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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!

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2 Answers 2

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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 $$\begin{aligned}E[|\overline{X}_n-\mu|]&=\frac{1}{n}E[|{S}_n-n\mu|]\\ &=\frac{1}{n}\int_0^\infty P(|{S}_n-n\mu|>t)dt\\ &\leq \frac{2}{n}\int_0^\infty \exp\bigg(-\frac{2t^2}{nM^2}\bigg)dt\\ &=M\sqrt{\frac{\pi}{2n}} \end{aligned}$$

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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).

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