Let $X_i$ be i.i.d. random variable with mean 0 and finite variance $E(X_i^2)=σ^ 2 > 0.$ Let $S_n=X_1+...+X_n$. Prove that $\lim\limits_{n \to \infty}E\left(\frac{|S_n|}{\sqrt n}\right)=\sqrt {\frac {\pi}{2}}σ$.
Since $X_i$ with mean 0 and finite variance, $E(\frac{Sn}{n})$ converges to 0 in probability. Since $E(Xi^2)=σ^2$, we know that $E(\frac{X_1^2+...+X_n^2}{n})$ converges to σ in probability. But how about $E(\frac{|S_n|}{\sqrt n})$? Since I want to be able to prove this rigorously, I am not sure where to go next.
Any help or comment will be greatly appreciated. Thank you.
By the central limit theorem, we know that $|S_n |/(\sqrt n \sigma) \to |N|$ in distribution, where $N$ is standard and normal. Unfortunately, in general, we may have a sequence of non-negative integrable random variables $(Y_n)_{n\geqslant 1}$ such that $Y_n\to Y$ in distribution (with $Y$ integrable) but the sequence $(\mathbb E[Y_n])_{n\geqslant 1}$ does not converge to $\mathbb E[Y ]$. However, when $(Y_n)_{n\geqslant 1}$ is uniformly integrable, we do have $\mathbb E[Y_n ]\to \mathbb E[Y]$. Indeed, for each $R$, $$ |\mathbb E[Y_n ]-\mathbb E[Y ] |=\left|\int_0^{+\infty}\left(\mu\{Y_n\gt t\}-\mu\{Y\gt t\}\right)\right|\mathrm dt \leqslant \int_0^R \left|\mu\{Y_n\gt t\}-\mu\{Y\gt t\}\right|\mathrm dt+\\ \int_R^{+\infty}\mu\{Y_n\gt t\}\mathrm dt+\int_R^{+\infty}\mu\{Y\gt t\}\mathrm dt.$$ Now, for a given $\varepsilon$, choose $R$ such that the last two terms are smaller than $\varepsilon$ for each $n$ (this is possible thanks to the uniform integrability). Once this $R$ is fixed, the first term goes to $0$ by dominated convergence.