Sigma-Algebra generated by a random vector let X,Y be real-valued random variables on some probability space. Then (X,Y) is random variable with values in $(\mathbb{R}^2, \mathcal{B}(\mathbb{R}^2))$(Borel-$\sigma$-Algebra). Do we have
$ \mathbb{F}^{(X,Y)}= \mathbb{F}^{X} \vee \mathbb{F}^{Y}$,
where $\mathbb{F}^X$ generates the sigma-algebra generated by X and so on?
 A: Firstly, the codomain of the random vector is $\mathbb{R}^{2}$ (rather
than $(\mathbb{R}^{2},\mathcal{B}(\mathbb{R}^{2}))$. More precisely,
given a measurable space $(\Omega,\mathcal{F})$, a random vector
$(X,Y)$ is a map $(X,Y):\Omega\rightarrow\mathbb{R}^{2}$, $\omega\mapsto(X(\omega),Y(\omega))$
such that for any $B\in\mathcal{B}(\mathbb{R}^{2})$, $(X,Y)^{-1}(B)=\{\omega\in\Omega\mid(X(\omega),Y(\omega))\in B\}\in\mathcal{F}$.
It is true that the $\sigma$-algebra generated by $(X,Y)$, denoted
by $\mathcal{F}_{(X,Y)}$, coincides with $\sigma(\mathcal{F}_{X}\cup\mathcal{F}_{Y})$,
where $\mathcal{F}_{X}$ and $\mathcal{F}_{Y}$ are the $\sigma$-algebras
generated by $X$ and $Y$ respectively.
Proof: Recall that $\mathcal{F}_{X}=\{X^{-1}(A)\mid A\in\mathcal{B}(\mathbb{R})\}$
and $\mathcal{F}_{Y}=\{Y^{-1}(B)\mid B\in\mathcal{B}(\mathbb{R})\}$.
Let $C\in\mathcal{F}_{X}$, then $C=X^{-1}(A)$ for some $A\in\mathcal{B}(\mathbb{R})$.
Note that $C=(X,Y)^{-1}(A\times\mathbb{R})\in\mathcal{F}_{(X,Y)}$.
Therefore $\mathcal{F}_{X}\subseteq\mathcal{F}(X,Y)$ . Similarly,
we can prove that $\mathcal{F}_{Y}\subseteq\mathcal{F}_{(X,Y)}$.
Hence $\mathcal{\mathcal{F}}_{X}\cup\mathcal{F}_{Y}\subseteq\mathcal{F}_{(X,Y)}$
and it follows that $\sigma\left(\mathcal{F}_{X}\cup\mathcal{F}_{Y}\right)\subseteq\mathcal{F}_{(X,Y)}$.
On the other hand, if $A,B\in\mathcal{B}(\mathbb{R})$, then $(X,Y)^{-1}(A\times B)=X^{-1}(A)\cap Y^{-1}(B)\in\sigma\left(\mathcal{F}_{X}\cup\mathcal{F}_{Y}\right)$.
That is, $(X,Y)^{-1}(C)\in\sigma\left(\mathcal{F}_{X}\cup\mathcal{F}_{Y}\right)$
for any $C\in\mathcal{C}:=\{A\times B\mid A,B\in\mathcal{B}(\mathbb{R})\}$.
Recall that $\sigma(\mathcal{C})=\mathcal{B}(\mathbb{R}^{2})$, by
the theorem below, we have $(X,Y)^{-1}(C)\in\sigma\left(\mathcal{F}_{X}\cup\mathcal{F}_{Y}\right)$
for any $C\in\sigma(\mathcal{C})=\mathcal{B}(\mathbb{R}^{2})$. That
is, $\mathcal{F}_{(X,Y)}\subseteq\sigma\left(\mathcal{F}_{X}\cup\mathcal{F}_{Y}\right)$.
Theorem: Let $(X,\mathcal{F})$ and $(Y,\mathcal{G})$ be measurable
spaces and let $f:X\rightarrow Y$ be a map. Let $\mathcal{C}\subseteq\mathcal{G}$
be such that $\sigma(\mathcal{C})=\mathcal{G}$. If $f^{-1}(A)\in\mathcal{F}$
for each $A\in\mathcal{C}$, then $f^{-1}(A)\in\mathcal{F}$ for each
$A\in\mathcal{G}$.
Proof of the theorem: Let $\mathcal{M}=\{A\subseteq Y\mid f^{-1}(A)\in\mathcal{F}\}$.
It can be verified directly that $\mathcal{M}$ is a $\sigma$-algebra
on $Y$ and $\mathcal{C}\subseteq\mathcal{M}$. Therefore, $\sigma(\mathcal{C})\subseteq\mathcal{M}$.
In particular, $f^{-1}(A)\in\mathcal{F}$ for each $A\in\mathcal{G}$.
