# $\lim\sup$ and $\lim\inf$ of a sequence of independent random variables with two states

Let $$(X_n)_{n\geq1}$$ be a sequence of random variables such that $$\mathbb{P}(X_n= n^{2/3})=1-\mathbb{P}(X_n=0)=\frac{1}{3n}$$ Find $$\lim\sup X_n$$ and $$\lim\inf X_n$$.

We know that

\begin{align} \{\omega:\limsup X_n(\omega) = +\infty\} \end{align}

is equivalent to \begin{align} \forall A >0, \exists n_0\in\mathbb{N} \quad&\text{s.t}\quad \forall n\geq n_0, \quad\sup_{k\geq n}X_k >A \\ \forall A >0, \exists n_0\in\mathbb{N} \quad&\text{s.t}\quad \forall n\geq n_0, \;\exists k\geq n, \quad X_k >A \end{align} and thus we have that

\begin{align} \{\omega:\limsup X_n(\omega) = +\infty\} = \bigcap_{A>0}\lim\sup\{X_k>A\} \end{align} and we actually take a sequence $$A_k$$ of rationals such that $$A_k \uparrow \infty$$ to make sure that the event is measurable. Finally we check that, $$\forall A_k>0$$,

$$\sum_{n\geq 1}\mathbb{P}(X_n > A_k) = \sum_{n\geq A^{3/2}}\mathbb{P}(X_n = n^{2/3}) = \sum_{n\geq A^{3/2}}\frac{1}{3n}$$ which diverges because it's basically the harmonic series less a finite set of terms and multiplied by a constant and we conclude by the second Borel-Cantelli lemma that the event occur almost-surely and as all the events of the intersection happen almost-surely then $$\mathbb{P}\left(\bigcap_{A_k>0}\lim_n\sup\{X_n>A_k\}\right)=1$$

The $$\lim\inf X_n =0$$ goes about the same way as I found in other posts, however I don't know if the above answer I wrote is correct. I found many questions showing the calculations for the case where limit superior and limit inferior are equal to a constant, but didn't find any about the sequence diverging.

Let $$B_n$$ be the event $$\{X_n=n^{2/3}\}$$. Then $$\sum_{n\geqslant 1}\mathbb P(B_n)$$ diverges hence, as you noticed, by the second Borel-Cantelli lemma, $$\limsup_{n\to \infty}B_n$$ has probability one hence for almost every $$\omega$$, there exists an infinite set $$I(\omega)\subset\mathbb N$$ such that for each $$n\in I(\omega)$$, $$\omega\in B_n$$ (that is, $$X_n(\omega)=n^{2/3}$$) hence we reach your conclusion in a bit shorter way.