Prove that fractional Brownian motion is not a semimartingale using the p-variation

Question

What follows, up to the horizontal line, is taken from Rogers "Arbitrage with fractional Brownian motion".

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 $H\ne1/2$.

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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.

Answer 1

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.

Answer 2

Let $B^{H}_{\cdot}$ denote the fBM with Hurst parameter $H\in (0,1)$. This is a centered Gaussian process with Covariance function (i.e.: $\Sigma(t,s) =\mathbb{E}[B_t^HB_s^H] $): $$ \Sigma(t,s)= 2^{-1}\left( t^{2H} - s^{2H} - |t-s|^{2H} \right)\qquad \boldsymbol{(1)}. $$ We may therefore apply this Komologorov-Chestov argument (i.e.: Theorem 1 in these lecture notes) to (1) deduce that $B_{\cdot}^H$ has a version with $H$-Hölder-continuous paths. Therefore, it must have a version with finite $H^{-1}$-variation.

Since we know that every semi-martingale has finite quadratic variation then if $H\neq 2^{-1}$ $B_{\cdot}^H$ is not a semi-martingale.

Note: This doesn't show that it is a (semi-)martingale for $H=2^{-1}$.