TITLE: partial sums of the sequences $(X_n)$ and $(Y_n)$
QUESTION [3 upvotes]: Let $(X_n : n = 1,2, \ldots )$ be a sequence of identically distributed random variables having finite mean. For $n=1,2,\ldots$, let $Y_n = X_nI_{\{w:|X_n(w)|\leq n\}}$, let $(S_n : n = 1,2, \ldots)$ and $(T_n : n = 1,2, \ldots)$ denote the sequences of partial sums of the sequences $(X_n)$ and $(Y_n)$, respectively. Then
$$\lim_{n \rightarrow \infty}(\frac{T_n}{n}-\frac{S_n}{n})=0 \ $$

The hint for this problem said proving first this result:

Let $(X_n : n = 1,2, ... )$ be a sequence of identically distributed
random variables having finite mean. Then $$ P(\limsup_{n \rightarrow \infty }\{w: |X_n(w)|> n \}) = 0$$

My attempt for this is:
Let $F$ denote the common distribution function of the random variables
$|Xn|$. Using the expression for the mean of a nonnegative random variable, we obtain $$\infty > \int_0^{\infty} [1-F(x)]dx \geq \sum_{n=1}^{\infty}[1-F(x)] = \sum_{n=1}^{\infty} P(\{w: |X_n(w)|> n \})$$ and then appealing to the Borel Lemma we have the probability is zero.
Is it correct?
but How can I apply this result to solve this problem?
Could someone help me pls. Thanks for your time and help.

REPLY [3 votes]: It's straightforward when you follow the hint. Note first that $$\frac{S_n}{n}-\frac{T_n}{n}=\frac 1n \sum_{k=1}^n X_k 1_{|X_k|>k}$$ 
Since $P(\limsup_{n \rightarrow \infty }\{w: |X_n(w)|> n \}) = 0$, we have $P(\liminf_{n \rightarrow \infty }\{w: |X_n(w)|\leq  n \}) = 1$.
Thus, for almost all $w$, there exists $N(w)$ such that $n\geq N(w)\implies |X_n(w)|\leq  n $. For $n\geq N(w)$, $$\begin{align}\frac{S_n(w)}{n}-\frac{T_n(w)}{n}&=\frac 1n \sum_{k=1}^{N(w)} X_k(w) 1_{|X_k|>k}(w)+\frac 1n \sum_{k=N(w)+1}^{n} X_k(w) 1_{|X_k|>k}(w)\\ &=\frac 1n \sum_{k=1}^{N(w)} X_k(w) 1_{|X_k|>k}(w)\end{align}$$
Since $\sum_{k=1}^{N(w)} X_k(w) 1_{|X_k|>k}(w)$ is a fixed quantity, $\lim_n  \frac{S_n(w)}{n}-\frac{T_n(w)}{n}=0$.
This holds for almost all $w$, hence the claim.
{"set_name": "stack_exchange", "score": 3, "question_id": 2538424}