%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%spring98 final #10 (added by Ivan)
\problem\points{25}
Let $R_1,R_2$ be random variables such that $\expect{R_1R_2} =
\expect{R_1}\expect{R_2}$.
Prove that
\[\variance{R_1 + R_2} = \variance{R_1} + \variance{R_2}.\]

\medskip
\solution{
\textbf{Solution:}
Let $m_i = \expect{R_i}$ and $R=R_1+R_2$.  Then
\begin{eqnarray*}
\variance{R_1 + R_2}  & = &  \expect{R^2} - \expectsq{R} \quad \mbox{[by formula for $\variance{R}$]}\\
 & = &  \expect{R_1^2 + 2R_1R_2 + R_2^2} - (m_1 + m_2)^2 \quad \mbox{[by
 linearity of $\expect{}$ and algebra]]}\\
 & = &  \expect{R_1^2} + 2\expect{R_1R_2} + \expect{R_2^2} - (m_1 + m_2)^2
 \quad \mbox{[by linearity of $\expect{}$]}\\
 & = &  \expect{R_1^2} + 2m_1m_2 + \expect{R_2^2} - (m_1 + m_2)^2
 \quad \mbox{[by hypothesis]}\\
 & = &  \expect{R_1^2} -m_1^2 + \expect{R_2^2} -m_2^2 \quad \mbox{[by algebra]]}\\
 & = &  \variance{R_1} + \variance{R_2} \quad \mbox{[by formula for $\variance{R_i}$]}.\\
\end{eqnarray*}
}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Fall 2000 Final exam
% Added by George
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% PROBLEM 12
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem\pts{8} \textbf {Overloading Circuits}

A memory card produced by NP Computer Hardware, Inc., consists of
infinitely many memory cells in series.  The card is attached to a
computer at one end.  If there is a power surge, the first memory cell
burns out.  Furthermore, if a memory cell burns out, the next memory
cell after that one burns out with probability $p$.  Otherwise, memory
cells have no chance of burning out in a power surge.

\problempart {\bf (5 points)} Express the expected number of memory
cells that burn out in a power surge, in terms of $p$.  (Hint: use
linearity of expectation.)

\solution{\bigskip {\bf Solution:}

Let $X_i$ be an indicator variable which is 1 if during a power surge,
memory cell $i$ burns out, and is a 0 otherwise.

We are asked to find Ex$(\sum_{i=0}^{\infty} X_i) =
\sum_{i=0}^{\infty}$ Ex($X_i$).

The probability that $X_i$ is 1 is equal to $p$ times the probability
that $X_{i-1}$ is 1.  This gives the simple recurrence relation $X_n =
pX_{n-1}$ with initial condition $X_0 = 1$, which simplifies to $X_n =
p^n$.  Thus, the value we want is $\sum_{i=0}^{\infty} p^n =
\frac{1}{1-p}$, provided that $p < 1$.  If $p=1$ all the memory cells
burn out.}

\mediumspace

\problempart {\bf (3 points)} It is discovered
that the mean number of memory cells that burn out from a power surge is
10.  What is $p$?

\solution{\bigskip {\bf Solution:}

If $10 = \frac{1}{1-p}$ then $10 - 10p = 1$ so $10p = 9$ so $p = 9/10$.
}
