussed how to statistically correct interpolate Brownian motion. You have probably noticed that the correct interpolation formula has one peculiarity:

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

Why $$\sqrt{0.25}$$ and not $$\sqrt{0.5}$$?

The answer is quite intuitive. Part of the variance we expect to obtain is already present in the linear interpolation part of the expression:

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

In the above $$\xi$$ is a sample from standard normal distribution. Thus $$X$$ will be normally distributed with mean of $$W(0)$$ and variance of $$1/4$$. To have $$W(0.5)$$, which would have expected variance of $$1/2$$ we need to add random sample, which would contribute variance of $$1/4$$. This is why $$\xi$$ is multiplied by $$\sqrt{0.25}$$ (square root was kept for clarity).

## 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. This app is almost exactly the same as the app from the previous post. The difference is that after the trajectory is generated it is differentiated

$$\Delta x(t) = x(t) - x(t-1) .$$

Then the distribution of increments, $$\Delta x$$, is plotted and compared against the normal distribution with $$\frac{x(256) - x(0)}{256}$$ mean and unit variance.

Everything seems fine doesn't it?