Hewitt–Savage zero–one law

The Hewitt–Savage zero–one law is a theorem in probability theory, similar to Kolmogorov's zero–one law and the Borel–Cantelli lemma, that specifies that a certain type of event will either almost surely happen or almost surely not happen. It is sometimes known as the Savage-Hewitt law for symmetric events. It is named after Edwin Hewitt and Leonard Jimmie Savage.^[1]

Statement of the Hewitt-Savage zero-one law

Let ${\displaystyle \left\{X_{n}\right\}_{n=1}^{\infty }}$ be a sequence of independent 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 finite permutations of the indices, has probability either 0 or 1 (a “finite” permutation is one that leaves all but finitely many of the indices fixed).

Somewhat more abstractly, define the exchangeable sigma algebra or sigma algebra of symmetric events ${\displaystyle {\mathcal {E}}}$ to be the set of events (depending on the sequence of variables ${\displaystyle \left\{X_{n}\right\}_{n=1}^{\infty }}$) which are invariant under finite permutations of the indices in the sequence ${\displaystyle \left\{X_{n}\right\}_{n=1}^{\infty }}$. Then ${\displaystyle A\in {\mathcal {E}}\implies \mathbb {P} (A)\in \{0,1\}}$.

Since any finite permutation can be written as a product of transpositions, 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 ${\displaystyle (i,j)}$, ${\displaystyle i,j\in \mathbb {N} }$.

Example

Let the sequence ${\displaystyle \left\{X_{n}\right\}_{n=1}^{\infty }}$ 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:

• ${\displaystyle S_{N}=0}$
• ${\displaystyle S_{N}\to \infty }$
• ${\displaystyle S_{N}\to -\infty }$
• ${\displaystyle \liminf S_{N}=-\infty }$ and ${\displaystyle \limsup S_{N}=\infty }$.

Since S_N are not independent the Kolmogorov's zero–one law is not directly applicable.

First consider the case when X₁ is a.s. constant. Then with probability 1 we have that either (${\displaystyle S_{N}=0}$), (${\displaystyle S_{N}\to \infty }$) or (${\displaystyle S_{N}\to -\infty }$).

Now consider the case, when X₁ is not a.s. constant. Then for any ${\displaystyle t\in [-\infty ,\infty ]}$ the event ${\displaystyle \left\{\limsup S_{N}\geq t\right\}}$ is in the exchangeable sigma algebra. That is because limit supremum does not change with finite permutation of the indices. From Hewitt-Savage zero-one law we have that

${\displaystyle \mathbb {P} \left(\limsup S_{N}\geq t\right)\in \{0,1\}}$.

There has to exist t, where probability switches from 0 to 1 i.e. exists ${\displaystyle t^{*}\in [-\infty ,\infty ]}$ such that ${\displaystyle \limsup S_{n}=t^{*}}$ almost surely. Similarly exists ${\displaystyle t_{*}\in [-\infty ,\infty ]}$ such that ${\displaystyle \liminf S_{n}=t_{*}}$ almost surely.

Since almost surely

${\displaystyle t^{*}=\limsup S_{N}=X_{1}+\limsup \sum _{n=2}^{N}X_{n}=X_{1}+t^{*}}$

and X₁ is not a.s. 0, then ${\displaystyle t^{*}}$ is not finite. Similarly ${\displaystyle t_{*}}$ in not finite.

Therefore, with probability 1 either (${\displaystyle S_{N}\to \infty }$), (${\displaystyle S_{N}\to -\infty }$) or (${\displaystyle \liminf S_{N}=-\infty }$ and ${\displaystyle \limsup S_{N}=\infty }$).^[2]

References