• $(\Omega,\mathcal A,\operatorname P)$ be a complete probability space
  • $(\mathcal F_t)_{t\ge0}$ be a complete and right-continuous filtration on $(\Omega,\mathcal A)$
  • $U,H$ be infinite-dimensional separable $\mathbb R$-Hilbert spaces
  • $(e_n)_{n\in\mathbb N}$ and $(f_n)_{n\in\mathbb N}$ be orthonormal bases of $U$ and $H$, respectively
  • $M$ and $N$ be $U$-valued and $H$-valued local $\mathcal F$-martingales on $(\Omega,\mathcal A,\operatorname P)$, respectively
  • $M^n:=\langle M,e_n\rangle_U$ and $N^n:=\langle N,f_n\rangle_H$ for $n\in\mathbb N$

By the Kunita-Watanabe inequality, we obtain $$\left|[M^m,N^n]_t-[M^m,N^n]_s\right|\le\sqrt{[M^m]_t-[M^m]_s}\sqrt{[N^n]_t-[N^n]_s}$$ for all $t\ge s\ge0$ almost surely for all $m,n\in\mathbb N$. Hence, $$[\![M,N]\!]_t:=\sum_{(m,\:n)\in\mathbb N^2}[M^m,N^n]_te_m\otimes f_n$$ exists for all $t\ge0$ almost surey in the sense of summability in $\operatorname{HS}(U,H)$ (Hilbert-Schmidt operators from $U$ to $H$).

$[M^m,N^n]$ has paths of locally bounded variation for all $m,n\in\mathbb N$. Are we able to show that the paths of $[\![M,N]\!]$ have locally bounded variation too?

Note that $(e_m\otimes f_n)_{(m,\:n)\in\mathbb N^2}$ is an orthonormal basis of $\operatorname{HS}(U,H)$ and hence the variation $\operatorname{Var}_\varsigma[\![M,N]\!]$ of $[\![M,N]\!]$ along a division $\varsigma=(t_0,\ldots,t_k)$ of $[s,t]$ is given by \begin{equation}\begin{split}\operatorname{Var}_\varsigma[\![M,N]\!]&=\sum_{i=1}^k\left\|[\![M,N]\!]_{t_i}-[\![M,N]\!]_{t_{i-1}}\right\|_{\operatorname{HS}(U,\:H)}\\&\le\sum_{(m,\:n)\in\mathbb N^2}\sum_{i=1}^k\left|[M^m,N^n]_{t_i}-[M^m,N^n]_{t_{i-1}}\right|\\&=\sum_{(m,\:n)\in\mathbb N^2}\operatorname{Var}_\varsigma[M^m,N^n]\;.\end{split}\end{equation} My problem is to find a bound of $\operatorname{Var}_\varsigma[M^m,N^n]$ which is summable over $(m,n)$. By the Kunita-Watanabe inequality, we obtain $$\operatorname{Var}_\varsigma[M^m,N^n]\le\sum_{m\in\mathbb N}\sqrt{[M^m]_t-[M^m]_s}\sum_{n\in\mathbb N}\sqrt{[N^n]_t-[N^n]_s}\;.$$ But are these sums finite?

Note that $[M^m]$ and $[N^n]$ are summable over $m$ and $n$, respectively, with $$\sum_{m\in\mathbb N}[M^m]=[M]$$ and $$\sum_{n\in\mathbb N}[N^n]=[N]\;.$$


Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.