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Given an Ito-process $X(t)$, $t\in[0,T]$

$$X(t)=X_{0}+\int_{0}^{t}F(s)ds + \int_{0}^{t}G(s)dW(s)$$

with $F\in \mathbb{L}^{1}(0,T)$ and $G\in\mathbb{L}^{2}(0,T)$. It is now often claimed that this representation is unique (up to indistinguishability of the processes $F$ and $G$). By linearity it is enough to assume the case $X=0$ and to show that $X_{0}=0$, $F=0$ and $G=0$. Taking $t=0$, it follows immediately that $X_{0}=0$. I now thought one could just use simply the fact that

$$\int_{0}^{t}G(s)dW(s)$$

is a martingale as well as to use Ito-isometry to prove this. If we have shown that $F=0$. Then $$ 0 = \mathbb{E}\left[\left(\int_{0}^{t}G(s)dW(s) \right)^{2}\right]=\mathbb{E}\left[\int_{0}^{t}G^{2}(s)ds \right]$$ by Ito-isometry. Thus $$\int_{0}^{t}G^{2}(s)ds=0$$ almost surely. Can I now already conclude that $G=0$ up to indistinguishability?

Additionally, how do I show that $F=0$?

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1 Answer 1

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Since

$$0 = \int_0^t F(s) \, ds + \int_0^t G(s) \, dW_s$$

we have

$$M_t := \int_0^t G(s) \, dW_s = - \int_0^t F(s) \, ds.$$

In particular, $(M_t)_{t \geq 0}$ is a martingale with continuous sample paths of bounded variation. This implies that $M_t = M_0 =0$ almost surely. Now it follows from Itô's isometry that

$$0 = \int_0^t G(s)^2 \, ds =0$$

and so $G=0$ (up to an exceptional null set). On the other hand,

$$\int_0^t F(s) \, ds=0 \quad \text{for all $t \geq 0$}$$

clearly implies $F=0$ (up to an exceptional null set).

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  • $\begingroup$ Obviously, I am missing something: but why is a martingale with continuous sample paths of bounded variation constant almost surely? $\endgroup$
    – Strickland
    Commented Oct 24, 2017 at 17:56
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    $\begingroup$ @Strickland Well, that's a known fact; see e.g. corollary 4 here almostsure.wordpress.com/2010/04/01/… $\endgroup$
    – saz
    Commented Oct 24, 2017 at 18:44

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