Smallest $\sigma$-algebras generated by finite Cartesian products of sets from the coordinate $\sigma$-algebras. Which is the meaning

independenceprobability theoryrandom variablestensor-products

Suppose we want to construct an infinite sequence of independent random variables of given distributions. Specifically, for each $n$ let $X_n$ be defined on $\left(\Omega_n\text{, }\mathcal{A}_n\text{, }\mathbb{P}_n \right)$ and let us set
$$
\Omega=\prod\limits_{n=1}^{\infty}\Omega_n\hspace{0.8cm}(\text{Countable Cartesian product})
$$

and
$$
\mathcal{A}=\displaystyle\otimes_{n=1}^{\infty}\mathcal{A_n}
$$

where $\displaystyle\otimes_{n=1}^{\infty}\mathcal{A_n}$ denotes the smallest $\sigma$-algebra on $\Omega$ generated by all sets of the form
$$A_1\times A_2\times \ldots\times A_k\times \Omega_{k+1}\times\Omega_{k+2}\times\ldots\text{,}\hspace{0.4cm}A_i\in\mathcal{A}_i\text{;}\hspace{0.4cm}k=1,2,3,\ldots$$
That is, $\mathcal{A}$ is the smallest $\sigma$-algebra generated by finite Cartesian product of sets from the coordinate $\sigma$-algebras

$\bf{Questions:}$

  1. why is $\displaystyle\otimes_{n=1}^{\infty}\mathcal{A_n}$ generated by finite product of sets $A_i$ with $i$ ranging from $1$ to $k$ times product of $\Omega_i$ with $i$ ranging instead from $k+1$ to $\infty$?
  2. In other terms, what do they mean when stating that $\mathcal{A}$ is the smallest $\sigma$-algebra generated by finite Cartesian products of sets from coordinate $\sigma$-algebras? What are coordinate $\sigma$-algebras?

Best Answer

The definition is weird at first sight. Let's have a closer look.

On the product space $\Omega:=\prod_{n =1}^\infty \Omega_n$, we want to put a useful $\sigma$-algebra, call it $\bigotimes_n \mathcal{A}_n$. Definitely, we want to ask that the projections (= coordinate maps)

$$\pi_n: \Omega \to \Omega_n: n \geq 1$$

become measurable maps with respect to the $\sigma$-algebra $\bigotimes_n \mathcal{A}_n$. So let's just define $\bigotimes_n \mathcal{A}_n$ as the smallest $\sigma$-algebra making these projections measurable:

$$\bigotimes_n \mathcal{A}_n:= \sigma(\{\pi_n:n \geq 1\})$$

Now, what does it have to do with your question? You can check (exercise for you!) that this definition is equivalent with the one you are given. I.e. prove that

$$\bigotimes_n \mathcal{A_n}= \sigma\left(\left\{A_1\times A_2 \times \dots \times A_k \times \Omega_{k+1} \times \dots\mid A_i \in \mathcal{A_i}, i=1, \dots, k, k \geq 1\right\}\right) \quad (*)$$

Now, getting to your questions:

(1) Working with finitely many sets is definitely easier in practise. What you wrote down is a generating set, you don't want that to be complicated, but you want the "basis" sets to be as easy as possible, hence we choose them finite.

(2) Means exactly what I wrote down in $(*)$. The coordinate $\sigma$-algebras are the $\sigma$-algebras $\mathcal{A}_n, n \geq 1$.

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