For $T\in R$ given and fixed consider: $$ {\rm d}F(T-t,X_t)=g(T-t,X_t)\,{\rm d}W_t. $$ where $g(t,x)$ is a given functions and $X_t$ is a given process driven by a brownian motion ($dX_t=(...)dt+(...)dW_t$). How does the stochastic integral representation of $F$ look like?

Are any of these two representations correct?

$$F(T-t,X_t)=F(T-0,X_0)+\int^t_0g(T-u)dW_u (1)$$ $$F(T-t,X_t)=F(T-0,X_0)+\int^{T-t}_0g(T-u,X)dW_u (2)$$

  • $\begingroup$ I think the 2nd representation is correct. $\endgroup$
    – FunnyBuzer
    Feb 19, 2019 at 13:58


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