expectation value of angle from expectation value of trigonometric quantity

Question

Given $\cos(x)$, we can solve for x using the inverse trigonometric functions. Does this same logic apply if we instead know $\langle \cos(x) \rangle$ and want to find $\langle x \rangle$, or does the expectation value operator change things? That is, given $y = \langle \cos(x) \rangle$, does $\langle x \rangle = \arccos(y)$? For instance, when considering the autocorrelation between tangent vectors along a worm-like chain, the expectation value of $\cos(x)$ approaches zero as distance increases, corresponding to the loss of correlation at large separations. Does this then imply that $\langle x \rangle = 90^\circ$? Does it make sense to speak of the average angle in this case?

Answer 1

No. As a very simple example of why this wouldn't work, consider the difference between $\langle x^2 \rangle$ and $\langle x \rangle^2$ for the normal distribution.