If $M$ is a martingale, why $M\mapsto \left<M\right>$ is a quadratic form?

Question

In the book of Schilling and Partzsch (Brownian motion, an introduction to stochastic process), page 206, they say that $$M\mapsto \left<M\right>,$$ is a Quadratic form where $M$ is a martingale and $\left<M\right>=\left(\left<M\right>_n\right)_n $ where $\left<M\right>_n$ is the unique adapte increasing process in the Doob decomposition $$M^2=M_0^2+N_n+\left<M\right>_n,$$ where $(N_n)$ is a martingale.

The thing I don't get it's that a quadratic form $q:V\to \mathbb F$ is a homogeneous polynomial of degree 2 where $V$ is a $\mathbb F-$ vector space. Here, $\left<M\right>$ is a stochastic process not an element of a field... so it's a quadratic form in which sense ?

Answer 1

We have an explicit expression for $\langle M\rangle_n$, namely $$ \langle M\rangle_n=\sum_{k=1}^{n-1}\mathbb E\left[M_k^2\mid\mathcal F_{k-1}\right]-M_{k-1}^2. $$ We have to check that for all $n$, $\langle aM\rangle_n=a^2\langle M\rangle_n$ and that $(M,N)\mapsto \langle M+N\rangle_n-\langle M\rangle_n-\langle N\rangle_n$ is bilinear. The first condition is easy to check. For the second, we look that $\langle M+N\rangle_n$, expand the square and notice that the terms $\mathbb E\left[M_k^2\mid\mathcal F_{k-1}\right]$ and $\mathbb E\left[N_k^2\mid\mathcal F_{k-1}\right]$ disappear after having substracted $\langle M\rangle_n$ and $\langle N\rangle_n$. Therefore, $$ \langle M+N\rangle_n-\langle M\rangle_n-\langle N\rangle_n=2 \left(\sum_{k=1}^{n-1}\mathbb E\left[M_kN_k\mid\mathcal F_{k-1}\right]-M_{k-1}N_{k-1}\right), $$ which is bilinear.

Answer 2

"Quadratic form" is to be understood in the following sense: $Q$ is a quadratic form on a $\mathbb{K}$-vector space $V$ if

$$B(M,N) := \frac{1}{4} (Q(M+N) - Q(M-N))$$

defines a bilinear form on $V \times V$. It is equivalent to saying that $$Q(\alpha M) = \alpha^2 Q(M) \quad \text{and} \quad Q(M+N)+Q(M-N) = 2Q(M) + 2Q(N) $$ for all $\alpha \in \mathbb{K}$, $M,N \in V$. Note that any bilinear form $B$ gives rise to a quadratic form $Q$; just set $Q(M) := B(M,M)$. See this article by Gleason for some further material on the topic.

See this question here for a proof that the quadratic variation $\langle M \rangle$ is a quadratic form on the space of square integrable martingales (in the sense of the above definition).