Given a filtered probability space $(\Omega, \mathbb{F}, \{\mathcal{F}_t\}_{0\le t\le 1}, \mathbb{P})$, let $X$ be a cadlag martingale and $V$ be cadlag supermartingale. Suppose $V$ has the following decomposition: there exists a $X$-integrable predictable process $H$ s.t.

$$V_t=V_0+\int_0^tH_sdX_s-C_t,~ \forall t\in [0,1]$$

where $C$ is an adapted increasing process (not necessarily predictable!) with $C_0=0$. Now if we have another predictable process $H'$ s.t.

$$V_t\le V_{t-}+H'_t(X_t-X_{t-}), \forall t\in [0,1]$$

where $V_{t-}$ and $X_{t-}$ denote the left limit. Could we say that $H=H'$ a.s. or

$$\int_0^tH_sdX_s\le \int_0^tH'_sdX_s, \forall t\in [0,1]$$

Many thanks for the reply!


1 Answer 1


Hi you can't say that $H=H'$ a.s. in general, indeed the jumps of $V_t$ can be written as :

$$\Delta V_t=H_t.\Delta X_t - \Delta C_t$$

So we have $\Delta V_t \leq H_t.\Delta X_t$ a.s. which is equivalent to :

$$V_t\leq V_{t-}+ H_t.(X_t-X_{t-}), \forall t\in [0,1]$$

Now suppose that $X$ is a positive martingale, then for any constant $c>0$, taking $H'=H+c$ is enough to exhibit a counterexample where your condition holds a.s..

For your second point, it is not true neither. For example if $X$ is a Brownian motion and $C_t=t$

Because in this case any process $H'$ satisfy your condition as $\Delta X_t=\Delta V_t=0$ a.s. but taking $H'_t=c$ where $c$ is a positive constant you don't have :

$$B_t=\int_0^tH_sdX_s\le \int_0^tH'_sdX_s=2.B_t, \forall t\in [0,1]$$ a.s.


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