Suppose $X_1, X_2, \ldots, X_n $ are iid RVs with $E|X_1| < \infty$. Let $M_n = \max_{1\leq i \leq n}|X_i|$. Use the following inequality to prove that $M_n/n \to_p 0$.


$$ P(|X| \geq \epsilon) \leq \frac{E[g(X)1_{[|X|\geq \epsilon]}]}{g(\epsilon)}$$ for a non-negative, even function $g$ increasing on $[0,\infty)$ and any $\epsilon > 0$.


We want to show that $P(|\frac{M_n}{n}| \geq \epsilon) \to 0$ as $n \to \infty$.

We know from the inequality that $$ P\left(|\frac{M_n}{n}| \geq \epsilon\right) \leq \frac{E[g(M_n/n)1_{[|X|\geq \epsilon]}]}{g(\epsilon)}.$$

So I must somehow show that

$$ \frac{E\left[g(M_n/n)1_{[|X|\geq \epsilon]}\right]}{g(\epsilon)} \to_{n\to \infty}0.$$

I thought of using the Markov inequality to show that $|X_i| < \infty$ but I'm not sure it's valid. My thought process:

$$ P(|X|= \infty) = P\left(\bigcap_{n=1}^\infty |X| \geq n\right) = \lim_{k\to\infty} P(|X|\geq k) \leq \lim_{k\to\infty}\frac {E|X|} k = 0$$

By Markov Inequality and because $E|X|<\infty$, so $|X|<\infty$.

Since $X$ was arbitrary, we can generalize to all $X_i$ which implies $M_n < \infty.$

Then it follows that

$$ P\left(\left|\frac{M_n} n\right| \geq \epsilon\right) \leq \lim_{n\to\infty} \frac{E[g(M_n/n) 1_{[|M_n/n| \geq \epsilon]}}{g(\epsilon)} = 0 $$

I just feel really iffy about $ P(|X|= \infty) = 0$ implying that $|X|< \infty$, even though it seems to make sense.


1 Answer 1


Actually the formula that was given to you is a form of Markov's inequality. Perhaps the question has a slicker solution using some special $g$, but $g(x)=|x|$ works: $$P(|M_n/n|>\epsilon)=P(|M_n|>n\epsilon)\le \sum_{i=1}^nP(|X_i|>n\epsilon)=nP(|X_1|>n\epsilon)\le nE(|X_1|\{|X_1|>n\epsilon\})/(n\epsilon)=E(|X_1|\{|X_1|>n\epsilon\})/\epsilon\to 0 $$ as $n\to\infty$.

  • $\begingroup$ So this is saying that the expectation approaches 0 because E|X| approaches 0 and 1{|X| > n*epsilon} approaches 0 correct? Does this follow because |X| is finite? $\endgroup$
    – Karen M.
    Oct 6, 2016 at 16:50
  • $\begingroup$ Given a random variable $X$ that is integrable ($E|X|<\infty$), you have $E(|X_1|\{|X_1|>n\})\to 0$ as $n\to\infty.$ If you're familiar with the dominated convergence theorem, this implication follows quickly from it. $|X|$ itself need only be finite almost surely. $\endgroup$
    – snarfblaat
    Oct 7, 2016 at 2:57

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