A variation on the Borel–Cantelli lemma theme

Question

Let $X,X_0,X_1,\dots$ be nonnegative independent identically distributed (i.i.d.) random variables. Let \begin{equation*} E:=\bigcap_{n\ge0}B_n, \end{equation*} where \begin{equation*} B_n:=\bigcup_{k\ge n}\{X_k>k-n\}. \end{equation*}

It was asked (see this post and this comment) if then necessarily $P(E)=0$.

It was shown that actually $P(E)=1$ if $EX=\infty$. Soon after that, the question, and with it the answer, were deleted by the OP.

The question remained as to what happens otherwise, when $EX<\infty$.

To wrap up this matter, below it will also be shown that $P(E)=0$ if $EX<\infty$.

Thus, we have a nice zero–one dichotomy.

However, the Kolmogorov zero–one law does not seem to be directly applicable here: even though the event $B_n$ depends only on the tail $(X_n,X_{n+1},\dots)$ of the sequence $(X_0,X_1,\dots)$, the sequence $(B_0,B_1,\dots)$ of the events is not nonincreasing, so that it is unclear if the event $E$ is in the terminal $\sigma$-algebra generated by $(X_0,X_1,\dots)$.

Answer 1

$\newcommand{\ep}{\varepsilon}$To begin, note that \begin{equation*} P(B_n)=1-\prod_{k\ge n}P(X_k\le k-n) =p:=1-\prod_{j\ge0}(1-P(X>j)). \tag{1}\label{1} \end{equation*} Next, \begin{equation*} \sum_{j\ge0}P(X>j)=E\sum_{j\ge0}1(X>j)\in[EX,EX+1), \end{equation*} since $\sum_{j\ge0}1(X>j)\in[X,X+1)$. So, \begin{equation*} p<1\iff EX<\infty. \end{equation*}

In particular, we have $P(B_n)=p=1$ for all $n$ if $EX=\infty$, and then clearly $P(E)=1$.

The case when $EX<\infty$, and hence $p<1$, is more interesting. To consider this case, take a natural $d$ and and let \begin{equation*} C_i:=\bigcup_{(i+1)d>k\ge id}\{X_k>k-id\}\quad\text{and}\quad D_i:=B_{id}\setminus C_i \end{equation*} for $i=0,1,\dots$, so that the events $C_0,C_1,\dots$ are independent, \begin{equation*} P(C_i)\le P(B_{id})=p, \end{equation*} and (cf. \eqref{1}) \begin{equation*} P(D_i)\le P\Big(\bigcup_{k\ge(i+1)d}\{X_k>k-id\}\Big) =1-\prod_{j\ge d}(1-P(X>j))=:\ep_d. \end{equation*} Then, for any natural $q$, \begin{equation*} P(E)\le P\Big(\bigcap_{i=0}^{q-1}B_{id}\Big) =P\Big(\bigcap_{i=0}^{q-1}(C_i\cup D_i)\Big) \\ \le P\Big(\bigcap_{i=0}^{q-1}C_i\Big)+\sum_{i=0}^{q-1}P(D_i) \le\prod_{i=0}^{q-1}P(C_i)+q\ep_d \le p^q+q\ep_d. \end{equation*} Note now that $\ep_d\to0$ as $d\to\infty$, because the product $\prod_{j\ge 0}(1-P(X>j))$ converges (to $1-p>0$). So, letting $d\to\infty$ and then $q\to\infty$, we get $P(E)\le0$. $\quad\Box$