% scribe: Lauren Deason
% lastupdate: Oct. 2, 2005
% lecture: 4
% title: Product spaces and independence
% references: Durrett, section 1.4
% keywords: product space, product measure, Fubini's theorem, projection map, independence, joint distribution, product sigma-field, 
% end

\documentclass[12pt,letterpaper]{article}

\include{macros}

\newcommand{\FF}{\mathcal F}
\newcommand{\CC}{\mathcal C}

\begin{document}

\lecture{4}{Product spaces and independence}{Lauren Deason}
{laurend@math.berkeley.edu}

\section{Product spaces and Fubini's Theorem}
% keywords: product space, product measure, Fubini's theorem, projection map, product sigma-field
% end

\begin{definition}
If $(\Omega_i,\FF_i)$ are measurable spaces, $i\in I$ (index set),
form $\prod_i\Omega_i$.  For simplicity, $\Omega_i=\Omega_1$.

$\prod_i\Omega_i$ (write $\Omega$ for this) is the space of all maps: 
$I\to \Omega_1$.
For $\omega\in\prod_i\Omega_i$, $\omega=(\omega_i: i\in I, \omega_i\in\Omega_i)$.  $\Omega$ is
equipped with \emph{projections}, $X_i:\Omega\to\Omega_i$,
$X_i(\omega)=\omega_i$.
\end{definition}

[picture of square, $\Omega_1$ on one side, $\Omega_2$ on other, point
$\omega$ in middle, maps under projection to each side]

\begin{definition}
A product $\sigma$-field on $\Omega$ is that generated by the
projections: $\FF=\sigma\left((X_i\in F_i)\mid F_i\in\FF_i\right)$.

$F_1\times F_2 = (\Omega\mid \Omega_1\in F_1\land\Omega_2\in
F_2)\in\FF = (X_1\in F_1)\cap(X_2\in F_2)$.
\end{definition}

We now seek to construct the \emph{product measure}.  We start with the 
$n=2$ case.
$(\Omega,\FF)=(\Omega_1,\FF_1)\times(\Omega_2,\FF_2)$.  Suppose we
have probability measures $P_i$ on $(\Omega_i,\FF_i)$, $i=1,2$.  Then
there is a natural way to construct $P=P_1\times P_2$ (product
measure) on $(\Omega,\FF)$.

\emph{Key idea:} $F_1\in\FF_1$, $F_2\in\FF_2$, $P(F_1\times
F_2)=P_1(F_1)\times P_2(F_2)$.

\begin{theorem}
[Existence of product measure and Fubini theorem]: (notation
$\omega=(\omega_1,\omega_2)$) There is a unique probability measure
$P$ on $(\Omega_1,\FF_1)\times(\Omega_2,\FF_2)$ such that for every
non-negative product-measurable [measurable w.r.t. the product
$\sigma$-field] function $f:\Omega_1\times\Omega_2\to[0,\infty)$,
\begin{align*}
&\int_{\Omega_1\times\Omega_2} f(\omega_1\times\omega_2)P(d\omega) =
\int_{\Omega_1}\left[\int_{\Omega_2}
  f(\omega_1,\omega_2)P_2(d\omega_2)\right]P_1(d\omega_1) = \\
&\int_{\Omega_2}\left[\int_{\Omega_1}
  f(\omega_1,\omega_2)P_1(d\omega_1)\right]P_2(d\omega_2).
\end{align*}
\end{theorem}

Note that this really tells very explicitly what $P$ is:
\begin{equation*}
\tag{$*$}\label{star} P(A) = \int_{\Omega_2} 1_A(\omega)P(d\omega)
\end{equation*}

Fix $\omega_2$, look at $A\omega_2=\{\omega_1\mid (\omega_1,\omega_2)\in A\}$.
\begin{equation*}
\int P_1(A\omega_2)P_2(d\omega_2)
\end{equation*}

Look at the level of sets.  $f=1_A$.  Look at formula \eqref{star}.
Look at the collection $\CC$ of all $A\subset\Omega_1\times\Omega_2$
such that $*$ makes sense, i.e.:
\begin{enumerate}
\item $A\omega_2\in\FF_1$ for every $\omega_2\in\Omega_2$.
\item $\omega_2\to P_1(A\omega_1)$ is measurable.
\end{enumerate}

Observe:
\begin{enumerate}
\item $\CC$ contains all $A_1\times A_2$ ($\pi$-system, closed under
intersection), and get $P(A_1\times A_2)=P_1(A_1)P_2(A_2)$
\item $\CC$ is a $\lambda$-system.
\end{enumerate}
Thus, $\CC\supset\sigma(A_1\times A_2)$, which is the product of
$\sigma$-fields.

(Checking $\lambda$-system uses monotone convergence theorem.)

We know that the order of integration was not relevant, because the
other way gives us a measure which agrees on rectangles (by
commutativity of addition), and rectangles generate (and we get a
$\pi$-system out of them), so we can apply the $\pi$-$\lambda$-theorem.

Extend our measure on indicator functions to simple functions,
additively, and so on.

\section{Independence}
% keywords: independence, joint distribution
% end


Random variables $X_1$ and $X_2$ with values in $(\Omega_1,\FF_1)$,
$(\Omega_2,\FF_2)$ are called \emph{independent} iff $\P(X_1\in
F_1,X_2\in F_2)=\P(X_1\in F_1)\P(X_2\in F_2)$ for all $F_1\in\FF_1$,
$F_2\in \FF_2$.

Observe:
\begin{enumerate}
\item If we take $\Omega=\Omega_1\times\Omega_2$ and
$\FF=\FF_1\times\FF_2$ and $\P=P_1\times P_2$, and
$X_i(\omega)=\omega_i$ projections as before, then $X_1$ and $X_2$
are independent random variables with distributions $P_1$ and $P_2$.
\item If $X_1$ and $X_2$ are independent random variables, defined on
  any background space $(\Omega,\FF)$, then the \emph{joint distribution}
  of $(X_1,X_2)$ is the product measure $P_1\times P_2$, $P_i=P_{X_i}$
  which is the $\P$ distribution of $X_i$.
\end{enumerate}
$\omega\to X_1(\omega)$, $\omega\to X_2(\omega)$.  Look at $\omega\to(X_1(\omega),X_2(\omega))$ as
a map from $\Omega$ to $\Omega_1\times \Omega_2$.  

We check that this map is product-measurable.  If $A_1\in\FF_1$ and 
$A_2\in\FF_2$, then

$$\left\{(X_1,X_2)\in A_1\times A_2\right\} = (X_1\in A_1)\cap(X_2\in A_2)\in\FF.$$

Now $P_{(X_1,X_2)}=$ $\P$ distribution of $(X_1,X_2)$ makes sense.


\begin{corollary}[Fubini] If $X_1$ and $X_2$ are independent random
variables,
\begin{equation*}
\E\left[f(X_1,X_2)\right]=\int\left[\int f(x_1,x_2)P_1(dx_1)\right]P_2(dx_2).
\end{equation*}
\end{corollary}

This justifies formulas for distribution of $X_1+X_2$, $X_1X_2$ for
real r.v.'s.  Example: If $X_1$ has density $f_1$, $X_2$ has density
$f_2$, $\P(X_1\in A)=\int_{A_1} f(x_1)dx_1$.  Then $X_1+X_2$ has a
density $$f(z)=(f_1\times f_2)(z) = \int_{\R} f_1(x)f_2(z-x)dx.$$
Check this by application of Fubini theorem.

$\E[g(X_1+X_2)]=\int g(z)f(z)dz.$


\emph{Useful fact:} For real random variables, $X_1$ and $X_2$ are
independent if and only if $\P(X_1\le x_1,X_2\le x_2)=\P(X_1\le
x_1)\P(X_2\le x_2)$ for all real $x_1,x_2$.

Fix $F_1=(-\infty,x_1]$ first.  Consider all sets with
$\P(X_1\in F_1,X_2\in F_2)=\P(X_1\in F_1)\P(X_2\in F_2)$.

Extend to $n$ variables.  $X_1,\ldots,X_n$ are independent iff
$$\P\left(\bigcap_i(X_i\in F_i)\right) = \prod_i \P(X_i\in F_i).$$
Same discussion with product spaces.  Most proofs work by induction on
$n$, reduces to $n=2$.  For example, $X_1,X_2,X_3$ are independent iff $X_1$
and $X_2$ are independent and $(X_1,X_2)$ and $X_3$ are independent.

Intuitive properties: e.g. if $X_1,\ldots,X_5$ are independent, then
$X_1+X_3+X_5$ and $X_2+X_4$ are independent.

To check this, we need to check that we can factor the probabilities as above, 
which goes for simple
functions, and then extends.

\begin{proposition}
If $X$ and $Y$ are independent and $\E(|X|)<\infty$ and $\E(|Y|)<\infty$
then $\E(XY)=\E(X)\E(Y)$.
\end{proposition}

\begin{proof} Use Fubini.  Be careful: we only showed Fubini for non-negative
functions.

First, check in case $X\ge 0$, $Y\ge 0$.  $\E(XY)=\int_{\R}\int_{\R}
xy\P(X\in dx)\P(Y\in dy)=\E(X)\E(Y)$, by Fubini.

In general, $X=X^+-X^-$, $Y=Y^+-Y^-$, and get four pieces, then put
back together.
\end{proof}

Note that the most general form of Fubini's theorem gives 
$$\E(f(X,Y)) = \int\int f$$ 
provided this $\int\int f$ formula is finite when $f$ is replaced by $|f|$.]


Recall $\E(X+Y)=\E(X)+\E(Y)$ always, provided they are finite.
$\var(X+Y)=\var(X)+\var(Y)+2\E\left[(X-\E(X))(Y-\E(Y))\right]$ (last term is
\emph{covariance}), and last term is $0$ if $X$ and $Y$ are independent.



\end{document}