random 2-dimensional vectors, $X_n$ is uniformly distributed

Question

Let $(X_1 ,X_2, \ldots)$ be an independent sequence of random 2-dimensional vectors, where for each $n$, $X_n$ is uniformly distributed on the squares with vertices $[\pm n,\pm n]$.

How can I calculate the probability $$ P(\{w: |X_n(w)| \rightarrow \infty \ \ as \ n \rightarrow \infty\}).$$

My attempt:

Let $A_n=\{ w \in \Omega | |X_k| \geq k\}$ with $k \in \mathbb{Z}^{+}$, then $$ P(A_n)= \left\{ \begin{array}{lcc} 0 & if & k \geq \sqrt{2}n \\ \\ 1-\frac{k^2 \pi}{n^2} & if & k \leq n \\ \\ c(k) & if & n < k < \sqrt{2}n \end{array} \right. $$

for $c(k)$ a certain value depending of $k$.

I that correct? Or How can I compute the probability on this $A_n=\{|X_n(w)| < n^\alpha\}$ ?

Could someone help pls? Thanks for your time and help.

Answer 1

Why not use the same approach I showed you in your previous question ?

Since all norms on $\mathbb R^2$ are equivalent, let $\|\cdot\|$ denote the maximum norm (its unit ball is a square, how convenient...).

Note that $\displaystyle P(\|X_n\|\to \infty)= P\left(\bigcap_N\bigcup_n \bigcap_{k\geq n}(\|X_k\| \geq N)\right)$.

Since the sequence of sets (indexed by $N$) $\bigcup_n \bigcap_{k\geq n}(\|X_k\| \geq N)$ is decreasing, $$P\left(\bigcap_N\bigcup_n \bigcap_{k\geq n}(\|X_k\| \geq N)\right)=\lim_N P\left(\bigcup_n \bigcap_{k\geq n}(\|X_k\| \geq N)\right)$$

so it suffices to compute $P\left(\bigcup_n \bigcap_{k\geq n}(\|X_k\| \geq N)\right)$ for each $N$.

But $P\left(\bigcup_n \bigcap_{k\geq n}(\|X_k\| \geq N)\right)$ is nothing other than $$P(\liminf_n \|X_n\|\geq N) = 1-P(\limsup_n \|X_n\|<N)$$

Since $\sum_{n=1}^\infty P(\|X_n\|< N)=\sum_{n=1}^N 1 + \sum_{n=N+1}^\infty \frac{N^2}{n^2}< \infty$, Borel-Cantelli lemma yields $P(\limsup_n \|X_n\|<N) = 0$, hence $P(\liminf_n \|X_n\|\geq N)=1$ for all $N$.

Therefore, $P(\|X_n\|\to \infty)=1$.

Answer 2

If $X_n\not\to\infty$, then some subsequence must remain bounded, which for large $n$ is unlikely. To make this idea a little more formal, take a sequence $(a_n)$ that goes to infinity slowly ($a_n=\log n$ will do) and show that $P(|X_n|\le a_n) $ is summable. You can then apply Borel-Cantelli.