# Question on the application of Kolmogorov Zero-One law

Let $X_1$, $X_2$,... be independent random variables on ($\Omega$, $\mathcal{F}$, $\mathbb{P}$). Suppose that the $X_i$ are symmetric (i.e. $X_i$ and $-X_i$ have the same distribution) and that there exists $c \in \mathbb{R}$ such that $\mathbb{P} ( | X_i| \leq c ) =1$ for all $i$. Show that $$\mathbb{P}( |S_n| \leq \frac{c}{2} \text{ i.o.}) =1.$$ (Kolmogorov zero-one law is needed.)

• There seems to be no direct relationship between the fact that $|X_i| \leq c$ and $|S_n| \leq \frac{c}{2}$. – Richard Sep 5 '14 at 14:37
• This isn't true without more assumptions. Take $X_1 = \pm c$ with probability $1/2$, and take $X_2 = X_3 = \dots = 0$. – Nate Eldredge May 5 '17 at 17:47

Not sure if it is right (amended from the solution given above):

WLOG, let $c>0$. (The case where $c=0$ is trivial.) \begin{eqnarray} &&\{ |S_n| > \frac{c}{2} \text{ev.} \} \nonumber\\ & \subseteq &\{ S_n > \frac{c}{2} \text{ev.} \} \cup \{ - S_n > \frac{c}{2} \text{ev.} \} \quad \quad \text{ (explanation by above)} \nonumber\\ & \subseteq & \{ \text{ liminf } S_n \geq \frac{c}{2} \} \cup \{ \text{ liminf } - S_n \geq \frac{c}{2} \} \end{eqnarray} By Kolmogorov zero-one law, $\mathbb{P} \{ \text{ liminf } S_n \geq \frac{c}{2} \} = 0$ or $1$. Suppose, for a contradiction, $\mathbb{P} \{ \text{ liminf } S_n \geq \frac{c}{2} \} = 1$. Then, since $S_n$ and $-S_n$ are identically distributed, $\mathbb{P} \{ \text{ limsup } S_n \leq -\frac{c}{2} \} = \mathbb{P} \{ \text{ liminf } - S_n \geq \frac{c}{2} \} =1$. This implies that $\mathbb{P}\{ \emptyset \} = \mathbb{P} ( \{\text{ liminf } S_n \geq \frac{c}{2} \} \cap \{ \text{ limsup } S_n \leq -\frac{c}{2} \} ) = 1$. This is a contradiction. Hence, $\mathbb{P} \{ \text{ liminf } S_n \geq \frac{c}{2} \} = \mathbb{P} \{ \text{ liminf } - S_n \geq \frac{c}{2} \} =0$. This shows the result.

• This is not valid. $\liminf S_n$ and $\limsup S_n$ are not tail random variables so the Kolmogorov 0-1 law does not apply. – Nate Eldredge May 5 '17 at 17:46

Since the event $A:=\{|S_n|\leqslant c/2 \,\mathrm{i.o.}\}$ is exchangeable, it is probably better to consider the Hewitt–Savage zero–one law instead of Kolmogorov's one.

Assume that $\mathbb P(A)=0$. Then for almost every $\omega$, there is a $N=(\omega)$ such that if $n\geqslant N$, then $|S_n|>c/2$. Notice that if $S_n>c/2$ where $n\geqslant N$, then we have $|S_{n+1}|>c/2$ and $|X_{n+1}|\leqslant c$, hence we necessarily have $S_{n+1}>c/2$. In other words, we have $S_n>c/2$ for $n$ large enough or $S_n<-c/2$ for $n$ large enough.

Notice that the random variables $\limsup_n S_n$ and $\liminf_n S_n$ are almost surely constant, possibly infinite. Since we have the equality in distribution $$\limsup_n S_n=\limsup_n S_{n+1}-X_1,$$ we have $\limsup_n S_n\in\{-\infty,\infty\}$ and similarly for the $\liminf$ (unless $X_1$ is degenerated, but in this case the problem is trivial). By symmetry, $\limsup_n S_n=+\infty$ and $\liminf_nS_n=-\infty$ andthis contradicts the fact that $S_n\in (c/2,\infty)$ for $n$ large enough or $S_n\in (-\infty,c/2)$ for $n$ large enough.

• What is the sequence of i.i.d. variables you apply the Hewitt-Savage law to? – tomasz Sep 7 '14 at 10:20
• Opps, I missed the fact that the $X_i$ are not necessarily identically distributed. – Davide Giraudo Sep 7 '14 at 10:22