Difference Between Local Martingale and Martingale

Question

I would like to try and distinguish between the two concepts of martingale and local martingale. I have read this answer in Martingale / local martingale : some confusion which was a good start and wanted to check the following idea:

If we have a local martingale $(M_t)$ then by definition there exists a sequence of stopping times $\tau_k$ such that $M_t^{\tau_k}$ is a martingale, with $\tau_k \rightarrow \infty$ almost surely and in an increasing manner. Given this fact, can we not argue that for any local martingale the process $M_t$ is a martingale on $[0,\infty)$.
$\textbf{Why?}$ Well for any finite time $t$, I can find a $k$ such that the stopping time $\tau_k$ with $t<\tau_k$ a.s., since the stopping times increase to $\infty$, so then $M_t^{\tau_k}=M_t$, and hence my process is a martingale $\textbf{for all finite times}.$

Now my attempt to explain the distinction between a martingale and local martingale is that the limiting behaviour of the process $(M_t)$ at $t=\infty$ is where the problem arises. So we cannot talk about the process having some 'ending value' $M_{\infty}$. I have heard the famous martingale property of expectation can fail with local martingales, so does this also relate to some weird behaviour of the local martingale process $(M_t$) at the end.

Answer 1

Your observation is correct - martingales are actually very similar to local martingales (and I would say that martingales are closer to local martingales than they are to uniformly integrable martingales). Before proceeding further, I would first recommend looking at the (typically easier to grasp) notion of a $\sigma$-finite measure.

Let $\mu$ be a measure on a measurable space $(X,\mathscr{A})$. $\mu$ is said to be $\sigma$-finite if there is an increasing sequence $\{X_{n}\}_{n}\subset\Sigma$ such that $\mu|_{X_{n}}$ is a finite measure, and $X=\bigcup_{n=1}^{\infty}X_{n}$. The idea is that, although $\mu(X)$ may fail to be finite, it is still possible to treat $\mu$ like a finite measure.

Hopefully, you are at least familiar with the notion of stochastic integration. Given a sufficiently regular process $X$, one may define the stochastic integral $(H\cdot X)$ for a large class of integrands $H$ (assumed to be measurable with respect to the predictable $\sigma$-algebra). Now, the stochastic integral allows us to define something like a measure on the predictable $\sigma$-algebra as follows \begin{equation*} D\longmapsto (\mathbf{1}_{D}\cdot X), \end{equation*} where $D\subset [0,\infty)\times\Omega$ is predictable. Inspired by the above definition, we may be tempted to say that $X$ is "like" a martingale if there is an increasing sequence $\{D_{n}\}_{n}$ of predictable sets (which cover $[0,\infty)\times\Omega$) such that $(\mathbf{1}_{D_{n}}\cdot X)$ is a martingale for each $n$. Now, if $D_{n}=\{(t,\omega):0\leq t\leq T_{n}(\omega)\}$ for some sequence $\{T_{n}\}_{n}$ of stopping times, then one recovers the notion of a local martingale from this heuristic discussion. Thus, local martingales are like $\sigma$-finite measures, in that they are essentially martingales, but not forced to hold on all (some notion of measurable) subsets of $[0,\infty)\times\Omega$.

Thus, the importance of local martingales comes from exactly the same reason as $\sigma$-finite measures. Although they are not in general martingales, they have enough of their essential properties to be treated as martingales in the proofs of theorems (exactly like $\sigma$-finite measures compared to finite measures).