# Fractional Brownian motion of Riemann-Liouville type is not a semimartingale

Given a filtered probability space $$(\Omega,\mathcal{F},\mathbb{F},\mathbb{P})$$ satisfying the usual conditions, $$B$$ a standard one-dimensional Brownian motion and $$H\in(0,1/2)$$. Consider the process $$X$$ given by $$X_t = \int_0^t \frac{(t-s)^{H-1/2}}{\Gamma(H+1/2)} d B_s,$$ where $$\Gamma$$ denotes the gamma function. The process $$X$$ is called fractional Brownian motion of Riemann-Liouville type with Hurst parameter $$H.$$ I have read in various articles that it is well-known that $$X$$ is not a semimartingale. I am trying to understand why this statement is true. However, I find it difficult to show this. I tried approximating the non-differentiable kernel $$K(t)=\frac{t^{H-1/2}}{\Gamma(H+1/2)}$$ by continuously differentiable kernels, $$\varepsilon>0,$$ $$K_\varepsilon(t)= \frac{(t+\varepsilon)^{H-1/2}}{\Gamma(H+1/2)}.$$ Then, an application of the stochastic Fubini theorem shows that $$X_t^\varepsilon = \int_0^t K_\varepsilon(t-s) dB_s$$ is a semimartingale with quadratic variation $$\langle X^\varepsilon\rangle_t = K_\varepsilon(0)^2 t$$ and $$\langle X^\varepsilon\rangle_t\to\infty$$ as $$\varepsilon\downarrow0.$$ Moreover, I was able to show that $$\mathbb{E}(|X_t^\varepsilon-X_t|^2)\to0$$ as $$\varepsilon\downarrow0.$$ Now, I aim to conclude that $$\langle X\rangle_t=\infty$$ which would yield the claim.

I would be grateful for any suggestion on how to finish my proof or any other way that shows the assertion.

It is well known that the fractional Brownian motion is not a semi martingale unless H=1/2; see for instance this post or Section 2 in this paper. Then, from Proposition 9 in the paper Equivalence of Volterra processes the Riemann-Liouville process is locally equivalent to the fractional Brownian motion. Therefore the Riemann-Liouville process is not a semi martingale unless H=1/2.

• Many thanks for your answer and drawing my attention to this interesting paper about the equivalence of Volterra processes! I have to think about your "therefore". In particular, is it clear that two continuous processes that are locally equivalent are neither both a semimartingale nor both not a semimartingale? Commented May 8, 2023 at 22:26
• The "therefore" follows from Girsanov theorem, but essentially if P(<X>_T <+\infty)=0 for one process then it should also be zero for the other process since the laws of the two processes are equivalent on the time interval [0,T] Commented May 8, 2023 at 23:23

$$\newcommand{\al}{\alpha}\newcommand{\Ga}{\Gamma}\newcommand{\be}{\beta}\newcommand\ip[1]{\langle #1\rangle}\newcommand\R{\mathbb R}$$This property is briefly proved in Section 5.1 of this paper. The proof is based on Basse's characterization of the spectral representation of Gaussian semimartingales, namely, Theorem 4.6.

Here are details on this. It follows from Basse's theorem that, if $$(X_t)_{t\ge0}$$ were a semimartingale, then we would have $$\begin{equation*} (t-s)^\al=g(s)+\int_s^t\Psi_r(s)\mu(dr) \tag{1}\label{1} \end{equation*}$$ for all real $$t\ge0$$ and almost all (a.a.) $$s\in[0,t]$$, where $$\al:=H-1/2\in(-1/2,0)$$, $$g\colon\R_+\to\R$$ is square integrable on $$[0,t]$$ for all real $$t\ge0$$, $$\mu$$ is a Radon measure on $$\R_+$$, and $$\R_+\times\R_+\ni(t,s)\mapsto\Psi_t(s)\in\R$$ is a measurable mapping such that $$\|\Psi_r\|_{L^2(\R_+)}=1$$.

In view of the Tonelli theorem, it follows from \eqref{1} that for a.a. triples $$(s,c,t)$$ such that $$0 we have $$\begin{equation*} \int_c^t\al(r-s)^{\al-1}\,dr=\int_c^t\Psi_r(s)\mu(dr). \tag{2}\label{2} \end{equation*}$$ Since $$\al(r-s)^{\al-1}<0$$ for $$r\in(c,t)$$, we see that the Lebesgue measure on $$(0,\infty)$$ is absolutely continuous w.r.t. $$\mu$$, with some density $$h$$, so that $$h(r)=\frac{dr}{\mu(dr)}$$ for real $$r>0$$. So, \eqref{2} implies $$\begin{equation*} \int_c^t\al(r-s)^{\al-1}\,h(r)\mu(dr)=\int_c^t\Psi_r(s)\mu(dr) \end{equation*}$$ for a.a. triples $$(s,c,t)$$ such that $$0. So, $$\begin{equation*} \Psi_r(s)=\al(r-s)^{\al-1}\,h(r) \tag{3}\label{3} \end{equation*}$$ for $$\mu$$-a.a. pairs $$(s,r)$$ such that $$0. Since the Lebesgue measure on $$(0,\infty)$$ is absolutely continuous w.r.t. $$\mu$$, we get \eqref{3} for a.a. pairs $$(s,r)$$ such that $$0.

For any real $$r>0$$ with $$h(r)\ne0$$, by \eqref{3},
$$\begin{equation*} 1=\|\Psi_r\|_{L^2(\R_+)}^2\ge\al^2 h(r)^2\int_0^r(r-s)^{2\al-2}\,ds=\infty, \end{equation*}$$ a contradiction.

Finally, if $$h(r)=0$$ for a.a. real $$r>0$$, then, by \eqref{3}, $$\Psi_r(s)=0$$ for a.a. pairs $$(s,r)$$ such that $$0, which contradicts \eqref{1}. $$\quad\Box$$

Details on \eqref{2}: Recall that \eqref{1} holds for all real $$t\ge0$$ and a.a. $$s\in[0,t]$$. Replace each of the two entries of $$t$$ in \eqref{1} by $$c\in(s,t)$$ and refer to this modification of \eqref{1} as \eqref{1}$$_c$$. By the Tonelli theorem, the conjunction of \eqref{1} and \eqref{1}$$_c$$ holds for a.a. triples $$(s,c,t)$$ such that $$0. For any such $$(s,c,t)$$, subtract the left-hand side of \eqref{1}$$_c$$ from that of \eqref{1}, and also subtract the right-hand side of \eqref{1}$$_c$$ from that of \eqref{1}. Then ($$g(s)$$ disappears and) we get \eqref{2} for a.a. triples $$(s,c,t)$$ such that $$0.

Further details, on the use of the Tonelli theorem, to show that the conjunction of \eqref{1} and \eqref{1}$$_c$$ holds for a.a. triples $$(s,c,t)$$ such that $$0: Let $$T\subset\R^3$$ denote the set of triples $$(s,c,t)$$ such that $$0 and the conjunction of \eqref{1} and \eqref{1}$$_c$$ fails to hold. We want to show that the Lebesgue measure $$|T|$$ of $$T$$ is $$0$$. Let $$S\subset\R^2$$ denote the set of pairs $$(s,t)$$ such that $$t\ge0$$, $$s\in[0,t]$$, and \eqref{1} fails to hold. Then, in view of the Tonelli theorem, \begin{equation*} \begin{aligned} |S|&=\iint_{\R^2}ds\,dt\,1(t\ge0, s\in[0,t], \text{\eqref{1} fails to hold}) \\ &=\int_\R dt\,1(t\ge0)\int_\R ds\,1(s\in[0,t], \text{\eqref{1} fails to hold}) \\ &=\int_\R dt\,1(t\ge0)0=0, \end{aligned} \end{equation*} since \eqref{1} holds for all real $$t\ge0$$ and a.a. $$s\in[0,t]$$. Note that $$\begin{equation*} T\subseteq\{(s,c,t)\colon(s,t)\in S\text{ or }(s,c)\in S\}. \end{equation*}$$ So, again in view of the Tonelli theorem, \begin{equation*} \begin{aligned} |T|&=\iiint_{\R^3}ds\,dc\,dt\,(1(s,c,t)\in T)) \\ &\le\iiint_{\R^3}ds\,dc\,dt\,(1((s,t)\in S)+1((s,c)\in S)) \\ &=\int_\R dc\,\iint_{\R^2}ds\,dt\,(1((s,t)\in S) +\int_\R dt\,\iint_{\R^2}ds\,dc\,(1((s,c)\in S) \\ &=\int_\R dc\,|S| +\int_\R dt\,|S|=0. \end{aligned} \end{equation*} Thus, $$|T|=0$$. $$\quad\Box$$

• Thank you very much for the effort you have put into your answer. I really appreciate it. Your reasoning is clear and I understand the idea. The only thing I do not get is how you obtained (2) from Tonelli's theorem. In particular, what did happen to $g$? Commented May 8, 2023 at 22:06
• @El_mago : I have added details on this. Please let me know if further details are needed. Commented May 8, 2023 at 23:31
• Many thanks again for your explanation. I do not get, how Tonelli's theorem is used? I only know it as a special case of Fubini's theorem. Commented May 9, 2023 at 8:24
• @El_mago : I have added further details on the use of the Tonelli theorem. (BTW, it is not a special case of the Fubini theorem -- because, in contrast with the Fubini theorem, the Tonelli theorem holds for all nonnegative measurable functions.) Commented May 9, 2023 at 19:20
• Many thanks for adding further details and pointing out my mistake. Commented May 17, 2023 at 13:27