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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions for Problem Set 8}
\centerline{November 5, 2009}
\vskip 10pt
\noindent
4.129.  Instead of computing the expected values directly 
from the integral formulas, we can use the general results
for $\beta$-distributed random variables from Theorem 4.11.
We have 
$$E(C) = E(10 + 20Y + 4Y^2) = 10 + 20E(Y) + 4E(Y^2).$$
From Theorem 4.11, we have $E(Y) = \frac{\alpha}{\alpha+\beta} = \frac{1}{3}$.
Then 
$$
E(Y^2) = V(Y) + E(Y)^2 = \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)} +
\frac{1}{9} = \frac{1}{18} + \frac{1}{9} = \frac{1}{6}.
$$
So $E(C) = 10 + \frac{20}{3} + \frac{4}{6} = \frac{52}{3}$.
To get $V(C) = E(C^2) - E(C)^2$, we can proceed as follows:
$E(C^2) = 100 + 400E(Y) + 480 E(Y^2) + 160 E(Y^3) + 16 E(Y^4)$.
We already know $E(Y),E(Y^2)$ from the above.  To get $E(Y^3)$
and $E(Y^4)$ proceed like this:
$$
E(Y^3) = 2\int_0^1 y^3 (1 - y)\ dy.
$$
Apart from the constant multiple, this is the integral of a 
beta density function with $\alpha = 4$ and $\beta = 2$.  
So 
$$
E(Y^3) = 2B(4,2) = 2\frac{\Gamma(4)\Gamma(2)}{\Gamma(6)} = 
2\frac{3!1!}{5!} = \frac{1}{10}.
$$
Similarly,
$$
E(Y^4) = 2B(5,2) = 2\frac{\Gamma(5)\Gamma(2)}{\Gamma(7)} = 
2\frac{4!1!}{6!} = \frac{1}{15}.
$$
So then
$$
V(C) = 10 + \frac{400}{3} + \frac{480}{6} + \frac{160}{10} + \frac{16}{15} - 
\left(\frac{52}{3}\right)^2 = \frac{1348}{45}.
$$
\vskip 10pt
\noindent
4.130. We use the same ``trick'' as in the last problem. 
We know $E(Y) = \frac{\alpha}{\alpha+\beta}$ from Theorem 4.11
and work in class. 
\begin{align*}
E(Y^2) &= \frac{1}{B(\alpha,\beta)} \int y^2\cdot y^{\alpha-1} (1 - y)^{\beta - 1}\ dy\\
&= \frac{1}{B(\alpha,\beta)} \int y^{\alpha+1} (1 - y)^{\beta - 1}\ dy\\
&= \frac{B(\alpha+2,\beta)}{B(\alpha,\beta)}\\
&= \frac{\Gamma(\alpha+2)\Gamma(\beta) \Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta) \Gamma(\alpha+\beta+2)}.
\end{align*}
 Then using the functional equation for
the $\Gamma$-function
\begin{align*}
&=\frac{(\alpha+1)\alpha\Gamma(\alpha)\Gamma(\beta) \Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta) (\alpha+\beta+1)(\alpha+\beta)\Gamma(\alpha+\beta)}\\
&=\frac{(\alpha+1)\alpha}{(\alpha+\beta+1)(\alpha+\beta)}.
\end{align*}
So then
\begin{align*}
V(Y) &= \frac{(\alpha+1)\alpha}{(\alpha+\beta+1)(\alpha+\beta)} - 
\left(\frac{\alpha}{\alpha+\beta}\right)^2\\
&= \frac{(\alpha + \beta)(\alpha + 1)(\alpha) - (\alpha + \beta + 1)(\alpha^2)}{(\alpha+\beta)^2(\alpha + \beta + 1)}\\
&= \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha + \beta + 1)},
\end{align*}
which is what we wanted to show.  
\vskip 10pt
\noindent
4.133.  (a)  This is the form for a beta density function with
$\alpha = 3$ and $\beta = 5$, so $c = \frac{1}{B(3,5)} = \frac{7!}{2!4!} = 105$.

(b)  $E(Y) = \frac{3}{8} = .375 $  from Theorem 4.11.

(c)  The standard deviation is 
$$\sigma = \sqrt{\frac{15}{64\cdot 9}} \doteq .1614.$$

(d)  We have:
$$P(Y > .375 + 2(.1614)) = \int_{.6978}^1 105 y^2 (1-y)^4\ dy \doteq .02972.$$
\vskip 10pt
\noindent
4.136. (a) $Y$ has an exponential distribution with $\beta =\theta$.
So by Example 4.13 with $\alpha = 1$ and $\beta = \theta$,
the moment generating function is $m_Y(t) = \frac{1}{1 - \theta t}$.

(b) Then 
$$E(Y) = m_Y'(0) = \left.\frac{\theta}{(1 - \theta t)^2}\right|_{t=0} = \theta$$
and 
$$E(Y) = m_Y''(0) = \left.\frac{2\theta^2}{(1 - \theta t)^3}\right|_{t=0} = 2\theta^2$$
Hence $V(Y) = E(Y^2) - E(Y)^2 = 2\theta^2 - \theta^2 = \theta^2$.   
\vskip 10pt
\noindent
4.137.  Letting $U = aY + b$, since $U$ is a function of $Y$,
we find its moment-generating function by integrating
$e^{t(ay + b)}$ times the density function $f(y)$:
\begin{align*}
m_U(t) 
&= \int_{-\infty}^\infty e^{t(ay + b)} f(y)\ dy \\
&=\int_{-\infty}^\infty e^{(at)y}\cdot e^{tb} f(y)\ dy\\
&=e^{tb}\int_{-\infty}^\infty e^{(at)y} f(y)\ dy\\
&=e^{tb}m_Y(at)
\end{align*}

Then to find $E(U)$ and $V(U)$, we compute as usual, 
using the product and chain rules (note:  $\frac{d}{dt}m_Y(at) = m_Y'(at)\cdot a$)  to find the necessary 
derivatives:
\begin{align*}
E(U) &= m_U'(0)\\
&= \left.\frac{d}{dt}(e^{tb}m_Y(at))\right|_{t=0}\\
&= \left.(be^{tb}m_Y(at) + e^{tb}m_Y'(at)a)\right|_{t=0}\\
&= bm_Y(0) + am_Y'(0)\\
&=b + a\mu,
\end{align*}
since $m_Y(0) = 1$ and $m_Y'(0) = E(Y) = \mu$.  Then
\begin{align*}
E(U^2) &= m_U''(0)\\
&= \left.\frac{d}{dt}(be^{tb}m_Y(at) + e^{tb}m_Y'(at)a)\right|_{t=0}\\
&= \left.(b^2e^{tb}m_Y(at) + 2b e^{tb}m_Y'(at)a + e^tb m_Y''(t)a^2)\right|_{t=0}\\
&= b^2 + 2ab\mu + a^2 E(Y^2).
\end{align*}
Hence 
$$V(U) = b^2 + 2ab\mu + a^2 E(Y^2) - (b + a\mu)^2 = 
a^2(E(Y^2) - \mu^2) = a^2\sigma^2.$$
\vskip 10pt
\noindent
4.139.   Using the formula of Exercise 4.138 and the moment-generating
function for a normal random variable, we have
$$m_X(t) = e^{4t}m_Y(-3t) = e^{4t}\cdot e^{\frac{\sigma^2(-3t)^2}{2} + \mu (-3t)}
= e^{\frac{(3\sigma)^2 t^2}{2} + (4 - 3\mu)t}$$
This is the moment-generating function for a normal random variable
with mean $4 - 3\mu$ and variance $9\sigma^2$.  Hence by 
the uniqueness theorem we quoted in class, $X$ is {\it normally} distributed.
\vskip 10pt
\noindent
4.140.  Using the standard forms (for instance from the table inside
the back cover of the textbook!),  (a)  is  the moment-generating
function for a gamma-distributed random variable with 
$\alpha = 2, \beta = 4$.  (b) is  the moment-generating
function for a gamma-distributed random variable with 
$\alpha = 1, \beta = 3.2$.  Recall that the gamma-distributed 
variables with $\alpha = 1$ are also called {\it exponential}.
(c)  is normal with $\mu = -5$ and $\sigma^2 = 12$.
\vskip 10pt
\noindent
4.141. The moment generating function is 
\begin{align*}
m_Y(t) &= E(e^{ty})\\
&= \int_{\theta_1}^{\theta_2} \frac{e^{ty}}{\theta_2 - \theta_1}\ dy\\
&= \frac{e^{t\theta_2} - e^{t\theta_1}}{t(\theta_2 - \theta_1)}.
\end{align*}
\vskip 10pt
\noindent
4.147. By Tchebysheff's Theorem, 
$P(|Y - \mu| \le k\sigma) \ge 1 - 1/k^2$.  To 
get $1 - 1/k^2 = .75$, we take $k = 2$.  
$2\sigma = 1$oz implies $\sigma = .5$oz. 
\vskip 10pt
\noindent
4.149.  We can shift and scale to consider a uniform random variable on $[0,1]$,
where we know $\mu = \frac{1}{2}$ and 
$\sigma^2 = \frac{1}{12}$.  
Hence $\mu \pm 2\sigma = \frac{1}{2} \pm {1}{\sqrt{3}}$.
Since $\frac{1}{\sqrt{3}} > \frac{1}{2}$, 
$\mu + 2\sigma > 1$ and $\mu - 2\sigma < 0$.  
Hence
$P(|Y - \mu| \le 2\sigma) = 1$.
This compares with $P(|Y - \mu| \le 2\sigma) \ge \frac{3}{4}$
by Tchebysheff.  It is greater than the
$P(|Y - \mu| \le 2\sigma) \doteq .95$ according to 
the empirical rule.
\vskip 10pt
\noindent
4.150.  For an exponential random variable, $\mu = \beta$ and
$\sigma = \beta$, so $P(|Y - \beta| \le 2\beta) = P(-\beta < Y < 3\beta) = 
P(0 < Y < 3\beta)$.  This is 
\begin{align*}
\int_0^{3\beta} \frac{1}{\beta} e^{\frac{-y}{\beta}}\ dy
&= 1 - e^{-3}\\
&\doteq .9502.
\end{align*}
This compares with $P(|Y - \mu| \le 2\sigma) \ge \frac{3}{4}$
by Tchebysheff, and is almost exactly the same as the 
$P(|Y - \mu| \le 2\sigma) \doteq .95$ predicted by the empirical
rule.
\vskip 10pt
\noindent
5.4.   (a) is clear.  

(b)  $F(1,2) = .05$.  This is the probability that the child
did not survive if a car-set was used.
\vskip 10pt
\noindent
5.6.   (a)  The region described by the inequality 
$Y_1 - Y_2 > .5$ is the region in the unit square
below the line $Y_2 = Y_1 - .5$, which 
intersects the edges at $(Y_1,Y_2) = (.5,0)$ and
$(1,.5)$.  The given probability is 
$$\int_{.5}^1\int_0^{y_1 - .5} \ dy_2\ dy_1 = \left. \frac{(y_1 - .5)^2}{2}\right|_{.5}^1
= .125.$$

(b)  The region is now the part of the unit square
underneath the hyperbola $Y_1Y_2 = .5$.  Note
that to set up the double integrals, we must split
the region of integration in this case.
\begin{align*}
P(Y_1Y_2 < .5) &= \int_0^{.5} \int_0^1 \ dy_2\ dy_1  + \int_{.5}^1 \int_0^{.5/y_1}\ dy_2\ dy_1\\
&= .5 + \left. .5\ln(y_1)\right|_{.5}^1\\
&= .5(1 - \ln(.5))\\
&\doteq .8465.
\end{align*}
\vskip 10pt
\noindent
5.7. The density is only nonzero on the first quadrant of the $(y_1,y_2)$-plane.
(a)  The inequalities $Y_1 < 1$, $Y_2 > 5$ describe an infinite vertical strip.
\begin{align*}
P(Y_1 < 1, Y_2 > 5) &= \int_0^1 \int_5^\infty e^{-y_1 - y_2}\ dy_2\ dy_1\\
&=\int_0^1 \left. -e^{-y_1 - y_2}\right|_{y_2=5}^{y_2=\infty}\ dy_1\\
&=\int_0^1 e^{-y_1 - 5}\ dy_1\\
&= \left. -e^{-y_1 - 5}\right|_0^1\\
&= e^{-5} - e^{-6}\\
&\doteq .004259.
\end{align*}

(b)  The inequality $Y_1 + Y_2 < 3$ describes the region in the the
first quadrant below the line $Y_2 = -Y_1 + 3$:
\begin{align*}
P(Y_1 + Y_2 < 3) &= \int_0^3 \int_0^{3 - y_1} e^{-y_1 - y_2}\ dy_2\ dy_1\\
&= \int_0^3 e^{-y_1} - e^{-3}\ dy_1\\
&= 1 - 4e^{-3}\\
&\doteq .8009.
\end{align*}
\vskip 10pt
\noindent
5.8.  (a)  We must have 
$1 = \int_0^1 \int_0^1 ky_1 y_2\ dy_2\ dy_1 = \frac{k}{4}$.
So $k = 4$.

(b)  b)  The joint cdf is 
$$F(y_1,y_2) = P(Y_1 \le y_1, Y_2 \le y_2) =
\begin{cases} y_1^2y_2^2 & \text{ if } 0\le y_1, y_2 \le 1 \\
          y_2^2 & \text{ if }  y_1 \ge 1, 0\le y_2 \le 1 \\
          y_1^2 & \text{ if }  y_2 \ge 1, 0\le y_1 \le 1 \\
          1 & \text{ if } y_1,y_2 \ge 1\\
          0 & \text{ otherwise. }
\end{cases}
$$
c)  $P(Y_1\le 1/2, Y_2\le 3/4) = \int_0^{1/2}\int_0^{3/4} 4y_1y_2\ dy_2\ dy_1
= (1/2)^2 (3/4)^2 = 9/64$.


\end{document}
