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:

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

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:

\begin{equation} X = \frac{W(0) + W(1)}{2} = \frac{2 W(0) + \xi}{2} = W(0) + \frac{\xi}{2} . \end{equation}

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

\begin{equation} \Delta x(t) = x(t) - x(t-1) . \end{equation}

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?