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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions -- Problem Set 10}
\centerline{November 30, 2005}
\bigskip
\noindent
5.87/ By linearity of expected values, 
$E(3Y_1 +4Y_1 - 6Y_3) = 3E(Y_1) + 4E(Y_2) - 6E(Y_3) = 6 - 4 - 24 = -22$.
We use Theorem 5.12b to find 
$$\eqalign{
V(3Y_1+4Y_1 - 6Y_3) &= 9V(Y_1) + 16V(Y_2)+36V(Y_3)\cr
&\qquad +24 Cov(Y_1,Y_2) -18 Cov(Y_1,Y_3) -48 Cov(Y_2,Y_3)\cr
&= 36 + 96 + 288 + 24  + 36 + 0\cr
&=480.\cr}$$

\bigskip
\noindent
5.92/  By computing marginal densities here, we
see $Y_1$ has a Gamma distribution with $\alpha = 2$ and $\beta = 1$.
$Y_2$ is exponentially distributed.  So $E(Y_1) = 2$, $E(Y_2) = 1$.
So $E(Y_1 - Y_2) = 1$.  An alternate way to see this: 
integrate against the joint density:
$$\eqalign{
E(Y_1 - Y_2) &= \int_0^\infty\int_0^{y_1} (y_1-y_2)e^{-y_1}\ dy_1\ dy_2\cr
&= {1\over 2}\int_0^\infty y_1^2 e^{-y_1}\ dy_1\cr
&= {1\over 2} \Gamma(3)\cr
&= 1\cr}$$
Then by Theorem 5.12 b,
$$\eqalign{
V(Y_1 - Y_2) &= V(Y_1) + V(Y_2) - 2Cov(Y_1,Y_2)\cr
&=2 + 1 - 2(E(Y_1Y_2) - 2\cdot 1)\cr
&=7 - 2\int_0^\infty \int_0^{y_1} y_1y_2e^{-y_1}\ dy_2\ dy_1\cr
&=7 - \int_0^\infty y_1^3e^{-y_1}\ dy_1\cr
&=7 - \Gamma(4)\cr
&=7 - 6 = 1.\cr}$$
It is unlikely that a customer spends 4 minutes or more
at the window, since $4 = \mu + 3\sigma$ (three standard
deviations above mean). 

\bigskip
\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
$$f(u) = F'(u) = \cases{(1-u)/2 & if $-1 \le u \le 1$\cr 0 & otherwise}$$

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

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  $f(u) = F'(u) = {1 - \sqrt{u}\over \sqrt{u}}$
(for $0 < u \le 1$, zero otherwise).

d)  Using these pdf's:  
$$\eqalign{
E(U_1) &= \int_{-1}^1 u(1-u)/2\ du = -1/3\cr
E(U_2) &= \int_{-1}^1 u(1+u)/2\ du = 1/3\cr
E(U_3) &= \int_0^1 u\cdot {1 - \sqrt{u}\over \sqrt{u}}\ du
=\int_0^1 \sqrt{u} - u\ du = 1/6\cr}$$

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!
 
\bigskip
\noindent
6.3/ a)  Since $U = 10Y - 4$, $U\le u \Leftrightarrow
Y \le (u+4)/10$.  So
$$F(u) = \int_{-\infty}^{(u+10)/4} f(y)\ dy
=\cases{0 & if $u < -4$\cr
        {1\over 2}((u+4)/10)^2 & if $-4 \le u \le 6$\cr
         - {1\over 2} + (u+4)/10 & if $6\le u \le 11$\cr
         1 & if $u \ge 11$\cr}$$
(Note: this is found by integrating each ``piece'', then
adjusting constants of integration to make the whole
function $F$ continuous.)  Finally, to get the pdf, we
take
$$f(u)= F'(u) = = \cases{0 & if $u < -4$\cr
        (u+4)/100 & if $-4 \le u \le 6$\cr
         1/10 & if $6\le u \le 11$\cr
         0 & if $u \ge 11$\cr}$$

b)  Integrate ``piecewise'':
$$E(U) = \int_{-4}^6 u\cdot (u+4)/100\ du + \int_6^{11} u\cdot (1/10)\ du
= 67/12.$$

c)  ``Using the methods of Chapter 4'' means finding $E(Y)$
then using linearity to get $E(U)$:
$$E(Y) = \int_0^1 y\cdot y\ dy + \int_1^{3/2} y\ dy = 23/24.$$
So $E(U) = 10E(Y) - 4 = (230/24) - 4 = 134/24 = 67/12$.
Check!
 
\bigskip
\noindent
6.8/ Note $U = Y_1 - Y_2 \le u \Leftrightarrow Y_2 \ge Y_1 - u$.
The region where the joint density is nonzero is the 
infinite ``wedge'' bounded by $y_2 = 0, y_2 = y_1$ and $y_1 \ge 0$.
Note that $y_2 = y_1 - u$ is a line parallel to the 
slanted edge of the wedge.  To find $F_U(u) = P(U \le u)$
we want to integrate over the portion of the wedge {\it above}
the line $y_2 = y_1 - u$.  For this, it is simplest to 
set up the iterated integral with $dy_1$ ``inside''
(since the left and right boundaries are given by the same
formulas all the way, unlike the bottom edge if we
integrate with $dy_2$ inside -- that way works too, of 
course, but it's significantly more complicated!).  
$$\eqalign{
F_U(u) &= \int_0^\infty \int_{y_2}^{y_2+u} e^{-y_1} \ dy_1 \ dy_2 \cr
&= \int_0^\infty - e^{-(y_2+u)} + e^{-y_2}\ dy_2\cr
&= \left. e^{-(y_2+u)}  - e^{-y_2} \right|_0^\infty\cr
&= 1 - e^{-u}\cr}$$
So $f_U(u) = F_U'(u) = e^{-u}$
(exponential with $\beta = 1$.)

\bigskip
\noindent
6.22/ a) Let $U = Y^m \Leftrightarrow Y = U^{1/m} = h^{-1}(U)$
Then ${d h^{-1}(u) \over du} = {1\over m} u^{1/m - 1}$
and 
$$\eqalign{
f_U(u) &= f_Y(h^{-1}(u))\left|{d h^{-1}(u)\over du}\right| \cr
&={1\over \alpha} m \left(u^{1/m}\right)^{m-1} e^{-u/\alpha}\cdot {1\over m}u^{1/m - 1}\cr
&= {1\over \alpha} e^{-u/\alpha}\cr}$$
(after some simplification).  

b)  If $U = Y^k$ ($k\ne m$), the same transformation gives
$f_U(u) = {m\over \alpha} u^{(m-k)/k} e^{-u^{m/k}/\alpha}$.
So 
$$E(U) = \int_0^\infty {m\over \alpha} u^{m/k} e^{-u^{m/k}/\alpha}\ du$$
The substitution $v = u^{m/k}$, $dv = {m\over k} u^{(m-k)/k}$
takes this to the form 
$$E(U) = {1\over \alpha} \int_0^\infty v^{k/m} e^{-v/\alpha}\ dv
= \alpha^{k/m} \Gamma\left({k\over m} + 1\right).$$

\bigskip
\noindent
6.24/ $U = -2\ln Y \Leftrightarrow Y = e^{-U/2} = h^{-1}(U)$.
Hence ${d h^{-1}(u)\over du} = -{1\over 2} e^{-u/2}$.  
So $f(u) = 1\cdot \left|{d h^{-1}(u)\over du}\right| =
{1\over 2}e^{-y/2}$.  (That is, $U$ is exponentially distributed
with $\beta = 1/2$.)

\bigskip
\noindent
6.33/ $U = Y_1/2 + Y_2/2$ where each $Y_i$ is exponentially
distributed and $Y_1,Y_2$ are independent.  Hence by
Theorem 6.2, the mgf for $U$ is 
$$m_U(t) = m_{Y_1} (t/2)\cdot m_{Y_2}(t/2) = {1\over (1-t/2)^2}.$$
Comparing with the table inside the back cover of the text, 
we see that $U$ has a Gamma distribution with $\alpha = 2$
and $\beta = 1/2$.

\bigskip
\noindent
6.34/ $Y_i$ are $N(0,1)$, so by Example 6.11 $Y_i^2$ are 
$\chi^2(1)$.  Then by Theorem 6.4, $U = Y_1^2 + Y_2^2$
has a $\chi^2(2)$ distribution.  Therefore
$f_U(u) = {1\over 2} e^{-u/2}$.  (Note: this is also same
as an exponential pdf with $\beta = 2$.)

\bigskip
\noindent
6.35/  By Theorem 6.3, $U$ is normally distributed with
$\mu_U = (\sum_i a_i)\mu$ and $\sigma_U^2 = (\sum_i a_i^2)\sigma^2$.
Hence
$$f_U(u) = {1\over \sqrt{2\pi \sigma_U^2}} e^{-(u - \mu_U)^2/(2\sigma_U^2)}$$

\bigskip
\noindent
6.36/  Using the result from Exercise 6.35, $U = Y_1 - Y_2$
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
$$P(Y_1 - Y_2 > 0) = P\left({(Y_1 - Y_2) + 1000\over 500} > 2\right) = 
P(Z > 2) = .0228.$$


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