Particularly the sample mean, which I think should be $O(1/n)$, no?
The definition I have says, $f(n)$ is of constant order ($O(1)$), when there exist some non-zero constant $c$ such that, $$\frac{f(n)}{c} \to 1, i.e. f(n) \to c, as \ n \to \infty. $$ So in this case $f(n) = n^{-1}\sum\limits_i^n{x_i}$, and $c = \mu$, the expected value of the sequence of random variables $x_i$. But then I get, $$f(n)\to \mu,\ as\ n \to \infty$$ $$f(n) = O(1),\ as\ n \to \infty$$
Shouldn't there be a way to show it is $O(1/n)$? For example, if $$h(n) = n^{-1}(h_1 + h_2 + h_3),$$ then $h(n) = O(1)$, right? Defining a another sequence $l(n) = h_1 + h_2 + h_3$, $l(n)$ should be $O(1)$ as well. But then, I can write $l(n) = nh(n)$. Then I could say, $$nh(n) = O(1), \ h(n) = O(1/n),$$ no? Why can't I do this with the sample mean? I am not sure I can say $\sum\limits_i^n{x_i}$ is $O(1)$ as to do the same as with $h(n)$. Or am I totally wrong?