• $T>0$
  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $(\mathcal F_t)_{t\in[0,\:T]}$ be a filtration on $(\Omega,\mathcal A)$
  • $M$ be an almost surely continuous local martingale on $(\Omega,\mathcal A,\mathcal F,\operatorname P)$ with $M_0=0$ almost surely and $$N:=\left\{\omega\in\Omega:X(\omega)\text{ is not continuous}\right\}\cap\left\{X_0\ne0\right\}$$
  • $n\in\mathbb N$, $\tau_0^n:=0$ and $$\tau_k^n:=\inf\left\{t\in(t_{k-1}^n,T]:\left|M_t-M_{\tau_{k-1}^n}\right|=\frac1{2^n}\right\}\wedge T\;\;\;\text{for }k\in\mathbb N$$ with $\inf\emptyset:=\infty$

We can show that $\tau_k^n$ is an $\mathcal F$-stopping for all $k\in\mathbb N$ with $$\tau_k^n\uparrow T\;\;\;\text{for }k\to\infty\tag1$$ on $\Omega\setminus N$. Now, let $$X_t:=\left\{\begin{array}{{{\displaystyle}}l}\displaystyle\sum_{k\in\mathbb N}M_{\tau_{k-1}^n}\left(M_{\tau_k^n\:\wedge\:t}-M_{\tau_{k-1}^n\:\wedge\:t}\right)&&\text{on }\Omega\setminus N\\0&&\text{on }N\end{array}\right\}\;\;\;\text{for }t\in[0,T]\;.$$

We can show that the sum in the definition of $X(\omega)$ has only finitely many terms, for all $\omega\in\Omega\setminus N$.

How can we show that $X$ is a martingale on $(\Omega,\mathcal A,\mathcal F,\operatorname P)$?

You may assume that $M$ is bounded (and hence a martingale), if it's not possible to show the statement otherwise.


Show that each term in the sum is a martingale. To wit, a stopped local martingale is a local martingale, and since $M_{\tau_k^n\wedge t}$ is a.s. bounded (by $k2^{-n}$), each $M_{\tau_k^n\wedge t}$ is a martingale, as is the difference $M_{\tau_k^n\wedge t}-M_{\tau_{k-1}^n\wedge t}$. As this difference vanishes on $[0,\tau_{k-1}^n]$, when you multiply it by a bounded $\mathcal F_{\tau_{k-1}^n}$-measurable random variable, it is still a martingale.

  • $\begingroup$ I'vse asked this question cause I want to unterstand the existence proof of the covariation presented in the book of Kallenberg (Google Books link). $X$ is called $V^n\cdot M$ there. In a first step, where he's showing that $V^n\cdot M$ is a martingale, he's explicitly assuming that $M$ is bounded. In light of your answer: Why? I could imagine that the problem is that he's considering the index set $[0,\infty)$ (instead of $[0,T]$) and hence the martingales in the definition of the martingale space (see Lemma 15.4) need to be $L^2$-bounded. Is that the reason? $\endgroup$ – 0xbadf00d Nov 1 '16 at 16:50
  • $\begingroup$ And please take a look at my other, related, question too. Thank you in advance for your efforts. $\endgroup$ – 0xbadf00d Nov 1 '16 at 17:12
  • $\begingroup$ It's hard to read his mind; perhaps the boundedness hypothesis is made to simplify the argument. $\endgroup$ – John Dawkins Nov 1 '16 at 20:11
  • $\begingroup$ For completeness: He uses that $M$ itself is a martingale whenever $M$ is bounded. $\endgroup$ – 0xbadf00d May 18 '17 at 18:02

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