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

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?

• Hi first : "I know that every semimartingale has finite quadratic variation" I don't think that's true, only locally true. Jul 2, 2019 at 13:54
• Your question is a bit unclear. In the title is saying "finite nonzero quadratic variation" but in your question your are not mentioning the "nonzero" assumption.
– saz
Jul 2, 2019 at 17:30
– user658409
Jul 2, 2019 at 17:30

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.