3
$\begingroup$

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!

$\endgroup$

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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 $A_2 = \{|X_2|<k,|X_3|<k\}$. $\endgroup$
    – Marc
    Aug 10, 2016 at 13:19
  • $\begingroup$ 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! $\endgroup$
    – Marc
    Aug 10, 2016 at 16:11
1
$\begingroup$

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 $A_t=B_t=1$.

$\endgroup$
2
  • 1
    $\begingroup$ 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 $\ge 1-2\delta$. $\endgroup$
    – fedja
    Aug 10, 2016 at 23:52
  • $\begingroup$ Thank you very much for the addition. That should complete the general problem. $\endgroup$
    – Marc
    Aug 11, 2016 at 8:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.