I am following Terrence Tao notes on concentration inequalities that can be found here.
Define $S_n = X_1 + \dots + X_n$ where $X_i, 1 \le i \le n$ are random variables. From the famous Markov inequality we have
$$P( |S_n| \geq \lambda ) \leq \frac{1}{\lambda} \sum_{i=1}^n {\bf E} |X_i|.$$
Tao states that one can informally view this as "the assertion that ${S_n}$ typically has size ${S_n = O( \sum_{i=1}^n |X_i| )}.$" What is the reason for this?
I realize that if $\lambda > \sum_{i=1}^n |X_i|$ then
$$P( |S_n| \geq \lambda ) < \frac{1}{\sum_{i=1}^n |X_i|} \sum_{i=1}^n {\bf E} |X_i|.$$
So the probability that $|S_n| \ge \lambda$ is smaller than a random variable that has expected value equal to one half. Is this the reason or something else?