I would like to know if my proof of this fact is correct : A bounded local martingale is a martingale.

The sequence of stopping time has the following property : it is increasing and it converges to $+\infty$ such that for all $n$ $1_{\tau_n}X_{t}^{\tau_n}$ is a martingale.

Attempt : We notice that for all $t\geq 0$ we have $X_{t}(\omega) = \lim_{n\to\infty}X_{t}^{\tau_n}(\omega)$ almost surely. Moreover there exists $M>0$ such that $\forall t\geq 0$ we have $X_t(\omega)\leq M$ almost surely. Also $lim_{n\to\infty}1_{\tau_n>0}=1$ almost surely. Thus we have that $1_{\tau_n>0} X_{t}^{\tau_n}$ converges almost surely to $X_t$ and it is dominated, so we have the $L_1$ convergence also. Now by the continuity of the operator conditional expectation we get for $s>t$

$$ \mathbb{E}(X_s | \mathcal{F}_t) = \mathbb{E}(\lim_{n\to\infty}1_{\tau_n>0} X_{s}^{\tau_n} | \mathcal{F}_t) = \lim_{n\to\infty}\mathbb{E}(1_{\tau_n>0} X_{s}^{\tau_n} | \mathcal{F}_t) = **\lim_{n\to\infty} 1_{\tau_n>0}\mathbb{E}( X_{s}^{\tau_n} | \mathcal{F}_t)**= \lim_{n\to\infty} 1_{\tau_n>0}X_{t}^{\tau_n} = X_t $$

Is this seems correct please ?

Thank you a lot !

The step with double stars is in fact false !


1 Answer 1


I would define what $\tau_n$ is (i.e. a localizing sequence for $X$), even though it can be inferred from the context. It's also not obvious to me why you take $1_{\tau_n > 0}$ out of the conditional expectation in the penultimate inequality. The definition of localizing sequence I'm familiar with is that $1_{\tau_n > 0} X_t^{\tau_n}$ is a martingale, so you can directly state $$\mathbb{E}[1_{\tau_n > 0} X_s^{\tau_n}|\mathcal F_t] = 1_{\tau_n > 0} X_t^{\tau_n}.$$

But aside from these small points about the writing, everything is correct.

  • $\begingroup$ Hello, thank you a lot for this answer. For the first point I will add it. For the second point it is a (big) mistake from myself ! $\endgroup$
    – coboy
    Oct 18, 2023 at 13:36

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