Why is a fractional Brownian motion not a semi-martingale? I am wondering what the simplest explanation is of why a fractional Brownian motion is not a semi-martingale. Also, it would be great if the answer could explain which of the assumptions going into Doob-Meyer's decomposition theorem are violated in this case.
many thanks!
 A: Let $K_{S}^{n}=\sum_{k=1}^{k_{n}} \alpha_{k}^{n} 1_{\left(t_{k-1}, t_{k}\right]}(s)$ be a simple predictable process, $\alpha_{k}^{n} \in F_{t_{k-1}}  \text { and } 0=t_{0}<t_{1}<\cdots<t_{k_{n}}=T$.
According to the Bichteler–Dellacherie–Mokobodski- theorem the process X is a semimartingale if and only if for any sequence of predictable simple processes $k_{n}$ with the property $\sup _{s \leq T}\left|K_{s}^{n}\right| \stackrel{P}{\rightarrow} 0$. we also have
$$\left(K^{n} \circ X\right)_{T}=\sum_{k=1}^{k_{n}} \alpha_{k}^{n}\left(X_{t_{k}}-X_{t_{k-1}}\right) \stackrel{P}{\rightarrow} 0.$$
Take now $ X=B^{H}$ and put $t_{k}=\frac{k T}{n}, k=0,1, \ldots, n .$ In the definition of $ K^{n}$ take now $\alpha_{1}^{n}=0$ and for $k=2, \ldots, n $ put $\alpha_{k}^{n}=n^{2 H-1}\left(B_{k-1}^{H}-B_{k-2}^{H}\right) .$ 
The Hölder continuity of $B^{H} $ gives now that our special $K^{n}$ satisfies
$$\sup _{s \leq T}\left|K_{s}^{n}\right| \stackrel{P}{\rightarrow} 0.$$
On the other hand the self-similarity of $B^{H} \text { gives }\left[b_{k}^{H}=B_{k}^{H}-B_{k-1}^{H}\right]$.
\begin{array}{l}{\text { But } 2^{2 H-1}-1 \neq 0 \text { for } H \in(0,1) \backslash \frac{1}{2}, \text { and the BDM-theorem gives }} \\ {\text { that } B^{H} \text { is not a semimartingale for } H \neq \frac{1}{2} \text { . }}\end{array}
