Proof that if $X$ and $Y$ are independent, then $f(X)$ and $g(Y)$ are also independent

Solution 1:

The definition of independence states that $X$ and $Y$ are independent if and only if for sets of the form $E_1 = (-\infty, a)$, $E_2 = (-\infty, b)$, $P(X \in E_1) \cdot P(Y \in E_2) = P(X \in E_1 \land Y \in E_2)$.

Recall we can show that $X$ and $Y$ are independent if and only if for all measurable sets $E_1, E_2 \subseteq \mathbb{R}$, $P(X \in E_1) \cdot P(Y \in E_2) = P(X \in E_1 \land Y \in E_2)$. This characterisation of independence is harder to prove that the definition of independence but easier to apply, so we will use it here.

Suppose that $f, g$ are measurable functions.

Note that $f(X) \in E_1$ iff $X \in f^{-1}(E_1)$, and $g(Y) \in E_2$ iff $Y \in g^{-1}(E_2)$. Note that both $f^{-1}(E_1)$ and $g^{-1}(E_2)$ are measurable.

Then we see that

$\begin{equation} \begin{split} P(f(X) \in E_1) \cdot P(g(Y) \in E_2) &= P(X \in f^{-1}(E_1)) \cdot P(y \in g^{-1}(E_2)) \\ &= P(X \in f^{-1}(E_1) \land Y \in g^{-1}(E_2)) \\ &= P(f(X) \in E_1 \land g(Y) \in E_2) \end{split} \end{equation}$

as required.