TheHewitt–Savage zero–one law is atheorem inprobability theory, similar toKolmogorov's zero–one law and theBorel–Cantelli lemma, that specifies that a certain type of event will eitheralmost surely happen or almost surely not happen. It is sometimes known as theSavage-Hewitt law for symmetric events. It is named afterEdwin Hewitt andLeonard Jimmie Savage.^[1]

Let${\displaystyle {\left\{X_n\right\}}_{n=1}^{\mathrm{\infty }}}$ be asequence ofindependent and identically distributed random variables taking values in a set${\displaystyle \mathbb{X}}$. The Hewitt-Savage zero–one law says that any event whose occurrence or non-occurrence is determined by the values of these random variables and whose occurrence or non-occurrence is unchanged by finitepermutations of the indices, hasprobability either 0 or 1 (a “finite” permutation is one that leaves all but finitely many of the indices fixed).

Somewhat more abstractly, define theexchangeable sigma algebra orsigma algebra of symmetric events${\displaystyle \mathcal{E}}$ to be the set of events (depending on the sequence of variables${\displaystyle {\left\{X_n\right\}}_{n=1}^{\mathrm{\infty }}}$) which are invariant underfinitepermutations of the indices in the sequence${\displaystyle {\left\{X_n\right\}}_{n=1}^{\mathrm{\infty }}}$. Then${\displaystyle A\in \mathcal{E}⟹\mathbb{P}(A)\in \{0,1\}}$.

Since any finite permutation can be written as a product oftranspositions, if we wish to check whether or not an event${\displaystyle A}$ is symmetric (lies in${\displaystyle \mathcal{E}}$), it is enough to check if its occurrence is unchanged by an arbitrary transposition${\displaystyle (i,j)}$,${\displaystyle i,j\in \mathbb{N}}$.

Let the sequence${\displaystyle {\left\{X_n\right\}}_{n=1}^{\mathrm{\infty }}}$ of independent and identically distributed random variables taking values in${\displaystyle \mathbb{R}}$. Consider the random walk${\displaystyle S_N=\sum _{n=1}^N{X}_n}$. Then one of the following occurs with probability 1:

• ${\displaystyle S_N=0}$
• ${\displaystyle S_N\to \mathrm{\infty }}$
• ${\displaystyle S_N\to -\mathrm{\infty }}$
• ${\displaystyle lim inf{S}_N=-\mathrm{\infty }}$ and${\displaystyle lim sup{S}_N=\mathrm{\infty }}$.

SinceS_N are not independent theKolmogorov's zero–one law is not directly applicable.

First consider the case whenX₁ is a.s. constant. Then with probability 1 we have that either (${\displaystyle S_N=0}$), (${\displaystyle S_N\to \mathrm{\infty }}$) or (${\displaystyle S_N\to -\mathrm{\infty }}$).

Now consider the case, whenX₁ is not a.s. constant. Then for any${\displaystyle t\in [-\mathrm{\infty },\mathrm{\infty }]}$ the event${\displaystyle \left\{lim sup{S}_N\ge t\right\}}$ is in the exchangeable sigma algebra. That is becauselimit supremum does not change with finite permutation of the indices. From Hewitt-Savage zero-one law we have that

${\displaystyle \mathbb{P}\left(lim sup{S}_N\ge t\right)\in \{0,1\}}$.

There has to existt, where probability switches from 0 to 1 i.e. exists${\displaystyle t^{\ast }\in [-\mathrm{\infty },\mathrm{\infty }]}$ such that${\displaystyle lim sup{S}_n=t^{\ast }}$ almost surely. Similarly exists${\displaystyle t_{\ast }\in [-\mathrm{\infty },\mathrm{\infty }]}$ such that${\displaystyle lim inf{S}_n=t_{\ast }}$ almost surely.

Since almost surely

${\displaystyle t^{\ast }=lim sup{S}_N=X_1+lim sup\sum _{n=2}^N{X}_n=X_1+t^{\ast }}$

andX₁ is not a.s. 0, then${\displaystyle t^{\ast }}$ is not finite. Similarly${\displaystyle t_{\ast }}$ in not finite.

Therefore, with probability 1 either (${\displaystyle S_N\to \mathrm{\infty }}$), (${\displaystyle S_N\to -\mathrm{\infty }}$)or (${\displaystyle lim inf{S}_N=-\mathrm{\infty }}$ and${\displaystyle lim sup{S}_N=\mathrm{\infty }}$).^[2]

• Theorems in probability theory
• Covering lemmas