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If I let $g(x)$ be a deterministic function of a real variable $x$ and define $X(t)$ as: $$X_T=\int_{0}^{T}f(u)dW_u$$ with $W_t$ being a wiener process. For $s<t$, Will $X_s$ and $X_s-X_t$ then be independent?

My intuition says it will be independent because the stochastic $W_t-W_s$ and $Ws$ is independent by definition of the Wiener proces. However, I can't prove this.

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  • $\begingroup$ Do you intend to calculate the Itô integral of $g$? $\endgroup$ – Raskolnikov Jan 27 '18 at 4:47
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It is equivalent to ask whether $X_s$ and $X_t-X_s$ are equivalent. Now $X_t-X_s=\int_s^t f(u)dW_u$ and $X_s=\int_0^s f(u)dW_u$. But we have a definition of the Itō integral like: $$\int_s^t f(u)dW_u = \lim_{n\to\infty} \sum_{i=1}^{n} f(x^{(n)}_i)(W_{x^{(n)}_{i+1}}-W_{x^{(n)}_{i}})$$ for suitable partitions $s=x_1^{(n)}<\dots<x_{n+1}^{(n)}=t$ and all these increments are $W_{x_{i+1}}-W_{x_{i}}$ are independent of $\{W_u\}_{u\le s}$.

So yes, they are independent.

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