Probability theory - n random variables independent, two vectors formed of those variables independent

Question

Use measure theory to prove: If random variables $X_1,..,X_n$ are independent, then two vectors $(X_1,..,X_k)$ and $(X_{k+1},..,X_n)$ are independent.

Answer 1

Let $\left(\Omega,\mathcal{A},P\right)$ be the underlying probability space for the $X_i$.

Further let $\mathcal{V},\mathcal{W}\subseteq\mathcal{A}$ such that $P\left(V\cap W\right)=P\left(V\right)P\left(W\right)$ whenever $V\in\mathcal{V}$ and $W\in\mathcal{W}$.

It is our aim to show that more broadly: $$P\left(V\cap W\right)=P\left(V\right)P\left(W\right)\text{ whenever }V\in\sigma\left(\mathcal{V}\right)\text{ and }W\in\sigma\left(\mathcal{W}\right)\tag0$$

If that has been reached then we are ready because we can apply it on $\mathcal{V}=\bigcup_{i=1}^{k}\sigma\left(X_{i}\right)$ and $\mathcal{W}=\bigcup_{i=k+1}^{n}\sigma\left(X_{i}\right)$.

Be aware that here $\sigma\left(\mathcal{V}\right)=\sigma\left(X_{1},\dots,X_{k}\right)$ and $\sigma\left(\mathcal{W}\right)=\sigma\left(X_{k+1},\dots,X_{n}\right)$

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Proof:

Fix some $V_{0}\in\mathcal{V}$ and let: $$\mathcal{W}\left(V_{0}\right):=\left\{ W\in\sigma\left(\mathcal{W}\right)\mid P\left(V_{0}\cap W\right)=P\left(V_{0}\right)P\left(W\right)\right\} $$

Then it is not difficult to prove that $\mathcal{W}\left(V_{0}\right)$ is a $\sigma$-algebra and this with $\mathcal{W}\subseteq\mathcal{W}\left(V_{0}\right)$. This justifies the conclusion that $\mathcal{W}\left(V_{0}\right)=\sigma\left(\mathcal{W}\right)$.

This can be done for every $V_{0}\in\mathcal{V}$ so actually we have proved that: $$\forall V\in\mathcal{V}\;\forall W\in\sigma\left(\mathcal{W}\right)\;P\left(V\cap W\right)=P\left(V\right)P\left(W\right)\tag1$$

Now secondly fix some $W_{0}\in\sigma\left(\mathcal{W}\right)$ and let: $$\mathcal{V}\left(W_{0}\right):=\left\{ V\in\sigma\left(\mathcal{V}\right)\mid P\left(V\cap W_{0}\right)=P\left(V\right)P\left(W_{0}\right)\right\} $$

Similarly it can be prove that $\mathcal{V}\left(W_{0}\right)$ is a $\sigma$-algebra and this with $\mathcal{V}\subseteq\mathcal{V}\left(W_{0}\right)$ as is proved above and is stated in $(1)$. This justifies the conclusion that $\mathcal{V}\left(W_{0}\right)=\sigma\left(\mathcal{V}\right)$.

This can be done for every $W_{0}\in\sigma\left(\mathcal{W}\right)$ so actually we have proved that: $$\forall V\in\sigma\left(\mathcal{V}\right)\;\forall W\in\sigma\left(\mathcal{W}\right)\;P\left(V\cap W\right)=P\left(V\right)P\left(W\right)$$ which is of course the same as $(0)$. So we reached our aim.