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Is a continuous supermartingale with vanishing drift already a martingale? In my concrete problem, I have a continuous nonnegative local martingale $ (X_t) $ on $ \left[0, T\right] $ which is bounded from below by 0 (and hence a supermartingale) and which in addition satisfies $ sup_{t \in \left[0,T\right]} \mathbb{E} \left[ X_t^p \right] < + \infty$, for some $p > 1$. When I apply Ito, I find

$dX_t = 0 dt + H dW_t$,

where W is a Brownian motion and H is some predictable process. Now, I am wondering if $(X_t)$ is in fact a true martingale (the reason is that I would like to apply the Doob inequality for martingales in $L^p$).

Thanks a lot for your help! Simon

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What is the drift term of a (general?) supermartingale? What does it mean that it is vanishing? Maybe the following fact is useful: if $X_t$ is a supermartingale and $\mathsf E X_t$ is constant in time, then $X_t$ is a martingale. Perhaps, some integrability condition are needed though - if this is something you are looking for, I'll fish them out. –  Ilya Jan 9 '13 at 13:36
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there is a standard example, which is either the reciprocal of a bessel 3 or of a squared bessel 3 in which routine ito formula calculations show that the drift term is 0, but the process is a strict local martingale, and a supermartingale (by virtue of positivity), but not a martingale. I'm told Chung & William's book works it out in great detail. –  mike Jan 9 '13 at 14:35
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1 Answer

Take $dX_t= f(t)dt+ dW_t$ such that :
-$F(t)=\int_0^t f(s)ds$ is strictly decreasing (determisitic)
-$f(t)\to 0$ as $t\to \infty$

Then $X_t= F(t)+ W_t$ and as the sum of a martingale term (the Brownian motion) plus a deterministic supermartingale $F(t)$, $X$ it is a supermaringale with a drift term that goes to 0, and $X$ is not be a martingale (unless F is null).

For example $f(t)=-\frac{1}{(1+t)^2}$ gives $F(t)=\frac{1}{1+t}-1$ seems to work fine.

Best regards

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