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For each $n$ let $M_{n}(t)$ be a martingale on $[0,\infty)$ and $\mathbb{E}(M_n(0))=0$. Also $\sigma(t)\geq 0$ be a continuous function such that $$ [M_n,M_n]_t \xrightarrow{p} \sigma(t), \,\,\&\,\, \sup_{n}\mathbb{E}([M_n,M_n]_t)<\infty, $$ as $n\rightarrow \infty$, where $[M_n,M_n]_t =\sum_{0<s\leq t}(M_n(s)-M_n(s-))^2$. Furthermore assume that $$ M_n(0)\xrightarrow{d} N(0,\sigma^2), $$ as $n \rightarrow \infty$. Can we prove that $M_n\xrightarrow{d}M$, in $D[0,\infty)$ where $M(t)$ is a Gaussian martingale with $\mathbb{E}(M(t))=0$ and $\mathbb{V}(M(t))= \sigma(t)+\sigma^2$? (we can assume all moments of $M_n$ are uniformly bounded). Otherwise, can the limit $M$ be different (or does not exists)?

Note: If $M_n(0)=0$ then the assertion holds with $\sigma=0$ (Proposition 9.1 in ``Functional limit theorems for multitype branching processes and generalized Pólya urn" by Svante Janson).

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    $\begingroup$ 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. $\endgroup$
    – Iosif Pinelis
    Commented Feb 11, 2021 at 20:44
  • $\begingroup$ Many thanks for pointing out this error in my question. $\sigma(t)$ is continuous. I am correcting the statement. Any help is appreciated. $\endgroup$
    – Suman Chakraborty
    Commented Feb 11, 2021 at 22:15
  • $\begingroup$ 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 $\tau:=\inf\{t\ge0\colon|M_1(t)|\ge1\}$. $\endgroup$
    – Iosif Pinelis
    Commented Feb 12, 2021 at 2:12

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