The following is not completely clear to me: Let $\mathcal{H}^2$ be the set of all RCLL martingales $ M$ which are bounded in $L^2$, i.e. $\sup_tE[M_t^2]<\infty$. Then it is a standard result that $\mathcal{H}^2$ becomes a Hilbert space with the norm $\|M_\infty\|_{L^2}$. One can show that the subspace of all continuous martingales in $\mathcal{H}^2$, denoted with $\mathcal{H}_c^2$, is a closed subspace. Then the space of $\mathcal{H}^2_d$ is called the space of all purely discontinuous martingales and it is defined to be the orthogonal complement of $\mathcal{H}^2_c$, i.e. all $M\in\mathcal{H}^2$ such that

$$E[M_\infty N_\infty]=0$$ for all $N\in\mathcal{H}^2_c$. Of course you can denote with $\mathcal{H}^2_{loc}$ the localized version of the above spaces.

Now a general local martingale $M$ is called purely discontinuous if $MN$ is a local martingale for very continuous local martingale $N$. I was wondering what is the connection between these definitions. I assume that all local martingales satisfy $M_0=0$. Why are there two different definitions for the same "name"?


1 Answer 1


Actually, both definitions are equivalent. Some preliminaries:

Definition A subspace $\mathcal{N} \subseteq \mathcal{H}^2$ (endowed with the scalar product $\mathbb{E}(\langle \cdot,\cdot \rangle_{\infty})$) is called a stable subspace if

  1. $\mathcal{N}$ is a closed subspace of $\mathcal{H}^2$.
  2. $\int_0^{\cdot} f(s) \, dX_s \in \mathcal{N}$ for all $X \in \mathcal{N}$ and $f \in L^2(\langle X \rangle)$ previsible.

Denote by $\mathcal{N}^{\bot}$ the orthogonal complement of $\mathcal{N}$, i.e.

$$\mathcal{N}^{\bot} := \{Y \in \mathcal{H}^2; \forall X \in \mathcal{N}: \mathbb{E}(X_{\infty} \cdot Y_{\infty}) = 0\}$$

Then we can prove the following

Proposition Let $\mathcal{N}$ a stable subspace of $\mathcal{H}^2$. Then $$\mathcal{N}^{\bot} = \{Y \in \mathcal{H}^2; \forall X \in \mathcal{N},t \geq 0: \langle X,Y \rangle_t = 0\}$$ where $\langle X,Y \rangle$ denotes the previsible quadratic co-variation of $X$ and $Y$.

The subspace $\mathcal{H}_c^2$ of all continuous martingales in $\mathcal{H}^2$ is a stable subspace (you already mentioned the closedness and the second property follows easily from the properties of the stochastic integral). By applying the previous theorem, we obtain that

$$\begin{align*} \mathcal{H}_d^2 &:= (\mathcal{H}_c^2)^{\bot} \\ &:= \{Y \in \mathcal{H}^2; \forall X \in \mathcal{H}_c^2: \mathbb{E}(X_{\infty} \cdot Y_{\infty}) = 0\} \\ &= \{Y \in \mathcal{H}^2; \forall X \in \mathcal{H}_c^2, t \geq 0: \langle X,Y \rangle_t = 0\}\end{align*}$$

It is known that $\langle X,Y \rangle_t = 0$ is equivalent to $X \cdot Y$ being a martingale. Thus, we obtain

$$\mathcal{H}_d^2 = \{Y \in \mathcal{H}^2; \forall X \in \mathcal{H}_c^2: X \cdot Y \, \text{is a martingale}\}$$

i.e. a $\mathcal{H}^2$-martingale $Y$ is purely discontinuous if and only if $X \cdot Y$ is a martingale for any continuous $L^2$-martingale $X \in \mathcal{H}_c^2$.

Stopping yields the corresponding statement for the localized versions of the defined spaces.


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