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Let $X_1, \ldots , X_n:\Omega \to \mathbb{R}$ be random variables such that $(X_1, \ldots , X_n)$ is a Gaussian vector in $\mathbb{R}^n$.

I want to show that $X_1$ and $\sigma(X_2, \ldots , X_n)$ are independent if and only if $\operatorname{Cov}(X_1,X_j) = 0$ for all $ 2 \leq j \leq n$.

At first, I am not totally sure what is meant with $\sigma$. From a measure theoretical point of view, it would be logical that it is the sigma-algebra generated by the r.v.'s $X_2, \ldots , X_n$. However, in the context of Gaussian vector and covariance matrices it could be something to do with variance. Who can help me out?

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    $\begingroup$ It means "sigma-algebra generated by $(X_2,\ldots,X_n)$". Also, the covariances should be $0$ not $1$, if you are aiming to capture independence. $\endgroup$ – John Dawkins Mar 19 '16 at 16:23
  • $\begingroup$ Yes, you are right! What happens if $X_1$ is independent of $\sigma(X_2, ... , X_n)$? $\endgroup$ – iJup Mar 19 '16 at 16:35
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Here $\sigma(X_2,\dots,X_n)$ is the $\sigma$-algebra generated $X_2,\dots,X_n$.

Suppose that $X_1$ and $\sigma(X_2,\dots,X_n)$ are independent. In particular, $X_1$ an $X_i$ are independent for any $i \ge 2$. Therefore

$$\mbox{cov} (X_1,X_i)= E [(X_1-E X_1) (X_i-E X_i) ] = E [X_1-E X_1]E [X_i-EX_i]=0.$$

Conversely, suppose that $\mbox{cov}(X_1,X_i)=0$ for all $i$. Since $(X_1,\dots,X_n)$ is a Gaussian vector, it follows from definition that any linear transformation of $(X_1,\dots,X_n)$ is a Gaussian vector. Let $Y=\sum_{j=2}^n \theta_j X_j$ be any linear combination of $X_2,\dots,X_n$. Then $(X_1,Y)$ is a linear transformation of $(X_1,\dots,X_n)$ and is therefore Gaussian. In particular, $X_1$ and $Y$ are independent if and only if $\mbox{cov}(X_1,Y)=0$. But this is equal to

$$ E[(X_1 -E X_1)\sum_{i=2}^n \theta_j (X_j -EX_j) ]=\sum_{j=2}^n \theta_j\mbox{cov} (X_1,X_j) =0. $$

Since $X$ is independent of any linear combination of $X_2,\dots,X_n$, $X_1$ is independent of $\sigma(X_2,\dots,X_n)$. Indeed, the joint distribution of $X_1$ and $(X_2,\dots,X_n)$ is determined by the characteristic function

$$\phi(\theta_1,\dots,\theta_n) = E [\exp (i \sum_{j=1}^n \theta_j X_j)],$$

but by the independence of $X_1$ and $\sum_{j=2}^n \theta_j X_j$, we have

$$\phi (\theta_1,\dots,\theta_n) = E [\exp(i \theta_1 X_1)] \times E [\exp( i\sum_{j=2}^n \theta_j X_j)],$$

that is the characteristic function is a product of the characteristic function of $X_1$ and of $(X_2,\dots,X_n)$. This is equivalent to $X_1$ independent of $\sigma(X_2,\dots,X_n)$, because, again, the joint distribution of $(X_2,\dots,X_n)$ is determined by the corresponding characteristic function.

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  • $\begingroup$ If $X_1$ and $\sigma(X_2, \ldots X_n)$ are independent, why does this imply that $X_1$ and $X_i$ are independent for $i \geq 2$? What is the translation between the sigma-algebra of the rv's and the rv themselves? $\endgroup$ – iJup Mar 22 '16 at 19:21
  • $\begingroup$ What do you mean by $X$? $\endgroup$ – iJup Mar 22 '16 at 19:23
  • $\begingroup$ $X_1$ independent of $\sigma(X_2,\dots,X_n)$ means that $X_1$ is independent of any function of $X_2,\dots,X_n$. $X_i$ is such a function. $\endgroup$ – Fnacool Mar 29 '16 at 4:16

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