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\begin{document}
\centerline{Mathematics 375 -- Probability Theory}
\centerline{Solutions for Problem Set 8}
\centerline{November 11, 2011}
\vskip 10pt
\noindent
5.1 (a)  Since the assignments are random, the probabilities are
\begin{align*}
P(Y_1 = 0, Y_2 = 0) &= \frac{1}{9}\\
P(Y_1 = 0, Y_2 = 1) &= \frac{2}{9}\\
P(Y_1 = 0, Y_2 = 2) &= \frac{1}{9}\\
P(Y_1 = 1, Y_2 = 0) &= \frac{2}{9}\\
P(Y_1 = 1, Y_2 = 1) &= \frac{2}{9}\\
P(Y_1 = 1, Y_2 = 2) &= 0\\
P(Y_1 = 2, Y_2 = 0) &= \frac{1}{9}\\
P(Y_1 = 2, Y_2 = 1) &= 0\\
P(Y_1 = 2, Y_2 = 2) &= 0.
\end{align*}
(b)  $F(1,0) = P(Y_1 \le 1, Y_2 \le 0) = P(Y_1 = 0, Y_2 = 0) + P(Y_1 = 1,Y_2 = 0) = \frac{1}{3}$.
\vskip 10pt
\noindent
5.4.   (a) All the entries are $\ge 0$ and they sum to one
so this is a legal joint probability function.  

(b) $F(y_1,y_2)$ is the joint cumulative distribution, so
$F(1,2) = P(Y_1 \le 1, Y_2 \le 2) = 1$.  This is the sum
of all the entries in the table, which works out to 1.
%\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
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)  The joint cdf is 
$$F(y_1,y_2) = P(Y_1 \le y_1, Y_2 \le y_2) =
\begin{cases} \int_0^{y_1}\int_0^{y_2} 4t_1t_2\ dt_2\ dt_1 = y_1^2y_2^2 & \text{ if } 0\le y_1, y_2 \le 1 \\
           \int_0^{y_1}\int_0^1 4t_1t_2\ dt_2\ dt_1 = y_2^2 & \text{ if }  y_1 \ge 1, 0\le y_2 \le 1 \\
 \int_0^1\int_0^{y_2} 4t_1t_2\ dt_2\ dt_1  = y_1^2 & \text{ if }  y_2 \ge 1, 0\le y_1 \le 1 \\
\int_0^1\int_0^1 4t_1t_2\ dt_2\ dt_1 =  1 & \text{ if } y_1,y_2 \ge 1\\
          0 & \text{ otherwise. }
\end{cases}
$$
c)  $P(Y_1\le 1/2, Y_2\le 3/4) = \int_0^{1/2}\int_0^{3/4} 4y_1y_2\ dy_2\ dy_1
= (1/2)^2 (3/4)^2 = 9/64$.
\vskip 10pt
\noindent
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
5.12.  Since the density is constant on the triangle with corners
at $(0,0), (1,0), (0,1)$, we can find the required probabilities
by computing areas.  (a) $P(Y_1\le 3/4,Y_2\le 2)$ is
twice the area of the pentagon with corners at $(0,0), (3/4,0),
(3/4,1/4), (1/4,3/4), (0,3/4)$.  This is $2\cdot \frac{7}{16} = \frac{7}{8}$.  
(b)  Twice the area of a square with side $\frac{1}{2}$, so equal
to $\frac{1}{2}$.  
\vskip 10pt
\noindent
5.16.  (a) The region where the joint density is nonzero is the unit
square.  So the required probability is computed by
\begin{align*}
P(Y_1 < 1/2, Y_2 > 1/4) &= \int_0^{1/2} \int_{1/4}^1 y_1 + y_2\ dy_2\ dy_1\\
&= \int_0^{1/2} \frac{3}{4} y_1 + \frac{15}{32}\ dy_1\\
&= \frac{21}{64}.
\end{align*}
(b) The required probability is:
\begin{align*}
P(Y_1 + Y_2 \le 1) &= \int_0^1 \int_0^{1 - y_1} y_1 + y_2\ dy_2\ dy_1\\
&= \int_0^1 y_1(1 - y_1) + \frac{(1 - y_1)^2}{2} \ dy_1\\
&= \frac{1}{3}.
\end{align*}
\vskip 10pt 
\noindent
5.18.  The region where the joint density is nonzero is the first
quadrant.  Since the region of integration is an infinite rectangle,
the required double integral can be computed by separating into 
a product of two one-variable integrals.  The first one can be done by 
parts:
\begin{align*}
P(Y_1 > 1, Y_2 > 1) &= \frac{1}{8} \int_1^\infty \int_1^\infty y_1 e^{-y_1/2} e^{-y_2/2} dy_2\ dy_1\\
&= \frac{1}{8}\int_1^\infty y_1 e^{-y_1/2}\ dy_1 \cdot \int_1^\infty e^{-y_2/2}\ dy_2\\
&= \frac{6 e^{-1/2}\cdot 2 e^{-1/2}}{8}\\
&= \frac{3 e^{-1}}{2} \doteq .5518.
\end{align*}
\vskip 10pt
\noindent
5.25.  (a)  The marginal densities are
$$f_1(y_1) = \int_0^\infty e^{-y_1 - y_2}\ dy_2 =  \left. -e^{-y_1 - y_2}\right|_0^\infty = e^{-y_1}$$
if $y_1 > 0$ and zero otherwise.  
Similarly, 
$$f_2(y_2) = \int_0^\infty e^{-y_1 - y_2}\ dy_1 =  \left. -e^{-y_1 - y_2}\right|_0^\infty = e^{-y_2}$$
if $y_2 > 0$ and zero otherwise.
These are both exponential densities with $\beta = 1$.

(b)  $P(1 < Y_i < 2.5) = \int_{1}^{2.5} e^{-y_i}\ dy_i = e^{-1} - e^{-2.5}$ for $i = 1,2$.

(c)  The conditional density is only defined for $y_2 > 0$ (since we must have $f_2(y_2) > 0$
in Definition 5.7).

(d)  The conditional density $f(y_1|y_2) = \frac{e^{-y_1-y_2}}{e^{-y_2}} = e^{-y_1}$.

(e) Similarly, $f(y_1|y_2) = e^{-y_2}$.

(f) The conditional and marginal densities for both variables are \emph{equal} for 
$y_i > 0$.

(g)  The marginal and conditional probabilities in any interval will also be
equal.
\vskip 10pt
\noindent
5.26.  (a)  The marginal densities are only nonzero if $0 \le y_i \le 1$ and then 
\begin{align*}
f_1(y_1) &= \int_0^1 4y_1y_2\ dy_2 = 2y_1, \text{ and similarly }\\
f_2(y_2) &= \int_0^1 4y_1y_2\ dy_1 = 2y_2.
\end{align*}
Looking ahead to the next section, note that these
satisfy the criterion of Theorem 5.4.  So $Y_1$ and $Y_2$
are independent.

(b)  To compute the denominator in the conditional probability, 
we can use the marginal density from part a:
\begin{align*}
P(Y_1 \le 1/2 | Y_2 \ge 3/4) 
&= \frac{P(Y_1 \le 1/2, Y_2 \ge 3/4)}{P(Y_2 \ge 3/4)}\\
&= \frac{\int_0^{1/2}\int_{3/4}^1 4y_1y_2\ dy_2\ dy_1}{\int_{3/4}^1 2y_2\ dy_2}\\
&= \frac{\int_0^{1/2} 2y_1\ dy_1 \int_{3/4}^1 2y_2\ dy_2}{\int_{3/4}^1 2y_2\ dy_2}\\
&= \int_0^{1/2} 2y_1\ dy_1\\
&= 1/4.
\end{align*}
(The cancellation works this way because of independence.) 

(c and d)  The conditional densities are nonzero in the same intervals as
the marginal densities, and are given there by
$$ f(y_1 | y_2) = \frac{4y_1y_2}{2y_2} = 2y_1$$
and 
$$ f(y_2 | y_1) = \frac{4y_1y_2}{2y_1} = 2y_2$$

(e)  Using the conditional density,
$$P(Y_1\le 3/4 \mid Y_2 = 1/2) = \int_0^{3/4} 2y_1\ dy_1 = \left. y_1^2\right|_0^{3/4} = \frac{9}{16}.$$ 
\vskip 10pt
\noindent
5.37.  The best way to do this is to compute the marginal density for
$Y_2$ first:
$$f_2(y_2) = \int_0^\infty \frac{1}{8} y_1 e^{-(y_1+y_2)/2}\ dy_1 = \frac{1}{2} e^{-y_2/2},$$
for $y_2 > 0$.  (Note this says $Y_2$ has an exponential distribution with mean 
$\beta = 2 \times 100 {\rm hours}$.)  Hence the probability that $Y_2$
is more than 200 hours is
$$\int_2^\infty \frac{1}{2}e^{-y_2/2}\ dy_2 = e^{-1} \doteq .3679.$$   

\end{document}
