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$X_1, X_2, X_3,\dots$ is a sequence of random signs, does this sequence converge almost surely?

I know the answer is no and I can use Cauchy in prob to prove it. However, I am very confused about the definition of convergence a.s.

Definition should be $P(w ∈ \Omega, lim X_n(w) \to X(w))=1$; now this is very trivial but obviously the sequence of random signs converges a.s. to a random sign using this definition: $P(w_1 ∈ \Omega, lim X_n(w_1) \to X(w_1))=1/2$, $P(w_2 \in \Omega, lim X_n(w_2) \to X(w_2))=1/2$ if $X(w_1)=1/2$, $X(w_2)=-1/2$. The question didn't specify what it converges to so I can specify it converges a.s. to X defined as another random sign. And I am using the definition of a.s. convergence here, not convergence in distribution.

Please tell me what the flaw in my logic is.

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  • $\begingroup$ (1) What do you assume for the joint distribution of $X_i$'s? In one extreme, if $X_i = X_1$ for all $i$ then $X_i$ converges almost surely, while if $X_i$'s are mutually independent, then $X_i$ does not converge a.s. (2) What are $\lim_n X_n$ and $X$? You are introducing them as if they both exist, but the point of the original question is to investigate whether $\lim_n X_n$ exists or not, letting alone how will the limit look like. $\endgroup$ – Sangchul Lee Apr 9 '18 at 22:15
  • $\begingroup$ oh so you're saying if indepdendent, then X1(w1) does not necessarily equal to X2(w1) they might be opposite? as in X1(w1)=1 while X2(w1)=-1 (since random sign only has two events). Also, what do you mean joint distribution of XI? $\endgroup$ – james black Apr 9 '18 at 22:19
  • $\begingroup$ and lim Xn and X is part of the definition? what do you mean by investigating it? $\endgroup$ – james black Apr 9 '18 at 22:24
  • $\begingroup$ Correct me if I am wrong, but it seems to me that you are assuming that $\Omega$ consists of two elements $\omega_1$ and $\omega_2$. In general, however, one needs a sufficiently large sample space to realize any sequence of random variables. For instance, in order to host a sequence of mutually independent random variables taking values in $\{-1, 1\}$, you need a sample space at least as large as the continuum $\simeq \{-1, 1\}^{\mathbb{N}}$. $\endgroup$ – Sangchul Lee Apr 9 '18 at 22:26
  • $\begingroup$ but random sign only has two elements since it only has two outputs so R={1,-1} for any random sign, then it has two have two elements in the prob space? Can you explain how i would investigate in this case if all signs are independent? thanks $\endgroup$ – james black Apr 9 '18 at 22:32
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Assume that $(X_n)$ is a sequence of i.i.d. random variables satisfying

$$\mathsf{P}[X_n = 1] = \mathsf{P}[X_n = -1] = \tfrac{1}{2}, \quad \forall \ n \geq 1.$$

Let $A$ be the event that $X_n$ converges. Then for each $\omega \in A$, there exists a real number $X(\omega) \in \mathbb{R}$ and a positive integer $N(\omega) \geq 1$ such that $|X_n(\omega) - X(\omega)| < \frac{1}{2}$ for all $n \geq N(\omega)$. This implies that $|X_n(\omega) - X_{N(\omega)}(\omega)| < 1$ for all $n \geq N(\omega)$, and since both $X_n$ and $X_{N(\omega)}$ take values in $\{-1, 1\}$, this forces that $X_n(\omega) = X_{N(\omega)}(\omega)$. So it follows that

$$ A \subseteq \bigcup_{N\geq 1} \{ \omega \in \Omega : X_n(\omega) = X_N(\omega) \text{ for all } n \geq N \}. $$

Splitting the RHS further depending on the value of $X_N(\omega)$, we get

\begin{align*} \mathsf{P}[A] &\leq \sum_{N\geq 1} \mathsf{P}[\{ \omega \in \Omega : X_n(\omega) = X_N(\omega) \text{ for all } n \geq N \}] \\ &\leq \sum_{N\geq 1} \mathsf{P}[\{ \omega \in \Omega : X_n(\omega) = 1 \text{ for all } n \geq N \}] \\ &\quad + \sum_{N\geq 1} \mathsf{P}[\{ \omega \in \Omega : X_n(\omega) = -1 \text{ for all } n \geq N \}] \end{align*}

But by the mutual independence, it follows that

$$ \mathsf{P}[\{ \omega \in \Omega : X_n(\omega) = 1 \text{ for all } n \geq N \}] = \prod_{n\geq N} \mathsf{P}[\{ \omega \in \Omega : X_n(\omega) = 1\}] = \prod_{n\geq N} \frac{1}{2} = 0 $$

and likewise for the second sum. So $\mathsf{P}[A] \leq 0$ and this implies that $\mathsf{P}[A] = 0$. Therefore $(X_n)$ does not converges a.s.

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