Functions of Independent Random Variables 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.   
 A: Yes, $g(X)$ and $h(Y)$ are independent for any functions $g$ and $h$ so long as $X$ and $Y$ are independent. It's a very well known results, which is studied in probability theory courses. I'm sure you can find it in any standard text like Billingsley's.
A: 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.
A: Not as an alternative, but as an addition to the previous brilliant answers, note that this result is in fact very intuitive. 
Usually, we think that $X$ and $Y$ being independent means that knowing the value of $X$ gives no information about the value of $Y$ and vice versa. This interpretation obviously implies that you can't somehow "squeeze" an information out by applying a function (or by any other means actually).
A: Consider this "less advanced" proof:
Let $X:\Omega_X\to\mathbb{R}^n,Y:\Omega_Y\to\mathbb{R}^m,f:\mathbb{R}^n\to\mathbb{R}^k,g:\mathbb{R}^m\to\mathbb{R}^p$, where $X,Y$ are independent random variables and $f,g$ are measurable functions. Then:
$$
P\{f(X)\leq x \text{ and } g(Y)\leq y\}\\=P(\{f(X)\leq x\}\cap\{g(Y)\leq y\})\\=P(\{X\in\{w\in\mathbb{R}^n:f(w)\leq x\}\}\cap\{Y\in\{w\in\mathbb{R}^m:g(w)\leq y\}\}).
$$
Using independence of $X$ and $Y$,
$$
P(\{X\in\{w\in\mathbb{R}^n:f(w)\leq x\}\}\cap\{Y\in\{w\in\mathbb{R}^m:g(w)\leq y\}\})=\\=P\{X\in\{w\in\mathbb{R}^n:f(w)\leq x\}\cdot P\{Y\in\{w\in\mathbb{R}^m:g(w)\leq y\}\}
\\=P\{f(X)\leq x\}\cdot P\{g(Y)\leq y\}.
$$
The idea is to notice that the set 
$$
\{f(X)\leq x\}\equiv\{w\in\Omega_X:f(X(w))\leq x\}=\{X\in\{w\in\mathbb{R}^n:f(w)\leq x\}\},
$$
so properties that are valid for $X$ are extended to $f(X)$ and the same happens for $Y$.
