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Given a sequence $X_k$ of $k$ independent random variables, the book am reading says I can form 'truncated' random variables, i.e.

$$X_k(n) = X_k \cdot \textbf{1}_{\{ \omega : |X_k(\omega)| \leq n\}}$$

where $\omega$ means an outcome from the sample space and $\textbf{1}$ is an indicator function.

Define $S_n$ and $\hat{S}_n$ as:

$$ S_n = X_1 + X_2 + ...+X_n $$

$$ \hat{S}_n = X_1(n) + X_2(n) + ... X_n(n) $$

so this means that $\hat{S}_n$ is the sum of $n$ truncated random variables.

We also define $m_n$ as:

$$ m_n = \mathbb{E}(X_1(n)) $$

Since the variable distributions (of $X_k$) are the same, we have $m_n=\mathbb{E}(X_k(n))$ for all $k \geq 1$.

The book am reading says that the following inequality is 'obvious', where given $\epsilon > 0$, we have:

$$ P\bigg(\big | \frac{S_n}{n} - m_n \big | \geq \epsilon\bigg) \leq P\bigg(\big | \frac{\hat{S}_n}{n} - m_n \big | \geq \epsilon\bigg) + P\bigg(\hat{S}_n \neq S_n\bigg) $$

But it's not at all obvious to me and I got lost. How can we have the above inequality? (i.e. we might get some large $X_k$ inside the $S_n$, such that the left term is much larger than the right term which is truncated).

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The event $[|S_n/n-m_n|\ge\epsilon]$ is contained in the union of the two events $[|\hat S_n/n-m_n|\ge\epsilon]$ and $[\hat S_n\ne S_n]$. Think: if some outcome is not in $[\hat S_n\ne S_n]$, then $\hat S_n - S_n$, and in that case for $|S_n/n-m_n|\ge\epsilon$ to occur it must also happen that $|\hat S_n/n-m_n|\ge\epsilon$.

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