\documentclass[11pt]{article}
\def\Z{\mathbb{Z}}
\def\Q{\mathbb{Q}}
\def\R{\mathbb{R}}
\def\C{\mathbb{C}}
\usepackage{latexsym,amsmath,amssymb}
\usepackage[all]{xy}
\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\textheight}{8.0in}
\begin{document}
\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions for Problem Set 10}
\centerline{November 30, 2009}
\vskip 10pt
\noindent
\vskip 10pt
\noindent
5.85.  Whenver we can, we use the results of Exercise 5.65 above (\emph{not} the method
of integration of the function times the joint density!!).  The point
of this problem is that some approaches to doing things can be 
much simpler than others!

(a) $Y_1,Y_2$ both have exponential distributions with $\beta = 1$, so 
$E(Y_1) = E(Y_2) = 1$.

(b) Similarly, $V(Y_1) = V(Y_2) = 1$.

(c)  $E(Y_1 - Y_2) = E(Y_1) - E(Y_2) = 0$.  

(d)  This is the only part of this question for which no obvious
``shortcut'' presents itself.  We must compute:
$$
E(Y_1Y_2) &= \int_0^\infty \int_0^\infty y_1y_2 \left(1 - \alpha(1 - 2e^{-y_1})(1 - 2e^{-y_2})\right)e^{-y_1-y_2}\ dy_2\ dy_1.$$
The best approach is to multiply out and split the integral into a sum of 
several terms where each is a product $g(y_1)h(y_2)$:
\begin{align*}
&=  \int_0^\infty \int_0^\infty y_1y_2 e^{-y_1-y_2}\ dy_2\ dy_1 \\
&= \quad - \alpha \int_0^\infty\int_0^\infty y_1y_2 \left(1 - \alpha(1 - 2e^{-y_1})(1 - 2e^{-y_2})\right)e^{-y_1-y_2}\ dy_2\ dy_1\\
&= (\int_0^\infty y_1 e^{-y_1}\ dy_1)(\int_0^\infty y_2e^{-y_2}\ dy_2) \\
&\quad -\alpha(\int_0^\infty y_1(1 - 2e^{-y_1})e^{-y_1}\ dy_1)(\int_0^\infty y_2(1 - 2e^{-y_2})e^{-y_2}\ dy_2).
\end{align*}
Integrating by parts, we have $\int_0^\infty xe^{-x}\ dx = 1$ and 
$\int_0^\infty x(1-e^{-x})e^{-x}\ dx = \frac{1}{2}$.  So $E(Y_1Y_2) = 1 - \frac{\alpha}{4}$.  

(e)  This requires the concept of covariance:  $V(Y_1 - Y_2) = V(Y_1) -2 {\rm Cov}(Y_1,Y_2) + V(Y_2)
= 1 + \frac{\alpha}{2}$.  The only way to estimate an interval is to use
Tchebysheff's theorem. 
\vskip 10pt
\noindent
5.91.  From Exercise 5.76 on the last problem set, $E(Y_1) = E(Y_2) = \frac{2}{3}$.
Then 
$$
E(Y_1Y_2) = \int_0^1\int_0^1 4y_1^2y_2^2\ dy_2\ dy_1 = \frac{4}{9}.
$$
Hence ${\rm Cov}(Y_1,Y_2) = E(Y_1Y_2) - E(Y_1)E(Y_2) = 0$.  This 
is not a surprise since we know $Y_1$ and $Y_2$ are independent (see solution 
to Exercise 5.26 on the last problem set).  
\vskip 10pt
\noindent
5.94.  ``Uncorrelated'' means that it is given that ${\rm Cov}(Y_1,Y_2) = 0$.  

(a) By the bilinearity of the covariance (part c of Theorem 5.12),
\begin{align*}
{\rm Cov}(U_1,U_2) &= {\rm Cov}(Y_1+Y_2,Y_1-Y_2)\\
&= {\rm Cov}(Y_1,Y_1) + {\rm Cov}(Y_2,Y_1) - {\rm Cov}(Y_1,Y_2) - {\rm Cov}(Y_2,Y_2)\\
&= V(Y_1) + 0 - 0 - V(Y_2)\\
&= V(Y_1) - V(Y_2).
\end{align*}

(b) $\rho = \frac{V(Y_1) - V(Y_2)}{V(Y_1)V(Y_2)}$.

(c)  It is possible for ${\rm Cov}(U_1,U_2)$ to be zero.  This happens
if and only if $V(Y_1) = V(Y_2)$.  
\vskip 10pt
\noindent
5.97.  (a) $Cov(Y_1,Y_1) = E(Y_1^2) - (E(Y_1))^2 = V(Y_1) = 2$
by the given information.  b,c,d)  If $Cov(Y_1,Y_2) = 7$, 
the correlation coefficient
$\rho = \frac{7}{\sigma_1\sigma_2} = \frac{7}{\sqrt{2}\cdot 2\sqrt{2}} = 7/4$.
That is not possible because $-1 \le \rho \le 1$ in all cases.
The largest and smallest values for $Cov(Y_1,Y_2)$ are $-4, 4$, corresponding
to cases where $Y_2$ is a decreasing or increasing linear function 
of $Y_1$.  
\vskip 10pt
\noindent
5.109.

\end{document}
