Proof of Independent Random Variables and Independent Standard Normal Coordinates on a Product Space

For $k\in[n]$ let $c_{k}:\mathbb{R}^{n}\to\mathbb{R}$, $c_{k}(z)=z_{k}$, be the $k$-th coordinate function. The identity of $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}),\gamma_{n})$ is a standard normal vector (Standard Normal Vectors: Density, Moments, Orthogonal Invariance and Concatenation (ii)), so its coordinates $c_{1},\dots,c_{n}$ are independent standard normal random variables on that space (Standard Normal Vectors and the Standard Normal Distribution on ℝᵏ (iii), Standard Normal Distribution (iii)); in particular $c_{k}^{-1}(B)\in\mathcal{B}(\mathbb{R}^{n})$ and $\gamma_{n}(c_{k}^{-1}(B))=\gamma(B)$ for every Borel set $B\subseteq\mathbb{R}$.

The product space. $P$ and $\gamma_{n}$ are probability measures (Standard Normal Vectors: Density, Moments, Orthogonal Invariance and Concatenation (i)), in particular $\sigma$-finite, so the product measure $\tilde P=P\otimes\gamma_{n}$ is defined (Product Measure (i)) and

$$\tilde P(\tilde\Omega)=P(\Omega)\,\gamma_{n}(\mathbb{R}^{n})=1 .$$

Let $k\in[n]$ and let $B\subseteq\mathbb{R}$ be a Borel set. Then

$$\tilde Y_{k}^{-1}(B)=Y_{k}^{-1}(B)\times\mathbb{R}^{n},\qquad\tilde Z_{k}^{-1}(B)=\Omega\times c_{k}^{-1}(B),$$

and both are measurable rectangles, since $Y_{k}^{-1}(B)\in\mathcal{F}$ and $c_{k}^{-1}(B)\in\mathcal{B}(\mathbb{R}^{n})$; so they lie in $\tilde{\mathcal{F}}$ (Measurable Rectangles and the Product σ-Algebra (ii)), and $\tilde Y_{k}$ and $\tilde Z_{k}$ are random variables (Probability Spaces, Events and Random Elements (iv)). This proves (i).

Laws. By Product Measure (i), the same two rectangles give

$$\tilde P(\tilde Y_{k}\in B)=P(Y_{k}\in B)\,\gamma_{n}(\mathbb{R}^{n})=P(Y_{k}\in B),\qquad\tilde P(\tilde Z_{k}\in B)=P(\Omega)\,\gamma_{n}\bigl(c_{k}^{-1}(B)\bigr)=\gamma(B).$$

So $\tilde Y_{k}$ has the law of $Y_{k}$ (The Law of a Random Element (ii)) and $\tilde Z_{k}$ has the law $\gamma$, that is, it is standard normal (Standard Normal Distribution (iii)). This proves (iii).

Rectangles. Let $C_{1},\dots,C_{2n}\subseteq\mathbb{R}$ be Borel sets. The event $\{W_{i}\in C_{i}\ (i\in[2n])\}$ is the rectangle $E\times F$ with

$$E=\bigcap_{k\in[n]}Y_{k}^{-1}(C_{k}),\qquad F=\bigcap_{k\in[n]}c_{k}^{-1}(C_{n+k}).$$

These are intersections over the nonempty finite, hence countable (Countable Sets: a Bijection from ℕ onto ℕ×ℕ, Finite Sets, Subsets, Surjective Images, Images, Products, and the Integers and Rationals (iii)), index set $[n]$ of sets in $\mathcal{F}$ and in $\mathcal{B}(\mathbb{R}^{n})$ respectively, so $E\in\mathcal{F}$ and $F\in\mathcal{B}(\mathbb{R}^{n})$ (Unions and Intersections over a Countable Index Set in a σ-Algebra (ii)). By Product Measure (i),

$$\tilde P\bigl(W_{i}\in C_{i}\ (i\in[2n])\bigr)=P(E)\,\gamma_{n}(F)=\prod_{k=1}^{n}P(Y_{k}\in C_{k})\prod_{k=1}^{n}\gamma(C_{n+k})=\prod_{i=1}^{2n}\tilde P(W_{i}\in C_{i}).\qquad(*)$$

Here $P(E)$ factors because the family $(Y_{k})_{k\in[n]}$ is independent, and $\gamma_{n}(F)$ factors because the family $(c_{k})_{k\in[n]}$ is independent, both by Independent Random Elements: Preimages, Functions and Blocks (i), with $\gamma_{n}(c_{k}^{-1}(C_{n+k}))=\gamma(C_{n+k})$; the last equality uses the laws just found and splits $[2n]$ into the disjoint sets $[n]$ and $\{n+1,\dots,2n\}$ (Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (ii)).

Independence. Let $m\in\mathbb{N}$, let $i_{1},\dots,i_{m}\in[2n]$ be distinct, let $I=\{i_{1},\dots,i_{m}\}$ and let $B_{1},\dots,B_{m}\subseteq\mathbb{R}$ be Borel sets. Let $(C_{i})_{i\in[2n]}$ be the union of the function $i_{r}\mapsto B_{r}$ on $I$ (well defined, the $i_{r}$ being distinct) and the constant function $\mathbb{R}$ on $[2n]\setminus I$, whose domains are disjoint; so $C_{i_{r}}=B_{r}$ for $r\in[m]$ and $C_{i}=\mathbb{R}$ for $i\in[2n]\setminus I$. Then $\{W_{i_{1}}\in B_{1},\dots,W_{i_{m}}\in B_{m}\}=\{W_{i}\in C_{i}\ (i\in[2n])\}$, and by $(*)$ its probability is

$$\prod_{i=1}^{2n}\tilde P(W_{i}\in C_{i})=\prod_{i\in I}\tilde P(W_{i}\in C_{i})\prod_{i\in[2n]\setminus I}\tilde P(W_{i}\in\mathbb{R})=\prod_{r=1}^{m}\tilde P(W_{i_{r}}\in B_{r}),$$

by splitting $[2n]$ into $I$ and its complement, finite as a subset of the finite set $[2n]$ (Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (iv), Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (ii)), the factors off $I$ being $\tilde P(\tilde\Omega)=1$, and reindexing along the bijection $r\mapsto i_{r}$ of $[m]$ onto $I$ (Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (iii)). By Independent Random Elements: Preimages, Functions and Blocks (i), the family $(W_{i})_{i\in[2n]}$ is independent, which is (ii).