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Suppose I have a sequence of independent random variables $\{X_n, n \in \mathbb N\}$.

How do I show formally that

$P((X_1,...,X_n)\in A, (X_{n+1},...)\in B) = P((X_1,...,X_n)\in A)P((X_{n+1},...)\in B)$

if $A$ is a rectangle in $\mathcal B^n$ and $B$ is a rectangle in $\mathcal B^{\infty}$?

I tried, but I'm stuck on this.

Thank you.

Benzio
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1 Answers1

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We want to deduce from the equality $$\tag{$\star$}P\left(\bigcap_{i\in I}\{X_i\in B_i\}\cap\bigcap_{j\in J}\{X_j\in B_j\} \right)=P\left(\bigcap_{i\in I}\{X_i\in B_i\}\right)\cdot P\left(\bigcap_{j\in J}\{X_j\in B_j\}\right)$$ for $I$ and $J$ finite disjoint subsets of $\mathbf N$ and $B_k$ Borel subsets that $$\tag{1}P((X_1,...,X_n)\in A, (X_{n+1},...)\in B) = P((X_1,...,X_n)\in A)P((X_{n+1},...)\in B).$$ Using $(\star)$, we obtain that $(1)$ holds if $A$ is a finite disjoint union of products of Borel sets and $B$ a finite union of set of the form $\prod_{j=1}^NB_j\times \mathbb R\times\dots $. Sets of this form generate respectively the Borel $\sigma$-algebra of $\mathbf R^n$ and $\mathbf R^\infty$ and are algebras. We can thus conclude by an approximation argument.

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Davide Giraudo
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  • Sorry, I don't understand well your answer. In particular why $A$ have to be a finite disjoint union of products of Borel sets? A product of borel sets isn't enough? Then what do you mean by $\prod_{j=1}^NB_j\times \mathbb R\times\dots $? Cartesian product of cartesian products? My mind is exploding! – Benzio Jun 20 '14 at 20:24
  • The set $A$ is not necessarily of the form $A_1\times \dots\times A_n$, for example if $A:={(x_1,\dots,x_n), x_1\dots x_n\geqslant 0}$: that's why we consider the unions. The notation $\prod_{j=1}^NB_j\times \mathbb R\times\dots$ means the subset of $\mathcal B^\infty$ such that the $N$ first factors are the $B_j$ and the others the whole real line. – Davide Giraudo Jun 20 '14 at 20:31
  • Ok thanks, now I should meditate a while on this :-) – Benzio Jun 20 '14 at 20:38