5
$\begingroup$

For a fractional Brownian motion $B_H$ consider the sequence for $p>0$ $$Y_{n,p}={1\over n}\sum\limits_{i=1}^n \left|B_H(i)-B_H(i-1)\right|^p.$$ By the Ergodic Theorem it is $$\lim\limits_{n\to\infty}Y_{n,p}=\mathbb{E}[|B_H(1)|^p] \ a.s.\text{ and in } L^1.$$ The Ergodic Theorem of Birkhoff says:

Let $(\Omega,\mathcal{F},\mathbb{P},\tau)$ be a measure-preserving dynamical system, $p>0$, $X_0\in\mathcal{L}^p$ and $X_n=X_0\circ \tau^n$. If $\tau$ is ergodic, then it holds $${1\over n}\sum\limits_{k=0}^{n-1}X_k\overset{n\to\infty}{\longrightarrow}\mathbb{E}[X_0]\ a.s.\text{ and in }L^1.$$

My problem is that I don't know how this theorem is used on the case described above, i.e. what is $\tau$ and what is $X_n$ in this case?

Another question is: Why do I have to use the ergodic theorem while by using the stationarity I have $$Y_{n,p}\sim {1\over n}\sum\limits_{i=1}^n |B_H(1)|^p=|B_H(1)|^p\ ?$$

I would be thankful for every help and explanation.

Maybe I should add saying that I am trying to understand how to prove that a fractional Brownian motion is not a semi-martingale for $H\neq {1\over2}$, by applying these statements.

$\endgroup$
  • 4
    $\begingroup$ I would be thankful for every help and explanation... Sure about that? $\endgroup$ – Did Jan 3 '13 at 0:29
5
$\begingroup$

1. The canonical shift $\tau$ is defined by the identity $$ (B_H\circ\tau)(t) =B_H(t+1)-B_H(1),\qquad t\geqslant0. $$ The covariance structure of $B_H$ then readily implies that $\tau$ leaves invariant the distribution of the process $B_H=(B_H(t))_{t\geqslant0}$.

2. The convergence of $Y_{n,p}$ is indeed a special case of the ergodic theorem for $$ X_0=|B_H(1)|^p, $$ since then, for every $k\geqslant0$ and every $t$, $(B_H\circ\tau^k)(t)=B_H(k+t)-B_H(k)$, hence $$ X_0\circ\tau^k=|(B_H\circ\tau^k)(1)|^p=|B_H(k+1)-B_H(k)|^p. $$

3. The last part of the post is wrong and this is due to a basic misunderstanding: while each random variable $$ X_k=|B_H(k+1)-B_H(k)|^p $$ is indeed distributed as $X_1=|B_H(1)|^p$, it does not follow that $Y_{n,p}=\frac1n\sum\limits_{k=1}^nX_k$ is distributed as $\frac1n\sum\limits_{k=1}^nX_1=X_1$, since the joint distributions of the random vectors $(X_k)_{1\leqslant k\leqslant n}$ and $(X_1)_{1\leqslant k\leqslant n}$ are quite different.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.