Let $\{X\_t\}\_{t\ge1}$ and $\{Y\_t\}\_{t\ge1}$ be two iid sequences of random variables that have full support. That is, if $A\subseteq\mathbb{R}$ has positive Lebesgue measure, then $P(X\in A) >0$ and $P(Y\in A)>0$. For such a sequence we almost surely have for any $M>0$ that $|X\_t|<M$ for infinitely many $t$. That means that with probability one
$$
\liminf\_{t\rightarrow\infty} |X\_t| = 0.
$$


**Question:** Is it possible to find a dependence structure between $\{X\_t\}\_{t\ge1}$ and $\{Y\_t\}\_{t\ge1}$ such that
$$
P\left(\liminf\_{t\rightarrow\infty} \,\,\max\{|X\_t|,|Y\_t|\} = \infty\right) > 0,
$$
that is, with positive probability there exists no $M>0$ such that $\max\{|X\_t|,|Y\_t|\}<M$ for infinitely many $t$.


**Thoughts:** I believe this is not possible, because for any subsequence of the $|Y\_t|$ that is bounded we need that same subsequence of the $|X\_t|$ to go to infinity. Let $t\_1,t\_2,\ldots$ denote the stochastic times that $|Y\_{t\_k}|\le M$. Then we must have $\lim\_{k\rightarrow\infty}|X\_{t\_k}| = \infty$. Moreover,
$$
P(|X\_t|\ge M \,\mid\,|Y\_t|< M) = \frac{P(|X\_t|\ge M \,;\,|Y\_t|< M)}{P(|Y\_t|< M)} \le \frac{P(|X\_t|\ge M)}{P(|Y\_t|< M)}
$$
can be made arbitrarily small by increasing $M$. I feel I should somehow be able to combine these results to get the desired answer.


Any help is highly appreciated. Thank you in advance!



Liminf of the maximum of two iid sequences
take a k so that $P(|X\_i | < k, | Y\_i | < k ) > \epsilon > 0$. They must exist because any k for $P(|X| < k) > \frac 3 4 $ and same for Y works, Let $A\_i = \{ |X\_i | < k, | Y\_i | < k \}$ By borel cantelli $A\_i$ happens infinitely often, and so the liminf is < k


sorry, had misinterpreted dependence structure. In that case pick any k for which $P( |X\_i | < k \} >\frac 12 $ and same for $Y\_i$, Then the density of $\{n> N\_0: |X\_i | < k \}$ is $ > \frac 12$ and similarly for Y. It is easy to argue that therefore they can't be disjoint, and there is and $ n > N\_0$ which is in both.


