Suppose $X\sim N(\mu,\sigma^2)$ and $X_1,\cdots,X_n$ are samples from $X$. Let $\bar X_n=\frac1n\sum_{i=1}^nX_i$. Then it is well known that
$$\bar X_n\overset{p}\to\mu\qquad\qquad(1)$$
and
$$\sqrt{n}(\bar X_n-\mu)\overset{d}\to N(0,\sigma^2)\qquad\qquad(2)$$
From $(1)$ we have $\bar X_n-\mu=o_p(1)$. From $(2)$ we have $\bar X_n-\mu=O_p(1/\sqrt{n})$.
My question is that can we derive the accurate rate of $\bar X_n-\mu$ convergence to $0$, say, $\bar X_n-\mu=o_p(n^\alpha)$?
Best Answer
I think what you’re looking for is the law of iterated logarithm, which says for finite variance iid variables, $\frac{\bar X_n-\mu}{\sigma}\sim \frac{\sqrt{2\log\log n}}{\sqrt{n}}$ in the sense that the $\limsup$ and $\liminf$ of the ratio are $\pm 1$ almost surely.