I noticed other similar questions about this, but I'm going in circles. Am I correct that the following are equivalent ways of showing $X_n$ a sequence of random variables converges almost surely to zero? Assume $X_n$ greater than or equal to $0$ for all $n$ for ease of notation, and limsup's and liminf's are taken as n goes to infinity.
- Show that $P(\limsup \{X_n > \epsilon\}) = 0$
- Show that $P(\liminf {X_n < \epsilon}) = 1$, this being the complement of (1), correct?
Is it correct to interpret this as: if the limit of $X_n$ exists, say $0$, then
$$\liminf X_n = \limsup X_n = \lim X_n = 0$$
so statements (1) and (2) above are really just saying that, for sufficiently large $n$, the probability of $X_n$ exceeding arbitrarily small $\epsilon$ is zero, or, equivalently, that the probability of $X_n$ being less than or equal to $\epsilon$ is one. But this sounds exactly like the definition of convergence in probability, but proving this will not guarantee almost sure convergence, hence, my issue.
I appreciate any help!
Best Answer
The complement of $\limsup \{X_n > \epsilon\}$ is $\liminf \{X_n \leqslant \epsilon\}$, which contains $\{\limsup X_n\lt\epsilon\}$ and is included in $\{\limsup X_n\leqslant\epsilon\}$. Equivalently, $$ \{\limsup X_n \gt \epsilon\}\subseteq\limsup \{X_n > \epsilon\}\subseteq\{\limsup X_n \geqslant \epsilon\}. $$ An example to show that the first inclusion may be strict is $\{\forall n,X_n=\epsilon+1/n\}$. An example to show that the second inclusion may be strict is $\{\forall n,X_n=\epsilon-1/n\}$.
Yes these are rather subtle interplays between the limsup of sets and the limsup of random variables, and between strict and large inequalities.