Is a continuous process with finite nonzero quadratic variation a semimartingale?

Question

I know that every semimartingale has finite quadratic variation but the converse is not true.

However, what if we assume continuity? If $X$ is a continuous process with finite nonzero quadratic variation is $X$ a semimartingale?

Answer 1

No, it's not true. There are continuous processes with finite quadratic variation which are not semimartingales:

Let $(B_t)_{t \geq 0}$ be a Brownian motion and consider $X_t := |B_t|^{\alpha}$, $ t \geq 0$. For $\alpha \in (0,1)$ the process $(X_t)_{t \geq 0}$ is not a semimartingale; this follows from the fact that $f(x):=|x|^{\alpha}$ cannot be written as the difference of two convex functions. On the other hand, it is possible to show that for $\alpha \in (1/2,1)$ the quadratic variation of $(X_t)_{t \geq 0}$ is zero. Hence, for any $\alpha \in (1/2,1)$ the continuous process $(|B_t|^{\alpha})_{t \geq 0}$ has zero quadratic variation but it is not a semimartingale.

For more details see the paper On functions transforming Brownian motion into a Dirichlet process by R. Chitashvili and M. Mania.

To get a continuous process with non-zero quadratic variation consider

$$Y_t := B_t + |B_t|^{\alpha}$$

for some $\alpha \in (1/2,1)$. Since the quadratic variation of $(B_t)_{t \geq 0}$ equals $\langle B \rangle_t =t$ and $(|B_t|^{\alpha})_{t \geq 0}$ has quadratic variation zero, it follows that the quadratic variation of $(Y_t)_{t \geq 0}$ equals $\langle Y \rangle_t = t$. On the other hand, $(Y_t)_{t \geq 0}$ cannot be a semimartingale because otherwise $|B_t|^{\alpha} = Y_t-B_t$ would be a semimartingale.