% scribe: Daisy Yan Huang & Jing Lei
% lastupdate: 05 December 2005
% lecture: 25
% references: Durrett, Section 4.6
% title: Reversed Martingales
% keywords: backwards martingales, reversed Martingales, exchangeable sequences, Hewitt-Savage 0-1 law, de Finetti's Theorem
% end

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\begin{document}

\lecture{25} {Reversed Martingales} {Daisy Huang, Jing
Lei}{yanhuang@stat,jinglei@stat}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

References: \cite{durrett}, section 4.6.

\section{Exchangeable $\sigma$-field}
% keywords: exchangeable sequences, exchangeable sigma fields
% end

Let $X_1, X_2,...$ be a sequence of real-valued random variables.
Define $\mathcal{E}_n=\sigma(f(X_1, X_2,...))$ where $f(X_1,
X_2,...)$ is symmetric with respect to the first n variables; i.e.,
$f(X_1, X_2,...)=f(X_{\pi(1)}, X_{\pi(2)},...)$ where $\pi \mbox{ is
a permutation such that }\pi:\{1,2,...n\} \rightarrow \{1,2,...n\}$
and $\pi(k)=k$ for $k>n$ and $f$ is product Borel measurable.
$\mathcal{E}_{\infty}=\bigcap_n\mathcal{E}_n$ is called the
\emph{exchangeable $\sigma$-field}.

\textbf{Remark}: \begin{enumerate}
\item
X is $\mathcal{E}_n$-measurable iff $X=f(X_1, X_2,...)$ for some
such $f$.

\item
$\mathcal{E}_n \supseteq \mathcal{E}_{n+1}$ because
$\mathcal{E}_{n+1}$ requires more symmetries.  Also, $\mathcal{E}_n
\downarrow \mathcal{E}_{\infty}$ since

\begin{eqnarray}
\label{exch-sigma-field} 
\mathcal{E}_{\infty} &\supseteq & \mbox{tail $\sigma$-field of } 
\{X_1, X_2,...\}\\
  & = & \bigcap_n \sigma(X_n, X_{n+1},...)
\end{eqnarray}
(noting that $\sigma(X_{n+1}, X_{n+2},...)\subseteq \mathcal{E}_n$ and
shifting the index $n$).

\end{enumerate}

The ``basic'' symmetric functions of $X_1,X_2,...X_n$ are the order
statistics
$$
\label{order-stat}
\min_{1\leq i \leq n}X_i= X_{n,1}\leq X_{n,2}
\leq ...\leq X_{n,n}=\max_{1\leq i \leq n}X_i.
$$

Obviously, each $X_{n,k}$ is a symmetric function of
$X_1,X_2,...X_n$.  You can check that $\mathcal{E}_n =
\sigma(X_{n,1},X_{n,2},...X_{n,n},X_{n+1},X_{n+2},...)$.

\begin{example}
$S_n=X_1+X_2+...+X_n$ is in $\mathcal{E}_n$
but not in the tail $\sigma$-field of $\{X_1, X_2,...\}$.
\end{example}

\section{Hewitt-Savage $0-1$ Law}
\begin{theorem}[Hewitt-Savage $0-1$ Law]
If $X_1, X_2,...$ is
i.i.d., then every event in $\mathcal{E}_\infty$ has probability
$0$ or $1$.
\end{theorem}

(Compare this with Kolmogorov's $0-1$ Law.)

Recall that if $X_1, X_2,...$ are i.i.d. and exchangeable (i.e.,
%
$$(X_1,X_2,...X_n) \overset{d}= (X_{\pi(1)},
X_{\pi(2)},...,X_{\pi(n)})$$ 
%
for all permutations $\pi$ on $n$ elements) and $\E{|X_1|}<\infty$,
then $(S_n/n, \mathcal{E}_n)_{n\geq 1}$ is a reversed
martingale.
If $S_n/n=\E{(X_1|\mathcal{E}_n)}$, then this is obvious
because $\mathcal{E}_n \downarrow$.

To see this: exchangeability implies that
$$
\label{claim-HS} \E{(X_1|\mathcal{E}_n)}
=\E{(X_k|\mathcal{E}_n)}\qquad \mbox{for every $1\leq k \leq n$}.
$$

This is because 
\begin{eqnarray*}
&& (X_1,f(X_1,\cdots,X_n,X_{n+1},X_{n+2}, \cdots)) \\
&& \overset{d}= \,
(X_k,f(X_k,\cdots,X_{k-1},X_1,X_{k+1}, \cdots ,X_n,X_{n+1},X_{n+2},\cdots))\\
&& = \,(X_k, f(X_1,X_2,...,X_n,X_{n+1},X_{n+2}...))
\end{eqnarray*}

Now check the definition of the claim:
$$
n \E{(X_1|\mathcal{E}_n)}=\E{(S_n|\mathcal{E}_n)}=S_n  \mbox{ since } 
S_n\subseteq \mathcal{E}_n.
$$

Now, we prove the theorem.  The plan is to show that
for every event $F\in \sigma(X_1,...,X_{n})$,
$\P(F|\mathcal{E}_\infty)=\P(F)$.  This says that
$\sigma(X_1,...,X_{n})$ is independent of $\mathcal{E}_\infty$. To
see this, let $n \rightarrow \infty$ and learn that
$\sigma(X_1,X_2,...)$ is independent of $\mathcal{E}_\infty$. But
this implies that $\mathcal{E}_n$ is independent of
$\mathcal{E}_\infty$, which leads to the result of the $0-1$ Law.

\begin{proof} By the Martingale Convergence Theorem,
$$
\label{HS-proof}
\P(F|\mathcal{E}_\infty)=\lim_{n\rightarrow\infty}{\P(F|\mathcal{E}_n)}
$$

Let $F=\{ X_1 \in \hat{F} \}$ for some $\hat{F}\subseteq \R$.  Then,
\begin{eqnarray*}
\P(F \mid \mathcal{E}_n)
  &=& \E(\1_{\hat{F}(X_1)}|\mathcal{E}_n)\\
  &=& \E(\1_{\hat{F}(X_k)}|\mathcal{E}_n)\\
  &=& \E\!\left(\frac{1}{n}\sum_{k=1}^{n} \1_{\hat{F}(X_k)} 
| \mathcal{E}_n\right)\\
  &=& \frac{1}{n}\sum_{k=1}^{n} {\1_{\hat{F}(X_k)}}\\
  &\rightarrow & \E{(\1_{\hat{F}(X_k)})}\\
  &=& \P(X_1\in \hat{F})\\ 
  &=& \P(F).
\end{eqnarray*}

This proves for the case $F \in \sigma(X_1)$.
To deal with the case $F \in \sigma(X_1, X_2)$, we do the same thing.
Let $F=\{ (X_1, X_2) \in \hat{F} \}$ for some $\hat{F}\subseteq
\R^2$ and $\varphi(X_1,X_2)=\1((X_1,X_2)\in\hat{F})$. Then,
%
\begin{eqnarray*}
\P(F \mid \mathcal{E}_n)
  &=& \E(\varphi{(X_1,X_2)}|\mathcal{E}_n)\\
  &=& \E(\varphi{(X_i,X_j)}|\mathcal{E}_n) \mbox{ for } i\neq j\\
  &=& \E\!\left({1\over {n(n-1)}}\sum_{1\leq i \neq j\leq n} \ {\varphi{(X_i,X_j)}} |
\mathcal{E}_n\right)\\
  &=& {1\over {n(n-1)}}\sum_{1\leq i \neq j\leq n} \ {\varphi{(X_i,X_j)}}.
\end{eqnarray*}

Consider $\varphi{(X_i,X_j)}=f(X_i)g(X_j)$, i.e.\ $\hat{F}$ is
rectangular; then
%
\begin{eqnarray*}
\P(F \mid \mathcal{E}_n)
  &=& \frac{1}{n(n-1)}\sum_{1\leq i \neq j\leq n}  f(X_i) g(X_j)\\
  &=& \frac{1}{n(n-1)}\left(\sum_{1\leq i,j \leq n} f(X_i) g(X_j) - \sum_{i=1}^{n} f(X_i) g(X_i)  \right)\\
  &=& \frac{1}{n(n-1)}\left(\sum_{i=1}^{n}f(X_i) \sum_{i=1}^{n} g(X_i)-\sum_{i=1}^{n} f(X_i) g(X_i) \right)\\
  &\ascv & \E f(X_1) \E g(X_2)\\
  &=& \E \varphi(X_1,X_2)\\
  &=& \P(F).
\end{eqnarray*}

To finish, for $\hat{F}\in \borel(\R ^2)$ we just use the $\pi-\lambda$
theorem. Similarly for $\hat{F}\in \borel(\R ^k)$, $k\geq 3$.

So far, we've shown that
$$
\P(F \mid \mathcal{E}_{\infty})=\P(F) \mbox{ for all } F\in
\sigma(X_1,\cdots,X_n),
$$
i.e.\ $\mathcal{E}_{\infty}$ is independent of
$\sigma(X_1,\cdots,X_n)$. Similarly as in the proof of Kolmogorov's 0-1 Law, we
can learn that $\mathcal{E}_{\infty}$ is independent of
$\sigma(X_1,X_2,\cdots)$ by sending $n\rightarrow \infty$. Thus
$\mathcal{E}_{\infty}$ is independent of itself, which completes the proof.
\end{proof}

\section{de Finetti's Theorem}
\begin{theorem}
If $X_1,X_2,\cdots,$ are exchangeable, and
$F_n(x):=\frac{1}{n}\sum_{1\leq i\leq n}\1(X_i\leq x)$, then
%
$$\lim_{n\toinf}\sup_x|F_n(x)-F(x)|=0$$ 
for some random CDF $(F(x), x\in \R)$.

Also, given $\mathcal{E}_{\infty}$, the $X_1,X_2,\cdots,$ are i.i.d.\ with 
common distribution $F$; i.e.
\begin{equation}\label{exch_indpt}
  \P(X_1\leq x_1,X_2\leq x_2,\cdots,X_k\leq
  x_k|\mathcal{E}_{\infty})=F(x_1)F(x_2)\cdot\cdots F(x_k).
\end{equation}
\end{theorem}

\textbf{Note:} Conceptually, it's as if $F$ were first picked at
random in some way from the set of probability CDF's on $\R$, and then we
sample from $F$.

\begin{proofsketch}
(This sketch proof is incomplete -- see the text book for details.)
\begin{enumerate}
\item
Look at \eqref{exch_indpt} for $k=1$, and by MGCT and
exchangeability we have
%
\begin{eqnarray*}
F(x)&=&\P(X_1\leq x|\mathcal{E}_{\infty})\\
    &=&\lim_{n\toinf}\P(X_1\leq x|\mathcal{E}_n) \mbox{ a.s.}\\
    &=&\lim_{n\toinf}F_n(x) \mbox{ a.s.}
\end{eqnarray*}

\item Using the fact that $F_n(x)$ is non-decreasing in $x$ for
fixed $n$, we learn that $F(x)$ is non-decreasing in $x$ almost
surely.

\item Clean up over rationals: let
$$F^{\star}(x):=\lim_{q\downarrow x}F(q),$$ 
where $q\in \Q$, the set of all rational numbers.
Then argue that $F^{\star}(x)$ is a CDF and replace $F(x)$ by
$F^{\star}(x)$.
\end{enumerate}
\end{proofsketch}


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