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Let $X_1, X_2, \ldots$ be i.i.d. random variables with mean $\mu$ and variance $\sigma^2$. How does

\begin{equation} \mathbb E\left[\,\left|\frac{1}{N} \sum_{i=1}^n X_i - \mu\, \right|\,\right] \to O\left(\frac{1}{\sqrt N}\right) \end{equation}

follow from the central limit theorem? We easily get

\begin{equation} \mathbb E\left[\left(\frac{1}{N} \sum_{i=1}^n X_i - \mu\, \right)^2\right] = \frac{\sigma^2}{N}, \end{equation}

but how to get the first one?

EDIT: Actually, any proof would do $-$ does not have to use the central limit theorem.

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Since $r\mapsto(\mathsf{E}|\cdot|^r)^{1/r}$ is nondecreasing,

$$ \mathsf{E}|\bar{X}_N-\mu|\le \frac{\sigma}{\sqrt{N}}. $$

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  • $\begingroup$ This follows from what? $\endgroup$ – user3749105 Feb 7 at 3:48
  • $\begingroup$ For $0<r\le s$, $(\mathsf{E}|\cdot|^r)^{1/r}\le(\mathsf{E}|\cdot|^s)^{1/s}$. $\endgroup$ – d.k.o. Feb 7 at 3:57
  • $\begingroup$ But this only implies $\lim_{N \to \infty} \sqrt N\;\mathbb E|\bar X_N - \mu| \leq \sigma$. This does not prevent the limit from being $0$. In particular, this means that the case $\mathbb E|\bar X_N - \mu| \to O(1/N)$ is not excluded. We need to bound $\lim_{N \to \infty} \mathbb E|\bar X_N - \mu|$ strictly away from $0$. $\endgroup$ – user3749105 Feb 7 at 7:01
  • $\begingroup$ This is what you need. Big $O(N^{-1/2})$ notation means that there is a constant $M$ s.t. $\sqrt{N}\mathsf{E}|\bar{X}_N-\mu|\le M$ for all $N$ large enough. $\endgroup$ – d.k.o. Feb 7 at 7:18
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Use Jensen's inequality: $\phi(\mathbb{E}[Y])\le \mathbb{E}[\phi(Y)]$ for any convex function $\phi$. Take $Y=|\overline{X}_N-\mu|$ and $\phi(x)=x^2$.

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