According to Wikipedia https://en.wikipedia.org/wiki/Convergence_of_random_variables

$X_n \to X$ almost surely if and only if $\forall\epsilon>0$ $$P(\liminf_{n\to\infty}\{{\lvert X_n-X\rvert\lt \epsilon\}})=1$$

While, according to Almost sure convergence and lim sup

$X_n \to X$ almost surely if and only if $\forall\epsilon>0$ $$P(\limsup\limits_{n\to \infty}\{|X_n-X|\geq \epsilon\})=0$$

This seems to make sense to me because the events in the brackets are complements. However I am not sure if we had to swap $\forall\epsilon>0$ with $\exists\epsilon>0 $

However my lecturer gave us this characterisation $$P(\cup_{m\ge1}\cap_{n\ge1}\cup_{k\ge n}\{|X_k-X|\gt\frac{1}{m}\})=0 $$

Is the characterisation given by my lecturer correct? I am of the opinion that my lecturer's characterisation is $\exists\epsilon>0$

$$P(\limsup\limits_{n\to \infty}\{|X_n-X|\gt \epsilon\})=0$$.

My lecturer's characterisation seems correct intuitively. There is an epsilon such that for all n, there is a k greater than n such that $|X_k-X|>\epsilon $. This seems like divergence.


1 Answer 1


you don't need to switch $\forall$ and $\exists$, because given $\varepsilon >0$, those statements are, as you said, equivalent by taking complements; for fixed $\varepsilon >0$:$$1-P\left(\liminf\limits_{n\to \infty}\{|X_n-X|< \epsilon\}\right)=P\left(\limsup\limits_{n\to \infty}\{|X_n-X|\geq \epsilon\}\right)=0$$because $\left(\limsup_{n\to \infty} A_n \right)^c = \liminf_{n\to \infty}A_n^c$ for some events $A_n$, where $A^c$ means taking complements. So whenever one of those is fulfilled for all $\varepsilon >0$, the other one is, too.
The characterisation given by your lecturer is just the same; note that$$\bigcap_{n\ge1}\bigcup_{k\ge n}\left\{|X_k-X|\gt\frac{1}{m}\right\}=\limsup_{n\to \infty} \left\{ \vert X_k - X \vert > \frac{1}{m} \right\}.$$Now assume $$P\left(\bigcup_{m\ge1}\bigcap_{n\ge1}\bigcup_{k\ge n}\left\{|X_k-X|\gt\frac{1}{m}\right\}\right)=0$$ and take $\varepsilon>0$. Let $l \in \mathbb{N}$ be big enough, s.t. $\frac{1}{l}<\varepsilon$. Then $$0=P\left(\bigcup_{m\ge1}\bigcap_{n\ge1}\bigcup_{k\ge n}\left\{|X_k-X|\gt\frac{1}{m}\right\}\right)\overset{m=l}{\geq} P\left(\bigcap_{n\ge1}\bigcup_{k\ge n}\left\{|X_k-X|\gt\frac{1}{l}\right\}\right)\ge P\left(\limsup_{n\to\infty}\left\{|X_k-X|\gt\varepsilon\right\}\right).$$The other way round, if it holds for every $\varepsilon>0$, then it holds for all $\frac{1}{m}$ with $m \in \mathbb{N}$ and therefore $$P\left(\bigcup_{m\ge1}\bigcap_{n\ge1}\bigcup_{k\ge n}\left\{|X_k-X|\gt\frac{1}{m}\right\}\right)\leq\sum_{m=1}^\infty \underbrace{P\left(\bigcap_{n\ge1}\bigcup_{k\ge n}\left\{|X_k-X|\gt\frac{1}{m}\right\}\right)}_{=0}=0.$$

  • $\begingroup$ Would you mind submitting an answer to my question here too. math.stackexchange.com/questions/3271795/…. More people have commented there and some have said things a little different than what you said here, though I have not the time to really read and understand your answer today. $\endgroup$
    – matt
    Jun 23, 2019 at 17:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .