Example of a random variable $X$ that is an $\mathscr{F}_t$-local martingale, but not an $\mathscr{F}_t^X$-local martingale.

Question

This is a problem from Ethier and Kurtz' Markov Processes. The book introduces some theorems on local martingales but they all involve the process being right continuous. I think this problem must be solved using the definitions of a local martingale, that is, there is $\mathscr{F}_t$ stopping times $\tau_1\le \tau_2 \le \cdots$ with $\tau_n \to \infty$ a.s. such that $X^{\tau_n}$ is a martingale. However, I am new to this concept and am lost on how to solve this problem. I would greatly appreciate some help.

Let $\eta$ and $\xi$ be independent random variables with $P(\eta=1)=P(\eta=-1)=\frac{1}{2}$ and $E|\xi|=\infty$. Define

$$X(t) = \begin{cases} 0, & 0\le t<1, \\ \eta \xi, & t\ge 1, \end{cases}$$ and $$ \mathscr{F}_t = \begin{cases} \sigma(\xi), & 0\le t<1 \\ \sigma(\xi,\eta), & t\ge 1. \end{cases}$$

Show that $X$ is an $\mathscr{F}_t$-local martingale, but that $X$ is not an $\mathscr{F}_t^X$-local martingale.

Answer 1

As noticed by @JohnDawkins $\{X_t\}$ is a local martingale w.r.t. to the sequence of stopping times $\{n1_{\{|\xi|\le n\}}\}$. However, $\{X_t\}$ cannot be a local martingale w.r.t. the filtration $(\mathcal{F}_t^X)$ because for any $(\mathcal{F}_t^X)$-stopping time $\tau$, $\{\tau<1\}\in \bigvee_{t<1}\mathcal{F}_t^X=\{\emptyset,\Omega\}$. Thus, for $t<1$, $$ \mathsf{E}|X_{1\wedge\tau}|=\mathsf{E}[|X_1|1\{\tau\ge 1\}]=\mathsf{E}[|X_1|]1\{\tau\ge 1\}, $$ which implies that there is no a sequence of $(\mathcal{F}_t^X)$-stopping times $\tau_n\uparrow\infty$ a.s. such that $X^{\tau_n}$ is a martingale for each $n$.

Answer 2

Because $E|X_t|=E|\xi|=\infty$ for $t\ge 1$, $\{X_t\}$ is not a martingale (with respect to any filtration).

Define $\tau_n:=n$ on $\{|\xi|\le n\}$ and $\tau_n=0$ on $\{|\xi|>n\}$. Then $\{\tau_n\}$ is an increasing sequence of $(\mathcal F_t)$ stopping times (note that $\xi$ is $\mathcal F_0$-measurable) with limit $\infty$ on $\{|\xi|<\infty\}$. I assume that $P(|\xi|<\infty)=1$. Because, for $t\ge 1$, $X_{t\wedge \tau_n} = \eta\xi1_{\{|\xi|\le n\}}$, the stopped process $X^{\tau_n}$ is an $(\mathcal F_t)$-martingale. This shows that $X$ is an $(\mathcal F_t)$ local martingale. But, for $n$ sufficiently large, the random variable $\tau_n$ is not an $({\mathcal F}^X_t)$ stopping time, because ${\mathcal F}^X_0=\{\emptyset,\Omega\}$ but $\{\tau_n=0\}=\{|\xi|\le n\}$.

As @d.k.o shows, $(X_t)$ does not admit a localizing sequence of $({\mathcal F}^X_t)$ stopping times, and so $(X_t)$ cannot be an $({\mathcal F}^X_t)$ local martingale.