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I have the following problem: We have to sigma algebras $\sigma (u)$ and $\sigma (X_{i})$ which are not independent. $u$ and $X_{i}$ are random variables.

Now my question: Why is it possible, in some cases, to find $X_{k}, X_{k+1} .....$ so that $\sigma (u)$ is independent from $\sigma (X_{i}, X_{k}, X_{k+1} .....)$.

The only thing I know is that the sigma Algebra $\sigma (X_{i}, X_{k}, X_{k+1} .....)$ would be finer than $\sigma (X_{i})$.

Finn
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  • This is not possible. If $\sigma (X_{i}, X_{k}, X_{k+1} .....)$ is independent from $\sigma(u)$ then every sub sigma algebra must also be independent from $\sigma(u)$. In particular $\sigma (X_{i})$. – Kurt G. Mar 11 '22 at 15:34
  • ah thanks! I was so confused – Finn Mar 11 '22 at 15:51

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