# Bound on expected norm of the difference between the sample mean $\bar{X_n}$ and population mean $\mu$ as a function of the sample size $n$ for LLN?

My question is motivated by this question:

Does law of large numbers converge in $L^1$?

that asks about the the convergence in $$L^1$$-norm of the sample mean $$\bar{X_n}$$ to the population mean $$\mu.$$

I looked at the answers and comments and I understood that the answer was in the affirmative, and the proof uses the fact that the sample means $$\bar{X_n}$$ are uniformly integrable(UI). The two steps a) and b) outlined in the previous link follow immediately using:

triangle inequality for norms, exchanging finite sum and integrals and then using that $$X_i\sim_{iid}X,$$ and finally $$\mathbb{E}||X-\mu||< \infty.$$

While the use of UI does indirectly show that the Law of Large Numbers (LLN) is true in $$L^1,$$ I still wonder if there's a bound like the below where $$\{X_i\}\sim_{iid} X: \Omega \to \mathbb{R}^d, \bar{X_n}:=\frac{1}{n}\sum_{i=1}^{n}X_i, \mu:=\mathbb{E}[X]:$$

$$\mathbb{E}||\bar{X_n}-\mu||\le C_X \frac{1}{\phi(n)}, \phi(n)\to \infty, n \to \infty?$$

Here $$C_X$$ is a constant depending only upon $$X,$$ and my guesses for possible choices for $$\phi(n)$$ would be $$n$$ or $$\sqrt{n}$$ etc. but not sure. But anyway, is there such an explicit bound? Resources, links or deductions are highly appreciated!

Here is an example of such bound in the case of $$P(X_1\in [a,b])=1$$. Set $$M=b-a$$ and $$S_n=X_1+...++X_n$$ and by Hoeffding's inequality we get \begin{aligned}E[|\overline{X}_n-\mu|]&=\frac{1}{n}E[|{S}_n-n\mu|]\\ &=\frac{1}{n}\int_0^\infty P(|{S}_n-n\mu|>t)dt\\ &\leq \frac{2}{n}\int_0^\infty \exp\bigg(-\frac{2t^2}{nM^2}\bigg)dt\\ &=M\sqrt{\frac{\pi}{2n}} \end{aligned}
Assuming that $$X \in L^2$$ then by Jensen's inequality and the fact that $$\sqrt{x}$$ is concave we have
$$E(|\bar X_n - \mu|) = E\left(\sqrt{(\bar X_n - \mu)^2}\right) \le \sqrt{\text{Var}(\bar X_n)} = \frac{\sigma}{\sqrt{n}}.$$
The $$\sqrt n$$ rate is sharp, as it is obtained when the $$X_i$$'s are iid from a normal distribution. As mentioned here, my belief is that the rate can be arbitrarily poor for $$X \in L^1$$, and in particular the density $$f(x) \propto 1 / |x|^{2 + \epsilon} I(|x| > 1)$$ seems to have a rate $$n^{\psi(\epsilon)}$$ for some $$\psi(\epsilon) \to 0$$ as $$\epsilon \to 0$$ (I did not attempt to determine what $$\psi(\epsilon)$$ is).