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Suppose $M(t)$ is a continuous local martingale. That is, there exists a sequence of stopping times $T_n$ which almost surely increase to $\infty$, and such that $M(t\wedge T_n)$ is a martingale for all $n$. By continuous I mean that it is almost surely continuous.

Suppose that $M(t)$ is almost surely bounded from below by some fixed constant $c$ (to clarify, we choose $c$ before we "roll"). Does it follow that $M(t)$ is a supermartingale?

Edit: it should probably be assumed in addition that $\mathbb{E}(M(0))<\infty$.

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    $\begingroup$ I think that the assumption $\mathbb{E}(M(0))<\infty$ can be relaxed as by definition, $M(t\hat{}T_n)$ is a martingale so $\forall t>0, M(t\hat{}T_n)$ is integrable. If one takes $t=0$ you get that $M(0\hat{}T_n)=M(0)$ is integrable. $\endgroup$
    – Tom Mike
    Commented Nov 25, 2016 at 7:25

2 Answers 2

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Here's my solution, based on a hint by @saz. I am not 100% percent it is true.

Let $M'(t) = M(t) + c$. Since $M'(t)$ is a (continuous) local martingale, there exists a sequence $T_n$ of stopping times such that $M'_n(t) = M'(t\wedge T_n)$ is a martingale for each $n$. That is, with respect to a filteration $\mathcal{F}(t)$, $t\ge 0$, it holds that for all $0\le s\le t$, $$\mathbb{E}\left(M'_n(t)\mid\mathcal{F}(s)\right)=M'_n(s).$$

By Fatou's lemma, \begin{eqnarray*} \mathbb{E}\left(\left|M'(t)\right|\right) &=& \mathbb{E}\left(M'(t)\right)\\ &=& \mathbb{E}\left(\liminf M'_n(t)\right)\\ &\le& \liminf\mathbb{E}\left(M'_n(t)\right) = \mathbb{E}\left(M'(0)\right) < \infty \end{eqnarray*}

By Fatou's lemma again, $$\mathbb{E}\left(M'(t)\mid\mathcal{F}(s)\right) =\mathbb{E}\left(\underline{\lim} M'_n(t)\mid\mathcal{F}(s)\right) \le \underline{\lim}\mathbb{E}\left(M'_n(t)\mid\mathcal{F}(s)\right) =\underline{\lim} M'_n(s) = M'(s)$$ and it follows that $M'$ is a supermartingale, hence $M$ is a supermartingale.

Note: I have edited my answer to include proof of integrability. While doing that, I figured out that one should assume $\mathbb{E}(M'(0))<\infty$. I'll update the question.

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  • $\begingroup$ What's may be missing here is that Fatou's lemma can be applied to nonnegative random variables, while here all I have is that it is almost surely positive. $\endgroup$
    – Bach
    Commented May 24, 2014 at 13:13
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    $\begingroup$ By assumption, we have $\mathbb{P}(\Omega_0)=1$ for $$\Omega_0 := \{\omega; \forall t: M_t(\omega) \geq c\}.$$ Define $$\tilde{M}_t(\omega) := \begin{cases} M_t(\omega) & \omega \in \Omega_0 \\ c & \omega \notin \Omega_0 \end{cases}$$ Now apply the above argumentation and note that, as $M_t = \tilde{M}_t$ a.s. we have $$\mathbb{E}(M_t \mid \mathcal{F}_s) = \mathbb{E}(\tilde{M}_t \mid \mathcal{F}_s).$$ $\endgroup$
    – saz
    Commented May 24, 2014 at 19:36
  • $\begingroup$ That's great, thanks! $\endgroup$
    – Bach
    Commented May 24, 2014 at 19:42
  • $\begingroup$ Why does $\liminf\mathbb{E}\left(M'_n(t)\right) = \mathbb{E}\left(M'(0)\right) < \infty$? $\endgroup$
    – alpastor
    Commented Jun 2, 2019 at 2:25
  • $\begingroup$ For the given sequence of stopping times as above, the stopped local martingale actually becomes a martingale and so the property follows as the expectation of a martingale at time $t$ is equal to the expectation at time zero @alpastor. $\endgroup$
    – user258521
    Commented Feb 5, 2020 at 17:53
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Hint Apply Fatou's lemma in order to prove $(M_t)_{t \geq 0}$ a supermartingale.

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  • $\begingroup$ Thanks for your hint, I believe that I have a full answer now (added below). $\endgroup$
    – Bach
    Commented May 24, 2014 at 13:49

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