# If $X$ and $Y$ are independent then $f(X)$ and $g(Y)$ are also independent.

Knowing that if you have two independent $X$ and $Y$, and $f$ and $g$ measurable functions, how to show that then $U = f (X)$ and $V = g (Y)$ are still independent.

• Jonas: This is a typical homework question, so please see these guidelines that apply to such questions, and edit your question accordingly. Otherwise, your question may be closed. Jul 14, 2013 at 21:33
• A hand-waving proof starts with the observation that independence of $X$ and $Y$ means that knowing the value of $X$ tells you nothing that you did not already know about the value of $Y$. For example, the conditional distribution $F_{Y\mid X}(y\mid X=x)$ is the same as the unconditional distribution $F_Y(y)$. If $X$ and $Y$ are independent but $g(X)$ and $h(Y)$ were dependent, then knowing the value of $X$ (and therefore $g(X)$) would mean that we could make some inference about $h(Y)$ and hence possibly about $Y$ which contradicts the independence of $X$ and $Y$. Jul 14, 2013 at 21:45
• math.stackexchange.com/q/8742/321264 May 7, 2020 at 20:53

You said measurable so I am going to assume you want a measure-theoretic answer, and that your definition of independence is $X$ and $Y$ are independent iff $\sigma(X)$ is independent of $\sigma(Y)$, i.e. that $P[X \in B_1, Y \in B_2] = P[X \in B_1]P[Y \in B_2]$ for all borel sets $B_1,B_2$. You must assume $f,g$ are borel functions (this is so that $f(X),g(Y)$ are measurable, so asking the question of independence makes sense). The $\sigma$-algebra generated by $f(X)$ is a sub-$\sigma$-algebra of the $\sigma$-algebra generated by $X$, and similarly for $g(Y)$ and $Y$. To see this note that for any borel set $B$ we have $(f\circ X)^{-1}(B) = X^{-1}(f^{-1}(B)) = X^{-1}(\text{some borel set}) \in \sigma(X)$. Since $\sigma(X) \perp \sigma(Y)$ it follows that $\sigma(f(X)) \perp \sigma(g(Y))$.