# Quadratic Covariation of an Increasing Process with another Process is 0

According to the book I'm reading on Option Pricing:

Since $$V$$ is an increasing process, $$\langle X, V \rangle_t = \langle V \rangle_t = 0$$

In this case $$X$$ is just a price process (according to the book the specific form shouldn't matter, but the process is assumed to be a continuous local martingale)

Does anyone see why this is true? Are there specific conditions under which it is true, or is it just for any increasing process?

I found a similar unanswered question, but wanted to try again: Quadratic Variation of Increasing Process?

Thanks a lot!

• The quadratic covariation of an increasing process with a continuous process is zero. That's a general statement which is not very difficult to show using the very definition of the quadratic covariation.
– saz
Jan 30, 2019 at 7:23
• Cross-posted: quant.stackexchange.com/questions/43789/… Jan 30, 2019 at 7:43

Let $$Y^1, Y^2$$ be semi-martingales with representation $$Y^i = X^i + B^i$$ where $$X^i$$ is a continuous local martingale and $$B^i$$ is a process with paths of bounded variation. Then the quadratic variation of these processes is defined as the quadratic variation of the local martingales, namely $$\langle Y^1, Y^2\rangle := \langle X^1, X^2\rangle,$$ and write $$\langle Y^i \rangle := \langle Y^i, Y^i\rangle$$.

Using $$X$$ and $$V$$ from your example, note that they are both semi-martingales with finite variation part and local martingale part respectively being equal to $$0$$, so $$\langle X + V\rangle = \langle X \rangle$$ and $$\langle V \rangle = \langle 0 \rangle= 0$$. Then using a polarization identity for local martingales we obtain $$\langle X, V \rangle = \tfrac{1}{2}(\langle X + V\rangle - \langle X \rangle - \langle V \rangle) = \tfrac{1}{2}(\langle X \rangle - \langle X \rangle - \langle 0 \rangle) = 0.$$

So as far as I know $$\langle X, V\rangle = \langle V \rangle = 0$$ holds for $$V$$ being of finite variation and $$X$$ being a continuous semi-martingale, but these conditions can probably be somewhat loosened.

Let $$(\pi_n)_{n\geq 0}$$ be a sequence of partitions of $$[0,T]$$ whose mesh is going to zero (that is, if the partition is $$\pi_i:0=t_0\leq t_1 \leq ... t_{N(i)} = T$$, then $$mesh(\pi_i)=min_p(t_{p+1}-t_p)$$).

Then a/the definition of quadratic covariation of processes X, Y over $$[0,T]$$ is: $$\langle X,Y \rangle_T = lim_{n \to \infty} \sum_{i=0}^{N(n)-1}|X_{t_{i+1}}-X_{t_i}||Y_{t_{i+1}}-Y_{t_i}|$$ (see Revuz & Yor, Continuous Martingales and Stochastic Calculus, Chapter IV on Stochastic Integration, Theorem 1.9 for the equivalence of this definition with the other main definition; that quadratic covariation is the unique increasing stochastic process such that $$(X_tY_t-\langle X,Y \rangle_T)_{t\geq 0}$$ is a martingale) and $$\langle X \rangle_T=\langle X,X \rangle_T$$.

Then if V is some general increasing process, and X some general continuous process, then we prove that $$\langle X,V \rangle_T=0$$ for all T: $$\langle X,V \rangle_T = lim_{n \to \infty} \sum_{i=0}^{N(n)-1}|X_{t_{i+1}}-X_{t_i}||V_{t_{i+1}}-V_{t_i}| \\ \leq lim_{n \to \infty} \sum_{i=0}^{N(n)-1}(max_j|X_{t_{j+1}}-X_{t_j}|)|V_{t_{i+1}}-V_{t_i}| \\ = lim_{n \to \infty} (max_j|X_{t_{j+1}}-X_{t_j}|)\sum_{i=0}^{N(n)-1}|V_{t_{i+1}}-V_{t_i}|\\ = lim_{n \to \infty} (max_j|X_{t_{j+1}}-X_{t_j}|)\sum_{i=0}^{N(n)-1}(V_{t_{i+1}}-V_{t_i})$$ since V is increasing, and then as the sum is telescoping: $$= lim_{n \to \infty} (max_j|X_{t_{j+1}}-X_{t_j}|)(V_T-V_0)=0$$ since X is continuous, and the mesh of the partition is tending to zero, so $$max_j|X_{t_{j+1}}-X_{t_j}|\to 0$$

Since our proof here is for general processes X, V, with V increasing and X continuous, we thus have $$\langle X,V \rangle_T = \langle V \rangle_T = 0$$ for all T.

As for the conditions required, the proof should inform what conditions we need to impose. So provided is $$V_T-V_0$$ is finite, we should not even need to impose continuity on V (although for the proof we clearly do need continuity of X)