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If $X_n$ converges to $X$ in distribution, does it imply $$\lim_{n \rightarrow \infty }E[|X_n|] = E[|X|]$$ If not, suppose, $X_1, X_2, .. X_n $ are i.i.d. random variables with mean $\mu$ and finite variance, what is $$\lim_{n \rightarrow \infty}E\left[\left|\frac{1}{N}\sum_{i=1}^{N}X_i - \mu\right| \right]$$

What I think? Finite variance, so Central limit theorem is valid, i.e. convergence in distribution of "debiased scaled variables" to $N(0,1)$, but I cannot relate this convergence in distribution to the first moment!

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Convergence in distribution does not imply convergence of the first moment. For example consider $X_n$ defined by \begin{align}P(X_n = 0) &= 1-\frac{1}{n} \\ P(X_n = n) &= \frac{1}{n} \end{align} Then $X_n \Rightarrow 0$ in distribution but $E[X_n] = 1 \nrightarrow 0$. To prove convergence in distribution, take $f$ a bounded real function, then $$E[f(X_n)] = \big(1-\frac{1}{n}\big)\,f(0) + \frac{1}{n}\,f(n) \longrightarrow f(0) = E[f(0)]$$ To get the convergence you want, you should first use the law of large numbers: $$\frac{1}{n}\sum_{i=1}^{n}X_i \longrightarrow \mu \ \ \text{a.s.}$$ then argue that the $\overline{X}_n = \frac{1}{n}\sum_{i=1}^{n}X_i$ are uniformly integrable (for instance because they are bounded in $L^2$). Now applying Vitali convergence theorem proves that $\overline{X}_n \to \mu$ in $L^1$.

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