Applying Central Limit Theorem to show that $E\left(\frac{|S_n|}{\sqrt{n}}\right) \to \sqrt{\frac{2}{\pi}}\sigma$ In the book Probability Essentials, by Jacod and Protter, the following question has bugged me for a long while and I'm wondering if it is bugged. The question is an application of Central Limit Theorem:
Let $(X_j)_{j\geq1}$ be iid with $E[X_j] =0$ and $\sigma_{X_j}^2 = \sigma^2 < \infty$. Let $S_n = \sum_{j=1}^n X_j$. Show that
$$ \lim_{n\rightarrow \infty} E\left\{\frac{|S_n|}{\sqrt{n}}\right\} = \sqrt{\frac{2}{\pi}}\sigma$$
What I tried and am aware of:
Interchanging E and lim is wrong as $\frac{|S_n|}{\sqrt{n}}$ does NOT converge a.s to $|Z|$ where $ Z \sim \mathcal{N}(0,\sigma^2)$, but does so in distribution. Note the use of continuous mapping theorem.
Weak convergence aka convergence in distribution implies $E[f(X_n)] \rightarrow E[f(X)]$ when $X_n \rightarrow^d X$ for f continuous and bounded. $f(x)=|x|$ is not bounded.
I tried using a truncated version of $|X|$. But at one stage I had to swap limits and couldn't justify the steps.
I appreciate any help. Also kindly avoid Skorokhod's theorem if possible as it has not been covered. Although if you have an idea using that, I'm all ears.
Note: The RHS is $E|Z|$.
 A: We assume $\sigma=1$.


*

*Show that for all $K$ and all $n$, 
$$P\left(\frac{|S_n|}{\sqrt n}\geqslant K\right)\leqslant\frac 1{K^2}.$$ 

*This implies 
$$\int \frac{|S_n|}{\sqrt n}dP\leqslant \int_{|S_n|<K\sqrt n} \frac{|S_n|}{\sqrt n}dP+\frac 1{K^2}.$$

*Use the definition of weak convergence together with $x\mapsto \min\{|x|,K\}$, a continuous bounded map.



Actually, there is a deeper result (not needed here) called the invariance principle (see Billingsley's book Convergence of probability measures) which stated the following. Take $\{X_n\}$ a sequence of i.i.d. centered random variables, with $EX_n^2=1$, and define for each $n$, a random function $f_n(\omega,\cdot)$ in the following way:


*

*$f_n(\omega,kn^{-1})=\frac 1{\sqrt n}\sum_{j=1}^kX_j(\omega)$ for $0\leqslant k\leqslant n$. 

*$f_n(\omega,\cdot)$ is piecewise linear. 


The measures associated with these functions converge in law to a Brownian motion. 
Now, to get the result, take the continuous bounded functional $F\colon C[0,1]\to \Bbb R$ given by $F(f)=|f(1)|$. The first version of the invariance principle, found by Kac and Erdős, was about the functional $F(f):=\sup_{0\leqslant x\leqslant 1}f(x)$. Then Donsker generalized it. 
A: In a nutshell, you consider some random variables $(Y_n)_n$ and $Y$ defined on a probability space $(\Omega,\mathcal F,\mathbb P)$, such that $Y_n\to Y$ in distribution and you wonder whether $\mathbb E(Y_n)\to \mathbb E(Y)$ (in your case, $Y_n=|S_n|/\sqrt{n}$ and $Y=|Z|$). Obviously, this is not true in general and Fatou lemma only guarantees that $\liminf \mathbb E(Y_n)\geqslant \mathbb E(Y)$. But here, you also know that $(Y_n)_n$ is bounded in $L^2(\mathbb P)$ since $\mathbb E(Y_n^2)=\sigma^2$ for every $n$. Hence $(Y_n)_n$ is uniformly integrable, and, together with the convergence in distribution, this guarantees that $\mathbb E(Y_n)\to\mathbb  E(Y)$.
One can prove the last assertion as follows. First there exists some random variables $(\bar Y_n)_n$ and $\bar Y$, possibly defined on another probability space $(\bar\Omega,\bar{\mathcal F},\bar{\mathbb P})$, such that each $\bar Y_n$ is distributed like $Y_n$, $\bar Y$ is distributed like $Y$, and $\bar Y_n\to\bar Y$ almost surely. In particular, (1) $\bar Y_n\to\bar Y$ in probability and (2) $(\bar Y_n)_n$ is uniformly integrable. For any integrable $\bar Y$, properties (1) and (2) together are equivalent to the convergence $\bar Y_n\to\bar Y$ in $L^1(\bar{\mathbb P})$, that is, to the fact that $\bar{\mathbb E}(|\bar Y_n-\bar Y|)\to0$. As a consequence, $\mathbb E(Y_n)=\bar{\mathbb E}(\bar Y_n)\to \bar{\mathbb E}(\bar Y)=\mathbb E(Y)$.
A: There are two parts to this problem:
1) $\limsup_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}] \leq E[|Z|] $
Proof: (Credits to DavideGiraudo)
$$E[\frac{|S_n|}{\sqrt{n}}] = E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} \leq K}] + E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} > K}]$$
$$\leq E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} \leq K}] + \sqrt{E[\frac{S_n^2}{n}]}.\sqrt{P(\frac{|S_n|}{\sqrt{n}} > K) } $$ (Cauchy Schwarz)
$$\leq E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} \leq K}] + \sigma^2.\frac{1}{K}  $$
(Used the Markov Inequality)
Take limsup on both sides
$$\limsup_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}] \leq \limsup_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} \leq K}] + \sigma^2.\frac{1}{K}$$
$$ \leq \limsup_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} \leq K}] + K.P(\frac{|S_n|}{\sqrt{n}} > K) + \sigma^2.\frac{1}{K} \quad (\times) $$
I have added that extra term for a reason...
Now use the following: 
(a) $f(x)=\min(|x|,K)$ is continuous and bounded
(b) $E[f(X_n)] \rightarrow E[f(X)] $ if $X_n \rightarrow^w X$. f as in (a)
Note that $ E[f(\frac{|S_n|}{\sqrt{n}})] = E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} \leq K}] + K.P(\frac{|S_n|}{\sqrt{n}} > K)$
Now apply (a) and (b) to $(\times)$ to get
$$\limsup_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}] \leq  E[|Z|1_{|Z| \leq K}] + K.P(|Z| > K) + \sigma^2.\frac{1}{K}$$
LHS is independent of K. Take $K \rightarrow \infty$. Note that $K.P(|Z| > K) \rightarrow 0$ and MCT applied to the first term yields
$$\limsup_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}] \leq E[|Z|]  $$
(2) $\liminf_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}] \geq E[|Z|] $
Proof:
$$ E[\frac{|S_n|}{\sqrt{n}}] = E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} \leq K}] + E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} > K}] $$
$$ \geq E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} \leq K}] + K.P(\frac{|S_n|}{\sqrt{n}} > K)$$
Take $\liminf_{n \rightarrow \infty}$ to get
$$\liminf_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}] \geq E[|Z|1_{|Z| \leq K}] + K.P(|Z| > K)$$ (I used (b))
Take $K \rightarrow \infty$, use MCT on the first term on RHS to get
$\liminf_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}] \geq E[|Z|] $
Thus we get $\lim_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}] = E[|Z|] $
QED
