Is a square-integrable continuous local martingale a true martingale?

Question

I am wondering, if the following is true without any other assumptions and if so, how to prove it:

Let $(M_t)_{t \geq 0}$ be a continuous local martingale on a filtered probability space $(\Omega, \mathcal{F}, (\mathcal{F}_t)_{t\geq 0}, P)$ that satisfies the usual conditions.

If $(M_t)_{t\geq 0}$is square integrable (that is, for all $t\geq 0$ we have $E[M_t^2] < \infty$) then $(M_t)_{t\geq 0}$ is a true martingale.

Thanks for any advice.

Answer 1

The BDG inequalities give the result if instead of $E[X^2_t]<+\infty$ you have $E[X^{*2}_t]<+\infty$ ( the star stands for the sup) indeed the result hold by the lemma 3 below.

If we look at George Lowther's blog, we have the following characterisation that is of interest :

Lemma 3 : A local martingale $X$ is a square integrable martingale if and only if ${{\mathbb E}[X_0^2]<\infty}$ and $[X]$ is integrable, in which case ${X^2-[X]}$ is a martingale.(note : the last claim is a bonus)

So if we have a local martingale square integrable at every time $t$, and which quadratic variation isn't integrable we have a counterexample. This I believe probably exists even if I don't have a concrete construction to exhibit right now.

Here is a counterexample, this is a continuous local martingale it is such that $E[X^2_t]$ is bounded and tends to 0as $t\to \infty$ (look at comet after equation (4)) yet it is not martingale as proven in the post.

NB: This by the way of lemma 3 proves that $E[[X]_t]=\infty$ for some $t$ which is not trivial.

Regards

Answer 2

I have thought about this problem for quite a while. The following is my result so far. However I'm not sure if my attempt is correct. Clarification and improvements are appreciated.

attempt: Since we have a continuous local martingale, we know that $\sup_{0\le s\le t}|M_s|$ is attained for every $\omega$ in some $t'$ (depending on $\omega$). Therefore $\sup_{0\le s\le t}|M_s(\omega)|=|M_{t'}(\omega)|$. Integrating yields $$E[\sup_{0\le s\le t}|M_s(\omega)|]=E[|M_{t'}(\omega)|]$$ Using $|M_{t'}(\omega)|=|M_{t'}(\omega)|(\mathbf1_{|M|_{t'}<1}+\mathbf1_{|M|_{t'}\ge 1})\le1+|M_{t'}(\omega)|^2$. After all, we get

$$E[\sup_{0\le s\le t}|M_s(\omega)|]\le 1+E[|M_{t'}(\omega)|^2]$$

Since $t'$ is depending on $\omega$ (but finite!) I think we can still use square integrability for the second term. Hence, if everything is correct, we would have found an integrable bound.

As I said, I'm not entirely sure about the above.

Answer 3

Unfortunately, the answer to your question is that it cannot be done, since square-integrable strict local martingales do in fact exist. The most famous example, due to Johnson and Helms (1962), is the inverse of a Bessel process of dimension three. The Bessel process $\rho$ of dimension three is determined by the SDE $$d\rho_t=\frac{1}{\rho_t}\,dt+dB_t,$$ where $B$ is a standard one-dimensional Brownian motion. Its inverse $X\equiv 1/\rho$ is the local martingale determined by $$dX_t=-X_t^2\,dB_t.$$ Revuz and Yor (2000) provide a lot of information about Bessel processes, including transition densities, which you can use to verify the square-integrability of $X$. You might also find the following paper I wrote useful: http://www.qfrc.uts.edu.au/research/research_papers/rp279.pdf.

Regards, Hardy