\problemdata      
        {?} % latex-friendly label for the prob.
        {Probability-Set Theory} % The topic(s) of the problem content.
        {?} % Source (if known)
        {Fall 2001, final} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).


\begin{problem}
Let $X,Y$ be real-valued random variables defined on the same
sample space, and let $[X=Y]$ be the event that the value of $X$ equals
the value of $Y$.  Define a binary relation, $\rho$, on such random
variables by the rule that $X \rho Y$ iff $\prob{X = Y} = 1$.

It is obvious from the definition that $\rho$ is \emph{reflexive} and
\emph{symmetric}, so only one more property is needed to prove that $\rho$
is an equivalence relation.
\bparts

\ppart 
State the remaining property of $\rho$ that confirms it is an
equivalence relation.

\solution{
$\rho$ must be  \emph{transitive}: if $X \rho Y$ and $Y \rho 
Z$m then $X \rho Z$.
}

\ppart 
Prove that $\rho$ has the property.

\solution{
Note that $X \rho Y$ iff $X(s) = Y(s)$ for every sample point, $s$, with
positive probability, \ie such that $\pr{s}>0$.

So let $s$ any sample positive with positive probability and suppose $X
\rho Y$ and $Y \rho Z$.  Now since $X \rho Y$, we know that $X(s) = Y(s)$,
and likewise since $Y \rho Z$ also $Y(s) = Z(s)$.  Therefore $X(s) =
Z(s)$, which implies that $X \rho Z$, as required.

\textbf{Alternative proof:}  
We will prove
\begin{lemma*}
For any event, $W$, of probability one, and any event, $A$,
\[
\pr{A \intersect W} = \pr{A}.
\]
\end{lemma*}

Assuming the Lemma for the moment, we prove transivity as follows: if
$X\rho Y$ and $Y \rho Z$, then the events $[X=Y]$ and $[Y=Z]$ each have
probability 1 by definition.  By the Lemma, their intersection also has
probability 1.  But this intersection in the same as the event $[X = Y =
Z]$, which is a subset of the event $[X=Z]$.  Hence,
\[
\pr{X=Z} \geq \pr{X=Y=Z} = \pr{[X=Y] \intersect [Y=Z]} = 1,
\]
so $X\rho Z$.

To prove the Lemma, note that $1-\pr{B} \geq 0$ for any event, $B$, so
\[
\pr{A} \geq \pr{A \intersect W} = \pr{A} + \pr{W} - \pr{A \union W} = 
\pr{A} + (1 - \pr{A \union W}) \geq \pr{A}.
\]
}

\eparts
\end{problem}
