If $n^{p}P(|X_1|>n) \to 0$ then there exist $y_n$ such that $\frac{1}{n^{1/p}}\sum_{k=1}^n X_k-y_n \to 0$ in probability Let $p>1,(X_n)_n$ a sequence of i.i.d random variables such that $\lim_nnP(|X_1|>n^{1/p})=0.$ Let $Y_n=\sum_{k=1}^n X_k.$
Prove that there exist $y_n$ such that $\frac{1}{n^{1/p}}Y_n-y_n$ converges in probability to $0.$
Hint: Consider $Y'_n=\sum_{k=1}^n X_k1_{|X_k| \leq n^{1/p}}$ and prove that $Y_n-Y_n'$ converges in probability to $0.$
Considering $Y_n'=\sum_{k=1}^n X'_k,$ $\forall \epsilon>0,P(|Y_n-Y_n'|>\epsilon)\leq P(Y_n \neq Y_n') \leq nP(|X_1|>n^{1/p}) \to0.$ Clearly $X_1 \in L^1$ since $n^{p}P(|X_1|>n) \to 0.$ Then $\frac{1}{n^{1/p}}Y_n-y_n=\frac{1}{n^{1/p}}(Y_n-Y_n')+\frac{1}{n^{1/p}}Y_n'-y_n,$ so we must find a suitable $y_n$ such that $\frac{1}{n^{1/p}}Y_n'-y_n$ to $0$ in probability.
What is the form of $y_n$? Is it $y_n=n^{1-1/p}E[X_11_{|X_1| \leq n^{1/p}}]$? Or $y_{n}=n^{1-1/p}E[X_1]$?
 A: Firstly, notice that form $ \lim_n n\mathsf{P}(|X_1|>n^{1/p})=0 $ it is easy to get
\begin{gather*}
 \lim_{x\to\infty}x^p\mathsf{P}(|X_1|>x)=0,\\
 \lim_{x\to\infty}\sup_{y\ge x}[y^p\mathsf{P}(|X_1|>y)]=0\qquad \tag{1}
\end{gather*}
and for $ c>0, 0<A<x $,
\begin{align*}
 &\frac1{x^c}\mathsf{E}[|X_1|^{p+c}1_{\{|X_1|\le x}\}]\\
 &\quad=\frac{p+c}{x^c}\mathsf{E}\Big[\int_0^x y^{p+c-1}1_{\{y<|X_1|\le x\}}dy\Big]\\
 &\quad\le \frac{p+c}{x^c}\Big[\int_0^A+\int_A^x\Big]y^{c-1}y^p\mathsf{P}(|X_1|>y)dy\\
 &\quad \le \frac{p+c}{x^c}\int_0^Ay^{p+c-1}\mathsf{P}(|X_1|>y)dy
      +\sup_{y\ge A}[y^p\mathsf{P}(|X_1|>y)]\frac{p+c}{c}.
\end{align*}
Letting $ x\to\infty, A\to\infty $ successively, get
\begin{gather*}
 \varlimsup_{x\to\infty}\frac1{x^c}\mathsf{E}[|X_1|^{p+c}1_{\{|X_1|\le x}\}]
 \le \sup_{y\ge A}[y^p\mathsf{P}(|X_1|>y)]\frac{p+c}{c},\\
 \lim_{x\to\infty}\frac1{x^c}\mathsf{E}[|X_1|^{p+c}1_{\{|X_1|\le x}\}]=0.\tag{2}
\end{gather*}
Taking $ c=2-p,x=n^{1/p} $ in (2),
\begin{equation*}
 \lim_{n\to\infty}\frac{n}{n^{2/p}}\mathsf{E}[|X_1|^21{\{|X_1|\le n^{1/p}\}}]=0. \tag{3}
\end{equation*}
Similarly, for $ 0<c\le p $,
\begin{gather*}
 \lim_{x\to\infty}x^c\mathsf{E}[|X_1|^{p-c}1_{\{|X_1|>x\}}]=0,\\
 \lim_{n\to\infty}n^{1-\frac1p}\mathsf{E}[|X_1|1_{\{|X_1|>n^{1/p}\}}]=0. \tag{4} 
\end{gather*}
As $ n\to\infty $,
\begin{equation*}
 \mathsf{Var}\Big[\frac{Y^\prime_n}{n^{1/p}}\Big]\le
 \frac{n}{n^{2/p}}\mathsf{E}[|X_1|^21{\{|X_1|\le n^{1/p}\}}]\rightarrow 0
\end{equation*}
from (3). Hence
\begin{equation*}
 \frac{Y^\prime_n}{n^{1/p}}-\frac{n}{n^{1/p}}\mathsf{E}[X_11_{\{|X_1|\le n^{1/p}\}}]
 \stackrel{pr}{\longrightarrow}0.
\end{equation*}
Furthermore,
\begin{gather*}
 \frac{Y_n}{n^{1/p}}-\frac{n}{n^{1/p}}\mathsf{E}[X_11_{\{|X_1|\le n^{1/p}\}}] 
 \stackrel{pr}{\longrightarrow}0.  \\
 \frac{Y_n}{n^{1/p}}-\frac{n}{n^{1/p}}\mathsf{E}[X_1]\stackrel{pr}{\longrightarrow}0.
 \tag{5}
\end{gather*}
from $ \mathsf{P}(Y^\prime_n\ne Y_n)\le n\mathsf{P}(|X_1|>n^{1/p}) $ and (4). For all $ y_n=n^{1-\frac1p}(\mathsf{E}[X_1]+o(1)) $,
\begin{equation*}
 \frac{Y_n}{n^{1/p}}-y_n\stackrel{pr}{\longrightarrow}0.
\end{equation*}
from (5) also.
