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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.

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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.

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  • $\begingroup$ Sorry I did a mistake is $\frac{T_n}{n}-\frac{S_n}{n}$ this limit must be zero... Is it the same with your argument? $\endgroup$ – Rachel Nov 26 '17 at 19:50
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    $\begingroup$ @Knight A sequence $u_n$ goes to $0$ if and only if $-u_n$ goes to $0$, there is no issue. $\endgroup$ – Gabriel Romon Nov 26 '17 at 19:56

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