Independence of Brownian Stochastic Integrals over Disjoint Intervals

Summary

The question asks whether a stochastic integral accumulated up to an earlier time is independent of the integral accumulated over the following interval. The answer rewrites the later increment as an Itō integral from the earlier time to the later time, while the earlier value is the integral over the preceding interval.

It appeals to the construction of the Itō integral as a limit of weighted sums of Brownian increments. Increments on the later interval are independent of the Brownian path up to the earlier time, so the integrals over the disjoint intervals are independent. The same conclusion applies if the later difference is written with the opposite sign. The explanation relies on the stated deterministic integrand and Brownian-motion setting; it does not establish the result for arbitrary stochastic integrands or processes with dependent increments.

Key ideas

• A stochastic integral over a later time interval is the increment of the accumulated integral across that interval.
• An Itō integral with a deterministic integrand can be represented through limits of weighted Brownian increments.
• Brownian increments after a fixed time are independent of the Brownian path up to that time.
• Therefore, integrals over the two disjoint intervals are independent in the stated setting.

Tags

# Stochastic Calculus: How to test for dependency of random variables


# Stochastic Calculus: How to test for dependency of random variables












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.

## Answer by Bjørn Kjos-Hanssen (score 2)

https://quant.stackexchange.com/a/37944

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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