1
$\begingroup$

Let

  • $T>0$
  • $(\Omega,\mathcal A,\operatorname P)$ 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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.