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Feb 12, 2021 at 2:12 comment added Iosif Pinelis The continuity condition does not help at all -- just take $M_n=M_1$, where $M_1$ is any continuous non-Gaussian martingale with $M_1(0)\sim N(0,\sigma^2)$ and a finite (and hence zero) quadratic variation. For instance, you may take $M_1(t):=Z+B(t\wedge\tau)$, where $B$ is a standard Brownian motion independent of $Z\sim N(0,1)$ and $\tau:=\inf\{t\ge0\colon|M_1(t)|\ge1\}$.
Feb 11, 2021 at 22:50 history edited YCor CC BY-SA 4.0
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Feb 11, 2021 at 22:16 history edited Suman Chakraborty CC BY-SA 4.0
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Feb 11, 2021 at 22:15 comment added Suman Chakraborty Many thanks for pointing out this error in my question. $\sigma(t)$ is continuous. I am correcting the statement. Any help is appreciated.
Feb 11, 2021 at 20:44 comment added Iosif Pinelis Some conditions are missing here. Indeed, suppose e.g. that for all $n$ we have $M_n=M_1$ and $M_1$ is a non-Gaussian martingale with $M_1(0)\sim N(0,\sigma^2)$. Then all the conditions hold but the conclusion does not. The same comment applies to Proposition 9.1 by Svante Janson.
Feb 11, 2021 at 20:18 history asked Suman Chakraborty CC BY-SA 4.0