\magnification=\magstep1
\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions -- Problem Set 9}
\centerline{November 25, 2003}
\bigskip
\noindent
5.4/ a)  It is clear from the table that $p(y_1,y_2) \ge 0$ and 
$\sum_{y_1,y_2} p(y_1,y_2) = 1$.  Hence this is a valid 
probability function.  b) $F(1,2) = 1$ since all possible combinations
of values of $Y_i$ satisfy $Y_1 \le 1$ and $Y_2 \le 2$.
\bigskip
\noindent
5.5/ The region where the joint density is nonzero is a triangle
with corners at $(0,0)$, $(1,0)$, $(1,1)$.
a) $P(Y_1\le 1/2, Y_2\le 1/3)$ is found by integrating the joint density
over the region shaded in this diagram:
\vglue 2.0truein
The simplest way to set up the double integral here is to 
integrate with respect to $y_1$ first (we don't need to split
the region):
$$\int_0^{1/3} \int_{y_2}^{1/2} 3y_2\ dy_1\ dy_2 = {23\over 216} \doteq .1065$$

b)  $P(Y_2\le Y_1/2)$ is found by integrating the joint density
over the region shaded in this diagram:
\vglue 2.0truein
We get
$$\int_0^1\int_0^{y_1/2} 3y_1\ dy_1\ dy_2 = {1\over 2}$$
\bigskip
\noindent
5.7/  The region where the joint density is nonzero is a triangle
with corners at $(0,0)$, $(1,1)$, $(0,1)$ (above the line $y_1 = y_2$).
\vglue 2.0truein
a)  We must have 
$$1 = \int_0^1 \int_{y_1}^1 k(1 - y_2)\ dy_2\ dy_1 \Rightarrow k = 6.$$
b) $P(Y_1\le 3/4, Y_2\ge 1/2)$ is found by integrating the joint density
over the region shaded in this diagram:
\vglue 2.0truein
This is one where either way of setting up the iterated integrals
leads to a case where you must split the region.  For instance,
setting up the integrals to integrate with respect to $y_2$ first,
we get:
$$\int_0^{1/2}\int_{1/2}^1 6(1-y_2)\ dy_2\ dy_1 + 
\int_{1/2}^{3/4}\int_{y_1}^1 6(1-y_2)\ dy_2\ dy_1 = {31\over 64}.$$ 
\bigskip
\noindent
5.13.  The region where the joint density is nonzero is the infinite
``wedge'' in the first quadrant between the $y_1$ axis and 
the  below the line $y_1 = y_2$.  a)  $P(Y_1< 2, Y_2 > 1$
is found by integrating the density over the region shaded in this diagram:
\vglue 2.0truein
We get
$$\int_1^2\int_1^{y_1} e^{-y_1}\ dy_2\ dy_1 = \int_1^2\int_{y_2}^2 e^{-y_1}\ dy_1\ dy_2$$
The second form is easier (don't need to integrate by parts!):
$$\int_1^2\int_{y_2}^2 e^{-y_1}\ dy_1\ dy_2 = \int_1^2 -e^{-2} + e^{-y_2}\ dy_2 = e^{-1} - 2e^{-2}.$$
b)  $P(Y_1\ge 2Y_2)$ is found by integrating the density over
the smaller ``wedge'' between the $y_1$-axis and the line $y_2 = y_1/2$.
$$\int_0^\infty\int_{2y_2}^\infty e^{-y_1}\ dy_1\ dy_2 = {1\over 2}$$
c)  $P(Y_1 - Y_2\ge 1)$ is found by integrating the density over
the shifted ``wedge'' in the first quadrant between the $y_1$-axis and the 
line $y_2 = y_1 - 1$:
\vglue 2.0truein
$$\int_0^\infty\int_{y_2+1}^\infty e^{-y_1}\ dy_1\ dy_2 = e^{-1}$$
\bigskip
\noindent
5.16/The region of integration here is the shifted quadrant bounded
by the lines $y_1 = 1$ and $y_2 = 1$.
$$P(Y_1 >1, Y_2 > 1) = \int_1^\infty \int_1^\infty {1\over 8}y_1 e^{-(y_1+y_2)/2}\ dy_1\ dy_2$$
The inner integral requires integration by parts with $u = y_1$ and
$dv = e^{(-y_1+1)/2}$  the final result is ${3\over 2e}$.
\bigskip
\noindent
5.20 a)  The marginal probabilities are the row and column totals given
in the table.
b)  The conditional probability function $p(y_2|y_1=0)$
is 
$$\eqalign{P(y_2 = 0 |y_1=0) &= .38/.76 = .5\cr
 P(y_2=1|y_1=0) &= .14/.76 = .184\cr
P(y_2 = 2| y_1 = 0) &= .24/.76 = .316\cr}$$
c)  It's $.24/.29 = .8276$
\bigskip
\noindent
5.21/a)
$$f_2(y_2) = \int_{y_2}^1 3y_1\ dy_1 = {3\over 2}(1 - y_2^2)$$
if $0 \le y_2 \le 1$ and 0 otherwise.  b)  The conditional
density $f(y_1 | y_2 ) = f(y_1,y_2)/f_2(y_2)$ will be undefined
when the denominator is zero -- at $y_2 = \pm 1$.  
c)  The conditional density $f(y_2 | y_1) = 1/y_1$ so 
at $y_1 = 3/4$ we get $f(y_2 | y_1 = 3/4) = 4/3$ if 
$0\le y_2\le 3/4$ and zero otherwise (i.e. a uniform density).
Then 
$$P(Y_2 > 1/2 | Y_1 = 3/4) = \int_{1/2}^{3/4} 4/3\ dy_2 = 1/3$$
(Note: we stop at the upper limit $y_2 = 3/4$ since the conditional
density is zero for $y_2 > 3/4$.)
\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
5.42/  These random variables are not independent, since
for instance 
$$p_1(0)p_2(0) = (.55)(.76) = .418 \ne p(0,0) = .38$$
\bigskip
\noindent
5.45/ The marginal densities are
$$f_1(y_1) = \int_{y_1}^1 6(1-y_2)\ dy_2 = 3(1-y_1)^2$$
(if $0\le y_1 \le 1$, and zero otherwise) and
$$f_2(y_2) = \int_0^{y_2} 6(1-y_2)\ dy_1 = 6y_2 - 6y_2^2$$
(if $0\le y_1 \le 1$, and zero otherwise).
The product is not equal to $f(y_1,y_2)$, 
so $Y_1$ and $Y_2$ are not independent.
\bigskip
\noindent
5.53/$Y_1$ and $Y_2$ are independent by the criterion of Theorem 5.5
in the text:  The joint density
$${1\over 8}y_1 e^{-(y_1+y_2)/2} = {1\over 4}y^1 e^{-y_1/2} \cdot {1\over 2}e^{-y_2/2}$$
factors into a product of a function of $y_1$ times a function of $y_2$ and
the region where the density is nonzero is a ``rectangle''.
(Note that $Y_1$ has a gamma distribution with $\alpha = 2$ and 
$Y_2$ is exponential.)  It is important to realize that Theorem 5.5
also applies on ``infinite rectangles'' i.e. 
$a,c$ can be $-\infty$ and $b,d$ can be $+\infty$ in that statement).
\bye

