TITLE: Applying Central Limit Theorem to show that $E\left(\frac{|S_n|}{\sqrt{n}}\right) \to \sqrt{\frac{2}{\pi}}\sigma$
QUESTION [9 upvotes]: 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 $E|Z|$.

REPLY [3 votes]: 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