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Let $\mathscr{H}_0^2$ the space of all $L^2$ martingales $X$ starting at zero s.t. $\sup_{t\geq 0}E[X_t^2]<\infty$.

Two martingales $N,M$ are said to be strongly orthogonal, if $NM$ is a uniformly integrable martingale. In Protter, Philip: Stochastic Integration and Differential Equation, there is a characterization for strongly orthogonal martingales (in chapter IV.3 after the definition of strongly orthogonal martingales).

Let $N,M\in \mathscr{H}_0^2$. Then $N,M$ are strongly orthogonal iff $[N,M]$ is a uniformly integrable martingale.

$"\Rightarrow"$ By definition, $[N,M]=NM-N_{-}\cdot M-M_{-}\cdot N$. $NM$ is a martingale by assumption, but are $N\cdot M$ and $M\cdot N$ martingales (they are clearly local martingales)?

$[N,M]$ is the unique process s.t. $NM-[N,M]$ is a local martingale. By assumption, we have $[N,M]=-(NM-[N,M]-NM)$ is a local martingale. Protter says, it is a uniformly integrable martingale by the Kunita-Watanabe inequality. But i don't get it?

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1 Answer 1

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To answer your question, mention several facts.

1. Let $X$ be a local martingale. If $\mathsf{E}[X^\ast_\infty]<\infty$, then $X$ is uniformly integrable martingale. (ibid. Th.I.51, p.38)

2. Suppose $X,Y\in\mathcal{H}^2$, then(ibid. Th.II.25 and Cor., p.69-70. and take H=K=1) \begin{gather} \mathsf{E}[([X,Y])^\ast_\infty]\le\mathsf{E}\biggl[\int_0^\infty|d[X,Y]_s|\biggr]\le (\mathsf{E}[X,X]_\infty)^{1/2}(\mathsf{E}[Y,Y]_\infty)^{1/2} =\|X\|_{\mathcal{H}^2}\|Y\|_{\mathcal{H}^2}<\infty.\tag{1}\\ \mathsf{E}[X^\ast_{\infty}Y^\ast_{\infty}]\le \{\mathsf{E}[(X^\ast_{\infty})^2]\}^{1/2}\{\mathsf{E}[(Y^\ast_{\infty})^2]\}^{1/2} \le 4\|X\|_{\mathcal{H}^2}\|Y\|_{\mathcal{H}^2}<\infty. \tag{2} \end{gather}

3. Let $X,Y\in\mathcal{H}^2$ be two martingale, then using integration by parts we have (ibid. Cor.II.2, p.68) $$ XY=X_-\centerdot Y+Y_-\centerdot X+[X,Y].$$ and $XY$ is a local martingale iff $[X,Y]$ is a local martingale, since the stochastic integrals $X_-\centerdot Y,Y_-\centerdot X$ are local martingale.

$\impliedby$ Since $[X,Y]$ is a local martingale, then $XY$ is a local martingale and uniformly integrable(by (2)).

$\implies$ Since $XY$ is a local martingale, then $[X,Y]$ is a local martingale and uniformly integrable(by (1)).

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  • $\begingroup$ For $"\Rightarrow"$, how do we get that $E[([X,Y])^*_\infty]<\infty$ in order to show the uniform integrability of $[X,Y]$ using (1) and 1.? Is it true to say $E[([X,Y])^*_\infty]=E[\int_0^\infty|d[X,Y]_s|]$? $\endgroup$
    – peer
    Commented Jun 19, 2017 at 17:45
  • $\begingroup$ @peer, Thank you for your replication. I add a expression in the (1). Indeed, $\endgroup$
    – JGWang
    Commented Jun 20, 2017 at 0:37
  • $\begingroup$ @peer, Thank you for your replication. I add a expression in the (1). Indeed, $|[X,Y]_t|\le \int_0^t|d[X,Y]_s| \le \int_0^\infty|d[X,Y]_s|$. $\endgroup$
    – JGWang
    Commented Jun 20, 2017 at 0:43
  • $\begingroup$ Thanks. Is the uniform integrability of $[N,\tilde{N}]$ and the martingale proberty for $[N,\tilde{N}]$ for $N,\tilde{N}\in\mathcal{H}^2_0$ equivalent to $<N,\tilde{N}>=0$? $\endgroup$
    – peer
    Commented Jun 22, 2017 at 13:02
  • $\begingroup$ @peer: Indeed: If $M,N\in\mathcal{H}^2_{\text{loc}}$, then $\langle M,N\rangle=0 \iff [M,N]\in \mathcal{M}_{\text{loc}}$; If $M,N\in\mathcal{H}^2_0$, then $(MN)^\ast_{\infty}, ([M,N])^\ast_{\infty}, \int_0^\infty|d[M,N]|\in L^1$. What you want is put above statements togather. $\endgroup$
    – JGWang
    Commented Jun 23, 2017 at 1:28

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