# Proving convergence in probability using basic inequality

Problem:

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 $M_n/n \to_p 0$.

Inequality:

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

Attempt:

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.

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 $n\to\infty$.
• 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. Oct 7, 2016 at 2:57