Is the claim that functions of independent random variables are themselves independent, true? 

I have seen that result often used implicitly in some proofs, for example in the proof of independence between the sample mean and the sample variance of a normal distribution, but I have not been able to find justification for it. It seems that some authors take it as given but I am not certain that this is always the case.   

The most general and abstract definition of independence makes this assertion trivial while supplying an important qualifying condition: that two random variables are independent means the sigma-algebras they generate are independent.  Because the sigma-algebra generated by a measurable function of a sigma-algebra is a sub-algebra, a fortiori any measurable functions of those random variables have independent algebras, whence those functions are independent.

(When a function is not measurable, it usually does not create a new random variable, so the concept of independent wouldn't even apply.)



Let's unwrap the definitions to see how simple this is.  Recall that a random variable $X$ is a real-valued function defined on the "sample space" $\Omega$ (the set of outcomes being studied via probability).


A random variable $X$ is studied by means of the probabilities that its value lies within various intervals of real numbers (or, more generally, sets constructed in simple ways out of intervals: these are the Borel measurable sets of real numbers).
Corresponding to any Borel measurable set $I$ is the event $X^{*}(I)$ consisting of all outcomes $\omega$ for which $X(\omega)$ lies in $I$.
The sigma-algebra generated by $X$ is determined by the collection of all such events.
The naive definition says two random variables $X$ and $Y$ are independent "when their probabilities multiply."  That is, when $I$ is one Borel measurable set and $J$ is another, then

$\Pr(X(\omega)\in I\text{ and }Y(\omega)\in J) = \Pr(X(\omega)\in I)\Pr(Y(\omega)\in J).$
But in the language of events (and sigma algebras) that's the same as

$\Pr(\omega \in X^{*}(I)\text{ and }\omega \in Y^{*}(J)) = \Pr(\omega\in X^{*}(I))\Pr(\omega\in Y^{*}(J)).$


Consider now two functions $f, g:\mathbb{R}\to\mathbb{R}$ and suppose that $f \circ X$ and $g\circ Y$ are random variables.  (The circle is functional composition: $(f\circ X)(\omega) = f(X(\omega))$.  This is what it means for $f$ to be a "function of a random variable".)  Notice--this is just elementary set theory--that

$$(f\circ X)^{*}(I) = X^{*}(f^{*}(I)).$$

In other words, every event generated by $f\circ X$ (which is on the left) is automatically an event generated by $X$ (as exhibited by the form of the right hand side).  Therefore (5) automatically holds for $f\circ X$ and $g\circ Y$: there's nothing to check!



NB You may replace "real-valued" everywhere by "with values in $\mathbb{R}^d$" without needing to change anything else in any material way.  This covers the case of vector-valued random variables.
