Expectations values of position in quantum mechanics In quantum mechanics, we can show that $$ \langle r \rangle^{-1} \neq \langle r^{-1} \rangle $$
I can understand this mathematically as the integrals are different but can anyone explain physically - in the context of quantum mechanics and its interpretation - why this is the case?
 A: This is nothing to do with QM per se, it's simply comes from the definition of the expectation value and the fact that the expectation operator doesn't commute with an arbitary function of a random variable.
To justify, consider that the expectation value is the average value we would expect to see after many trials. The average value also has the property of not commuting with a function of a random variable, for example:
Lets say we perform an experiment 100 times and 50 times we get the result '1', 25 times we get the result '5' and 25 times we get the result '10'. The average value is 4.75 (whose reciprocal is ~0.235).
Now let's say we perform an experiment 100 times and 50 times we get the result '1', 25 times we get the result '0.2' and 25 times we get the result '0.1'. The average value is 0.575.
A: It seems worth mentioning that the inequality 
$$\langle r\rangle ~\langle\frac{1}{r}\rangle  ~>~1$$ 
follows directly from the Cauchy-Schwarz inequality 
$$|| \sqrt{r}\psi ||~ || \frac{1}{\sqrt{r}}\psi ||~>~ || \psi ||^2~=~1$$ 
when the two functions $\sqrt{r}\psi $ and $\frac{1}{\sqrt{r}}\psi$ are not proportional. Here we have used the fact that an expectation value 
$$\langle f\rangle~=~\int_{\mathbb{R}^3}\! d^3r~f|\psi|^2=||\sqrt{f} \psi ||^2$$ 
of a non-negative function $f$ is related to the 2-norm 
$$ || \psi ||~:=~\sqrt{\int_{\mathbb{R}^3}\! d^3r~|\psi|^2} . $$
A: For the average $\langle r \rangle$, the function $r$ becomes large for larger $r$ and thus more dominantly samples the distribution over which you average at large distances. For the average $\langle r^{-1} \rangle$, the function gives its largest contribution at small $r$ and thus dominantly samples the distribution over small distances. So dependent on the distribution over which you average this is not the same.
A: As correctly indicated by @John Davis, this is a statistical problem, however one may try to connect a little more with physics.
Imagine a central potential (hydrogen atom) : 
 $V(r) = -\dfrac{\gamma}{r}$
Then, your question is equivalent to ask why the following equality is not true : 
$\langle V(r) \rangle \langle r \rangle = - \gamma = \langle V(r) \, r\rangle $ 
Clearly, this equality is only (obviously) true for a (deterministic) classical particle, but it is false for every statistical theory, including quantum mechanics (except if $V$ is a constant).
A: The transformation ($r^{-1}$) of a random variable $r$ is non-linear, as such it gives different results in taking the expectation of the transformation and the transformation of the expectation. In other words the operations do not commute. 
The correspondnce principle (or the Ehrenfest theorem of mean values) is supposed to be for the exact same observables and linear combinations of them.
Mathematicaly (in probability calculus) the transformation of this kind should take into account the zeros of the variable/operator etc..
