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\centerline{Mathematics 375 -- Probability Theory}
\centerline{Solutions for Problem Set 9}
\centerline{December 2, 2011}
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\noindent\vskip 10pt
\noindent
5.48.  No $Y_1$ and $Y_2$ are not independent.  If they
were, then for instance $P(Y_1 = 0, Y_2 = 0) = P(Y_1 = 0)P(Y_2 = 0) = (.76)(.55) = .418$.
However, that does not match the given value.  
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\noindent
5.51.  (a)  By Theorem 5.4, $Y_1$ and $Y_2$ are independent here. 
Each has an exponential distribution with mean $\beta = 1$ and the joint density
is the product $e^{-(y_1 + y_2)} = e^{-y_1} \cdot e^{-y_2}$.
(b)  Yes, this explains the results in 5.25.
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\noindent
5.55.  No they are not independent.  ``Uniformly distributed'' just
means that the joint density is constant on the triangle in the 
figure from 5.11.  But computing marginal densities we have
$$f_1(y_1) = \begin{cases} y_1 + 1 &  -1 \le y_1 \le  0\\
                           -y_1 + 1 & 0 \le y_1 \le 0\\
                                0 & \text{ otherwise.}
             \end{cases} $$
(These are separate cases based on the equation of the top
edge of the triangle on different intervals
of $y_1$-values; they are not added together.)
$$f_2(y_2) = \int_{y_2-1}^{1-y_2}\ dy_1 = 2 - 2y_2$$
for $0 \le y_2 \le 1$.  The product of the marginal
densities is not equal to the constant joint density.
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\noindent
5.61.  Yes they are independent by Theorem 5.5: take 
$g(y_1) = \frac{1}{4} y_1 e^{-y_1/2}$ and $h(y_2) = \frac{1}{2} e^{-y_2/2}$.
(These are the marginal densities in fact -- $Y_1$ has a $\Gamma(2,2)$
distribution and $Y_2$ has an exponential distribution with $\beta = 2$.)
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\noindent
5.70.  As in the Hint, let $Y_1$ = bus arrival time and $Y_2 = $ passenger
arrival time, both uniformly distributed on $[0,1]$.  We want 
$$P(Y_2 \le Y_1 \le Y_2 + 1/4).$$
This can be set up as a double integral, but the most direct way
is to think in terms of areas.  The area we want is what is left after
two triangles are cut out of the unit square.  The first triangle is 
the one above the line $y_2 = y_1$, so its area is $\frac{1}{2}$.  The
second triangle is below the line $y_2 = y_1 - \frac{1}{4}$, so it
has area $\frac{1}{2}\cdot \left(\frac{3}{4}\right)^2 = \frac{9}{32}$.
The area we want is then 
$$1 - \frac{1}{2} - \frac{9}{32} = \frac{7}{32}.$$
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\noindent
5.76.  This can be done most easily by using the
marginal densities for the $Y_i$ (although there
are several other valid methods too).  Note that 
$f_1(y_1) = \int_0^1 4y_1y_2\ dy_2 = 2y_1$ if $0 \le y_1 \le 1$, and zero otherwise.
Similarly for $f_2(y_2)$ by the symmetry of the joint
density.  

(a)  $E(Y_1) = \int_0^1 y_1 f_1(y_1)\ dy_1 = \int_0^1 2y_1^2\ dy_1 = \frac{2}{3}$.

(b)  Then the variance is:
$$V(Y_1) = E(Y_1^2) - \left(\frac{2}{3}\right)^2 = \int_0^1 2y_1^3\ dy_1 - \frac{4}{9}
= \frac{1}{2} - \frac{4}{9} = \frac{1}{18}.$$

(c)  By linearity, $E(Y_1 - Y_2) = E(Y_1) - E(Y_2) = \frac{2}{3} - \frac{2}{3} = 0$.
This can also be computed, for instance, as the integral:
$$\int_0^1 \int_0^1 (y_1 - y_2)\cdot 4y_1 y_2\ dy_2\ dy_1.$$
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\noindent
5.81.  There are again a multitude of different correct ways to 
approach this.  The most direct way is to compute as follows:
\begin{align*}
E(Y_1/Y_2) &= \int_0^\infty \int_0^\infty \frac{y_2}{y_1} 
\cdot \frac{1}{8} y_1 e^{-(y_1+y_2)/2}\ dy_2 dy_1\\
&= \int_0^\infty \int_0^\infty \frac{1}{8} y_2 e^{-(y_1+y_2)/2}\ dy_2 dy_1\\
&= \frac{1}{8} \int_0^\infty e^{-y_1/2}\ dy_1 \cdot \int_0^\infty y_2 e^{-y_2/2}\ dy_2\\
&= \frac{1}{8} \cdot 2 \cdot \Gamma(1) \cdot 4 \cdot \Gamma(2)\\
&= 1.
\end{align*}
To derive the values of the integrals, I substituted $u = y_1/2$ in the first
and recognized that the value was just $2 \Gamma(1)$.  Similarly, the second
integral is evaluated by setting $t = y_2/2$. 

Another, somewhat more clever, way is as follows.  Theorem 5.12 implies
that $Y_1$ and $Y_2$ are independent.  Note that independence
of $Y_1$ and $Y_2$ implies that $E(g(Y_1)h(Y_2)) = E(g(Y_1))E(h(Y_2))$ 
for any functions $g,h$ (why?).  Hence $E(Y_2/Y_1) = E(Y_2)E(1/Y_1)$.  
By examining the joint density and factoring, we can see $Y_2$ has an 
exponential distribution with $\beta = 2$, so $E(Y_2) = 2$.  
 We also see $Y_1$ has a $\Gamma(2,2)$
distribution.  By Exercise 4.111 from a previous problem set,
$$E(1/Y_1) = \frac{1}{2(2-1)} = \frac{1}{2}.$$ 
Therefore, $E(Y_2/Y_1) = 1$ as before(!)
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\noindent
5.92.   We have
$$E(Y_1) = \int_0^1 \int_{y_1}^1 6 y_1(1-y_2)\ dy_2\ dy_1 = \frac{1}{4}$$
and 
$$E(Y_2) = \int_0^1 \int_{y_1}^1 6 y_2(1-y_2)\ dy_2\ dy_1 = \frac{1}{2}.$$
Therefore
\begin{align*}
{\rm Cov}(Y_1,Y_2) &= E(Y_1Y_2) - \frac{1}{8}\\
&= \int_0^1 \int_{y_1}^1 6 y_1y_2(1-y_2)\ dy_2\ dy_1 - \frac{1}{8}\\
&= \frac{3}{20} - \frac{1}{8}\\
&= \frac{1}{40}.
\end{align*}
Since ${\rm Cov}(Y_1,Y_2) \ne 0$, they are not independent.
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\noindent
5.93. (a)  We have
$$E(Y_1) = \int_0^1 \int_{y_2-1}^{1-y_2} y_1 dy_1\ dy_2 = 0$$
and 
$$E(Y_2) =\int_0^1 \int_{y_2-1}^{1-y_2} y_2 dy_1\ dy_2 = \frac{1}{3}.$$
Therefore
\begin{align*}
{\rm Cov}(Y_1,Y_2) &= E(Y_1Y_2) - 0\\
&= \int_0^1 \int_{y_2-1}^{1-y_2} y_1 y_2 dy_1\ dy_2\\
&=  0.
\end{align*}
(b)  In Exercise 5.55 earlier, we saw that $Y_1$ and $Y_2$ 
are \emph{not} independent.  (This example shows that the
implication in Theorem 5.11 goes only one way.)
(c)  The correlation coefficient is also $0$.
(d)  No, it is possible for two dependent random 
variables to have zero covariance.  
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\noindent
5.96. (a)  By the definition, 
$${\rm Cov}(Y_1,Y_2) = E((Y_1 - \mu_1)(Y_2 - \mu_2)) = E((Y_2 - \mu_2)(Y_1 - \mu_1)) = {\rm Cov}(Y_2,Y_1).$$
(b)  Similarly,
$${\rm Cov}(Y_1,Y_1) = E((Y_1 - \mu_1)^2) = V(Y_1) = \sigma_1^2.$$
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\noindent
5.97.  (a) ${\rm Cov}(Y_1,Y_1) = E(Y_1^2) - (E(Y_1))^2 = V(Y_1) = 2$
by the given information.  b,c,d)  If ${\rm Cov}(Y_1,Y_2) = 7$, 
the correlation coefficient
$\rho = \frac{7}{\sigma_1\sigma_2} = \frac{7}{\sqrt{2}\cdot 2\sqrt{2}} = 7/4$.
That is not possible because $-1 \le \rho \le 1$ in all cases.
The largest and smallest values for ${\rm Cov}(Y_1,Y_2)$ are $-4, 4$, corresponding
to cases where $Y_2$ is a decreasing or increasing linear function 
of $Y_1$. 
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\noindent
5.103.  By the general formulas for expected value and variance of a
linear combination of random variables,
$$E(3Y_1 + 4Y_2 - 6Y_3) = 3E(Y_1) + 4E(Y_2) - 6E(Y_3) = (3)(2) + (4)(-1) - (6)(4) = -22$$
and 
\begin{align*}
V(3Y_1 + 4Y_2 - 6Y_3) &= 9V(Y_1) +16V(Y_2) + 36V(Y_3)\\
&\qquad + 24 {\rm Cov}(Y_1,Y_2) - 36 {\rm Cov}(Y_1,Y_3) - 48 {\rm Cov}(Y_2,Y_3)\\
&= 36 + 96 + 288 + 24 + 36 + 0\\
&= 480.
\end{align*}  
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\noindent
5.105.  From Exercise 5.76, we know $V(Y_i) = \frac{1}{18}$.  So 
we have $V(Y_1 - Y_2) = V(Y_1) + V(Y_2) - 2{\rm Cov}(Y_1,Y_2) = \frac{1}{9}$.  
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\noindent
5.114.  (a)  We recognize that $Y_1$ has a $\Gamma(4,1)$ distribution
and $Y_2$ has an exponential distribution with $\beta = 2$.  Hence
from our general formulas we know $E(Y_1) = 4$ and $E(Y_2) = 2$.
So $E(U) = E(Y_1) - E(Y_2) = 2$.  
(b)  Similarly, $V(Y_1) = 4$ and $V(Y_2) = 2$, so $V(U) = V(Y_1) + V(Y_2) + 2 {\rm Cov}(Y_1,Y_2)
= 6$, since independence implies the covariance is zero.  
(c)  $0$ is a bit more than one standard deviation below the mean for $U$.
The daily profit could drop below zero, but it would  not be extremely common.
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\noindent

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