In some sources (for example here: Why is a fractional Brownian motion not a semi-martingale?) you can find a proof why the fractional Brownian motion is not a semimartingale via the Bichteler-Dellacherie theorem.
There are two things I do not understand.
Firstly, the simple predictable processes $K^n$ are used with $$ K^n:=\sum_{j=1}^n n^{2H-1} \left(B^H_{\frac{jT}{n}}-B^H_{\frac{(j-1)T}{n}}\right)1_{\left[ \frac{jT}{n}, \frac{(j+1)T}{n}\right]}. $$
Secondly, it is mentioned that according to the Bichteler-Dellacherie Theorem a process $X$ is a semimartingale if and only if for any sequence of predictable simple processes $K^n$ with the property $\displaystyle\sup_{s≤T}|K^n_s|$ converge in probability to 0.
My understanding is:
- These processes are not simple predictable processes as they are not bounded.
- The Bichteler-Dellacherie theorem requires the essential supremum of simple processes converge to 0 and not the supremum in probability.
Is my understanding correct and, if yes, can we somehow still use this argument to prove that the fractional Brownian motion is not a semimartingale? For example, would it work to define a sequence like $H^{n,m}:=K^n\wedge \frac 1m$ for which you can choose an appropriate diagonal sequence which converges to 0 but the integral wrt the fractional Brownian motion does not?
Using p-variation
First for a short proof. We 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.
Using The Bichteler-Dellacherie Theorem
From The Bichteler-Dellacherie Theorem and Existence of the Stochastic Integral we use the definition/theorem
Theorem 1 A cadlag adapted process X is a semimartingale if and only if, for each {t\ge 0}, the set
$$\displaystyle \left\{\int_0^t\xi\,dX\colon \xi{\rm\ is\ elementary}, \vert\xi\vert\le 1\right\} (3)$$
is bounded in probability. This is equivalent to the statement that, for any sequence ${\xi^n}$ of bounded predictable processes converging uniformly to zero, ${\int_0^t\xi^n\,dX} $ converges to zero in probability.
As mentioned in Why is a fractional Brownian motion not a semi-martingale?, it suffices to take
$$\xi^{n}(t):=\sum^{2^{n}}_{k=1}\frac{1}{2^{n(1-2H)}}(B_{H}(t^n_{k+1})-B_{H}(t^n_k))1_{(t^n_{k},t^n_{k+1}]}(t)$$
because then by Hölder continuity of fBM we have $|B_{(k+1)/2^{n}}-B_{(k)/2^{n}}|\leq c/2^{n(H-\epsilon)}$ and so
$$\sup_{t\leq 1}|1_{|\xi^{n}(t)|\leq 1}\xi^{n}(t)|\leq \frac{c}{2^{n(1-H-\epsilon)}}\to 0,$$ for small $\epsilon>0$ and random constant $c$. On the other hand,
$$\int 1_{|\xi^{n}(t)|\leq 1}\xi^{n}(t)dB_{H}(t)=\frac{V_2^{n}(B_{H})}{2^{n(1-2H)}}-\int 1_{|\xi^{n}(t)|\geq 1}\xi^{n}(t)dB_{H}(t).$$
We already have $\frac{V_2^{n}(B_{H})}{2^{n(1-2H)}}\to c_{2}\neq 0.$ We also have
$$\int 1_{|\xi^{n}(t)|\geq 1}\xi^{n}(t)dB_{H}(t)\approx \sum^{2^{n}}_{k=1}\frac{1}{2^{n(1-2H)}}(B_{H}(t^n_{k+1})-B_{H}(t^n_k))^{2}1\{|\frac{1}{2^{n(1-2H)}}(B_{H}(t^n_{k+1})-B_{H}(t^n_k))|\geq 1\}$$
$$\leq1\{\frac{c}{2^{n(1-H-\epsilon)}}\geq 1\}\sum^{2^{n}}_{k=1}\frac{1}{2^{n(1-2H)}}(B_{H}(t^n_{k+1})-B_{H}(t^n_k))^{2}.$$
Here we can apply Cauchy-Schwartz to bound by $P[\frac{c}{2^{n(1-H-\epsilon)}}\geq 1](E[(\frac{V_2^{n}(B_{H})}{2^{n(1-2H)}})^{2}])^{1/2}$. The ergodic theorem also works for $L_{p},p>1$ convergence to turn the second factor into $c_{2}$ again. For the first factor, we use that the Hölder-norm has finite moments see "Kolmogorov-Chentsov" in Schilling
$$E[c^{\alpha}]=E[\sup\frac{|\xi(x)-\xi(y)|^{\alpha}}{|x-y|^{\alpha\gamma}}]<\infty$$
and hence $P[\frac{c}{2^{n(1-H-\epsilon)}}\geq 1]\to 0$.