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

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 $$H\ne1/2$$.

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 $$p_1 then $$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.

• Why do you think that the quadratic variation of BM is infinite? For $H=1/2$ we have $V_p(B)=\infty$ for $p \in (1,2)$ but this does not imply $V_2(B)=\infty$ (i.e. non-finiteness of the quadratic variation). In fact, Brownian motion has finite quadratic variation.
– saz
Commented Mar 12, 2020 at 6:41
• @saz I was trying to apply (but without having understood) the reasoning of Rogers, who says that if p-variation is infinite (finite) then the quadratic variation is infinite (finite) too. But sincerely, I don't understand how to deduce what the quadratic variation is, given that we know the p-variation. Commented Mar 12, 2020 at 6:57
• I'm reading that if $p_1 < p_2$ then $V_{p_2} \le V_{p_1}$, I think we have to use this fact Commented Mar 12, 2020 at 9:38

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.