Showing Stronger result of Weak Law of Large Numbers

Question

So, Khintchine's form of the Weak Law of Large Numbers asserts that $i) E(X_1)=0 \Rightarrow (S_n/n) \rightarrow 0$

The stronger result is: $ii) E(X_1)=0 \Rightarrow E(\|S_n\|)=o(n)$

Now ii) is shown to be stronger than i) using the Markov inequality:

$ii) P(\|Y\|\ge c)\le E(\|Y\|)/c$

I just don't understand how the Markov Inequality shows that ii) is stronger than i) could someone explain this in steps please?

Answer 1

Define $Y_n:=S_n/n$. The weak law of large numbers states that $Y_n\to 0$ in probability, while (ii) says that $\mathbb E|Y_n|\to 0$.

In general, convergence in $\mathbb L^1$ implies the convergence in probability. This is indeed a consequence of Markov's inequality: for each positive $\varepsilon$, $$\mathbb P\{|Y_n|\gt\varepsilon\}\leqslant \frac 1{\varepsilon}\mathbb E|Y_n| .$$