Consider a good integrator $X$ (semi-martingale) and the relative quadratic variation process indicated by: $Y_t:=[X,X]_t$.

Why is that:

$$[Y,Y]_t=0 \ \ \ \ \ and \ \ \ \ \ \ [X,Y]_t=0 \ \ ?$$

EDIT: I believe that for general good integrators the previous statement is not true. Indeed:

If $Y$ is adapted, cadlag, with paths of finite variation on compacts, then Y is a quadratic pure jump semimartingale. Therefore: $[Y,Y]_t=∑_{0<s<t}(ΔY_s)^2$.

• Hint: For which processes $U$ can one be sure that $[U,V]=0$ for every process $V$? The answer should be very near to the beginning of your notes.
– Did
Jul 15, 2015 at 14:14
• @Did I was thinking about $U$ being of Finite Variation but the Q.V. would be 0 just in the case of $U$ being continuous as well. Jul 15, 2015 at 14:30
• just out of curiosity: do you mean the predictable version of the quadratic variation? Jul 15, 2015 at 15:03
• @user190080 no, I mean the quadratic variation for general good integrators Jul 15, 2015 at 15:08

Quadratic variation of a semi-martingale is non-decreasing and right-continuous. (Or at least there exists such a version of it). You know that a function is of bounded variation if and only if it is the difference of two non-decreasing functions. So $[X,X]_t$ is of bounded variation and right-continuous. Then its quadratic variation must be the sum of the jumps squared over the interval the Q.V. is computed.

If you impose further that the semi-martingale is continuous, then its Q.V. is continuous as well. The argument above gives in that case a Q.V. of $0$.

• I follow the reasoning and it is what I was thinking after the hint gave by @Did, but: If $Y$ is adapted, cadlag, with paths of finite variation on compacts, then $Y$ is a quadratic pure jump semimartingale. Therefore: $[Y,Y]_t = \sum_{0<s<t}(\Delta Y_s)^2$. So why should this be 0 ? Jul 15, 2015 at 14:25
• @mastro You are absolutely right. One should impose continuity on $X$ to make sure $Y$ is continuous as well. Then the QV of $Y$ is $0$. Jul 15, 2015 at 14:32
• but why is $[X,X]$ difference of increasing functions? Jul 15, 2015 at 15:00
• $[X,X]$ is non-decreasing itself. Then just write it as $[X,X] - 0$ where $0$ is a stochastic process that is almost surely $0$ at all time instants, which is clearly also non-decreasing. Jul 15, 2015 at 15:02
• @mastro to see that $[X,X]$ is non-decreasing is, since you use the sum definition of the quadratic variation, not hard. Maybe a small remark, functions which are non-decreasing are of finite variation Jul 15, 2015 at 15:12

Quadratic variation of a quadratic variation is quartic variation, i.e., $$[[X,X],[X,X]] = \sum_{t\leq\cdot}(\Delta X_t)^4.$$

More generally, Emery, Michel, Stabilité des solutions des équations différentielles stochastiques. Application aux intégrales multiplicatives stochastiques, Z. Wahrscheinlichkeitstheor. Verw. Geb. 41, 241-262 (1978). ZBL0351.60054 has shown that for a $$\mathcal{C}^2$$ function $$\xi:\mathbb{R}\to\mathbb{R}$$ with $$\xi(0)=0$$, the quantity $$\int_0^\cdot \xi(dX_t) := \xi'(0)(X_t-X_0)+\frac{1}{2}\xi''(0)[X,X]^c+\sum_{t\leq\cdot}(\xi(\Delta X_t)-\xi'(0)\Delta X_t)$$ is the $$\xi$$-variation of $$X$$.

For nice enough $$\psi$$ and $$\xi$$, the variations compose, i.e., the $$\psi$$-variation of the $$\xi$$-variation of $$X$$ is equal to the $$\psi(\xi)$$-variation of $$X$$. The proof is given, for example, in Carr, Peter; Lee, Roger, Variation and share-weighted variation swaps on time-changed Lévy processes, Finance Stoch. 17, No. 4, 685-716 (2013). ZBL1275.91129.

One can extend the composition rule to predictable variations and build a simplified stochastic calculus as a result. For example, the Ito formula for $$f(X)$$ is a predictable variation of $$X$$ by means of the predictable function $$\xi(x) = f(X_-+x)-f(X_-)$$. For more details, see Černý, Aleš; Ruf, Johannes, Simplified stochastic calculus via semimartingale representations, Electron. J. Probab. 27, Paper No. 3, 32 p. (2022). ZBL1490.60103.