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\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).  
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\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 $V(Y_1) = V(Y_2)$.  
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\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 $Y_1$.  
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\noindent
5.109.  Using Theorem 5.12, 
$$V(30Y_1 + 25Y_2) = 900 V(Y_1) + 1500 {\rm Cov}(Y_1,Y_2) + 625 V(Y_2).$$
We compute:
\begin{align*}
E(Y_1) &= \int_0^1 \int_0^1 y_1(y_1+y_2)\ dy_1\ dy_2 = \frac{7}{12}\\
E(Y_2) &= \frac{7}{12} \text{ by symmetry }\\
E(Y_1^2) &= \int_0^1 \int_0^1 y_1^2(y_1+y_2)\ dy_1\ dy_2 = \frac{5}{12}\\
E(Y_1Y_2) &=  \int_0^1 \int_0^1 y_1y_2(y_1+y_2)\ dy_1\ dy_2 = \frac{1}{3}\\
E(Y_2^2) &= \frac{5}{12} \text{ by symmetry }.
\end{align*}
Then
$$V(Y_i) = E(Y_i^2) - E(Y_i)^2 = \frac{5}{12} - \left(\frac{7}{12}\right)^2 = \frac{11}{144}.$$
Similarly,
$${\rm Cov}(Y_1,Y_2) = E(Y_1Y_2) - E(Y_1)E(Y_2) = \frac{1}{3} - \frac{49}{144} = \frac{-1}{144}.$$
So finally, 
$$V(30Y_1 + 25Y_2) = \frac{1525\cdot 11 - 1500}{144} = \frac{15275}{144} \doteq 106.1.$$
By Tchebysheff's Theorem, and interval containing $30Y_1 + 25Y_2$ at least
$75\%$ of the time would be $[\mu - 2\sigma, \mu + 2\sigma] \doteq [11.48,52.68]$.  
\vskip 10pt
\noindent
6.1.  
(a) Letting $U_1 = 2Y-1$, $U_1 \le u \Leftrightarrow Y \le (u+1)/2$.
Then the cdf for $U_1$ is
$$F(u) = \int_0^{(u+1)/2} 2(1 - y)\ dy = -((1-u)/2)^2 + 1$$
(for $-1 \le u\le 1$, and zero otherwise).  To find the pdf for 
$U$, we differentiate
$$f(u) = F'(u) = (1-u)/2 \text{ if } -1 \le u \le 1;   0  \text{otherwise}$$

(b)  Derivation for $U_2$ is very similar to part (a).  Answer:
$$f(u) = (1+u)/2 \text{ if } -1 \le u \le 1, 0  \text{ otherwise }.$$

(c)  $U_3 = Y^2$.  We have $Y^2 \le u \Leftrightarrow =\sqrt{u}\le Y\le \sqrt{u}$.
Only positive values of $y$ occur with nonzero densities, though, so
the cdf for $U_3$ is 
$$F(u) = \int_0^{\sqrt{u}} 2(1-y)\ dy = -(1 - \sqrt{u})^2$$
(for $0 \le u\le 1$, and zero otherwise).  The pdf for 
$U_3$ is  $f(u) = F'(u) = \frac{1 - \sqrt{u}}{\sqrt{u}}$
(for $0 < u \le 1$, zero otherwise).

(d)  Using these pdf's:  
\begin{align*}
E(U_1) &= \int_{-1}^1 u(1-u)/2\ du = -1/3\\
E(U_2) &= \int_{-1}^1 u(1+u)/2\ du = 1/3\\
E(U_3) &= \int_0^1 u\cdot \frac{1 - \sqrt{u}}{\sqrt{u}}\ du
=\int_0^1 \sqrt{u} - u\ du = 1/6
\end{align*}

(e)  Using the methods from Chapter 4:  We know $Y$ has 
a Beta distribution with $\alpha = 1$, $\beta = 2$ from
the form of the pdf for $Y$.  So $E(Y) = 1/3$ and $V(Y) = 1/18$.
Hence $E(U_1) = 2(1/3) - 1 = -1/3$, 
$E(U_2) = 1 - 2(1/3) = 1/3$, and 
$E(U_3) = E(Y^2) = V(Y) + E(Y)^2 = 1/18 + (1/3)^2 = 1/6$.
Everything checks!
\vskip 10pt
\noindent
6.7. (a) Let $Z$ be a standard normal random variable and $U = Z^2$.  Then for any given $u > 0$
$U \le u \Leftrightarrow -\sqrt{u} \le Z \le \sqrt{u}$.  So 
$$F_U(u) = P(U \le u) = P(-\sqrt{u} < Z \le \sqrt{u}) 
= \int_{-\sqrt{u}}^{\sqrt{u}} \frac{1}{\sqrt{2\pi}} e^{-y^2/2}\ dy.$$
We will now differentiate this function of $u$ to find the density
function for $U$, using the Fundamental Theorem of Calculus:
\begin{align*}
\frac{d}{du}\int_{-\sqrt{u}}^{\sqrt{u}} \frac{1}{\sqrt{2\pi}} e^{-y^2/2}\ dy &= 
\frac{u^{-1/2} e^{-u/2}}{2\sqrt{2\pi}} +  \frac{u^{-1/2} e^{-u/2}}{2\sqrt{2\pi}}\\
&= \frac{u^{-1/2} e^{-u/2}}{\sqrt{\pi}\cdot 2^{1/2} }.
\end{align*}

(b) This is the form of a $\Gamma$ density function with $\alpha = 1/2$ and 
$\beta = 2$ because $\Gamma(1/2) = \sqrt{\pi}$.

(c) Another name for this is the $\chi^2$ distribution with one degree
of freedom.
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\noindent
6.8.  (a)  The inequality $U < u$ is equivalent to $1 - Y < u$ so $Y > 1 - u$.  Now,
$$F_U(u) = P(1 - Y < u) = \int_{1-u}^1 \frac{1}{B(\alpha,\beta)} y^{\alpha-1} (1 - y)^{\beta -1}\ dy.$$
To see what is happening here, make the substitution $z = 1 - y$, so $y = 1 - z$, and $dz = -dy$.
We get
$$F_U(u) = \int_u^0 \frac{1}{B(\alpha,\beta)} (1-z)^{\alpha-1} z^{\beta -1} (-dz) = 
\int_0^u \frac{1}{B(\alpha,\beta)} (1-z)^{\alpha-1} z^{\beta -1}\ dz.$$
Differentiating this with respect to $u$, using the FTC, we get:
$$f_U(u) = \frac{1}{B(\alpha,\beta)} u^{\beta-1} (1-u)^{\alpha-1}$$

(b) Since the beta function is symmetric in its arguments, this is the 
form for the density function of a variable with a beta distribution with
parameters $\beta$ and $\alpha$ (i.e. \emph{reversed} from the parameters of $Y$).

(c) $E(U) = \frac{\beta}{\alpha + \beta} = 1 - \frac{\alpha}{\alpha+\beta} = 1 - E(Y)$
as we expect from our general facts about linearity of expected values.

(d) We can see $V(U) = V(Y)$ because of the symmetry of 
$\frac{\alpha\beta}{(\alpha + \beta)^2(\alpha + \beta + 1)}$
in $\alpha,\beta$.  
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\noindent
6.38.  We have
\begin{align*}
m_U(t) &= E(e^{tU})\\
&= E(e^{t(a_1Y_1+a_2Y_2)})\\
&= E(e^{(a_1t)Y_1} \cdot e^{(a_2 t)Y_2})\\
&= E(e^{(a_1t)Y_1}) \cdot E(e^{(a_2 t)Y_2}) \text{ by Theorem 6.2 (independence)}\\
&= m_{Y_1}(a_1t) \cdot m_{Y_2}(a_2 t),
\end{align*}
which is what we wanted to show.  
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\noindent
6.39.  By the preceding exercise, if $Y_i$ are the lives of the two components and
$U = \frac{1}{2}Y_1 + \frac{1}{2}Y_2$ is the average life, then
$$m_U(t) = m_{Y_1}(\frac{1}{2} t) \cdot m_{Y_2} (\frac{1}{2} t) = \frac{1}{\left(1 - \frac{t}{2}\right)^2}.$$
By the uniqueness theorem, this implies that $U$ has a $\Gamma$ distribution with 
$\alpha = 2$ and $\beta = \frac{1}{2}$.  The density function is the $\Gamma$ density:
$$f(u) = \frac{1}{\Gamma(2) (1/2)^2} u e^{-\frac{u}{1/2}} = 4ue^{-2u}$$
(for $u > 0$, and $0$ otherwise).
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\noindent
6.40.  This is a $\chi^2$ distribution with 2 d.f. by Theorem 6.4.
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\noindent
6.41.  By Theorem 6.3, $U$ has a normal distribution with mean $\mu_U = (a_1 + \cdots + a_n)\mu$
and $\sigma_U^2 = (a_1^2 + \cdots + a_n^2)\sigma^2$.  The density function for $U$ is the
usual normal density:
$$ f_U(u) = \frac{1}{\sqrt{2\pi\sigma_U^2}} e^{\frac{(u - \mu_U)^2}{2\sigma_U^2}}.$$  
\vskip 10pt
\noindent
6.42.  Using the result from Exercise 6.41, $U = Y_2 - Y_1$
is normally distributed with $\mu_U = -1000$
and $\sigma_U^2 = 300^2 + 400^2 = 500^2$.  The probability
that the elevator is overloaded is found by standardizing
and using the table for the standard normal
$$P(Y_2 - Y_1 > 0) = P\left(\frac{(Y_2 - Y_1) + 1000}{500} > 2\right) = 
P(Z > 2) = .0228.$$
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\noindent
6.46.  Using Exercise 6.38 (with $a_2 = 0$) we have
$$m_W(t) = m_Y(\frac{2}{\beta} t) = \frac{1}{(1 - \beta \frac{2}{\beta} t)^{n/2}} = 
\frac{1}{(1 - 2t)^{n/2}}.$$
By the uniqueness theorem, this implies that $W$ has a $\chi^2$ distribution 
with $n$ d.f. 
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