Theorem 2.2.9 states the wlln in its most familiar form. Let $X_1, X_2...$ be i.i.d with $E|X_i|<\infty$. Let $S_n=X_1+...+X_n$ and let $\mu=EX_1$. Then $S_n/n\rightarrow \mu$ in probability. The proof first states $xP(|X_1|>x)\leq E(|X_1|I_{(|X_1|>x)})$. It seems using Markov's inequality here but not the exact form because of the indicator function inside the expectation. Can someone explain how this inequality holds?


1 Answer 1


We have

$$xP(|X_1| > x) = xE(I_{(|X_1| > x)}) = E(xI_{(|X_1| > x)}) \le E(|X_1|I_{(|X_1| > x)}).$$

  • $\begingroup$ how does the last inequality hold? $\endgroup$ Commented Nov 1, 2015 at 23:38
  • $\begingroup$ Hi @user3404321, in the event $(|X_1| > x)$, $x \le |X_1(\omega)|$ for all $\omega$ in the sample space. So then $xI_{(|X_1| > x)} \le |X_1(\omega)|I_{(|X_1| > x)}$ for all $\omega$. Therefore $E(xI_{(|X_1| > x)}) \le E(|X_1|I_{(|X_1| > x)})$. $\endgroup$
    – kobe
    Commented Nov 1, 2015 at 23:40
  • $\begingroup$ I see. but how can we assume x<|X_1| here? $\endgroup$ Commented Nov 1, 2015 at 23:41
  • $\begingroup$ I don't understand what you mean. Could you elaborate? $\endgroup$
    – kobe
    Commented Nov 1, 2015 at 23:42
  • $\begingroup$ Sure. I understand that in the event $(|X_1|>x)$, $x<|X_1(\omega)|$ for all $\omega$ in that event, but I do not see how come we know $x<|X_1|$ holds without specifying $x$? $\endgroup$ Commented Nov 2, 2015 at 0:04

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