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\overfullrule=0pt
\centerline{Mathematics 375 -- Probability and Statistics I}
\centerline{Solutions -- Problem Set 9}
\centerline{December 7, 2005}
\bigskip
\noindent
5.22/
a)  The marginal densities are 
$f_1(y_1) = \int_0^1 4y_1y_2\ dy_2 = 2y_1$ and similarly
$f_2(y_2) = \int_0^1 4y_1y_2\ dy_1 = 2y_2$.
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:
$$\eqalign{P(Y_1 \le 1/2 | Y_2 \ge 3/4) 
&= {P(Y_1 \le 1/2, Y_2 \ge 3/4)\over P(Y_2 \ge 3/4)}\cr
&= {\int_0^{1/2}\int_{3/4}^1 4y_1y_2\ dy_2\ dy_2\over \int_{3/4}^1 2y_2\ dy_2}\cr
&= {\int_0^{1/2} 2y_1\ dy_1 \int_{3/4}^1 2y_2\ dy_2\over \int_{3/4}^1 2y_2\ dy_2}\cr
&= \int_0^{1/2} 2y_1\ dy_1\cr
&= 1/4\cr}$$
(The cancellation works this way because of independence.)
\bigskip
\noindent
5.27/  The region where the density is nonzero here is the
interior of the triangle with vertices at $(0,1), (1,0), (0,-1)$.
As always, to find
the marginal density with respect to one of the
variables we compute the ``inside'' integral
of the iterated double integral of the joint density
with the ``other variable'' on the inside.  
The bottom edge of this triangle is the line $y_2 = y_1 - 1$ and
the top edge is $y_2 = 1 - y_1$.  
a)  Hence to find $f_1(y_1)$ we compute for $0 \le y_1 \le 1$:
$$\int_{y_1-1}^{1-y_1} 30y_1y_2^2\ dy_2 = 20y_1(1-y_1)^3$$
(Outside $0\le y_1\le 1$, the marginal density is zero.)
From Chapter 4, we recognize this as the form for a 
Beta pdf with $\alpha = 2, \beta = 4$.  

b)  The marginal density $f_2(y_2)$ is not as nice because the
right boundary curve of the triangle changes depending on 
the value of $y_2$.
If $0 \le y_2 \le 1$, then 
$f_2(y_2) = \int_0^{1-y_2} 30y_1y_2^2\ dy_1
= 15y_2^2(1-y_2)^2$.  On the other hand,
If $-1 \le y_2 \le 0$, then 
$f_2(y_2) = \int_0^{y_2+1} 30y_1y_2^2\ dy_1
= 15y_2^2(y_2+1)^2.$

\bigskip
\noindent
5.31/We take the factor depending only on $y_2$ outside the integral
with respect to $y_1$, then change variables to $u = y_1/2$:
$$\eqalign{f_2(y_2) &= \int_0^\infty {1\over 8} y_1 e^{-(y_1+y_2)/2}\ dy_1\cr
&=  {1\over 2}\int_0^\infty  u e^{-u}\ du \cdot e^{-y_2/2}\cr
&= {1\over 2}\Gamma(2)e^{-y_2/2}\cr
&= {1\over 2}e^{-y_2/2}\cr}$$
if $y_2 > 0$, and zero otherwise.  This gives the 
marginal density for $Y_2$. Then the desired probability
is found by integrating 
$$\int_2^\infty f_2(y_2)\ dy_2 = \int_2^\infty {1\over 2}e^{-y_2/2}\ dy_2 = 1/e.$$
(Note that the problem should say the life of the component is in 100's of 
hours to match the previous problem this one builds on.
So the lower limit of integration is 2 rather than 200.)

\bigskip
\noindent
5.32
a)  We claim 
$$f(y_1,y_2) = \cases{1/y_1 & if $0 < y_1 \le 1, 0 \le y_2 \le y_1$\cr
                       0 & otherwise}$$
The region where this is nonzero is the interior of the triangle 
with vertices $(0,0), (1,0), (1,1)$.  To convince ourselves this
is correct, we can compute the marginal densities and see
that they match what the problem says.  First, if $0 < y_1 \le 1$, 
$f_1(y_1) = \int_0^{y_1} 1/y_1\ dy_2 = y_2/y_1|_0^{y_1} = 1$.
(Outside $0 < y_1 \le 1$ we get zero.)
Hence we do get the uniform density on $(0,1)$.  Second,
If we fix a value for $y_1$, then the ``conditional''
density $f(y_2|Y_1 = y_1) = \cases{1/y_1 & if $0 \le y_2 \le y_1$\cr
0 & otherwise\cr}$ which is also correct for the uniform distribution
on $(0,y_1)$.  

b) $P(Y_2 \ge 1/4 | Y_1 = 1/2) = \int_{1/4}^{1/2} 2\ dy_2 = 1/2$.

\bigskip
\noindent
5.39/ Not independent.  For instance $P(Y_1 = 0, Y_2=0) = 
4/9 \ne P(Y_1 =0)P(Y_2=0) = 16/81$.  (We compute the marginal distributions
to find $P(Y_1 = 0) = 1/9 + 2/9 + 1/9 = 4/9$ and similarly
for $P(Y_2=0)$.  

%\bigskip
%\noindent
%5.44/  Yes, as noted in solution for 5.22 above.

\bigskip
\noindent
5.46/ No -- $Y_1, Y_2$ cannot be independent when 
the interval of $y_2$ values where the density is 
nonzero depends on $y_1$.  (You can also see this directly
by computing $f_1(y_1)$ and $f_2(y_2)$.  The region
where the joint density is nonzero is the triangle
with vertices at $(0,0), (2,0), (2,1)$.  So
$$f_1(y_1) = \int_0^{y_1/2} 1\ dy_2 = {y_1\over 2}$$
(when $0 \le y_1 \le 2$; 0 otherwise).
Similarly,
$$f_2(y_2) = \int_{2y_2}^2 1\ dy_1 = 2 - 2y_2$$
(when $0 \le y_2 \le 1$; 0 otherwise).
The product $f_1(y_1)f_2(y_2)$ is not the same as $f(y_1,y_2)$.)

%\bigskip
%\noindent
%5.51/ No -- same reason as 5.46.

%\bigskip
%\noindent
%5.61/  We assume that the times the calls arrive are
%uniformly distributed on $(0,1)$, and independent
%of each other.  That means the joint density
%is $f(y_1,y_2) = 1$ on the unit square $0 < y_1,y_2, < 1$
%and 0 elsewhere.  
%
%a)  $P(Y_1 \le 1/2, Y_2 \le 1/2) = \int_0^{1/2}\int_0^{1/2} 1\ dy_2\ dy_1 = 1/4$.
%(Note: this is just the area of the smaller square defined by 
%$0\le y_1, y_2 \le 1/2$.)
%
%b)  The probability of the event $E$ that the calls are made within 5 
%minutes of each other can be computed as follows.  $\overline{E}$ is the 
%event that the first call arrives more than 5 minutes = 1/12 hour before, or 
%more than 5 minutes after the second.  In the unit square, this
%corresponds to the union of two triangles, one below the line 
%$y_2 = y_1 - 1/12$, the other above the line $y_2 = y_1 + 1/12$.
%Each of these triangles has area $(1/2)(11/12)^2 = 121/288$.
%So $P(E) = 1 - (2)(121/288) = 23/144$.
%
\bigskip
\noindent
5.64/ When the function depends on only one of $Y_i$, we
can use the marginal densities computed in Exercise 5.22 above.
We will use that method in parts a and b.  For functions depending
on both variables, as in part c, we integrate against the joint density to 
find the expected value.  

\item{a)}  $E(Y_1) = \int_0^1 2y_1^2\ dy_1 = 2/3$.
\item{b)}  $E(Y_1^2) = \int_0^1 2y_1^3\ dy_1 - (2/3)^2 = 1/2 - 4/9 = 1/18$.
\item{c)}  $E(Y_1-Y_2) = \int_0^1\int_0^1 (y_1 - y_2) \cdot 4y_1y_2\ dy_2\ dy_1 
= 0$.

\noindent
(The last one can also be seen using the linearity 
$E(Y_1 - Y_2) = E(Y_1) - E(Y_2)$ then noting symmetry of 
the joint density.  The two terms must cancel.)
\bigskip
\noindent
5/70.  Instead of splitting up the expected value or computing
marginal densities, the fastest way to compute this is simply
to take
$$E(Y_1 - Y_2) = \int_0^1 \int_0^{y_1} (y_1-y_2){1\over y_1}\ dy_2\ dy_1 = 
\int_0^1 y_1/2\ dy_1 = 1/4.$$

\bigskip
\noindent
5.77/ $Cov(Y_1,Y_2) = E(Y_1Y_2) - E(Y_1)E(Y_2) = 0$
since we know $Y_1$ and $Y_2$ are independent.  (See Exercise 5.22
above and Theorem 5.11.)

\bigskip
\noindent
5.87/ By linearity of expected values, 
$E(3Y_1 +4Y_1 - 6Y_3) = 3E(Y_1) + 4E(Y_2) - 6E(Y_3) = 6 - 4 - 24 = -22$.
We use Theorem 5.12b to find 
$$\eqalign{
V(3Y_1+4Y_1 - 6Y_3) &= 9V(Y_1) + 16V(Y_2)+36V(Y_3)\cr
&\qquad +24 Cov(Y_1,Y_2) -18 Cov(Y_1,Y_3) -48 Cov(Y_2,Y_3)\cr
&= 36 + 96 + 288 + 24  + 36 + 0\cr
&=480.\cr}$$

\bigskip
\noindent
5.92/  By computing marginal densities here, we
see $Y_1$ has a Gamma distribution with $\alpha = 2$ and $\beta = 1$.
$Y_2$ is exponentially distributed.  So $E(Y_1) = 2$, $E(Y_2) = 1$,
and $E(Y_1 - Y_2) = 1$.  An alternate way to see this: 
integrate against the joint density:
$$\eqalign{
E(Y_1 - Y_2) &= \int_0^\infty\int_0^{y_1} (y_1-y_2)e^{-y_1}\ dy_2\ dy_1\cr
&= {1\over 2}\int_0^\infty y_1^2 e^{-y_1}\ dy_1\cr
&= {1\over 2} \Gamma(3)\cr
&= 1\cr}$$
Then by Theorem 5.12 b,
$$\eqalign{
V(Y_1 - Y_2) &= V(Y_1) + V(Y_2) - 2Cov(Y_1,Y_2)\cr
&=2 + 1 - 2(E(Y_1Y_2) - 2\cdot 1)\cr
&=7 - 2\int_0^\infty \int_0^{y_1} y_1y_2e^{-y_1}\ dy_2\ dy_1\cr
&=7 - \int_0^\infty y_1^3e^{-y_1}\ dy_1\cr
&=7 - \Gamma(4)\cr
&=7 - 6 = 1.\cr}$$
It is unlikely that a customer spends 4 minutes or more
at the window, since $4 = \mu + 3\sigma$ (three standard
deviations above mean). 


\bigskip
\noindent
6.33/ $U = Y_1/2 + Y_2/2$ where each $Y_i$ is exponentially
distributed and $Y_1,Y_2$ are independent.  Hence by
Theorem 6.2, the mgf for $U$ is 
$$m_U(t) = m_{Y_1} (t/2)\cdot m_{Y_2}(t/2) = {1\over (1-t/2)^2}.$$
Comparing with the table inside the back cover of the text, 
we see that $U$ has a Gamma distribution with $\alpha = 2$
and $\beta = 1/2$.

\bigskip
\noindent
6.34/ $Y_i$ are standard normal, so by Discussion 5 (or
Example 6.11) $Y_i^2$ are 
$\chi^2$ with $\nu = 1$.  They are independent as well.
So the mgf for $U$ is
$$m_U(t) = {1\over \sqrt{1 - 2t}}\cdot {1\over \sqrt{1 - 2t}} = {1\over 1 - 2t}$$
Then from the table,  $U = Y_1^2 + Y_2^2$
has a $\chi^2$ distribution with $\nu = 2$.  Therefore
$f_U(u) = {1\over 2} e^{-u/2}$.  (Note: this is also same
as an exponential pdf with $\beta = 2$.)

\bigskip
\noindent
6.36/  Since $Y_1$ is the capacity of the elevator and
$Y_2$ is the load, the elevator is overloaded when 
$Y_2 - Y_1 > 0$.
Using the result from Discussion 5, B, $U = Y_2 - Y_1$
is normally distributed with $\mu_U = -1000$
and $\sigma_U^2 = 300^2 + 400^2 = 500^2$.  The probability
that the elevator is overloaded is found by standardizing
and using the table for the standard normal
$$P(Y_2 - Y_1 > 0) = P\left({(Y_2 - Y_1) + 1000\over 500} > 2\right) = 
P(Z > 2) = .0228.$$

\bye

