Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I can prove it briefly, but I found a "counter" example. (There must be a mistake in the following words...)

I can prove: X is a continuous local martingale, with $X_0=0$ a.s, then X is $L_2$ bounded iff $E[X]_\infty<\infty$

I can prove: X is a continuous local martingale, if $E[X]_\infty<\infty$, then X is a true martingale.

It seems that we can get: any $L_2$ bounded continuous local martingale is a true martingale.

But there is a "counter" example: If X is a standard Brownian motion in $R^3$, started at $0$, I can prove (i) $Y_t =1/|X_{1+t}|$ is a local martingale; (ii) Y is bounded in $L^2$; (iii) Y is not a martingale.

Can I say: since Y isn't continuous, therefore Y is bounded in $L^2$, but not a martingale??? But I guess that we can prove Y is continuous almost surely. Is continuous L2 bounded local martingale a true martingale???

share|cite|improve this question
Help!!!!!!......... – XXX11235 Jan 8 '13 at 9:40
Why (iii)? $ $ $ $ – Did Jan 8 '13 at 10:49
So!!!!!!......... – Did Jan 10 '13 at 16:14
up vote 4 down vote accepted

No, not every continuous $L^2$ bounded local martingale is a true martingale (see your counterexample). However, every local martingale with $L^1$ bounded quadratic variation is a true martingale!

The problem is in your second statement:

If $X$ is a continuous martingale with $L^2$-bounded quadratic variation: $\mathbb{E}[X]_\infty <\infty$ THEN $X_t$ is $L_2$ bounded. The converse ('if f ') is not true in general (which is what is going wrong in your example).

Let $\tau_n$ be a localizing sequence. we then have:

$$ \mathbb{E}X_t^2 = \mathbb{E}\lim_{n \rightarrow \infty} X_{t \wedge \tau_n}^2 \leq \lim_{n \rightarrow \infty} \mathbb{E} X_{t \wedge \tau_n}^2 = \lim_{n \rightarrow \infty} \mathbb{E} [X]_{t \wedge \tau_n}= \mathbb{E} \lim_{n \rightarrow \infty} [X]_{t \wedge \tau_n} =E[X]_t $$

The inequality follows from Fatous lemma. The second exchange of limit and integration is, however, justified by the monotone convergence theorem.

In your specific example, Fatou's lemma results in a strict inequality:

$$ \frac{1}{1+t} = \mathbb{E} Y_t^2 < \mathbb{E}[Y]_t = \mathbb{E} \int_0^t \frac{1}{X_{1+t}^4} ds = \int_0^t \mathbb{E}Y_s^4 ds = \infty $$ The exchange of integrators is justified because $Y_s^4$ is positive. Further, while the second moment of $Y_s$ exists, the fourth moment does not. Hence the last value is infinity.

share|cite|improve this answer
Thank you very much! – XXX11235 Jan 23 '13 at 0:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.