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Let $\xi_{1},\xi_{2},\ldots$ be a sequence of independent random variables with mean 0. For simplicity, assume that each $\xi_{i}$ only takes two values in $[-1,1]$.

Let $\mathscr{F}$ denote the collection of all non-empty finite subsets of $\mathbb{N} = \{1,2,\ldots\}$. We think of $\mathscr{F}$ as a directed set, ordered by inclusion $\subseteq$. For $F \in \mathscr{F}$ define $$\zeta_{F} = \tfrac{1}{|F|}\sum_{i \in F}\xi_{i}.$$

Question: Is it true that the net $(\zeta_{F})_{F \in \mathscr{F}}$ converges to 0 almost surely?

"Converges to zero" means that for each $\epsilon > 0$ there is a $G_{\epsilon} \in \mathscr{F}$ such that $|\zeta_{F}| < \epsilon$ for each $F \in \mathscr{F}$ with $F \supseteq G_{\epsilon}$.

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    $\begingroup$ Certainly no: take $F$ equal to simething like the union $G_\epsilon$ and $\{n\leqslant N:\xi_n>1/2\} $, for very large $N$ $\endgroup$
    – Fedor Petrov
    Commented Oct 12 at 16:53

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The answer is no.

Consider any event $\omega$ such that $\xi_i(\omega) = 1$ for infinitely many $i$. Then for a given finite set $G$ one can find a finite superset $F$ such that $\frac{1}{\lvert F \rvert} \sum_{i \in F} \xi_i(\omega)$ is close to $1$ (to obtain $F$, simply add a lot of indices $i$ to $G$ for which $\xi_i(\omega) = 1$.

Similarly, if $\xi_i(\omega) = -1$ for infinitely many $i$, then for any given finite set $G$ we can find a finite superset $F$ such that $\frac{1}{\lvert F \rvert} \sum_{i \in F} \xi_i(\omega)$ is close to $-1$.

Hence, the net does not converge unless only the sequence $(\xi_i(\omega))$ is eventually constant. But the set of $\omega$ for which this happens has probability $0$.

I'm tempted to say that, intuitively speaking, a convergence property that relies in some way on harmonic analysis or maximal inequalities will, typically, depend on the order of the involved elements. (But experts in harmonic analysis might want to correct me if I'm oversimplifying things with this statement.)

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  • $\begingroup$ But the set of indices for which $\xi_i(\omega) = 1$ might depend on $\omega$. And so will our set $F$. Could you elaborate? Thanx. $\endgroup$
    – Arkadi Predtetchinski
    Commented Oct 12 at 17:16
  • $\begingroup$ @ArkadiPredtetchinski: I'm not sure I understand your comment. Of course all those objects depend on $\omega$. You asked about pointwise almost everywhere convergence (= almost sure convergence), so one has to consider every $\omega$ seperately, check whether the net converges for this $\omega$, and then check whether the set of all those $\omega$ for which the net converges has measure $1$. But my answer shows that this set actually has measure $0$. $\endgroup$
    – Jochen Glueck
    Commented Oct 12 at 17:40

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