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\begin{unit}{4}{Joint distributions and independence}\footnote{These notes
contain almost no proofs. For a complete set of notes, see
\cite{k}; alternatively, read the introductory book \cite{wil},
or the more advanced book \cite{fg}.}
% SYLLABUS:
%4.  Joint distributions and independence.                               TK
%      Joint distributions, 	DONE
% marginal 	DONE
% and conditional  distributions.	DONE
% Independence: DONE
% factorisation of joint distribution and moment generating function, DONE
% moment properties. [What does this mean???]
%      Covariance, correlation, DONE [L2 approach]
% multivariate normal distribution.	DONE

\section{Joint distributions}
Consider a random element $(X, Y): (\Omega, \FF) \to (\R^2, \BB(\R^2))$.
This random variable is called {\sc two random variables}. 
\begin{quote}
If
$\Pr$ is a probability of $(\Omega, \FF)$, the joint distribution of 
$(X,Y)$ is another name for the law $\Pr_{X,Y}$ of the
random variable $(X,Y)$. 
\end{quote}
The joint distribution function of $(X,Y)$
is the function
\begin{equation}
\label{jdf}
F_{X,Y}(x,y) := \Pr(X \le x, Y \le y) = \Pr_{X,Y}
((-\infty, x] \times (-\infty, y]), \quad (x,y) \in \R^2.
\end{equation}
The law $\Pr_X$ of $X$ is referred  to as the first marginal
of the law $\Pr_{X,Y}$.
The distribution function $F_X$ of $X$ is referred to as the
first marginal distribution function of the joint 
distribution function $F_{X,Y}$ and, of course,
\[
F_X(x) = \lim_{y \to \infty} F_{X,Y}(x,y).
\]
Note that we chose ti use $\le$ instead of $<$ in \eqref{jdf} for
no good reason other than a mere arbitrary convention.


\subsection{Knowledge of $F_{X,Y}$ implies knowledge of $\Pr_{X,Y}$}
Explaining why knowledge of the function $F_{X,Y}$
implies knowledge of $\Pr_{X,Y}(B)$ for all $B \in \BB(\R^2)$
is beyond the scope of these lectures. The explanation can be
found in \cite{k,wil,fg}.
But we give some intuition.
Consider a rectangle (with sides parallel to the axes--please think
geometrically)
\begin{equation}
\label{rrr}
(a_1, b_1] \times (a_2, b_2] :=
\{(x,y) \in \R^2:~ a_1 < x \le b_1, ~ a_2 < y \le b_2\}.
\end{equation}
We allow $a_1$, $a_2$ to take any value, including $-\infty$.
Since $(-\infty, b_1] \times (-\infty, b_2]$ is the
disjoint union of four rectangles, using additivity, we obtain
\[
\Pr_{X,Y}((a_1, b_1] \times (a_2, b_2])
= F_{X,Y}(b_1, b_2) 
- F_{X,Y}(b_1, a_2)
- F_{X,Y}(a_1, b_2)
+ F_{X,Y}(b_1, b_2).
\]
Rectangles of the form \eqref{rrr} have the following nice properties:
First, intersection of two of them is a rectangle
of the same form. Thus, if $\RR$ denotes the collection
of these rectangles we have that $\RR$ is closed under finite intersection,
i.e.\ it is a $\pi$-system. Second, the complement of
a rectangle from $\RR$ is a finite union of disjoint rectangles from $\RR$.
\begin{exe}
Write, explicitly, the complement of $(a_1, b_1] \times (a_2, b_2]$
as the disjoint union of elements of $\RR$.
Notice we can do that in at least two different ways.
Do so.
\end{exe}
Now consider the class
\[
\CC := \{\text{finite disjoint unions of elements of $\RR$} \}.
\]
It is easy to visualise, geometrically, what kind of elements
$\CC$ contains.
Then
\begin{exe}
Show that $\CC$ is a field, i.e.\ if $A \in \CC$ then $A^c \in \CC$ and
if $A, B \in \CC$ then $A \cup B \in \CC$.
\end{exe}

Hence, if $A \in \CC$, we can write $A = \cup_{i=1}^m B_i$, where
$B_i$ are disjoint rectangles from $\RR$, and, since for
each such rectangle $B_i$ we can use $F_{X,Y}$ to compute 
$\Pr_{X,Y}(B)$, we have that
\[
\Pr_{X,Y}(A) = \sum_{i=1}^m \Pr_{X,Y}(B), \quad A \in \CC,
\]
can be computed by using $F_{X,Y}$ only.

But there are many sets in $\BB(\R^2)$ that do not belong to $\CC$,
so we wish to continue our endeavour. 
If it were true that every set in $\BB(\R^2)$ was a limit of
sets of $\CC$ then we would be finished, by the continuity
property of a probability. However, there are many elements in
$\BB(\R^2)$ that are not limits of sets in $\CC$.
\begin{example}
The set of all points $(x,y)$ where $x, y$ are rationals cannot be obtained
as a limit of elements of $\CC^2$. To illuminate this point,
consider ``straightforward'' procedure that places a little rectangle
around each such point and then let the little rectangle
shrink. Specifically, let $\Q=\{q_1,q_2, \ldots\}$ be an enumeration
of the rationals. To each $(q_m, q_n)$ associate the rectangle
\[
I_{m,n}(\epsilon) := (q_m - \epsilon 2^{-m}, ~q_m + \epsilon 2^{-m}]
\times
(q_n - \epsilon 2^{-n}, ~q_n + \epsilon 2^{-n}],
\]
and let $I(\epsilon) := \cup_{m,n} I_{m,n}(\epsilon)$.
Show that $\cap_{\epsilon > 0, \epsilon \in \Q} I(\epsilon)$
is not equal to $\Q\times \Q$.\\
{\em Hint: The set $I(\epsilon)$ is uncountable.}
\end{example}

We will denote by $\Pr_0$ the function $\Pr_{X,Y}$ restricted
to $\CC$.
Clearly, $\Pr_0$ is uniquely specified by $F_{X,Y}$.
If $\Pr_0$ can be uniquely extended to
a probability on $(\RR^2, \BB(\R^2))$ then our claim that $F_{X,Y}$
completely specifies $\Pr_{X,Y}$ will be proved. 
The proof of this is contained in \cite{k}.
The most important step in the proof is to show that
$\Pr_0$ is countably additive on $\CC$.

\begin{exe}
Let $V, W$ be independent random variables with $F_V(t)= F_W(t) = 1-e^{-t}$,
$t > 0$. Let $X=2V$, $Y=V-W$. Compute the
joint distribution function $F_{X,Y}$ of the random element $(X,Y)$.
\end{exe}

\section{Independence}
Recall that $X, Y$ are independent random variables
on $(\Omega, \FF \Pr)$ if $\sigma(X)$, $\sigma(Y)$ are independent 
$\sigma$-fields.
We would like to show that
\begin{proposition}
\label{prodf}
$X, Y$ are independent on $(\Omega, \FF, \Pr)$ if and only if
$F_{X,Y}(x,y) := \Pr(X \le x, Y \le Y)$,
$F_X(x) := \Pr(X \le x)$, 
$F_Y(y) := \Pr(Y \le y)$ are related by
\[
F_{X,Y}(x,y) = F_X(x) F_Y(y).
\]
\end{proposition}
This is a consequence of the following:
\begin{lemma}
\label{indpi}
Let $(\Omega, \FF, \Pr)$ be a probability space.
Let $\RR_1, \RR_2$ be two classes of sets, each of which is closed
under the intersection operation.
Then $\sigma(\RR_1), \sigma(\RR_2)$ are independent if and only if
\[
\Pr(B_1 B_2) = \Pr(B_1) \Pr(B_2),
\quad
B_1 \in \RR_1, ~ B_2 \in \RR_2.
\]
\end{lemma}
For a proof see \cite{k}.

\no{\bf Proof of Proposition \ref{prodf}:}
Let $\RR_1 = \{X^{-1}(a,b]:~a < b\}$, $\RR_2 = \{Y^{-1}(a,b]:~a < b\}$.
These are both classes of sets closed under intersection.
It is easy to see that the assumption $F_{X,Y}(x,y) = F_X(x) F_Y(y)$
implies that $\Pr(B_1 B_2) = \Pr(B_1) \Pr(B_2)$, $B_1 \in \RR_1$,
$B_2 \in \RR_2$. Hence $\Pr(A_1 A_2) = \Pr(A_1) \Pr(A_2)$, $A_1 \in 
\sigma(\RR_1)= \sigma(X)$, $A_2 \in \sigma(\RR_2)=\sigma(Y)$.
\qed






\section{Joint density}
The concept of joint density requires understanding of the Lebesgue
integral of functions of two variables. You will not be asked to understand
this, but you may think of it, roughly, as a Riemann integral.
For a complete understanding, see \cite{wil,fg}, and also consult
the notes \cite{k}.

We do mention, without proof of course, the following:
\begin{theorem}
(i) If $h$ is Riemann integrable on a rectangle $R=[a,b] \times [c, d]$ then
its Lebesgue integral on $R$ coincides with its Riemann integral on $R$.\\
(ii) If $h$ is bounded and measurable then it is Riemann integrable on $R$
of the set of discontinuities $D$ of $h$ satisfies $\Pr((U_1, U_2) \in D) =0$.
\end{theorem}

Given two distribution functions $F_1, F_2$ on $\R$,
we can define the product measure $F_1 \times F_2$ on $(\R^2, \BB(\R^2))$
by defining it first on rectangles,
\[
(F_1\times F_2)(B_1 \times B_2) := \Pr(F_1^{-1}(U_1) \in B_1, ~
 F_2^{-1}(U_2) \in B_2),
\]
and then extending it to $\BB(\R^2)$ using the procedure explained in \cite{k}.
One could define the Lebesgue-Stieltjes integral by
\[
\int_{\R^2} h~ d(F_1 \times F_2) := \E h(F_1^{-1}(U_1), F_2^{-1}(U_2)),
\]
where $U_1, U_2$ are independent random variables, uniform in $[0, 1]$.


\begin{theorem}[Fubini]
If $h \ge 0$ or if $\int_{\R^2} |h|~ d(F_1 \times F_2) < \infty$, then
\[
\int_{\R^2} h~ d(F_1 \times F_2)
= \int_{\R} \left( \int_\R h(x,y) F_2(dy) \right) F_1(dx)
= \int_{\R} \left( \int_\R h(x,y) F_1(dx) \right) F_2(dy).
\]
\end{theorem}
\no{\bf Proof [sketch]:}
The statement is obvious when $h(x,y) = \1_{B_1}(x) \1_{B_2}(y)$.
Hence it is obvious for linear combinations of such functions.
The general statement follows from an approximation procedure
(see \cite{wil,fg} and the notes \cite{k}.)
\qed



Occasionally, it so happens that there exists a function
$f$ such that $F_{X,Y}$ can be written as a Lebesgue integral:
\[
F_{X,Y}(x,y) = \int_{(-\infty, x]\times (-\infty, y]} f(s,t) ds dt.
\]
In such a case, we say that $(X,Y)$ is absolutely continuous (jointly
absolutely continuous, I suppose, if you want to be pedantic)
and that $f_{X,Y}$ is a density.
Using Fubini's theorem, we see that
\[
f_X(x) = \int_{(-\infty, x]} f(s) ds
\]
is a density for $X$
(i.e.\ $X$ is absolutely continuous, and so is $Y$).

Another consequence of Fubini's theorem is:
\begin{lemma}
If $X, Y$ are independent then
\[
\E(XY) = (\E X )(\E Y),
\]
whenever the expectations are defined.
\end{lemma}

And another, useful consequence of Fubini's theorem is:
\begin{lemma}
If $X$ is a positive random variable then
\[
\E X = \int_0^\infty \Pr(X > x) ~ dx.
\]
\end{lemma}

A standard criterion for independence between $X, Y$, for 
absolutely continuous $(X, Y)$, is: 
\begin{lemma}
Suppose that $(X,Y)$ is absolutely continuous. Let $f_{X,Y}$
be a density of $(X,Y)$. Let $f_X, f_Y$ be densities of
$X, Y$, respectively. Then $X, Y$ are independent if and only if
\[
f(x,y) = f_X(x) f_Y(y),
\]
for all $(x,y)$ except, possibly, on a set of measure zero.
\end{lemma}

\section{Joint moment generating function}
When $(X,Y)$ is a random variable in $\R^2$ we can define its
{\sc joint moment generating function} by
\[
M_{X,Y}(\eta, \theta) := \E e^{\eta X + \theta Y}, \quad
\eta, \theta \in \R.
\]
whenever it exists.
Let also $M_X, M_Y$ be the moment generating functions of $X, Y$.
One can prove that:
\begin{lemma}
Suppose that $M_{X,Y}$ exists in a (one sided) neighbourhood of zero.
If $M_{X,Y}(\eta, \theta) = M_X(\eta) M_Y(\theta)$ then
$X, Y$ are independent.
\end{lemma}
\begin{exe}
Let $X, Y$ be independent random variables with
common moment generating function $M_X(\theta) = M_Y(\theta) =
e^{\theta^2}$. Let $X':= X+Y$, $Y':=X-Y$. Compute the joint moment
generating function of $(X', Y')$. Show that $X', Y'$ are
independent.
\end{exe}


\section{Correlations}
We now consider random variables $X$ on some probability space
$(\Omega, \FF, \Pr)$ with  $\E X^2 < \infty$. The aggregate of all 
these random variables will be denoted by 
\[
L^2(\Omega, \FF, \Pr).
\]
If $\E X^2 < \infty$, $\E Y^2 < \infty$, then $\E(X+Y)^2 < \infty$
(indeed, $(x+y)^2 \le 2x^2+2y^2$).
This means that if $X, Y \in L^2(\Omega, \FF, \Pr)$ then, for any 
$a, b \in \R$, $aX + bY \in L^2(\Omega, \FF, \Pr)$. Hence 
$ L^2(\Omega, \FF, \Pr)$ is a linear space (a linear subspace of
$\R^\Omega$).
The correlation between $X$ and $Y$ is defined by
\[
\E (X Y).
\]
The covariance between $X$ and $Y$ is defined by
\[
\cov(X,Y) = \E(X-\E X)(Y - \E Y).
\]
Since, for $X_1, X_2, Y \in L^2(\Omega, \FF, \Pr)$, $a_1, a_2 \in \R$,
\[
\cov(a_1 X_1 + a_2 X_2, Y) = a_1 \cov(X_1, Y) + a_2 \cov(X_2, Y),
\]
the covariance is linear in each of its arguments when the other is
kept fixed and it can thus be used to define an inner product:
\[
\langle X, Y \rangle := \cov (X,Y).
\]
We also define the semi-norm
\[
|| X || := \sqrt{\cov(X, X)} = \sqrt{\var(X, X)},
\]
where the word `semi-norm' means that it has the following properties:
\begin{itemize}
\item[1. ]
$|| X || \ge 0$
\item[2. ]
$|| a_1 X_1 + a_2 X_2 || = |a_1|~ ||X_1|| + |a_2|~ ||X_2||$.
\item[3. ]
$||X+Y|| \le ||X|| + ||Y||$.
\end{itemize}
Another name for $||X||$ is `standard deviation'.
We also note that
\begin{itemize}
\item[]
If $||X||=0$ then $\Pr(X=0)=1$.
\end{itemize}
We \underline{cannot} deduce, from $||X||=0$ alone that
$X(\omega) =0$ for all $\omega \in \Omega$, but only
that $X(\omega)=0$ for all $\omega$ except those in a set
of probability zero. If we could deduce that $X(\omega)=0$ for all
$\omega \in \Omega$, we would say that $||\cdot||$ is a norm.
To get around this problem, we merely identify all random variables
in $L^2(\Omega, \FF, \Pr)$ which differ on a set of measure zero:
That is, we let $[X]$ be the set of all $Y$ such that $\Pr(X \not = Y)=0$,
and redefine $L^2(\Omega, \FF, \Pr)$ to be the collection of
all such $[X]$. It is not hard to see that this is still a linear 
space and if we let $||[X]|| := ||X||$ (which is well defined),
then this is a norm.

Being a normed space with an inner product, $L^2(\Omega, \FF, \Pr)$
has a structure much like the geometric structure of the usual
Euclidean space, for instance, Pythagoras' theorem holds:
\[
||X + Y||^2 = ||X||^2 + ||Y||^2
\quad \text{ if } \langle X, Y \rangle =0.
\]
For more properties, see \cite{k}.

\begin{exe}
Let $U_1, U_2$ be independent, both uniform in $[0,1]$. Compute
the correlation and covariance between 
$X_1 := U_1+e^{U_2}$, $X_2=U_1^2-U_2^3$.
\end{exe}




\section{Conditioning}
%We warn the reader that this section is quite informal. In other
%words, we will not prove anything, because, without actually
%defining the concepts properly, we cannot prove anything at all.
%
%The proofs and justifications will be given in Chapter 8.

\subsection{Na\"ive conditioning}
Recall that, if
$(\Omega, \FF, \Pr)$ is a probability space and $B \in \FF$,
$\Pr(B) \not = 0$, we define
\[
\Pr(A | B) = \Pr(AB)/\Pr(B), \quad A \in \FF.
\]
This is a new probability, called $\Pr$ conditional on $B$. We
easily see that $(\Omega, \FF, \Pr(\cdot|B))$ is
a new probability space.

The problem is that, in many cases, we want to condition
with respect to an event $B$ that has probability zero.
This can be done. Loosely speaking, what saves us is the fact that if $\Pr(B)=0$
then $\Pr(AB)=0$ as well. 

We can use many methods for defining conditioning in a more general
sense. There is a geometric approach based on the structure of
$L^2(\Omega, \FF, \Pr)$, and this can be found in any good book,
like \cite{wil,fg}, and also in the notes \cite{k}.
In Chapter 8 you will see another approach.

What we will do, instead, is to define conditioning in very 
special cases, using formulae.


\subsection{Conditional density}
Suppose that $(X,Y)$ is an absolutely continuous random element of $\R^2$
with some joint density $f: \R^2 \to \R$. Then $X$ is an absolutely continuous
random variable. Let $f_1: \R \to \R$ be a density for $X$.
Similarly, $Y$ is an absolutely continuous
random variable. Let $f_2$ be a density for $Y$.
We call $f_1$, $f_2$ {\sc marginal densities} of $f$.
Define {\sc the conditional density of $X$ given $Y$} by
\[
f_{X|Y}(x|y) := 
\begin{cases}
\displaystyle
\frac{f(x,y)}{f_2(y)},& \quad \text{if } f_2(y) \not = 0,
\\
0,& \quad \text{otherwise}.
\end{cases}
\]
\begin{lemma}
For $B, C \in \BB$, let
\[
g_B(y) := \int_B f_{X|Y}(x|y) dx.
\]
Then
\[
P(X \in B , Y \in C) = \E[g_B(Y) \1_C(Y)],
\]
\end{lemma}
This is hard to prove in general. But you can sort of see it is true
by considering
\begin{exe}
Convince yourselves that the lemma is true if
$B$, $C$ are intervals.
\end{exe}


\subsection{Conditional probability mass function}
Let $(X,Y)$ be a random variable on $(\Omega, \FF, \Pr)$
with values in the discrete set $(S_1\times S_2, 2^{S_1\times S_2})$.
As usual, define its 
{\sc joint probability mass function} and {\sc marginal probability
 mass functions} by
\[
p(x,y) := \Pr((X,Y)=(x,y)),
\quad
p_1(x) := \Pr(X=x),
\quad
p_2(y) := \Pr(Y=y),
\quad
(x,y) \in S_1 \times S_2.
\]
Define the {\sc conditional probability mass function of $X$ given $Y$} by
\footnote{
The notation $p_{X|Y}$ is terrible. We only use it out of some respect
to the undergraduate probability courses. The reason that the notation
is terrible is that in the subscript `$X|Y$' in $p_{X|Y}(x | Y)$ plays
a merely cosmetic r\^ole, as opposed to the essential r\^ole played by the
last variable $Y$ inside the parenthesis.
}
\begin{equation} \label{llll}
p_{X|Y}(x|y) := 
\begin{cases}
\displaystyle
\frac{p(x,y)}{p_2(y)},& \quad \text{if } p_2(y) \not = 0,
\\
0,& \quad \text{otherwise}.
\end{cases}
\end{equation}
\begin{lemma}
\[
\Pr(X=x | Y=y) = p_{X|Y}(x | y) p_2(y).
\]
\end{lemma}
\begin{proof}
If $p_2(y)=0$ then $\Pr(X=x, Y=y) = 0$.
If $p_2 (y) >0$ then multiply both sides of \eqref{llll} by $p_2(y)$ to
obtain the desired result.
\end{proof}



%\section{Lots of independent random variables}



\noindent
\hrulefill

\begin{thebibliography}{9}
\bibitem{fg}
{\sc B. Fristedt \& L. Gray},
{\em A Modern Approach to Probability Theory},
Birkh\"auser, 1997.
\bibitem{hs}
{\sc E. Hewitt \& K. Stromberg},
{\em Real and Abstract Analysis},
Springer, 1965.
%\bibitem{kf}
%{\sc A. N. Kolmogorov \& S. V. Fomin}, 
%{\em Introductory Real Analysis}, 
%Dover, 1975.
\bibitem{k}
{\sc T. Konstantopoulos},
Extended set of lecture notes, with proofs,\\
{\tt www.ma.hw.ac.uk/$\sim$takis}
%\bibitem{dm}
%{\sc D. Mumford},
%The dawning of the age of stochasticity.\\
%{\tt http://www.dam.brown.edu/people/mumford/Papers/Dawning.pdf}, 1999.
\bibitem{wil}
{\sc D. Williams},
{\em Probability with Martingales},
Cambridge, 1991.
\end{thebibliography}


\end{unit}

\end{document}
