# Liminf of the maximum of two iid sequences

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 $$|X_t| for infinitely many $$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 $$\max\{|X_t|,|Y_t|\} for infinitely many $$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!

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.

• If I'm not mistaken, you use the second Borel-Cantelli lemma, which assumes independence of the events $A_i$. I don't think this is guaranteed. Take for example $Y_k = X_{k+1}$, then $A_1 = \{|X_1|<k,|X_2|<k\}$ and $A_2 = \{|X_2|<k,|X_3|<k\}$.
– Marc
Aug 10, 2016 at 13:19
• Oh wow. Thank you, that is a great answer in the sense that your solution is so simple. My actual sequences are not iid, but stationary ergodic and this generalizes directly. I'm a first year PhD and this is a great lesson to always start thinking simple. Thanks again!
– Marc
Aug 10, 2016 at 16:11

This answer contains the details to the answer provided by Michael, all credits go to him.

Choose $M>0$ such that $P(|X_t|<M)>1/2$ and $P(|Y_t|<M)>1/2$. Define $A_t = 1$ if $|X_t|<M$ and 0 otherwise. Similarly, define $B_t = 1$ if $|Y_t|<M$ and 0 otherwise. Then $\{A_t\}_{t\ge1}$ and $\{B_t\}_{t\ge1}$ are iid sequences again. Therefore, $$\frac{1}{n}\sum_{t=1}^{n}A_t + B_t \rightarrow E(A) + E(B) = P(|X|<M) + P(|Y|<M) > 1.$$ We conclude that there must be infinitely many $t$ such that $A_t=B_t=1$.

• The law of large numbers does not apply to dependent sequences. What you need in general, however, is an even simpler lemma: if $A_j$ is a sequence of events such that $P(A_j)\ge p$ for all $j$, then $P(\limsup A_j)\ge p$. Now for every $\delta$ choose $k$ such that $P(X_j>k), P(Y_j>k)<\delta$ (this is possible due to the identical distribution). Then your $\liminf$ is $\le k$ with probability $\ge 1-2\delta$. Aug 10, 2016 at 23:52
• Thank you very much for the addition. That should complete the general problem.
– Marc
Aug 11, 2016 at 8:12