Fix a filtered probability space satisfying the usual conditions.

Let $\mathcal{M}^2_0$ be the vector space of cadlag martingales null at $0$ bounded in $L^2$.

We state without proof the following theorem.

THEOREM For all $M \in \mathcal{M}^2_0$ there exists a unique process $[M]$, called the quadratic variation of $M$, such that

  • $[M]$ is càdlàg increasing and null at 0
  • $M^2-[M]$ is a uniformly integrable martingale
  • $\Delta[M]_t = (\Delta M_t)^2 \quad \forall t > 0$

LEMMA $[cM] = c^2[M].$

DEFINITION For $M$ and $N$ in $\mathcal{M}^2_0$, we define $[M,N] := \frac{1}{4}([M+N]-[M-N])$

LEMMA $[M,M] = [M]$

LEMMA For all $M$ and $N$ $\in \mathcal{M}^2_0$, [M,N] is the unique process such that

  • $[M,N]$ is càdlàg of FV and null at 0
  • $MN-[M,N]$ is a uniformly integrable martingale
  • $\Delta [M,N]_t = \Delta M_t \Delta N_t \quad \forall t > 0$

QUESTION: How to prove that $(M,N)\mapsto[M,N]$ is bilinear?

  • $\begingroup$ btw: I am not able to write accents in LaTeX. "c\`adl\`ag" does not work. Also, how to use the amsthm package? $\endgroup$ – Tom Jan 22 '14 at 18:18

$[\lambda M,\mu N]$ is the unique process such that

  1. $[\lambda M,\mu N]$ is of finite variation and $[\lambda M, \mu N]_0=0$
  2. $(\lambda M) \cdot (\mu N) - [\lambda M,\mu N]$ is a (uniformly integrable) martingale
  3. $\Delta [\lambda M,\mu N]_t = (\Delta \lambda M)_t \cdot (\Delta \mu N)_t$$

In order to show $[\lambda M,\mu N] = \lambda \mu [M,N]$ it suffices to show that $\lambda \mu [M,N]$ satisfies (1)-(3).

  1. Obviously, $\lambda \mu [M,N]$ is of finite variation as $[M,N]$ is of finite variation.
  2. Multiplying with a constant preserves the martingale property; hence $\mu \lambda M \cdot N- \lambda \mu [M,N]$ is a (uniformly integrable) martingale.
  3. By definition, $$\Delta (\lambda \mu [M,N]_t) = \lambda \mu \Delta[M,N]_t = \Delta (\lambda M)_t \cdot \Delta (\mu N)_t$$

The proof of the additiy is similar, I leave it to you.

  • $\begingroup$ thanks a lot! Actually it was very easy, I don't know why I didn't see it. $\endgroup$ – Tom Feb 2 '14 at 15:35

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