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On wikipedia there is an example of a local martingale which is not a martingale, but I do not understand why it is a local martingale. We have the process

$ X_t = \begin{cases} W_{\min(t/(1-t),T)} &\text{for } 0 \le t < 1,\\ -1 &\text{for } 1 \le t < \infty. \end{cases}$

where $(W_t)$ is a standard Brownian motion and $T = \inf\{ t : W_t = −1 \}$.

The expectation is

$\mathbb{E} X_t = \begin{cases} 0 &\text{for } 0 \le t < 1,\\ -1 &\text{for } 1 \le t < \infty. \end{cases}$

This expectation is clearly discontinuous. So we have that it is not a martingale.

Now we will conclude that it is a local martingale with localizing sequence $ \tau_k = \min \{ t : X_t = k \}$ if there is such $t$, otherwise $τ_k = k$. However, I can not figure out why this is true.

I would appreciate some help.

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    $\begingroup$ Did you read the note "details 1" on the WP page? "Since the localizing sequence tends to infinity, it looks like the local martingale is in the limit the same as the martingale" Well it might "look like" this but what the example shows is that it is not like this. What kind of explanation are you expecting? $\endgroup$ – Did May 1 '15 at 13:20
  • $\begingroup$ Why X is not a martingale is fully explained on the WP page (hint: consider the identity just before the assertion that "This process is not a martingale."). My advice would be to reread fully and carefully the WP page and to come back here if some points remain unclear. Sorry but the ones you cite as unclear at present, are definitely not. $\endgroup$ – Did May 2 '15 at 18:37
  • $\begingroup$ @Did The point that it is a local martingale remains unclear. I would like to have some help on that. $\endgroup$ – user202723 May 4 '15 at 10:44
  • $\begingroup$ What did you try exactly to check that $(\tau_k)$ is a localizing sequence for $(X_t)$? $\endgroup$ – Did May 4 '15 at 11:09
  • $\begingroup$ @Did That it is a localizing sequence is clear (increasing and a.s. divergent), but not why the stopped processes are martingales. For the first case (the BM) it is clear, but not for $t\geq 1$. $\endgroup$ – user202723 May 4 '15 at 11:14
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I also think the wikipedia example a little bit confusing. It seems the localizing sequence can not give a martingale results since when $t\in(0,1), X_t^{\tau_n}$ is again a martingale with mean 0 and when $t\geq 1, $ it has mean -1. I have checked the reference book in that wikipedia page, and the book gives a different localizing sequence.

Details have been discussed in this thread, and I think it is quite clear.

How to show the following process is a local martingale but not a martingale?

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The point you are missing is that if the process is localized and you look at $t>1$. Then you have two possibilities: The process is stopped either at $-1$ or $K$. By the dominated convergence theorem, the expectation is equal to $E[X_0]$. If the process is not localized, then the process $X_t$ will take almost surely only value $-1$ for $t>1$.

How I understand this is the following:

The problem in this example is just that the large value with low probability will distort your expectation. Therefore, no matter for $t>1$ or $t<1$, you have to take these "large values" into account in order to obtain a martingale. In the case where the process is not localized, you just omit these "large values" for t>1 and think somehow that, for $t>1$, the process will hit -1 almost surely. And this cause a problem, since the process can also goes to minus infinity with positive possibly.(otherwise it is not a martingale) The localization sequence will correct the value of the process $X_t$ for $t>1$ by adding a certain "portion" of the process not stopping at $-1$.

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wrt your general question: A local martingale is not necessarily integrable, whereas a martingale is (by definition). This all boils down to the statement that a limit of a sequence of integrable random variables need not be integrable!

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