Random Vector is Random Variable

Theorem
Let $n \in \N$.

Let $\struct {X, \Sigma, \Pr}$ be a probability space.

Let $\struct {S_1, \Sigma_1}, \struct {S_2, \Sigma_2}, \ldots, \struct {S_n, \Sigma_n}$ be measurable spaces.

Let:


 * $\ds S = \prod_{i \mathop = 1}^n S_i$

Let:


 * $\ds \Sigma' = \bigotimes_{i \mathop = 1}^n \Sigma_i$

where $\ds \bigotimes_{i \mathop = 1}^n \Sigma_i$ denotes the product $\sigma$-algebra of $\Sigma_1, \Sigma_2, \ldots, \Sigma_n$.

For each $i \in \set {1, 2, \ldots, n}$:


 * let $X_i$ be a random variable on $\struct {\Omega, \Sigma, \Pr}$ taking values in $\struct {S_i, \Sigma_i}$.

Define a function $\mathbf X : \Omega \to S$ by:


 * $\map {\mathbf X} \omega = \tuple {\map {X_1} \omega, \map {X_2} \omega, \ldots, \map {X_n} \omega}$

for each $\omega \in \Omega$.

Then $\mathbf X$ is a random variable on $\struct {\Omega, \Sigma, \Pr}$ taking values in $\struct {S, \Sigma'}$.

Proof
Note that from the definition of product $\sigma$-algebra: finite case, we have:


 * $\ds \Sigma' = \map \sigma {\set {\prod_{i \mathop = 1}^n S_i : S_i \in \Sigma_i \text { for each } i \in \set {1, 2, \ldots, n} } }$

where $\sigma$ denotes the $\sigma$-algebra generated by a collection of subsets.

We aim to apply Mapping Measurable iff Measurable on Generator.

We therefore aim to show that:


 * ${\mathbf X}^{-1} \sqbrk B \in \Sigma$

for each:


 * $\ds B \in \set {\prod_{i \mathop = 1}^n E_i : E_i \in \Sigma_i \text { for each } i \in \set {1, 2, \ldots, n} }$

Let:


 * $\ds B \in \set {\prod_{i \mathop = 1}^n E_i : E_i \in \Sigma_i \text { for each } i \in \set {1, 2, \ldots, n} }$

Then, for each $i$ there exists $E_i \in \Sigma_i$ such that:


 * $\ds B = \prod_{i \mathop = 1}^n E_i$

Then, for $\omega \in \Omega$, we have $\map {\mathbf X} \omega \in B$ :


 * $\ds \map {X_i} \omega \in E_i$ for each $i \in \set {1, 2, \ldots, n}$.

That is, $\omega \in {\mathbf X}^{-1} \sqbrk B$ :


 * $\omega \in {X_i}^{-1} \sqbrk {E_i}$ for each $i \in \set {1, 2, \ldots, n}$.

So:


 * $\ds {\mathbf X}^{-1} \sqbrk B = \bigcap_{i \mathop = 1}^n {X_i}^{-1} \sqbrk {E_i}$

Since $X_i$ is $\Sigma/\Sigma_i$-measurable, we have:


 * ${X_i}^{-1} \sqbrk {E_i} \in \Sigma$

for each $i \in \set {1, 2, \ldots, n}$.

From Sigma-Algebra Closed under Finite Intersection, we therefore have:


 * $\ds \bigcap_{i \mathop = 1}^n {X_i}^{-1} \sqbrk {E_i} \in \Sigma$

so:


 * $\ds {\mathbf X}^{-1} \sqbrk B \in \Sigma$

which was the demand.

From Mapping Measurable iff Measurable on Generator, we then have:


 * $\mathbf X$ is $\Sigma/\Sigma'$-measurable.

So:


 * $\mathbf X$ is a random variable on $\struct {\Omega, \Sigma, \Pr}$ taking values in $\struct {S, \Sigma'}$

as required.

Also see

 * Vector is Random Vector iff Components are Random Variables shows that the converse is also true: if $\tuple {X_1, X_2, \ldots, X_n}$ is a random variable on $\struct {\Omega, \Sigma, \Pr}$ taking values in $\struct {S, \Sigma'}$ then each $X_i$ is a random variable taking values in $\struct {S_i, \Sigma_i}$.