# Applying Ergodic Theorem on fractional Brownian motion

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.

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I would be thankful for every help and explanation... Sure about that? –  Did Jan 3 '13 at 0:29
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.