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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions for Problem Set 8}
\centerline{November 16, 2009}
\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
4.139.   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
with mean $\mu_X = 4 - 3\mu_Y$ and variance $\sigma_X^2 = 9\sigma_Y^2$
\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
4.147. 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)  $F(1,2) = 1$.  This is sum of all of the entries in 
the table, since $F(1,2) = P(Y_1 \le 1, Y_2 \le 2)$.
\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)  $P(Y_1\le 1/2, Y_2\le 3/4) = \int_0^{1/2}\int_0^{3/4} 4y_1y_2\ dy_2\ dy_1
= (1/2)^2 (3/4)^2 = 9/64$.

5.11. 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 $1/4$.
\vskip 10pt
\noindent
5.14.  Given: $f(y_1,y_2) = 6y_1^2y_2$ if $0 \le y_1 \le y_2,
0 \le y_1 + y_2 \le 2$ and  $0$ 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 $1$.
\begin{align*}
\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\\
&= \int_0^1 3y_1^2((2-y_1)^2 - y_1^2)\ dy_1\\
&= \int_0^1 12y_1^2 - 12y_1^3\ dy_1\\
&= \left. 4y_1^3 - 3y_1^4\right|_0^1 \\
&= 1
\end{align*}
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:
\begin{align*}
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\\
&= \int_0^{1/2} 3y_1^2((1-y_1)^2 - y_1^2)\ dy_1\\
&= \int_0^{1/2} 3y_1^2 - 6y_1^3\ dy_1\\
&= \left. y_1^3 - (3/2)y_1^4\right|_0^{1/2} \\
&=1/32.
\end{align*}
\vskip 10pt
\noindent
5.15.  The region where the joint density is nonzero is an
infinite wedge-shaped region in the first quadrant bounded
by the $y_1$-axis ($y_2 = 0$) and the line $y_2 = y_1$.  

(a) The inequalities $Y_1 < 2$ and $Y_2 > 1$ define a 
small triangle within the wedge, with vertices at
$(1,1),(2,1),(2,2)$.  To get the desired probability here
it is somewhat simpler to integrate first with respect to 
$y_1$ (doing the other order leads to an integral by parts):
\begin{align*}
P(Y_1 < 2, Y_2 > 1) &= \int_1^2 \int_{y_2}^2 e^{-y_1}\ dy_1\ dy_2 \\ 
&= \int_1^2 e^{-y_2} - e^{-2}\ dy_2\\
&= e^{-1}-2e^{-2}\\
&\doteq .0972.
\end{align*}

(b)  The inequality $Y_1 \ge 2 Y_2$ defines the lower half of the 
wedge-shaped region -- that is, the region below the line
$Y_2 = Y_1/2$.  Again, we set up the integrals to integrate
with respedt to $y_1$ first:
\begin{align*}
P(Y_1 \ge 2Y_2) &= \int_0^\infty \int_{2y_2}^\infty e^{-y_1}\ dy_1\ dy_2\\
&= \int_0^\infty e^{-2y_2}\ dy_2\\
&= \left.\frac{-e^{-2y_2}}{2}\right|_0^\infty\\
&= \frac{1}{2}.
\end{align*}

(c)  The inequality $Y_1 - Y_2 \ge 1$ defines the portion of the 
wedge-shaped region below the line $Y_2 = Y_1 - 1$.  This is 
a line parallel to the upper edge of the wedge, passing through $(1,0)$
on the $Y_1$-axis.  The probability we want is
\begin{align*}
P(Y_1 - Y_2 \ge 1) &= \int_0^\infty \int_{y_2+1}^\infty e^{-y_1}\ dy_1\ dy_2\\
&= \int_0^\infty e^{-y_2 - 1}\ dy_2\\
&= \left. -e^{-y_2 - 1}\right|_0^\infty\\
&= e^{-1}\\
&\doteq .3679.
\end{align*}
\vskip 10pt
\noindent
{\it Additional Problem}
\vskip 10pt
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),
\begin{align}
\label{eq1}
\Gamma(p)\Gamma(q) 
&= \int_0^\infty s^{p-1} e^{-s}\ ds\cdot \int_0^\infty t^{q-1} e^{-t}\ dt\nonumber\\
&=\int_0^\infty s^{p-1} e^{-s}\cdot \int_0^\infty t^{q-1} e^{-t}\ dt\ ds\\
&=\int_0^\infty \int_0^\infty s^{p-1} t^{q-1} e^{-(s+t)}\ dt\ ds.\nonumber
\end{align}
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
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 (\ref{eq1}) is 
\begin{equation}
\label{eq2}
\int_0^\infty \int_0^u s^{p-1} (u-s)^{q-1} e^{-u}\ ds\ du.
\end{equation}

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 (\ref{eq2}) becomes:
\begin{equation}
\label{eq3}
\int_0^\infty \int_0^1 (\lambda u)^{p-1} (u-\lambda u)^{q-1} e^{-u} u\ d\lambda\ du.
\end{equation}

Then, working with the integrand in (\ref{eq3}), we can simplify,
then split the result back up into two one-variable integrals
(reversing the process we did above in the first step):
\begin{align*}
&\int_0^\infty \int_0^1 (\lambda u)^{p-1} (u-\lambda u)^{q-1} e^{-u} u\ d\lambda\ du\\
&= \int_0^\infty \int_0^1 \lambda^{p-1} (1-\lambda)^{q-1} u^{p+q-1} e^{-u} d\lambda\ du\\
&=\int_0^1 \lambda^{p-1} (1-\lambda)^{q-1}\ d\lambda\cdot \int_0^\infty u^{p+q-1} e^{-u}\ du\\
&=B(p,q)\Gamma(p+q)
\end{align*}
(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!!)

\end{document}
