I have problems to under stand the following thing:
I totally see that the fractional Brownian Motion (https://en.wikipedia.org/wiki/Fractional_Brownian_motion) is no semimartingale hor $H\neq \frac{1}{2}$, since increments are not independent. But, I often read that it is possible to represent the fractional BM as a stochastic integral in something like that form: $$ X_t = C \int (t-s)^{H-\frac{1}{2}}dW_s $$, where W is a standard BM. Why should that integral not be a local martingale?
thanks for any help
We let $B_{H}(t):= C \int (t-s)^{H-\frac{1}{2}}dW_s$ and follow Arbitrage with fractional Brownian motion. The key is that a semimartingale always has finite quadratic variation, and if its quadratic variation is zero, then it is of bounded variation.
Consider the interval $[0,1]$ on which is defined the fractional Brownian motion $B_{H}$, and consider its partitions $\pi_n = \{t^n_k = \frac{k}{2^{n}} : 0\le 1\le 2^{n}\},\ n\in\mathbb N$ and
$$ V_p^{n}(B_{H}) = \sum_{k=1}^{2^{n}} |B_{H}(t^n_{k+1})-B_{H}(t^n_k)|^p. $$
By the ergodic theorem applied to fBM
$$\frac{V_p^{n}(B_{H})}{2^{n(1-pH)}}\to E[|B_{H}(1)-B_{H}(0)|^{p}]=c_{p}\neq 0.$$
Therefore, $$ V_p(B_{H}) =\begin{cases} \infty, & \text{if }\ pH < 1, \\ 0, & \text{if }\ pH > 1. \end{cases} $$
If $H<\frac{1}{2}$, we get $V_{2}(B_{H})=+\infty$. If $H>\frac{1}{2}$, then we get that $V_{2}(B_{H})=0$ and $V_{1}(B_{H})=+\infty$ but this contradicts that if a semimartingale has zero quadratic variation, it least has finite 1-variation.