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The Original Question: Demonstrate that every martingale is a local martingale.

Attempt at a Solution:

Consider the standard setup of this problem: $\mathscr{F}_t$ is the filtration that satisfies the normal conditions, $(\sigma_n)_{n\in \mathbb{N}}$ is the monotone increasing sequence of stopping times with $\text{lim}_{n \to \infty}\sigma_n = \infty$ and $\{X_t\}_{t \in [0,\infty]}$ is a martingale.

To show that $X$ is a local martingale I will show that $E[X_\infty | \mathscr{F}_t] = X_{min(t , \sigma_n)}$.

We know that $X_\infty = X_{min(\infty,\sigma_n)}=X_{\sigma_n}$ because $\sigma_n \in \mathbb{R}_+$. It then follows that:

$E[X_{\sigma_n}|\mathscr{F}_t] = X_{\sigma_n} = X_{min(t,\sigma_n)}$.

However this incorrect for any $t < \sigma_n$, and so $\exists$ $\geq 1$ mistakes in my attempted solution.

MY Question: Am I correct? If not, what's wrong with my solution, and what's some good hints to a correct solution (or just post the correct solution).

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You just have to show the existence of one particular localization $(\sigma_n)_{n\in\mathbb{N}}$ such that $(X^{\sigma_n}_t)_{t\geq 0}$ is a martingale. It seems to me you're on the way to showing that $(X^{\sigma_n}_t)_{t\geq 0}$ is a martingale for every localization $(\sigma_n)_{n\in\mathbb{N}}$ (which does not hold). – Stefan Hansen Nov 8 '12 at 17:26
@StefanHansen Could you please dumb down your criticism of my attempt? I wikipedia'd localization but couldn't relate it to what you're saying. – Jase Nov 8 '12 at 17:45
Sorry, I think the correct term is localizing sequence, see e.g. this. It's just a term meaning any sequence $(\sigma_n)_{n\in\mathbb{N}}$ of stopping times increasing to $\infty$, which is used to define local martingales (and other local properties). – Stefan Hansen Nov 8 '12 at 18:00
@StefanHansen I think I don't understand what you mean by "one particular" sequence $(\sigma_n)_{n \in \mathbb{N}}$. Guessing what you mean; we can specify that each $\sigma_n = \text{inf}\{t \geq 0 : X_t = n\}$ where $X_{min(t,\sigma_n)}$ is the stopped process. However I can't see how this can help me solve the problem. – Jase Nov 9 '12 at 5:13

This is too long to be a comment, so I posted it as an answer instead. Let me first introduce the definitions I use of a martingale, local martingale and localizing sequence.

A sequence $(\sigma_n)_{n\in\mathbb{N}}$ is called a localizing sequence if $\sigma_n$ is a stopping time for all $n\in\mathbb{N}$ and $\sigma_n\uparrow \infty$ a.s.

A stochastic process $(X_t)_{t\geq 0}$ is called a martingale (wrt. $(\mathscr{F}_t)_{t\geq 0}$) if it is integrable, adapted and satisfies $$ E[X_t\mid\mathscr{F}_s]=X_s \quad \text{a.s.} $$ for every $0\leq s<t$.

A stochastic process $(X_t)_{t\geq 0}$ is called a local martingale if there exists a localizing sequence $(\sigma_n)_{n\in\mathbb{N}}$ such that the stopped process $(X_t^{\sigma_n})_{t\geq 0}=(X_{t\wedge \sigma_n})_{t\geq 0}$ is a martingale for every $n\in\mathbb{N}$.

Now, you're starting with a martingale $(X_t)_{t\geq 0}$ and in order to show that it is also a local martingale, you just have to find a localizing sequence $(\sigma_n)_{n\in\mathbb{N}}$ such that $(X_{t\wedge\sigma_n})_{t\geq 0}$ is a martingale. Your task is now to find such sequence using the fact that $(X_t)_{t\geq 0}$ already is a martingale. To this end, note that a stopping time is a mapping from $\Omega$ into $\mathbb{R}_+\cup \{\infty\}$ and not into $\mathbb{R}_+$.

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Hi I think that by taking $\sigma_n=n$ as a localizing sequence, it is clear from the fact that a stopped martingale by a bounded stopping time stays a martinagle, that every martingales are local martingales. Best regards – TheBridge Nov 9 '12 at 13:14
You could also just pick $\sigma_n=\infty$ for all $n$. – Stefan Hansen Nov 9 '12 at 15:24

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