
> 
> What follows, up to the horizontal line, is taken from [Rogers "Arbitrage with fractional Brownian motion"](http://www.long-memory.com/fractional-brownian-motion/Rogers1997.pdf).
> 
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> 


Consider an interval $[0,T]$ on which is defined the fractional Brownian motion $B$, and consider its partitions $\pi\_n = \{t^n\_k = \frac{kT}{n} : 0\le k\le n\},\ n\in\mathbb N$.


Let $p\ge1$, the $p$-variation of $B$ is
$$
 V\_p(B) = \lim\_{n\to\infty} \sum\_{k=0}^{n-1} |B(t^n\_{k+1})-B(t^n\_k)|^p = 
 \begin{cases}
 \infty, & \text{if }\ pH < 1, \\
 0, & \text{if }\ pH > 1.
 \end{cases}
$$
If $H>1/2$ we can choose $p\in(1,\frac1H)$ so that $pH<1$, then the $p$-variation is infinite, hence the quadratic variation of $B$ is infinite too.


If $H<1/2$ we can choose $p>2$ so that $pH<1$, then again we obtain that the $p$-variation and the quadratic variation of $B$ are infinite.


In both cases the quadratic variation of $B$ is not finite, hence the fBm is not a semimartingale for $H\ne1/2$.




---


Could somebody further explain the above reasoning? I don't fully get what has to be proved, is it related to the fact that a semimartingale has to have finite *variation*? But which *variation*: quadratic, p-variation or another one?


Moreover, I don't understand how to deduce what the quadratic variation is, given that we know the p-variation. Is it related to the fact that given $p\_1<p\_2$ then $V\_{p\_2}\le V\_{p\_1}$?


Fianlly, what about the case $H=1/2$, in which $B$ is the usual Brownian motion? If we take $p\in(1,2)$ then we are still in the case $pH<1$ and so the $p$-variation is infinite hence the quadratic variation of $B$ is infinite too, contradicting the fact that B is a martingale.



Prove that fractional Brownian motion is not a semimartingale using the p-variation
Assume $B$ is a semimartingale, then it has finite quadratic variation.


Recall that if $s < b$ then $V\_b \le V\_s$.


* If $H<1/2$ we can choose $p>2$ s.t. $pH<1 \implies V\_p = \infty \implies \infty\le V\_2 \implies V\_2 = \infty$, i.e. the quadratic variation ($p=2$) is infinite too: contradiction.
* If $H>1/2$ we can choose $p\in(\frac1H,2)$ s.t. $pH>1 \implies V\_p = 0 \implies V\_2 \le 0 \implies V\_2 = 0 \implies B$ must have finite variation. But on the other hand, for $p\in(1,\frac1H)$ we have $V\_p = \infty$, hence $B$ cannot have finite variation: contradiction.


Either way, if $H\ne\frac12$, the fBm is not a semimartingale.


