Local martingale that is not integrable

Question

I have just been introduced to the notion of a local martingale, and I have asked myself the question whether the fact that $(X_{t})_{t\geq 0}$ is a local martingale with associated $\{\tau_{n}\}_{n}$ stopping times that go to infinity implies that for every $X_{t}$ is indeed integrable.

I cannot prove it and thus I believe it is false but I am struggling to find counterexamples. Could anyone give me a hint as to what such a process $X$ and the stopping times would look like so that they are indeed a local martingale however not integrable.

Answer 1

This is probably not the most exotic example, but it is indeed an example of a non-integrable local martingale:

Let $X_0$ be any non-integrable random variable and consider the proces $$X_t = X_0 \quad \text{ for all $t\geq 0$}.$$ Now if we let $$\tau_n = \begin{cases} 0 &, \text{ if } X_0 > n \\ \infty & , \text{ if }X_0 \leq n \end{cases},$$ then $(\tau_n)_{n\in \mathbb{N}}$ is an increasing sequence of stopping times going to infinity and the stopped proces $$X_{\min(\tau,t)} 1_{\{\tau_n > 0 \}} = X_0 1_{\{X_0 \leq n\} },$$ is bounded, constant (with respect to $t$) and thus a martingale. We conclude, that the proces $(X_t)_{t\geq 0}$ is indeed a local martingale.

From this odd example we can actually construct a slightly more interesting example. Let $(X_t)_{t \geq 0}$ be as above and consider a true martingale $(Y_t)_{t\geq 0}$ and let $$Z_t = X_t + Y_t.$$ Then $Z_t$ is not integrable, but since the sum of local martingales is again a local martingale, we actually get that $(Z_t)_{t\geq 0}$ is a local martingale.

Answer 2

Suppose $(B_t)_{t\ge 0}$ is a standard Brownian motion, with its natural filtration $(\mathcal F_t)_{t\ge 0}$.

Example: Let $H$ be $\mathcal F_1$ measurable, but not integrable. R.M. Dudley has shown ["Wiener functionals as Itô integrals," Ann. Probability, 5 (1977), pp. 140–141); http://www.jstor.org/stable/pdf/2242810.pdf] that the $\mathcal F_1$ measurability of $H$ alone ensures the existence of a progressive (even predictable) process $(K_s)_{0\le s\le 1}$ such that the stochastic integral $\int_0^1 K_s\,dB_s$ exists and is equal to $H$ a.s. Define $$ L_s:=\cases{K_s,&$0\le s<1$;\cr 0,& $s\ge 1$.\cr} $$ The associated stochastic integral process $$ M_t:=\int_0^t L_s\,dB_s $$ is then a local martingale. By design, $M_1=H$ is not integrable. In particular, $(M_t)$ is not a martingale.