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???

  • 1
    $\begingroup$ Why (iii)? $ $ $ $ $\endgroup$ – Did Jan 8 '13 at 10:49

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.


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