Uniqueness of Drift and Diffusion in an Itô Process

Summary

The question asks how to show uniqueness of the drift and diffusion terms in an Itô process representation. In the zero-process case, the author sees that the time integral of the drift must equal the stochastic integral and asks how to prove both components vanish. The answer uses their distinct path properties: the drift integral has bounded variation, while the stochastic integral is a continuous local martingale starting at zero.

A continuous local martingale with bounded variation must be constant, so the two integral processes vanish. This gives uniqueness of the represented integral processes. To infer the integrands, the drift is determined only up to equality almost everywhere in time (with the relevant probability qualification), and the diffusion is similarly identified through its stochastic-integral equivalence, not as a pointwise indistinguishable process. The answer’s claim that the integrands are equal at all times is too strong without additional regularity assumptions.

Key ideas

• The time integral of an integrable drift has bounded variation.
• The stochastic integral in the representation is a continuous local martingale.
• A continuous local martingale with bounded variation is constant, which establishes equality of the integral processes.
• Uniqueness of the integral processes does not imply pointwise equality of integrands without further assumptions.

Tags

# Ito representation unique up to indistinguishability? Proof?


# Ito representation unique up to indistinguishability? Proof?












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 $F=0$?

## Answer by Calculon (score 1)

https://quant.stackexchange.com/a/38984

You want to show that

$$\int_0^tF(s)\,ds = \int_0^tG(s)\,dW_s$$

implies that both $F$ and $G$ are indistinguishable from $0$. The process on the left has bounded variation. The process on the right is a continuous local martingale that is equal to $0$ at time $0$. Continuous local martingales of bounded variation are constant processes. So both integrals are equal to $0$ at all times. Hence both $F$ and $G$ are equal to $0$ at all times.

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