\magnification=\magstep1
\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions -- Problem Set 8}
\centerline{November 14, 2003}
\bigskip
\noindent
4.104/a) By inspection, this is an exponential distribution 
with parameter $\beta = \theta$, so by Example 4.13 in the text
with $\alpha = 1$, $\beta = \theta$,
$$m(t) = {1\over 1 - \theta t}$$ 
(for $t < {1\over \theta}$).  This could also be done, of course,
by computing 
$$m(t) = \int_0^\infty e^{ty} \cdot e^{-y/\theta}\ dy$$
directly.

b)  We have 
$$E(Y) = m'(0) = {\theta\over (1 - \theta t)^2}|_{t=0} = \theta$$
Similarly,
$$E(Y^2) = m''(0) = {2\theta^2\over (1 - \theta t)^3}|_{t=0} = 2\theta^2,$$
so $V(Y) = E(Y^2) - (E(Y))^2 = 2\theta^2 - \theta^2 = \theta^2$.
\bigskip
\noindent
4.105/ a) 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 for $f(y)$:
$$\eqalign{m_U(t) 
&= \int_{-\infty}^\infty e^{t(ay + b)} f(y)\ dy \cr
&=\int_{-\infty}^\infty e^{(at)y}\cdot e^{tb} f(y)\ dy\cr
&=e^{tb}\int_{-\infty}^\infty e^{(at)y} f(y)\ dy\cr
&=e^{tb}m_Y(at)\cr}$$

b)  Then to find $E(U)$ and $V(U)$, we compute as usual, 
using the product and chain rules (note:  ${d\over dt}m_Y(at) = m_Y'(at)\cdot a$)  to find the necessary 
derivatives:
$$\eqalign{E(U) &= m_U'(0)\cr
&= {d\over dt}(e^{tb}m_Y(at))|_{t=0}\cr
&= (be^{tb}m_Y(at) + e^{tb}m_Y'(at)a)|_{t=0}\cr
&= bm_Y(0) + am_Y'(0)\cr
&=b + a\mu\cr}$$
since $m_Y(0) = 1$ and $m_Y'(0) = E(Y) = \mu$.  Then
$$\eqalign{E(U^2) &= m_U''(0)\cr
&= {d\over dt}(be^{tb}m_Y(at) + e^{tb}m_Y'(at)a)|_{t=0}\cr
&= (b^2e^{tb}m_Y(at) + 2b e^{tb}m_Y'(at)a + e^tb m_Y''(t)a^2)|_{t=0} \cr
&= b^2 + 2ab\mu + a^2 E(Y^2)\cr}$$
Hence 
$$V(U) = b^2 + 2ab\mu + a^2 E(Y^2) - (b + a\mu)^2 = 
a^2(E(Y^2) - \mu^2) = a^2\sigma^2$$
\bigskip
\noindent
4.107/  Using the formula of 4.106 and the moment-generating
function for a normal random variable, we have
$$m_X(t) = e^{4t}m_Y(-3t) = e^{4t}\cdot e^{{\sigma^2(-3t)^2\over 2} + \mu (-3t)}
= e^{{(3\sigma)^2 t^2\over 2} + (4 - 3\mu)t}$$
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.
\bigskip
\noindent
4.108/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 $\sigma^2 = 12$.
\bigskip
\noindent
4.109/This computation was also done in class.  For a uniformly
distributed $Y$ on the interval $(\theta_1,\theta_2)$, 
$$\eqalign{m(t) &= \int_{\theta_1}^{\theta_2} e^{ty} {1\over \theta_2 - \theta_1}\ dy\cr
&= {e^{\theta_2 t} - e^{\theta_1 t} \over (\theta_2 - \theta_1) t}\cr}$$
\bigskip
\noindent
4.113/Note that this is a ``backwards'' exponential density (nonzero
for negative $y$).  Part a) is a ``warm-up'' for computing the 
moment generating function (it's the computation of the value
$m(3/2)$):
$$\eqalign{
E(e^{3Y/2}) &= \int_{-\infty}^0 e^{3y/2}\cdot e^y\ dy\cr
&= \int_{-\infty}^0 e^{5y/2}\ dy\cr
&= {2\over 5} e^{5y/2} |_{-\infty}^0\cr
&= {2\over 5}\cr}$$
For part b), we do the general computation in exactly the same
way, noting that $t$ is constant with respect to the integration
variable $y$:
$$\eqalign{m(t) &= \int_{-\infty}^0 e^{ty}\cdot e^y\ dy\cr
&= \int_{-\infty}^0 e^{(1+t)y}\ dy\cr
&= {1\over 1+t} e^{(1+t)y} |_{-\infty}^0\cr
&= {1\over 1+t}\cr}$$
For c), $m'(t) = {-1\over (1+t)^2}$, so $m'(0) = E(Y) = 1$.
Then, $m''(t) = {2\over (1+t)^3}$, so $m''(0) = E(Y^2) = 2$.
Hence $V(Y) = E(Y^2) - (E(Y))^2 = 2 - 1 = 1$.
\bigskip
\noindent
4.114/Use Tchebysheff's Theorem:
$$P(|Y - \mu| \le k\sigma) \ge 1 - {1\over k^2}$$
To get an interval $(\mu - k\sigma,\mu + k\sigma)$ that is guaranteed to
{\it include at least 90\%} of the tire lifetimes, we 
make $1 - {1\over k^2} = .9$, or $k = \sqrt{10} \doteq 3.162$.
This gives 
$$(\mu - k\sigma,\mu + k\sigma) = (25000 - (3.162)(4000),25000 + (3.162)(4000))
= (12352,37648)$$
\bigskip
\noindent
4.116/To apply Tchebysheff's Theorem, we need to know
$\mu$ and $\sigma$, so we begin by computing those values.
With the given density function,
$$\eqalign{\mu = E(Y) &= \int_0^2 y\cdot {1\over 2}(2-y)\ dy\cr
&= \left({y^2\over 2} - {y^3\over 6}\right|_0^2\cr
&= {2\over 3}\cr}$$
Then 
$$\eqalign{
E(Y^2) &= \int_0^2 y^2\cdot {1\over 2}(2-y)\ dy\cr
&= \left({y^3\over 3} - {y^4\over 8}\right|_0^2\cr
&= {2\over 3}\cr}$$
So $\sigma = \sqrt{(2/3) - (2/3)^2} = \sqrt{2}/3$.  In this 
case, $\mu - 2\sigma = (2 - 2\sqrt{2})/3 < 0$, so 
$$P(|Y-\mu| < 2\sigma) = \int_0^{(2+2\sqrt{2})/3} {1\over 2}(2-y)\ dy \doteq .96$$
Compared with the empirical rule value $.95$ and the 
Tchebysheff lower bound $1 - {1\over 2^2} = .75$, we see that 
the empirical rule is much closer.
\bigskip
\noindent
4.117/From previous work we know that the mean and standard 
deviation for the uniform distribution on $(a,b)$ are 
$\mu = {a + b \over 2}$ (the midpoint of the interval), 
and $\sigma = {b - a\over 2\sqrt{3}}$, which equals
${1\over \sqrt{3}}$ times the distance from the midpoint
to either endpoint.  In this case 
$$\mu + 2\sigma = \left({a+b\over 2}\right) + 2\cdot \left({b - a\over 2\sqrt{3}}\right)
= \left({a+b\over 2}\right) + {2\over \sqrt{3}}\cdot \left({b - a\over 2}\right)$$
Since $2 > \sqrt{3}$, this takes us past the right endpoint 
of the interval.  Similarly, $\mu - 2\sigma$ takes us past
the left endpoint.  But the uniform density is only nonzero
between $a$ and $b$, so 
$$P(|Y - \mu| \le 2\sigma) = \int_a^b {1\over b - a}\ dy = 1$$
Compared with the empirical rule value $.95$ and the 
Tchebysheff lower bound $1 - {1\over 2^2} = .75$, we again see that 
the empirical rule is much closer, although it is actually 
an underestimate in this case(!).
\bigskip
\noindent
4.118/From previous work, we know that the exponential distribution
with parameter $\beta$ has mean $\mu = \beta$, and standard
deviation $\sigma = \beta$.  The interval
$(\mu - 2\sigma, \mu + 2\sigma) = (-\beta,3\beta)$ extends past the left endpoint
of the interval where the density is nonzero.  Hence
$$P(|Y - \mu| \le 2\sigma) = \int_0^{3\beta} e^{-y/\beta} {1\over \beta}\ dy
= 1 - e^{-3} \doteq .9502$$
Compared with the empirical rule value $.95$ and the 
Tchebysheff lower bound $1 - {1\over 2^2} = .75$, we again see that 
the empirical rule is much closer.
\bigskip
\noindent
Additional Problem/By the definition of the $\Gamma$ function
and the fact that the two integration variables are independent 
of each other (i.e. $s$ is a constant with respect to $t$ and 
vice versa),
$$\eqalign{\Gamma(p)\Gamma(q) 
&= \int_0^\infty s^{p-1} e^{-s}\ ds\cdot \int_0^\infty t^{q-1} e^{-t}\ dt\cr
&=\int_0^\infty s^{p-1} e^{-s}\cdot \int_0^\infty t^{q-1} e^{-t}\ dt\ ds\cr
&=\int_0^\infty \int_0^\infty s^{p-1} t^{q-1} e^{-(s+t)}\ dt\ ds\cr}
\leqno(1)$$
The region of integration here is the entire first quadrant in the 
$(s,t)$-plane ($s$= horizontal axis, $t$= vertical axis).

Now we do the first change of variables, letting 
$u = s+t$ and putting $u$ on the horizontal axis, $s$ on the 
vertical axis in the new coordinate system.  The $s$-axis
in the old coordinate system maps to the line through $(1,1)$ in
the new coordinates.  The $t$-axis in the old coordinates maps
to the $u$ axis in the new coordinates:
$$\pmatrix{1&1\cr 1&0}\cdot \pmatrix{ 1\cr 0\cr} = \pmatrix{1\cr 1\cr}  \qquad
\pmatrix{1&1\cr 1&0}\cdot \pmatrix{ 0\cr 1\cr} = \pmatrix{1\cr 0\cr}$$
and the first quadrant in the $(s,t)$ plane maps to the ``wedge''
bounded by $s = u$ and the $u$-axis in the new coordinates.
This means that the new form of the double integral (1) is 
$$\int_0^\infty \int_0^u s^{p-1} (u-s)^{q-1} e^{-u}\ ds\ du\leqno(2)$$

Next, we perform the second change of coordinates, letting
$\lambda = s/u$, $u = u$.  Note that the value of $\lambda$ at 
any point is the {\it slope of the line through that point and 
the origin}.  Hence the wedge corresponds to the infinite
``rectangle'' $0 \le \lambda \le 1$, $0 \le u \le \infty$ 
in the new coordinates, and with the given information
about how the area element transforms, we get that (2) becomes:
$$\int_0^\infty \int_0^1 (\lambda u)^{p-1} (u-\lambda u)^{q-1} e^{-u} u d\lambda\ du\leqno(3)$$

Then, working with the integrand in (3), we can simplify,
then split the result back up into two one-variable integrals
(reversing the process we did above in the first step):
$$\eqalign{&\int_0^\infty \int_0^1 \lambda^{p-1} (1-\lambda)^{q-1} u^{p+q-1} e^{-u} d\lambda\ du\cr
&= \int_0^\infty \int_0^1 \lambda^{p-1} (1-\lambda)^{q-1} u^{p+q-1} e^{-u} d\lambda\ du\cr
&=\int_0^1 \lambda^{p-1} (1-\lambda)^{q-1}\ d\lambda\cdot \int_0^\infty u^{p+q-1} e^{-u}\ du\cr
&=B(p,q)\Gamma(p+q)\cr}$$
(by the definitions of the beta and gamma functions).  We have shown that
$$\Gamma(p)\Gamma(q) = B(p,q)\Gamma(p+q)$$
from which we get the desired equality by dividing both sides
by $\Gamma(p+q)$.  (Again, very cool!!)
\bye

