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Let $(\Omega, \mathcal F, P)$ be a probability space with sigma algebras $\mathcal F_{n, t}$ for $n \in \mathbb N$, $t \in [0, 1]$, where for all $n$, $\mathcal F_{n, t} \subset \mathcal F_{n, h}\ $ whenever $h > t$. Assume also for each $n$, the filtration $\mathcal \{ \mathcal F_{n, t} \}$ is right continuous in $t$. Finally, assume $\mathcal F_{0, 0}\ $ contains all $P$-null sets.

Let $\mathcal G_t := \sigma( \{ \mathcal F_{n, t} \}_{n \in \mathbb N} \ ) $ be the sigma algebra generated by $ \mathcal F_{n, t}$ for fixed $t$.

Given a sequence $M_n$ of $\{ \mathcal F_{n, t} \}$-martingales such that $\sup_{n \in \mathbb N} \int |M_n (t, \omega)| dt dP < \infty$,

define for each $k \in \mathbb Z^+$, the process $S_k$ on $(0, 1]$ given by

$S_k (t) := \frac{1}{2^k t} \sum_{h = 0}^{2^k - 1} 1_{[h/2^k, (h+1)/2^k]} (t) \sum_{j = 0}^{h-1} M_j (t).$

As the $S_k$ are equibounded in $L^1 (\Omega \times (0, 1])$ , it follows from a result of Komlos that a subsequence of the $S_k$ Cesaro converges in $L^1 (\Omega \times (0, 1])$ to an “almost” $L_1$ process, say $X$ - in the sense that $X_t$ is in $L^1 (\Omega)$ for (Lebesgue) almost every $t$. In what follows we restrict the process $X$, and the filtration $\mathcal G_t$ to those times $t$.

Question: Is $X$ a $\mathcal G_t$-martingale?

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  • $\begingroup$ Your $S_k(t)$ does not depend on $t$. $\endgroup$ – Iosif Pinelis Apr 7 at 13:39
  • $\begingroup$ So, it’s left implicit in the indicator function, but it’s not obvious as is so I’ll change it. $\endgroup$ – Nate River Apr 7 at 15:09
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    $\begingroup$ If $M_j = M$ is independent of $j$, then $X(t) = t M(1)$ so no. It's not even adapted to ${\mathcal G}$... $\endgroup$ – Martin Hairer Apr 7 at 15:17
  • $\begingroup$ Sorry, why is $M_j$ independent of $j$? $\endgroup$ – Nate River Apr 7 at 23:14
  • $\begingroup$ Oh I understand, I guess one would also need that the $\mathcal F_{j, t}$ are independent of $\mathcal F_{h, t}$ for $h > j$? $\endgroup$ – Nate River Apr 7 at 23:29

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