This question comes from chapter 3 of "Concepts in Thermal Physics" by Blundell & Blundell. The common result for variance is derived as follows:
$$\sigma_x^2 = <(x-<x>)^2>=<(x^2+<x>^2-2x<x>)>=<x^2>+<x>^2-2<x><x>=<x^2>-<x>^2$$
This implies that $<2x<x>>=2<x><x>$, and $<<x>^2>=<x>^2$. Both of these seem to imply that averaging an already averaged value does nothing. Is this true? If so, why?
On
If $c$ is a constant, then
$$\langle c \rangle = c$$ and $$\langle c f(x) \rangle = c\ \langle f(x) \rangle$$
for any function of $x$.
And guess what? $\langle x \rangle$ is a constant! I know the notation is incredibly confusing, but that's really all that's going on here.
(And also the square of a constant is also a constant too, if that wasn't clear.)
Formally $<x>$ is the expectation of $x$, which is $\int x p(x) dx$, where $p(x)$ is the probability density function. But once calculated, $\mu=<x>$ is just a number, so $$<<x>>\equiv<\mu>=\int \mu p(x) dx=\mu \int p(x) dx=\mu.$$ Thus in something like $<x-<x>>$ it should really say $<x-\mu>$ to make it clear that it's a constant.