3
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Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $(\mathcal F_t)_{t\ge0}$ be a complete filtration of $\mathcal A$
  • $H$ be a separable $\mathbb R$-Hilbert space
  • $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $H$
  • $X$ be an $L^2$-bounded almost surely continuous $H$-valued $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$ with $X_0=0$ almost surely and $$X^n:=\langle X,e_n\rangle_H\;\;\;\text{for }n\in\mathbb N$$

We can show that $([X^m,X^n]_te_m\otimes e_n)_{(m,\:n)\in\mathbb N^2}\subseteq\text{HS}(H)$ is summable for all $t\ge0$ almost surely and $$A_t:=\sum_{(m,\:n)\in\mathbb N^2}[X^m,X^n]_te_m\otimes e_n\;\;\;\text{for }t\ge0$$ is self-adjoint for all $t\ge0$ almost surely.

How can we show that $A_t$ is a nonnegative operator, i.e. $$\langle A_tx,x\rangle_H\ge0\;\;\;\text{for all }x\in H\;,\tag1$$ for all $t\ge0$ almost surely?

Let $x\in H$. Basic results about summability and the quadratic covariation yield $$\langle A_tx,x\rangle_H=\sum_{m\in\mathbb N}\sum_{n\in\mathbb N}\left[\langle X,\langle x,e_m\rangle_He_m\rangle_H,\underbrace{\langle X,\langle x,e_n\rangle_He_n\rangle_H}_{=:\:Y^n}\right]_t\tag2$$ for all $t\ge0$ almost surely. Now, $$\sum_{n\in\mathbb N}Y^n=\langle X,x\rangle_H\tag3\;,$$ but I wasn't able to conclude something useful from $(3)$ for $(2)$.

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  • $\begingroup$ Will a mimic of Bochner's theorem work? Have you try? $\endgroup$ – Henry.L Jul 24 '17 at 12:29

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