Probability truncation method $P(|S_n| \ge \epsilon) \le P(|S_{n,\le N}| \ge \epsilon/2) + P(|S_{n,>N}| \ge \epsilon/2)$?

Question

Suppose $X_i$ are iid random variables and let split it as $$X_i = X_{i,\le N} + X_{i,>N}$$ where $X_{i,\le N} = X_i\textbf{I}(|X_i|\le N)$ and $X_{i,> N} = X_i\textbf{I}(|X_i|> N)$.

Let $S_n = \sum_{i=1}^n X_i$ and similarly split $S_n$ as $$S_n = S_{n,\le N} + S_{n,>N}$$ where $S_{n,\le N} = \sum_{i=1}^n X_{i,\le N}$ and $S_{n,>N} =\sum_{i=1}^n X_{i,> N}$.

I am not sure how to show $$ P(|S_n| \ge \epsilon) \le P(|S_{n,\le N}| \ge \epsilon/2) + P(|S_{n,>N}| \ge \epsilon/2) $$ for some $\epsilon > 0$.

In the book I read (Tao's random matrix theory, especially, pdf page p86, book page p78, in the proof of Thm 2.1.7.), I think this argument is used. As there is no explanation on this, I am assuming it is somewhat trivial. But I am not sure how this can be done.

Here is my trial. Certainly, $|S_n| \ge \epsilon$ implies $|S_{n,\le N}| + |S_{n,> N}| \ge \epsilon$, from the triangular inequality. Thus $$ P(|S_n| \ge \epsilon) \le P(|S_{n,\le N}| + |S_{n,> N}| \ge \epsilon)$$ However I am not sure how the probability in the right handside could be smaller than $P(|S_{n,\le N}| \ge \epsilon/2) + P(|S_{n,>N}| \ge \epsilon/2)$.

Any comments or answers will be very appreciated.

Answer 1

Note that $\{ |S_n| \ge \epsilon \} \subseteq \{ |S_{n, \le N}| \ge \epsilon/2 \} \cup \{ |S_{n, > N} | \ge \epsilon/2 \}$ . . . now use the following two facts

• if $A \subseteq B$ then $P(A) \le P(B)$
• $P(A \cup B) = P(A) + P(B) - P(A\cap B) \le P(A) + P(B)$