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!

  • $\begingroup$ 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. $\endgroup$
    – saz
    Jan 30, 2019 at 7:23
  • $\begingroup$ Cross-posted: quant.stackexchange.com/questions/43789/… $\endgroup$ Jan 30, 2019 at 7:43

2 Answers 2


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)


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