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\centerline{Mathematics 375 -- Probability Theory}
\centerline{Solutions for Problem Set 10}
\centerline{December 9, 2011}
\vskip 10pt
\noindent
%5.122.  (a) To match this up with the general description given in
%Definition 5.11, we want to think of one class consisting of 
%mice with weights between 80 and 100 grams, a second class
%consisting of mice with weight more than 100 grams, and a third
%class consisting of all others.  From the given information 
%$p_1 = \frac{1}{2}(1 - 2(.1587)) = .3413$ (probability that
%a normal r.v. is between $\mu - \sigma$ and $\mu$), $p_2 = .5$, and 
%$p_3 = .1587$.   Then $P(Y_1 = 2, Y_2 = 1, Y_3 = 1)$
%is $$\frac{4!}{2!1!1!}(.3413)^2(.5)^1(.1587)^1 \doteq .1109.$$
%(b)  Similarly, 
%$$P(Y_1=0,Y_2=4,Y_3=0) = \frac{4!}{0!4!0!}(.3413)^0(.5)^4(.1587)^0 = .0625.$$  
%\vskip 10pt 
%\noindent
6.1.  
(a) Letting $U_1 = 2Y-1$, $U_1 \le u \Leftrightarrow Y \le (u+1)/2$, and 
the interval of $u$ values corresponding to $0 \le y \le 1$ is $-1 \le u \le 1$.
Then the cdf for $U_1$ is
$$F(u) = \int_0^{(u+1)/2} 2(1 - y)\ dy = 1 -((1-u)/2)^2$$
(for $-1 \le u\le 1$; and $1$ for $u \ge 1$, $0$ for $u \le -1$).  
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)  Letting $U_2 = 1 - 2Y$, $U_2 \le u \Leftrightarrow Y \ge (1-u)/2$, and 
the interval of $u$ values corresponding to $0 \le y \le 1$ is again $-1 \le u \le 1$.
The cdf for $U_2$ is
$$F(u) = \int_{(1-u)/2}^1 2(1 - y)\ dy = ((1+u)/2)^2$$
(for $-1 \le u\le 1$; and $1$ for $u \le -1$, $0$ for $u \ge 1$).  
To find the pdf for 
$U$, we differentiate
$$f(u) = F'(u) = (1+u)/2 \text{ if } -1 \le u \le 1;   0  \text{\ otherwise}$$

(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.)  The expected values
above can be computed directly from the density function
for $Y$.  But there is also a more clever way.   We recognize $Y$ 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
6.5.  Note that $y = 1$ gives $u = 5$ and $y = 5$ gives $u = 53$.
The density for $U$ will be nonzero only on the interval $[5,53]$.
(Note that there was a small typo in the answer to this problem --
the correct form should be $5 \le u \le 53$.)
 Using the method of distribution functions, if $u \in [5,53]$,
then since $f(y)$ is nonzero only when $1 \le y \le 5$, 
\begin{align*}
F_U(u) &= P(U \le u)\\
       &= P(2Y^2 + 3 \le u)\\
       &= P\left(1 \le Y \le \sqrt{\frac{u-3}{2}}\right)\\
       &= \int_1^{\sqrt{\frac{u-3}{2}}} \frac{1}{4}\ dy\\
       &= \frac{1}{4} \sqrt{ \frac{u-3}{2} }  - \frac{1}{4}.
\end{align*}
Therefore the density function for $U$ is the derivative:
$$f_U(u) = \frac{1}{16\sqrt{\frac{u-3}{2}}}$$
on the interval $5 \le u \le 53$.  
\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.
\vskip 10pt
\noindent
6.11. (a) Since the components operate independently, the joint density
for the lifetimes is
$$f(y_1,y_2) = \begin{cases}  e^{-(y_1 + y_2)} & y_1,y_2 > 0\\
                                    0 & {\rm otherwise}.
\end{cases}$$
Let $U = (Y_1 + Y_2)/2$, the average.  Then 
\begin{align*}
F_U(u) &= P(U \le u)\\
&= P(Y_1 + Y_2 \le 2u)\\
&= \int_0^{2u} \int_0^{2u - y_1} e^{-(y_1+y_2)}\ dy_2\ dy_1\\
&= 1 - e^{-2u} - 2u e^{-2u}. 
\end{align*}
Hence
$$f_U(u) = 4 u e^{-2u}.$$
That is, $U$ has a $\Gamma(2,1/2)$ distribution. 
(b)  $U$ has mean $1$ and variance $1/2$, using the formulas
for $\Gamma$-distributed random variables.  Checking our work, using
the formulas for linear combinations,
$E(U) = \frac{1}{2}(E(Y_1) + E(Y_2)) = \frac{1 + 1}{2} = 1$ and 
$V(U) = \frac{1}{4} V(Y_1) + \frac{1}{4} V(Y_2) + 0 = \frac{1}{2}$.  Check!
\vskip 10pt
\noindent
6.12.  (a)  Since $c > 0$, we have for all $u > 0$:
\begin{align*}
F_U(u) &= P(U \le u)\\
&= P(Y \le u/c)\\
&= \int_0^{u/c} \frac{y^{\alpha-1} e^{-y/\beta}}{\beta^\alpha \Gamma(\alpha)}\ dy.
\end{align*}
We find $f_U(u) = F_U'(u)$ this time by using the first part of the 
Fundamental Theorem of Calculus:
$$f_U(u) = \frac{1}{c} \frac{(u/c)^{\alpha-1} e^{-u/(c\beta)}}{\beta^\alpha \Gamma(\alpha)}
= \frac{u^{\alpha - 1} e^{-u/(c\beta)}}{(c\beta)^\alpha \Gamma(\alpha)}.$$
(b)  We recognize this as the density for a $\Gamma(\alpha,c\beta)$-distributed
random variable.   
(c)  The shape parameter is unchanged.  The scale parameter is multiplied by $c$.  
\vskip 10pt
\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
$$P(Y_2 - Y_1 > 0) = P\left(\frac{(Y_2 - Y_1) + 1000}{500} > 2\right) = 
P(Z > 2) = .0228.$$
\vskip 10pt
\noindent
6.43. (a)  Since $\overline{Y}$ is normally distributed with mean $\mu$
and variance $\sigma^2/n$, the density function is 
$$\frac{\sqrt{n}}{\sqrt{2\pi\sigma^2}} e^{-n\frac{(u - \mu)^2}{2\sigma^2}}$$
(b)  We standardize with $\sigma$ for $\overline{Y}$:
$$P(|Y-\mu|\le 1) = P(|Z| \le 5/4) = 1 - 2(.1056) = .7888.$$
(c) Similarly with $n = 36$:
$$P(|Y-\mu|\le 1) = P(|Z| \le 6/4) = 1 - 2(.0668) = .8664,$$
with $n = 64$:
$$P(|Y-\mu|\le 1) = P(|Z| \le 8/4) = 1 - 2(.0228) = .9544,$$
and with $n = 81$:
$$P(|Y-\mu|\le 1) = P(|Z| \le 9/4) = 1 - 2(.0122) = .9756.$$
As $n$ increases, the variance of $\overline{Y}$ decreases, so
the probability that $Y$ is within 1 unit of $\mu$ increases. 
(Note: This observation will be fundamental in our study
of statistics next semester.)
\vskip 10pt
\noindent
6.45.  From our previous work, we know that $U$ will 
have a normal distribution with $\mu = 100 + 70 + 12 = 182$
and standard deviation $\sqrt{49 (.5)^2 + 9 (.2)^2 + 2\cdot 7\cdot 3\cdot 0} \doteq 3.55$ 
(we're using the independence assumption to derive this since that gives 
${\rm Cov}(Y_1,Y_2) = 0$)
Then we want a value for the bid, $B$ such that 
$P(U > B) = .01$.  This says
$P(Z > \frac{B - 182}{3.55}) = .01$, so 
$\frac{B - 182}{3.55} = 2.33$, and hence $B = 190.27$.  The 
independence assumption here is questionable because  
sand is used in construction to make 
concrete.  The amount of sand used and the amount
of cement mix used are probably related.
\end{document}

7.42.  (a) We do not know the distribution of the fracture
strengths individually, but the CLT says that the average
of 100 measurements will have an approximately normal
distribution with mean $\mu = 14$ and standard deviation $\sigma = 2\sqrt{100} = .2$.
So $P(\overline{Y} > 14.5) \doteq P(Z > .5/.2) = P(Z > 2.5) \doteq .0062$.  
(b)  Such an interval is $\mu \pm 1.96 \sigma = [13.608, 14.392]$.  
\vskip 10pt
\noindent
7.43.  Using the CLT, we have 
$$P(|\overline{Y} - \mu| < .5) \doteq P\left(|Z| < \frac{.5}{2.5/10}\right) = 1 - 2(.0228) = .9554.$$
\vskip 10pt
\noindent
7.44.  We want to take $n$ so that $\frac{.4}{2.5/\sqrt{n}} = 1.96$, so $n \doteq 150$. 
\vskip 10pt
\noindent
7.45.  The probability of obtaining a sample of size 64 with average wage
$\$ 6.90$ or less if the population mean is $\$ 7.00$ can be approximated
using the CLT:
$$P(\overline{Y}\le 6.90) = P\left(Z \le \frac{6.90 - 7.00}{.5/\sqrt{64}}\right) = 
P(Z \le -1.6) = .0548.$$
Of course interpreting what this should mean is harder than just 
computing the number!  One way to say this in words is that one out of
every 18 or so samples of size 64 will probably have an average wage
of $\$6.90$ or less even if the population average wage is $\$ 7.00$.
So there's not super-strong evidence that some sort of discrimination
was going on here.  On the other hand, getting an average wage that
low is not very common, either.   
\vskip 10pt
\noindent
7.49.  Even though the distribution of individual service times
is skewed, the CLT says that the average of 50 times will 
have an approximately normal distribution with mean $\mu = 1.4$
and $\sigma = .7/\sqrt{50} \doteq .099$ (units are hours for both).  
We want 
$P(\overline{Y} > 1.6) = P\left(Z > \frac{1.6 - 1.4}{.099}\right) = P(Z > 2.02) = .0217$.  
\vskip 10pt
\noindent
7.58.  Let $W_i = X_i - Y_i$.  Then $W_i$ are independent and 
identically distributed with mean $\mu_1 - \mu_2$ and standard
deviation $\sqrt{\sigma_1^2 + \sigma_2^2}$.  Then Theorem 7.4 (the CLT)
implies the desired statement.
\vskip 10pt
\noindent
7.60.  Using the more general version of 7.58, 
\begin{align*}
P(|(\overline{X} - \overline{Y}) - (\mu_1-\mu_2)| < .05) 
&= P\left(|Z| < \frac{.05}{\sqrt{.01/50 + .02/100}}\right)\\
&=P(|Z| < 2.5)\\
&= 1 - 2(.0062) = .9876.
\end{align*} 
\vskip 10pt
\noindent
7.73.  Let $Y = $ number who
show up (which is a binomial random variable with $n = 160$, $p = .95$.
We want $P(Y \le 155) = \sum_{y=0}^{155} \binom{160}{y}(.95)^y (.05)^{160-y}$.
This can be computed directly using software like R or Maple, but it is awkward
otherwise.  
Using the normal approximation to the binomial (with the continuity
correction) we can approximate this by 
$P(W \le 155.5)$ where $W ~ N(160\cdot .95, \sqrt{160\cdot(.95)(.05)}) = N(152,2.76)$.  
so 
\begin{align*}
P(Y \le 155) &\doteq P(W \le 155.5)\\
&= P\left(Z \le \frac{155.5 - 152}{2.76}\right)\\
&\doteq P(Z \le 1.27)\\
&=1 - .102\\
&=.898.
\end{align*}
\vskip 10pt
\noindent
7.74.  (a)  Letting $Y$ be the number of lawyers in the sample (binomial
with $n = 1500$, $p \doteq .00244$), 
$$P(Y = 0) = \binom{1500}{0}(.00244)^0(.99756)^{1500} \doteq .04.$$
(b)   $Y = $ the number of lawyers in the sample from Washington
is binomial with $n = 1500$, $p \doteq .01563$.  So
$Y$ has an approximate $N(23.44, 4.804)$ distribution.
So 
$$P(Y > 30) \doteq P\left(Z > \frac{30.5 - 23.44}{4.804}\right) = P(Z > 1.47) = .0708.$$
(c)  With $n = 1000$ and $p = .01563$, the expected number would be $15.63$ lawyers
out of the $1000$ and the standard deviation would be 
$\sqrt{1000(.01563)(.98437)} \doteq 3.922$.  So
it would be extremely unlikely to get as many as 30 lawyers
in the sample if the density was the same $p = 1/64$ as the whole city, since
30 is only slightly less than 4 standard deviations above the mean.    
\end{document}
