Hewitt–Savage zero–one law

Remarks

Proof idea: approximate $A$ by an event $A_n$ that depends only on $X_1, \dots, X_n$. Let $A_n'$ be the image of $A_n$ under the permutation that swaps the blocks $1, \dots, n$ and $n+1, \dots, 2n$. By symmetry $A_n'$ approximates $A$ as well, and $A_n, A_n'$ are independent with equal probability. So $P(A) = \lim P(A_n \cap A_n') = \lim P(A_n)^2 = P(A)^2$. Every tail event is symmetric, so the theorem extends Kolmogorov's zero–one law in the i.i.d. case. Example: let $X_i$ be real and $S_n = X_1 + \dots + X_n$. The event $\set{S_n \in I \text{ for infinitely many } n}$, for an interval $I$, is symmetric, because a finite permutation leaves $S_n$ unchanged for all large $n$. It is not a tail event in general, since changing $X_1$ shifts all $S_n$. The identical distribution is needed: if $X_1$ is a fair coin and $X_n = 0$ for $n \ge 2$, then $\set{\text{some } X_n = 1}$ is symmetric and has probability $\frac{1}{2}$.

Going deeper

For a student

The theorem is used to show that quantities such as $\limsup_n S_n$ for a random walk are almost surely constant, and to obtain the recurrence dichotomy for random walks. Compared with Kolmogorov's zero–one law, it needs identical distribution but covers more events, namely all events of the exchangeable σ-algebra. In the language of de Finetti's theorem, it says that an i.i.d. sequence is an exchangeable sequence whose mixing measure is a point mass.