Local martingale but not martingale

Question

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.

Answer 1

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?

Answer 2

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!

Answer 3

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 $-1$.