The Brownian motion component of the model can be replaced with fractional Brownian motion with parameter (0

Question:

The Brownian motion component of the model can be replaced with fractional Brownian motion with parameter \(0
\[
\operatorname{cov}(Y(s), Y(t))=s^{2 v}+t^{2 v}-|s-t|^{2 v},
\]

where \(s, t \geq 0\). The index \(v\) is called the Hurst coefficient, and \(v=1 / 2\) is ordinary Brownian motion. Show that the fit of the plant growth model can be appreciably improved by taking \(v \simeq 1 / 4\).

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Related Book For  book-img-for-question
Question Posted: