If $X$ and $Y$ are independent random variables, are $X^2$ and $Y^2$ also independent?

Question

If $X$ and $Y$ are independent random variables, are $X^2$ and $Y^2$ also independent?

Can this be proven?

Answer 1

To prove that 2 (discrete) random variables are independent, it boils down to showing that $P(A=a, B=b) = P(A=a)P(B=b)$ all the time.
If they are not independent, then we must find a counter example.

If we suspect that they are independent, then we want to show that:

(Case when $x\neq 0, y \neq 0$)
P( X = x^2, Y = y^2 ) = P( X = x OR X = -x, Y = y OR Y = -y)
= P( X = x, Y = y) + P( X = -x, Y = y ) + ...
= ?
= ? P(X = x^2 ) P(Y= y^2)

Can you take it from here?
Must this equality be true? How are we using the independence of $X, Y$?
What about the case when $ x =0$? What if the random variables are continuous, or a mix?

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More generally, we say that if $X, Y$ are independent, then for measurable functions $f, g$, we get that $f(X), g(Y)$ is also independent.
The proof is immediate from understanding measurable functions.

If you haven't learn about measurable functions, then just think of them "reasonable, nicely behaving functions" like $ f(x) = x^2$. The proof follows by a similar approach as above.

Answer 2

The technical way to prove this...

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, and suppose $X, Y: (\Omega, \mathcal{F}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$, where $\mathcal{B}(\mathbb{R})$ is the collection of Borel sets of $\mathbb{R}$, and $(\Omega, \mathcal{F})$ denotes the measurable space consisting of the sample space $\Omega$ with a $\sigma$-algebra of subsets of $\Omega$ given by $\mathcal{F}$.

Let $g: \mathbb{R} \to \mathbb{R}$ be given by $g(x) = x^2$. Because $g$ is continuous over $\mathbb{R}$, we have that $g: (\mathbb{R}, \mathcal{B}(\mathbb{R})) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$.

Thus, for any $A, B \in \mathcal{B}(\mathbb{R})$, $g^{-1}(A) \in \mathcal{B}(\mathbb{R})$ and $g^{-1}(B) \in \mathcal{B}(\mathbb{R})$. By definition of $X$ and $Y$ being independent, it follows that $X^{-1}(g^{-1}(A))$ and $Y^{-1}(g^{-1}(B))$ are independent. Thus, $(g \circ X)^{-1}(A)$ and $(g \circ Y)^{-1}(B)$ are independent for any $A, B \in \mathcal{B}(\mathbb{R})$.

It follows that $g(X) = X^2$ and $g(Y) = Y^2$ are independent random variables.

Answer 3

In fact, we have: For any Borel functions $f,g:\mathbb{R}\rightarrow\mathbb{R}$ , if $X,Y$ are independent, then $f(X),g(Y)$ are also independent.

The proof is just few lines: Observe that $\sigma(f(X)):=\{f(X)^{-1}(B)\mid B\in\mathcal{B}(\mathbb{R})\}=\{X^{-1}(f^{-1}(B))\mid B\in\mathcal{B}(\mathbb{R})\}\subseteq\sigma(X)$. Similary $\sigma(g(Y))\subseteq\sigma(Y)$. It is given that for any $A\in\sigma(X)$ and $B\in\sigma(Y)$, $P(A\cap B)=P(A)P(B)$. Therefore, it is trivial that for any $A\in\sigma(f(X))$ and $B\in\sigma(f(Y))$, $P(A\cap B)=P(A)P(B)$.