Pove that the summation of iid sequence satisfies $\frac{S_n}{n\log n}\rightarrow c\quad \text{in probability}$

Question

Suppose that $X_1,X_2,\cdots,X_n$ are iid sequence with pdf $\frac{2}{\pi (1+x^2)}\cdot 1_{(0,+\infty)}(x)$. Denote $S_n$ as $S_n:=X_1+X_2+\cdots+X_n$. Prove that there exits $c>0$ such that $$\frac{S_n}{n\log n}\rightarrow c\quad \text{in probability}$$ and calculate $c$.

My solution: Choose another iid sequence $Y_1,Y_2,\cdots,Y_n$ such that $X_n,Y_n$ are independent and have same distribution. Let $X_n^s:=X_n-Y_n$. I have proved that $$\frac{X_1^s+X_2^s+\cdots+X_n^s}{n\log n}\rightarrow 0\quad \text{in probability}.$$ Therefore, $$\frac{S_n}{n\log n}-m_n\rightarrow 0\quad \text{in probability}$$ where $m_n$ is the medin of $\frac{S_n}{n\log n}$. I want to show $m_n\rightarrow c$ but I can't.

Answer 1

Let $S_n' = \sum_{k=1}^n X_k 1_{\{|X_k|\le a_n\}}$. Clearly,

$$ \mathbb P\left(\left|\frac{S_n - b_n}{a_n}\right|>\varepsilon\right)\le\mathbb P(S_n\ne S_n') + \mathbb P\left(\left|\frac{S_n' - b_n}{a_n}\right|>\varepsilon\right). $$

With $a_n = n\log n$ and $b_n = \mathbb E S_n'$, we note that $$ \mathbb P(S_n\ne S_n')\le\sum_{k=1}^n\mathbb P(|X_k|>a_n)\to 0, $$ and that $$ \mathbb P\left(\left|\frac{S_n' - b_n}{a_n}\right|>\varepsilon\right)\le \frac{1}{\varepsilon^2 a_n^2}\text{Var}\left(S_n'\right)\le\frac{1}{\varepsilon^2 a_n^2}\sum_{k=1}^n\mathbb E(X_k^2 1_{\{|X_k|\le a_n\}})\to 0. $$ So we have $$ \frac{S_n - b_n}{a_n}\to 0\quad\text{in probability} $$ with $b_n/a_n\to 2/\pi$. Then it's easy to see $$ \frac{S_n}{n\log n}\to\frac{2}{\pi}\quad\text{in probability.} $$

Answer 2

Note that $\frac{S_n}{n\log n}$ can be divided into two parts: $$\frac{S_n}{n\log n}=\frac{\sum_{k=1}^nX_k 1_{X_k\le n\log n}}{n\log n}+\frac{\sum_{k=1}^nX_k 1_{X_k> n\log n}}{n\log n}$$ The second part converges to zero in probability, and one can verify the mean square of the first part minus $c$ converges to $(2/\pi-c)^2$. Therefore, $\frac{S_n}{n\log n}$ converges $c=2/\pi$ in probability.