Equivalence of $\sigma$-field $\sigma(X,Y)$ and $\sigma (X+Y,X-Y)$

Question

Consider any two random variable $X$ and $Y$. Is it correct to say that $\sigma$-field $\sigma(X,Y)$ and $\sigma (X+Y,X-Y)$ are equal?

In my logic this is correct, because,

$$ \sigma(X+Y,X-Y) = \sigma\{\omega:(X+Y, X-Y)(\omega)\in H\} $$ Where $H\in \mathbb{R}^2$

Now $$ \sigma\{\omega:(X+Y, X-Y)(\omega)\in H\} = \sigma\{(X(\omega),Y(\omega))\left[\begin{matrix} 1 & 1\\ 1 & -1\\ \end{matrix}\right] \in H\} \\ = \sigma\{\omega:(X(\omega), Y(\omega))\in H'\} = \sigma(X,Y) $$ Where $H'\in \mathbb{R}^2$, $H'$ is nothing but orthogonal transformation of $H$. Hence the proof.

Is my logic correct? If not what should I do?

Answer 1

Here is a more general result.

If $X : \Omega \rightarrow (E ,\mathcal E)$ and $Y : \Omega \rightarrow (F ,\mathcal F)$ are random variables such that $Y = f(X)$, where $f : (E ,\mathcal E) \rightarrow (F ,\mathcal F)$ is a measurable function. Then $\sigma(Y) \subset \sigma(X)$.

proof: By definition $\sigma(Y) = \{ Y^{-1}(B) : B \in \mathcal F\}$. Let $B \in \mathcal F$. Since $f$ is measurable, we have $A = f^{-1}(B) \in \mathcal E$. Moreover, $Y^{-1}(B) = (f \circ X)^{-1} (B) = X^{-1}(f^{-1}(B )) = X^{-1}(A)$ so $Y^{-1}(B) \in \sigma(X)$. Therefore $\sigma(Y) \subset \sigma(X)$.

Now if you take $(X,Y)$ and $(X+Y,X-Y)$ respectively instead of the $X$ and the $Y$ in the result above, you have

$$ (X+Y,X-Y) = f(X,Y) $$

where $f : (x,y) \in \mathbb R^2 \mapsto (x+y,x-y)$ is continuous and thus measurable so $$ \sigma(X+Y,X-Y) \subset \sigma(X,Y). $$ You also have $(X,Y) = g(X+Y,X-Y)$ with $g : (a,b) \in \mathbb R^2 \mapsto (\frac{a+b}{2},\frac{a-b}{2})$ so $$ \sigma(X,Y) \subset \sigma(X+Y,X-Y) $$ which gives you $$ \sigma(X,Y) =\sigma(X+Y,X-Y). $$

Answer 2

Let $(\Omega,\mathcal{F})$ be the underlying measurable space. Let $X,Y$ be random variables. Firstly, recall that the symbol $\sigma(X,Y)$ denotes the smallest $\sigma$-algebra $\mathcal{M}$ such that $X$ is $\mathcal{M}/\mathcal{B}(\mathbb{R})$-measurable and $Y$ is $\mathcal{M}/\mathcal{B}(\mathbb{R})$-measurable. Such $\sigma$-algebra always exist. For, let $\mathcal{C}$ be the collection of all such $\sigma$-algebras. Then $\cap\mathcal{C}$ is a $\sigma$-algebra. Clearly $X$ and $Y$ are $\cap\mathcal{C}/\mathcal{B}(\mathbb{R})$-measurable and $\cap\mathcal{C}$ is "smallest".

Claim 1: $\sigma(X+Y,X-Y)\subseteq\sigma(X,Y)$.

Proof: Denote $\mathcal{M}=\sigma(X,Y)$. Since $X,Y$ are $\mathcal{M}/\mathcal{B}(\mathbb{R})$-measurable, so $X+Y$ and $X-Y$ are also $\mathcal{M}/\mathcal{B}(\mathbb{R})$-measurable. It follows that $\sigma(X+Y,X-Y)\subseteq\mathcal{M}$.

Claim 2: $\sigma(X,Y)\subseteq\sigma(X+Y,X-Y)$.

Proof: Observe that $X=\frac{1}{2}\{(X+Y)+(X-Y)\}$, $Y=\frac{1}{2}\{(X+Y)-(X-Y)\}.$ Therefore $X,Y$ are $\sigma(X+Y,X-Y)/\mathcal{B}(\mathbb{R})$-measurable. Hence $\sigma(X,Y)\subseteq\sigma(X+Y,X-Y)$.