\magnification=\magstep1
\centerline{Mathematics 375 -- Probability and Statistics I}
\centerline{Solutions -- Problem Set 8}
\centerline{November 4, 2005}
\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
4.110/  (Refer to the general formula for the mgf
of a uniform random variable given inside the back cover of the text,
and the result derived in Exercise 4.105.)
a)  $m_Y(t) = {e^t - 1\over t}$.
b)  With $W = aY$,  $m_W(t) = m_{Y}(at) = {e^{at} - 1\over at}$.
By uniqueness, this is the mgf for a uniformly distributed
random variable on $(0,a)$.
c)  Similarly,
$X = -aY \Rightarrow m_X(t) = m_{Y}(-at) = {e^{-at} - 1\over -at} = 
{1 - e^{-at}\over at}$.  This is the mgf for a uniformly distributed
random variable on $(-a,0)$.
d)  $U = aY + b \Rightarrow m_U(t) = e^{bt} m_{Y}(at) = 
e^{bt}\cdot {e^{at} - 1\over at} = {e^{(a+b)t} - e^{bt}\over at}$.
This is the mgf for a uniformly distributed
random variable on $(b,a+b)$.

\bigskip
\noindent
4.115/ 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.  

\bigskip
\noindent
4.120/  From 4.87, we have $\mu = E(L) = 276$, $\sigma = \sqrt{47664}
\doteq 218.32$.  In Tchebysheff, the value of $k$ for which
$1 - 1/k^2 = .89$ is $k = 3.015$.  Then
$(\mu - 3.015\sigma, \mu + 3.015\sigma)$
is the interval $(-382.2, 934.24)$.  
(Since $L$ is the loss due to downtime of the machine,
negative values of loss mean positive profits.)

\bigskip
\noindent
4.122/ a) Recall that the $\chi^2$ distribution with 
$\nu$ degrees of freedom is another name for the
Gamma distribution with $\alpha = \nu/2$ and $\beta = 2$.
(We will be using these a lot next semester, because
they arise in understanding variances of samples from 
a normal distribution.)  From our general formulas $E(Y) = 7$ 
and $V(Y) = 14$ ($\sigma = \sqrt{14} \doteq 3.742$). b)  We could compute the corresponding 
probability numerically.  For a ``quick and dirty''
estimate, though, we can also apply Tchebysheff:
Note $23 = \mu + 4.3\sigma$.  
$P(Y > \mu + 4.3\sigma) \le P(|Y - \mu| > 4.3\sigma) \le {1\over (4.3)^2}
\doteq .054$, which is quite small.  And of course, this
also includes $P(Y < \mu - 4.3\sigma)$.) So it is unlikely 
to find such a large value for $Y$.  

\bigskip
\noindent
5.3/  If $Y_1$ is the number married, and $Y_2$ is the number
never married, then the joint probability function is
$$P(Y_1 = y_1, Y_2 = y_2) = {{4\choose y_1}{3\choose y_2}{2\choose 3-y_1-y_2}\over
{9\choose 3}}$$

\bigskip
\noindent
5.6/  a)  We must have $1 = \int_0^1\int_0^1 ky_1y_2\ dy_2\ dy_1 = k/4$.
so $k = 4$.
b)  The joint cdf is 
$$F(y_1,y_2) = P(Y_1 \le y_1, Y_2 \le y_2) =
\cases{ y_1^2y_2^2 & if $0\le y_1, y_2 \le 1$\cr
          y_2^2 & if $y_1 \ge 1, 0\le y_2 \le 1$\cr
          y_1^2 & if $y_2 \ge 1, 0\le y_1 \le 1$\cr
          1 & if $y_1,y_2 \ge 1$\cr
          0 & otherwise.\cr}$$
c)  $P(Y_1\le 1/2, Y_2\le 3/4) = \int_0^{1/2}\int_0^{3/4} 4y_1y_2\ dy_2\ dy_1
= (1/2)^2 (3/4)^2 = 9/64$.
          
           

\bigskip
\noindent
5.9/ Since $Y_1,Y_2$ are uniformly distributed on this
triangle of area 1, the probability that $(Y_1,Y_2)$ lies 
in any region in the triangle is just the {\it area} of that region.
(Careful -- this is definitely false when the distribution 
is not uniform!)
a) $P(Y_1\le 3/4, Y_2\le 3/4$  is the area of the 
remaining region when we ``cut off'' two small
triangles of areas $1/32$ and $1/16$ from the big triangle.
That leaves an area of $1 - 3/32 = 29/32$.
b)  $P(Y_1 - Y_2 \ge 0)$.  $Y_1 \ge Y_2$ below the line
$y_1 = y_2$ in the plane.  This gives the triangle
with vertices $(0,0), (1,0), (1/2,1/2)$, whose area
is $1/4$.

\bigskip
\noindent
5.12/ Given $f(y_1,y_2) = \cases{6y_1^2y_2 & if $0 \le y_1 \le y_2,
0 \le y_1 + y_2 \le 2$\cr 0 & otherwise}$.  The region 
where the density is nonzero is the interior of the triangle $T$
with vertices $(0,0), (1,1), (0,2)$.  
a)  We have $f(y_1,y_2) \ge 0$ for all $y_1,y_2$, so the thing
that needs to be checked is that the total integral is $1$.
$$\eqalign{
\int\int_T f(y_1,y_2)\ dA &=
\int_0^1\int_{y_1}^{2-y_1} 6y_1^2y_2\ dy_2\ dy_1\cr
&= \int_0^1 3y_1^2((2-y_1)^2 - y_1^2)\ dy_1\cr
&= \int_0^1 12y_1^2 - 12y_1^3\ dy_1\cr
&= (4y_1^3 - 3y_1^4|_0^1 \cr
&= 1\cr}$$
as required for a density function.

b)  Note that $Y_1 + Y_2 = 1$ defines a line that cuts through
the triangle $T$ from part a.  The region we want is the 
part of $T$ that lies below that line:
$$\eqalign{
P(Y_1 + Y_2 \le 1) &= \int_0^{1/2}\int_{y_1}^{1-y_1} 6y_1^2y_2\ dy_2\ dy_1\cr
&= \int_0^{1/2} 3y_1^2((1-y_1)^2 - y_1^2)\ dy_1\cr
&= \int_0^{1/2} 3y_1^2 - 6y_1^3\ dy_1\cr
&= (y_1^3 - (3/2)y_1^4|_0^{1/2} \cr
&=1/32.\cr}$$

\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

