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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).2012-11-08
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    @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.2012-11-08
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    Sorry, I think the correct term is localizing sequence, see e.g. [this](http://almostsure.wordpress.com/2009/12/23/localization/). 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).2012-11-08
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    @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.2012-11-09

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