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  \paragraph{Problem}}
\newcommand(\solution}{%
  \paragraph{Solution}}

\begin{document}


\begin{problems}

%BEGIN Kavita's problems
%END   Kavita's problems

%BEGIN Joe's problems
\problem %source: spring00 ps8
\problem

\begin{problemparts}
\problempart

Prove that $Pr(\overline{A}|C) = 1 - Pr(A|C)$ for any events $A$ and $C$.

\problempart

Is it possible for $Pr(A|B)$ and $Pr(A|C)$ to be very small, if 
$Pr(A|B\cap C)$ is large? Give a counterexample or a proof to justify 
your answer.
\end{problemparts}

\omitsolution{

\begin{enumerate}
\item
\begin{eqnarray*}
1 - Pr(A|C) &=& 1 - \frac{Pr(A\cap C)}{Pr(C)} \\
            &=& \frac{Pr(C) - Pr(A\cap C)}{Pr(C)} \\
            &=& \frac{Pr(\overline{A}\cap C)}{Pr(C)}\\
            &=& Pr(\overline{A}|C)
\end{eqnarray*}

\item
Yes, it is.  
Consider the uniform sample space $S=\{1,2,\ldots, 2 \cdot 10^6\}$. \\
Let $A = \{1\}, B = \{1,2,\ldots,10^6 \}, C = \{1,10^6+1,\ldots,2 \cdot 10^6 -1\}$. \\
Then $Pr(A \mid  B)  = Pr(A \mid C) = 10^{-6}$, while $Pr(A \mid B \cap C) = 1$.
\end{enumerate}
}

\problem %source: spring00 ps9
%(Probability identities)

\begin{problemparts}
\problempart
Carefully prove the following theorem:
\begin{quote}
If $A$ and $B$ are events in a probability space with
$\Pr(B) \neq 0$ and $\Pr(\bar{B}) \neq 0$,
then
$\Pr(A) = \Pr(A \mid B) \Pr(B) + \Pr(A \mid \bar{B}) \Pr(\bar{B})$.
\end{quote}
Show all steps, and justify them with basic facts about probability.

\solution{
\begin{eqnarray*}
\Pr(A) & = & \Pr  ((A \cap B) \cup (A \cap \bar{B})) \\
      & = & \Pr  (A \cap B) + \Pr (A \cap \bar{B}) \\
        &  & \longrightarrow \text{(Because they are disjoint.)} \\ 
      & = & \Pr  (A \mid B) \Pr(B) + \Pr (A \mid \bar{B}) \Pr(\bar{B}) \\
\end{eqnarray*}
}


\problempart
A space probe near Mars communicates with Earth using bit strings. 
Suppose that in its transmissions it sends a 1 one-third of the time and 
a 0 two-thirds of the time. When a 0 is sent, the probability that it is 
correct is 0.9, and the probability it is received incorrectly (as a 1) is
0.1. When a 1 is sent, the probability it is received correctly is 0.8, and
the probability it is received incorrectly (as a 0) is 0.2.

Use the formula above to find the probability that a 0 was transmitted,
given that a 0 was received.

\solution{
Let R be the event that a 0 was received. Let S be the event that a 0 was transmitted. 

So we are trying to find $\Pr(S \mid R)$.

Then we have:
\begin{eqnarray*}
\Pr(S) & = & 2/3 \\
\Pr(\bar{S}) & = & 1/3 \\
\\
\Pr(R \mid S) & = & 0.9 \\
\Pr(\bar{R} \mid S) & = & 0.1 \\
\\
\Pr(\bar{R} \mid \bar{S}) & = & 0.8 \\
\Pr(R \mid \bar{S}) & = & 0.2 \\
\end{eqnarray*}

Now rearranging:
\begin{eqnarray*}
\Pr(S \mid R) & = & \frac{\Pr (S \cap R) }{\Pr (R)} \\
              & = & \frac{\Pr (R \mid S) \Pr (S) }{\Pr(R)} \\
              & = & \frac{\Pr (R \mid S) \Pr (S) }{\Pr(R \mid S) \Pr (S) 
                                                 + \Pr(R \mid \bar{S}) \Pr(\bar{S})} \\
              & = & \frac{0.9 * 2/3}{0.9 * 2/3 + 0.2 * 1/3} \\
              & = & \frac{1.8}{1.8+.2}\\
              & = & .9
\end{eqnarray*}
}

\problempart
Generalize the theorem above. 
Namely, given an event $A$ and mutually exclusive events $B_1, B_2,
\ldots B_k$ with nonzero probability, prove that:
\[
Pr(B_1 \cup B_2 \cup B_3 \cup \ldots B_k ) = 1
\]
implies:
\[
\Pr(A) = \sum_{i=1}^k \Pr(A | B_i) \Pr(B_i)
\]

\solution{
\begin{eqnarray*}
\Pr(A) & = & \Pr (A \cup (B_1 \cap B_2 \cap \ldots \cap B_k  ) )  \\
       & = & \Pr((A \cap B_1) \cup (A \cap B_2) \ldots \cup (A \cap B_k)) \\
       & = & \Pr(A \cap B_1) + \Pr(A \cap B_2) + \ldots +  \Pr(A \cap B_k) \longrightarrow \text{Disjoint} \\ 
       & = & \Pr(A \mid B_1) \Pr(B_1) + \Pr(A \mid B_2) \Pr(B_2) + \ldots +  \Pr(A \mid B_k) \Pr(B_k) \\
       & = & \sum_{i=1}^k \Pr(A | B_i) \Pr(B_i)
\end{eqnarray*}
}
%[[[IThis doens't make sense either. I changed it to Pr(A). Otherwise you have Bayes law. 
% And we need some constraints on the B_i ]]]
\end{problemparts}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


%END   Joe's problems


%BEGIN Ray's problems
%END   Ray's problems

%BEGIN Ivan's problems (spring '98)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%pset9 problem6 spring98, (left for pset10)
\problem [10 pts]

$A$, $B$, $C$ are events of an experiment. If $C$ is independent of $A$ and $C$
is independent of $B$, is $C$ independent of $A\cap B$? Prove your answer or
give a counterexample. 

\solution{
{\bf Solution}

It is {\em not} true. A counterexample would be an experiment as follows.
Suppose we  are to draw a number from the set $\{1,2,\ldots,8\}$ randomly.
Let $A$ be the event that the number is even. Let $B$ be the event that the
number is prime. Let $C$ be the event that the number is at least 5. Then we
have $A=\{2,4,6,8\}, B=\{2,3,5,7\}, C=\{5,6,7,8\}$.

Then $P(A)=\frac{1}{2}$, $P(B)=\frac{1}{2}$, $P(C)=\frac{1}{2}$.

$C$ is independent of $A$ because
$P(C|A)=\frac{P(C\cap A)}{P(A)}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}=P(C).$

$C$ is independent of $B$ because
$P(C|B)=\frac{P(C\cap B)}{P(B)}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}=P(C).$

Note that the only number in $A\cap B$ is 2 (an even prime), which is not in
$C$. Therefore, $C$ is not independent of $A\cap B$ because
$P(C|A\cap B)=0\ne P(C)$.
}

%END   Ivan's problems

%BEGIN George's problems
%%%%%%%%%%%%%%  Fall 2000  PS 10   %%%%%%%%%%%%%%   %%%%%%%%%%%%%%%

\problem

Let $A,B,C$ be events of some experiment.  For each of the following
statements, prove it if it is true or give a counterexample if it is
false:

\bparts

\ppart If $C$ is independent of $A$, and $C$ is independent of $B$, then
$C$ is independent of $A\cup B$.

\solution{
\medskip
\textbf{Solution:}
\medskip
The statement is false.  For example, let $A$ and $B$ be
unbiased independent bits, and let $C = A \oplus B$.  Then $C$ is
independent from both $A$ and $B$, but it is not independent from $A
\cup B = A\, {\sf or}\, B$.}

\ppart If $C$ is independent of $A$, and $C$ is independent of $B$, and
$C$ is independent of $A\cap{B}$, then $C$ is independent of $A\cup{B}$.

\solution{
\medskip
\textbf{Solution:}
\medskip
It is true.  To prove it, observe by the Sum Rule that
\begin{eqnarray*}
\lefteqn{\prob{x\in(C\cap(A\cup B))}}\\
& = &  \prob{x\in(C\cap(A- B))} +\prob{x\in(C\cap(B- A))} + \prob{x\in(C\cap(A\cap B))},
\end{eqnarray*}
and then that 
\[
\prob{x\in(C\cap(A - B))} + \prob{x\in(C\cap (A \cap B))}
= \prob{x \in (C\cap A)}
\]
also by the Sum Rule.  Then use the fact that $C$ is independent of
both $A$ and $A \cap B$ to get that $C$ is independent from $A - B$, and
similarly that it is independent from $B - A$.

}
\eparts
 
%%%%%%%%%%%%%%  Fall 2000 PS 11    %%%%%%%%%%%%%%   %%%%%%%%%%%%%%%

\problem Several statements involving probability
are listed below.  Assume that $A$ and $B$ are events, $S$ and $T$ are
random variables, and the underlying sample space is countable.
Furthermore, assume that all expectations exist.  For each statement,
describe restrictions on $A$, $B$, $S$, and $T$ sufficient to ensure
that the statement is true.  Make your restrictions as minimal as you
can.  No proofs are necessary.

\begin{problemparts}
%\begin{enumerate}

\problempart $\Pr(A \cup B) = \Pr(A) + \Pr(B)$
%\item[\textbf{(a)}] $\Pr(A \cup B) = \Pr(A) + \Pr(B)$

\solution{
\textbf{Solution:}\\
This statement is true provided that the event $A \cap B$
has probability zero.}

\problempart $\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)$
%\item[\textbf{(b)}] $\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)$

\solution{
\textbf{Solution:}\\
This statement is always true.
}

\problempart $\Pr(A \cap B) = \Pr(A) \cdot \Pr(B)$
%\item[\textbf{(c)}] $\Pr(A \cap B) = \Pr(A) \cdot \Pr(B)$

\solution{
\textbf{Solution:}\\
This statement is true provided that events $A$ and $B$ are independent.
}

\problempart $\Pr(A \cap B) < \Pr(A)$
%\item[\textbf{(d)}] $\Pr(A \cap B) < \Pr(A)$

\solution{
\textbf{Solution:}\\
This statement is true provided $\Pr(A \cap \overline{B}) > 0$.
}

\problempart $\Ex(S + T) = \Ex(S) + \Ex(T)$
%\item[\textbf{(e)}] $\Ex(S + T) = \Ex(S) + \Ex(T)$

\solution{
\textbf{Solution:}\\
This statement is always true.
}

\problempart $\Ex(S \cdot T) = \Ex(S) \cdot \Ex(T)$
%\item[\textbf{(f)}] $\Ex(S \cdot T) = \Ex(S) \cdot \Ex(T)$

\solution{
\textbf{Solution:}\\
This statement is true provided that random variables $S$
and $T$ are independent.}

\end{problemparts}
%\end{enumerate}   
%%%%%%%%%%%%%%  Fall 2000  PS 11.5   %%%%%%%%%%%%%%   %%%%%%%%%%%%%%%
\ppart Let $X$ be a random variable whose value is an observation drawn
uniformly at random from the $\set{i\in\ints \mid -n \le i \le n}$.  Let
$Y = X^2$.  Show that $Ex[XY] = Ex[X] Ex[Y]$. 
Are $X$ and $Y$ independent?

\solution{
\medskip
\textbf{Solution:}

Since $Y$ is determined by $X$, there are many cases indicating where $X$
and $Y$ are not independent, e.g., 
\[\Pr(Y = 0) = 1/(2n+1),\mbox{  but } \Pr(Y=0 \mid X= 0) = 1.\]

To see that $Ex[XY] = Ex[X] Ex[Y]$, note that $Ex[X] = 0$ and $Ex[XY]
= Ex[X^3] = 0$ since $X$ and $X^3$ are both symmetrical.
}

\ppart 
Show that in general for any two random variables $X$ and $Y$
that if $Ex[XY] = Ex[X] Ex[Y]$ then $var(X+Y) =
var(X) + var(Y)$.  In particular, this shows that it is possible
for nonindependent variables to have variances that exhibit
linearity.

\solution{
\medskip
\textbf{Solution:}

We have that $var(X + Y) = Ex((X+Y)^2) - (Ex(X+Y))^2 = Ex(X^2 + 2XY +
Y^2) - (Ex(X)+ Ex(Y))^2 = Ex(X^2) - (Ex(X))^2 + Ex(Y^2) - (Ex(Y))^2 +
2Ex(XY) - 2Ex(X)Ex(Y) = var(X) + var(Y)$. 
}

\eparts
                                          
%%%%%%%%%%%%%%
%END   George's problems

%BEGIN Josh's problems
%END   Josh's problems

%BEGIN Tina's problems
%END   Tina's problems


\end{problems}

\end{document}



