Can the sample mean converge faster than $1/\sqrt{n}$?

Question

Consider a IID sequence $X_1,...,X_n$ of random variables with $\mathbb{E}[X_i] = 0$ and $\mathbb{E}[X_i^2] = 1$.

The central limit theorem implies that the sample mean converges in $L_1$ to $0$ at a rate $O(1/\sqrt{n})$. I am interested in an anti-concentration result showing that this is the fastest possible rate over all $X_i$s.

So far, using the Paley-Zygmund inequality, I have derived such a result assuming the $X_i$'s have sufficiently small $4^{th}$ moments; specifically, for some universal constant $c_0$, $$\mathbb{E} \left[ \left| \frac{1}{n} \sum_{i = 1}^n X_i \right| \right] \geq \frac{c_0}{\sqrt{n}} \frac{n}{\mathbb{E}[X_i^4] + n} \in \Omega \left( \frac{1}{\sqrt{n}} \right),$$ which implies my desired result as long as $\mathbb{E}[X_i^4] \in O(n)$. Previously, using Berry-Esseen, I was also able to get a similar result assuming $(\mathbb{E}[X_i^3])^{4/3} \in O(\sqrt{n})$.

However, it seems counterintuitive to me that having a large $3^{rd}$ or $4^{th}$ moment would speed up the rate of convergence, and I'm wondering if these assumption can be relaxed. Actually, in my application, it would be ok to assume $\mathbb{E}[X_i^p] \in O(n^{p/2})$ for any $p \geq 1$, although it would be simpler if I can avoid making any assumptions on $X_i$.

Question: Without any further assumptions on the $X_i$s, do there exist universal constants $c_0, n_0 > 0$ such that, for all integers $n > n_0$, $$\mathbb{E} \left[ \left| \frac{1}{n} \sum_{i = 1}^n X_i \right| \right] \geq \frac{c_0}{\sqrt{n}}.$$ Alternatively, is there a counterexample, i.e., a sequence of IID sequences $\{\{X_{n,i}\}_{i = 1}^n\}_{n = 1}^\infty$ such that $$\lim_{n \to \infty} \mathbb{E} \left[ \left| \frac{1}{\sqrt{n}} \sum_{i = 1}^n X_{n,i} \right| \right] \to 0.$$

Answer 1

There's a counterexample. Define IID sequences $X_{n,i}$ such that $X_{n,i} = \pm n$ each with probability $\frac12 n^{-2}$, and $X_{n,i} = 0$ otherwise. Then $\mathbb E X_{n,i} = 0$, $\mathbb E X_{n,i}^2 = 1$, and

$$\mathbb E\left[\left\lvert \frac{1}{\sqrt{n}} \sum_{i=1}^n X_{n,i} \right \rvert\right] \le \mathbb E \left[\frac{1}{\sqrt{n}}\sum_{i=1}^n \lvert X_{n,i} \rvert \right] = O(n^{-1/2}).$$

(Arguing similarly, you can show that your condition $\mathbb E[X_{n,i}^4] = O(n)$ cannot be weakened.)

Answer 2

This is not an answer (there is already one perfect answer by @Adam), just too long for a comment.

In essence, it depends on how you quantify the "rate of convergence". Here you are looking at the expected absolute value while fixing the standard deviation. And the counterexample @Adam proposes is precisely where the former is negligible compared to the latter.

And there is nothing counterintuitive with the phenomenon you observe. Once more, you fix the second moment, so thanks to the Hölder inequality, your first moment can be bounded from below by (inverse of) higher moments. For instance, $$ \mathbb{E}\big[X^2\big]^{3/2} = \mathbb{E}\big[|X|^{2/3}|X|^{4/3}\big]^{3/2}\le \mathbb{E}[|X|]\cdot\mathbb{E}[X^4]^{1/2}. $$ Consequently, for iid centered $X_{n,1},\dots,X_{n,i}$ with unit variance, $$ \mathbb E\left[\left| \sum_{i=1}^n X_{n,i} \right |\right] \ge \frac{\mathbb E\left[\left( \sum_{i=1}^n X_{n,i} \right )^2\right]^{3/2}}{\mathbb E\left[\left( \sum_{i=1}^n X_{n,i} \right )^4\right]^{1/2}} = \frac{n^{3/2}}{\mathbb E\left[\left( \sum_{i=1}^n X_{n,i} \right )^4\right]^{1/2}}. $$ In turn, $$ \mathbb E\left[\left( \sum_{i=1}^n X_{n,i} \right )^4\right] = n \mathbb E [X_{n,1}^4] + 3n(n-1) \mathbb E [X_{n,1}^2]^2 = n \mathbb E [X_{n,1}^4] + 3n(n-1), $$ so you indeed get the desired lower bound as long as $\mathbb{E}[X_{n,1}^4] = O(n), n\to\infty$.