Probability of $\limsup_{n\to \infty} \{X_n X_{n+1}>0\}$ where $\{X_n\}$ are independent Gaussian r.v.'s with mean 0

Question

Let $\{X_n\}$ be a sequence of independent Gaussian random variables with $\mathbb{E}\, X_n = 0$ for all $n \geq 1$. Find the probability of the event $$ \limsup_{n\to \infty} \big\{ X_n X_{n+1}> 0 \big\} $$

My first thought is that it should be 1 since Gaussians are always positive for a finite value. I was thinking of applying Borel-Cantelli and was trying something along the lines of \begin{align*} \mathbb{P} \big( \limsup_{n\to \infty} \big\{ X_n X_{n+1}> 0 \big\}\big) &= \mathbb{P}\big( X_n X_{n+1} > 0 \,\,\, i.o. \big) \\ &\leq \mathbb{P}\big( \big\{ X_n X_{n+1}> 0 \,\,\, i.o \big\} \cap \big\{ X_{n+1} > 0 \,\,\, i.o\big\} \big)\\ &= \mathbb{P}\big( \big\{ X_n X_{n+1}> 0 \,\,\, i.o \big\}\big) \,\,\mathbb{P}\big( \big\{ X_{n+1} > 0 \,\,\, i.o\big\} \big) \,\,\,\, \text{(by independence)} \end{align*} I'm not sure I'm thinking of this problem right, though.

Answer 1

Note that it is enough to consider only events $\{X_{2k}X_{2k+1} > 0 \}_{k \in \mathbb N}$ and by independence of $\{X_k\}_{k \in \mathbb N}$, those are independent as well. Moreover $\mathbb P(X_{2k}X_{2k+1} > 0 ) = \mathbb P(X_{2k},X_{2k+1} > 0) + \mathbb P(X_{2k},X_{2k+1}<0) = \frac{1}{2}$ by symmetry, so by Borel Cantelli.

$$ \sum_{k=1}^\infty \mathbb P(X_{2k}X_{2k+1} > 0) = \sum_{k=1}^\infty \frac{1}{2} = \infty$$ and since events are independent we get $\mathbb P(\limsup \{X_{2k}X_{2k+1} > 0 \}) = 1$, so in particular $\mathbb P (\limsup \{X_{k}X_{k+1} > 0 \}) = 1$, since $\limsup \{X_{2k}X_{2k+1} >0 \} \subset \limsup \{X_kX_{k+1} > 0 \} $

Answer 2

For a direct approach with no reference to the Borel–Cantelli lemma, observe that for any $n\in\mathbb N$, $X_n X_{n+1}>0$ if and only if $X_n$ and $X_{n+1}$ have the same sign. (For the sake of simplicity, assume that $X_n\neq0$ for every $n\in\mathbb N$; this event has probability $1$.)

Now think about what the complementary event is. The complementary event is that successive realizations of $X_n$ and $X_{n+1}$ have the same sign for only finitely many $n\in\mathbb N$. Therefore, the sequence eventually switches to an alternating sign pattern. That is, there exists some $m\in\mathbb Z_+$ such that the sign pattern looks either like this: \begin{align*} \underbrace{?\,?\,?\,?\,?\,?\,?\,?\,?}_{\text{$m$ times}}\,|\,\underbrace{{+}\,{-}\,{+}\,{-}\,{+}\,\cdots}_{\textit{ad infinitum}} \end{align*} or like this: \begin{align*} \underbrace{?\,?\,?\,?\,?\,?\,?\,?\,?}_{\text{$m$ times}}\,|\,\underbrace{{-}\,{+}\,{-}\,{+}\,{-}\,\cdots}_{\textit{ad infinitum}} \end{align*} For each fixed $m$, either of these two events has probability $(1/2)^{\infty}=0$ (look at the tails and exploit independence), and thus so does the union (over $m\in\mathbb Z_+$) of these events. The conclusion is that the original event involving the set-theoretic limit superior has probability $1$.