In my textbook on Advanced Probability it reads
"Definition: $X_n$ converges in probability to X if for all $\epsilon >0$, $\lim_{n\rightarrow \infty} P(|X_n-X| \geq \epsilon) = 0$"
Now in a lemma, we set out to prove that $X_n$ also converges in probability to $X$ if and only if for all $\epsilon > 0$, $\lim_{n\rightarrow \infty} P(|X_n-X| > \epsilon) = 0$.
Now in the proof, proving the "if" direction, they fix $\epsilon > 0$ and reason (without further argument)
$${\lim \sup}_{n \rightarrow \infty} P(|X_n -X| \geq \epsilon) \leq {\lim \sup}_{n \rightarrow \infty}P(|X_n – X| > \epsilon/2) $$
and from here conclude that $\lim_{n\rightarrow \infty} P(|X_n-X| \geq \epsilon) = 0$.
I have two questions:
- Why use Lim Sup, rather than ordinary Lim, when the definition only concerns the limit?
- Why is this inequality "obviously" true?
Best Answer
The inequality follows from the fact that the event $|X_n-X| \geq \epsilon$ is contained in the event $|X_n-X| >\epsilon/2$. Recall that $A \subseteq B$ implies $P(A) \leq P(B)$.
When you want to prove that $a_n \to 0$ (where $a_n \geq 0$ ) but you don't know that $\lim a_n$ exists you start with $\lim \sup a_n$. But once you show that $\lim \sup a_n=0$ you can conclude that $\lim a_n$ exist and is $0$.