When is the quadratic variation of a process continuously differentiable?

Question

When is the quadratic variation of a continuous square integrable process continuously differentiable for almost all $\omega$?

For the Brownian motion this is obviously true. It is also true for an continuous square integrable martingale with stationary independent increments . But can we say this for any square integrable martingale ? Any hints towards an answer would be appreciated .

Answer 1

This is only a partial answer, but we know that a necessary condition for differentiability is continuity. Hence, for the quadratic variation of a process (*) to be differentiable, the quadratic variation must be continuous. In this answer, I will outline some conditions for the continuity of the quadratic variation.

Consider the semimartingale $X$, and denote its optional quadratic variation as $[X]$. If $X$ is a local martingale, denote its predictable quadratic variation process as $\langle X \rangle$.

Consider first the case where $X$ is a local martingale, with $\langle X \rangle$ its predictable quadratic variation. We will need the following definition, where we use the notation $X_{t-} = \lim_{s\uparrow t} X_s$.

Definition. A process $Y$ is regular if, for every predictable stopping time $T$, we have that $$ E[Y_{T-}] = E[Y_T]. $$

Recalling our local martingale $X$, we will then have that $\langle X \rangle$ is continuous if and only if $X^2$ is regular. A sufficient condition for $X^2$ to be regular is for $X$ to be continuous.

Suppose now that $X$ is a general semimartingale. We know that its optional quadratic variation can be written as $$ [X]_t = \langle X^c \rangle_t + \sum_{0<s\le t} \left( \Delta X_s \right)^2, $$ where $X^c$ is the continuous (local) martingale part of $X$, and $\Delta X_s = X_s - X_{s-}$. From the discussion above, we know that $\langle X^c \rangle$ is continuous. From the expression giving the optional quadratic variation above, $[X]$ has jumps whenever $X$ has jumps, so that $[X]$ is continuous if and only if $X$ is continuous.

Hence, a general square-integrable martingale need not have a differentiable quadratic variation process. In particular, if the martingale is not continuous, then it cannot have a differentiable optional quadratic variation. If it has predictable jump times, then it will also not have a differentiable predictable quadratic variation.

Reference: Cohen, S. N., & Elliott, R. J. (2015). Stochastic calculus and applications (Vol. 2). New York: Birkhäuser.

(*) In this answer, I assume that processes are semimartingales to guarantee that the quadratic variation is well-defined. I am also assuming a filtration satisfying the usual conditions and take martingales to be their càdlàg versions.