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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions for Problem Set 6}
\centerline{October 29, 2009}
\vskip 10pt
\noindent
%3.145.   By the binomial theorem, 
%\begin{align*}
%m(t) = E(e^{tY}) &=\sum_{y=0}^n e^{ty}p^y q^{n-y}\\
%                 &=\sum_{y=0}^n (e^tp)^y q^{n-y}\\
%                 &=(pe^t + q)^n.
%\end{align*}
%This is what we wanted to show.
%\vskip 10pt
%\noindent
%3.146.  The idea of this problem is to use the moment 
%generating function for a binomial $Y$ to give a different
%derivation of the
%formulas we know for $E(Y)$ and $V(Y)$.
%By the chain rule for derivatives,
%$m'(t) = n(pe^t + q)^{n-1}pe^t$.  So $E(Y) = m'(0) = np$.
%Then 
%$$m''(t) = n(n-1)(pe^t + q)^{n-2}p^2e^{2t} + n(pe^t + q)^{n-1}pe^t$$
%so 
%$E(Y^2) = m''(0) = n(n-1)p^2 + np$.  Then 
%$$V(Y) = E(Y^2) - E(Y)^2 = n(n-1)p^2 + np - n^2p^2 = -np^2 + np = 
%np(1 - p) = npq.$$
%\vskip 10pt
%\noindent
%3.150.  The uniqueness theorem says that if two 
%random variables have the same moment generating function,
%then they have the same distribution.  This given formula
%has the form of the moment generating function for a geometric
%random variable with $p = .3$ (this was proved in class; also 
%see the indicated problem 3.147).  So the random variable is geometric
%with $p = .3$.  
%\vskip 10pt
%\noindent
%3.153.  By the Uniqueness Theorem for the moment generating function,
%and the formulas from class and the text for moment generating
%functions for standard types of random variables,
%\begin{enumerate}
%\item[(a)] Binomial, $n = 5$, $p = 1/3$.
%\item[(b)] Geometric, $p = 1/2$.
%\item[(c)] Poisson, $\lambda = 2$.
%\end{enumerate}
%\vskip 10pt
%\noindent
%3.158.  Since $e^{bt}$ is not random (it's a fixed function of $t$),
%$$m_{aY+b}(t) = E(e^{(aY+b)t}) = E(e^{bt} e^{(at)Y}) = 
%e^{bt}E(e^{(at)Y}) = e^{bt}m_Y(at).$$
%\vskip 10pt
%\noindent
4.6.  (a) The probability function for the (discrete) geometric
random variable is $p(y) = q^{y-1}p$.  Hence if $y = i \ge 1$,
\begin{align*}
F(i) &= P(Y \le i) = P(Y = 1) + \cdots + P(Y = i)\\
     &= p(1 + q + q^2 + \cdots + q^{i-1})\\
     &= p\cdot \frac{1 - q^i}{1 - q}\\
     &=1 - q^i
\end{align*}
(where we used the formula for the sum of a finite geometric series
to get the next to last line, then simplified using $q = 1 - p$).
Now, if $y > 0$ is not an integer, we get $F(y) = F(i)$
where $i$ is the largest integer $\le y$.  Hence 
$$
F(y) = \begin{cases} 0  & \text{ if } y < 0\\
                       1-q^i & \text{ if } i \le y < i+1 \text{ for } i = 0, 1, 2, \ldots.
         \end{cases}
$$

b)  $\displaystyle{\lim_{y\to -\infty}F(y) = 0}$ by the definition.  Then,
$$\lim_{y\to +\infty}F(y) = \lim_{i\to\infty} 1 - q^i = 1$$
since $q < 1$.  Finally, $F(y)$ is non-decreasing since
$$i_1 > i_2 \Rightarrow q^{i_1} < q^{i_2} \Rightarrow
-q^{i_1} > -q^{i_2} \Rightarrow 1-q^{i_1} > 1-q^{i_2}.$$

\noindent
4.8.  (a)  We must have 
$\int_{-\infty}^{+\infty} f(y)\ dy = \int_0^1 k y(1 - y)\ dy = 1$.
This gives the equation $k \left.\left(\frac{y^2}{2} - \frac{y^3}{3}\right)\right|_0^1 = 1$, so $\frac{k}{6} = 1$, and $k = 6$.  

(b)  
\begin{align*}
P(.4 \le Y\le 1) &= \int_{.4}^1 6y(1-y)\ dy\\
&= \left. 3y^2 - 2 y^3\right|_{.4}^1\\
&= .648.
\end{align*}

(c)  This is the same as (b) because $Y$ is a continuous
random variable, so including the endpoints or not makes
no difference.

(d)  By the definition of conditional probability, 
$$
P(Y\le .4 | Y \le .8) = \frac{P(Y\le .4)}{P(Y\le .8)} = 
\frac{\int_0^{.4} 6y(1 - y)\ dy}{\int_0^{.8} 6y(1 - y)\ dy} \doteq .393.
$$

(e)  This is the same as (d) for the same reason as in
(c).
\vskip 10pt
\noindent
4.13.  (a)  $F(y) = \int_{-\infty}^y f(z)\ dz$ is a continuous
antiderivative of $f(y)$.  Hence 
$$
F(y) = \begin{cases}  
           0 & \text{ if } y < 0\\
          \frac{y^2}{2} & \text{ if } 0 \le y < 1\\
          y - \frac{1}{2} & \text{ if } 1 \le y < 1.5\\
              1 & \text{ if } y \ge 1.5.
       \end{cases}
$$

(b)  By the FTC, $P(0 \le Y \le .5) = F(.5) - F(0) = \frac{1}{8} = .125$.

(c)  By the FTC, $P(.5 \le Y \le 1.2) = F(1.2) - F(.5) = .7 - .125 = .575$.
\vskip 10pt
\noindent
4.17. (a)  We must have $1 = \int_0^1 cy^2 + y\ dy$.  So 
$1 = \frac{c}{3} + \frac{1}{2}$, and $c = \frac{3}{2}$.

(b)  As in 4.13 (b), 
$$
F(y) = \begin{cases}  
           0 & \text{ if } y < 0\\
          \frac{y^3}{2} + \frac{y^2}{2} & \text{ if } 0 \le y < 1\\
              1 & \text{ if } y \ge 1.
       \end{cases}
$$

(c)  plots omitted.  

(d)  $F(-1) = 0$,  $F(0) = 0$,  $F(1) = 1$.

(e)  $P(Y < 1/2) = F(1/2) = \frac{1}{16} + \frac{1}{8} = \frac{3}{16} = .1875$.

(f)  $P(Y \ge 1/2 | Y \ge 1/4) = \frac{P(Y\ge 1/2)}{P(Y\ge 1/4)} = 
\frac{1 - F(1/2)}{1 - F(1/4)} \doteq \frac{.8125}{.9609} \doteq .8455.$ 

\vskip 10pt
\noindent
4.21.  The mean is 
\begin{align*}
E(Y) &= \int_{-\infty}^\infty yf(y)\ dy\\
&= \int_0^1 \frac{3 y^3}{2} + y^2\ dy\\
&= \left.\frac{3y^4}{8} + \frac{y^3}{3}\right|_0^1\\
&= \frac{3}{8} + \frac{1}{3} = \frac{17}{24} \doteq .708.
\end{align*}
Then the variance is $V(Y) = E(Y^2) - E(Y)^2$.
\begin{align*}
E(Y^2) &= \int_{-\infty}^\infty y^2 f(y)\ dy\\
&= \int_0^1 \frac{3 y^4}{2} + y^3\ dy\\
&= \left.\frac{3y^5}{10} + \frac{y^4}{4}\right|_0^1\\
&= \frac{3}{10} + \frac{1}{4}\\ 
&= \frac{11}{20}.
\end{align*}
So $V(Y) = \frac{11}{20} - \left(\frac{17}{24}\right)^2 = \frac{139}{2880}\doteq .0483$.
\vskip 10pt
\noindent
4.26. (a) By Theorem 4.5, 
$E(aY + b) = aE(Y) + b = a\mu + b$.

(b)  Then, by the definition of the variance
and Theorem 4.5 again, 
\begin{align*}
V(aY + b) &= E(((aY + b) - (a\mu + b))^2)\\
&= E(a^2 (Y^2 - \mu^2))\\
&= a^2 (E(Y^2) - \mu^2)\\
&= a^2 (E(Y^2) - E(Y)^2) \text{ by one of our standard formulas }\\
&= a^2 V(Y).
\end{align*}  
\vskip 10pt
\noindent
4.27.  Using the result of 4.26,  $E(W) = 5 - .5E(Y) = 5 - \frac{17}{48} 
\doteq 4.646$ (dollars).  Then $V(W) = \frac{1}{4}V(Y) = \frac{139}{4\cdot 2880} \doteq
.0121$.   
\vskip 10pt
\noindent
4.32.  (a)  $E(Y) = \frac{3}{64}\int_0^4 y^3(4 - y)\ dy =
\frac{3}{64}\int_0^4 4y^3  - y^4\ dy = 2.4$.
Then $E(Y^2) = \frac{3}{64}\int_0^4 4y^4  - y^5\ dy = 6.4$
and $V(Y) = E(Y^2) - E(Y)^2 = .64$.

(b)  Easiest method here is to use general facts
about expected values of linear combinations as in 4.26 above:
$E(C) = 200(2.4) = \$480$.  $V(C) = 200^2(.64) = 25600$
(units of variance are dollars${}^2$), so $\sigma = 160$.

(c) $P(C > 600) = P(Y > 3) = \frac{3}{64}\int_3^4 y^2(4-y)\ dy \doteq .26$, 
so this can certainly happen.
\vskip 10pt
\noindent
4.33.  (a)  We have $E(Y) = \int_5^7 \frac{3}{8}y (7 - y)^2\ dy = \frac{11}{2} = 5.5$.
Then 
\begin{align*}
V(Y) &= E(Y^2) - \mu^2\\
&= \int \frac{3}{8} y^2 (7 - y)^2\ dy - \left(\frac{11}{2}\right)^2\\
&= \frac{3}{20} = .15.
\end{align*}

(b)  This part can be done via Tchebysheff's Theorem (which also 
applies for continuous random variables).  But we have not
talked about this yet.  So, another way (yielding
a different interval, of course!) is to 
find a $z$ such that $F(z) = .75$, where $F$ is the 
distribution function.  We have
$F(y) = 0$ for all $y \le 5$, $F(y) = 1 - \frac{1}{8} (7-y)^3$ for 
$5 \le y \le 7$, and $F(y) = 1$ for all $y \ge 7$.  Hence we have
$$
\frac{3}{4} = 1 - \frac{1}{8}(7 - z)^3 \Leftrightarrow
z = 7 - 2^{1/3} \doteq 5.74.
$$
The interval would be $5 \le y \le 5.74$.

(c)  $P(Y < 5.5) = \int_5^{5.5} \frac{3}{8}y (7 - y)^2\ dy \doteq .5781$,
so this is actually very common.
4.41.  From Theorem 4.6 (and work in class), we know that
$E(Y) = \frac{\theta_1 + \theta_2}{2}$.  Now compute
\begin{align*}
E(Y^2) &= \int_{\theta_1}^{\theta_2} y^2 \cdot \frac{1}{\theta_2 - \theta_1}\ dy\\
&= \frac{1}{\theta_2 - \theta_2} \left.\frac{y^3}{3}\right|_{\theta_1}^{\theta_2}\\
&= \frac{\theta_2^3 - \theta_1^3}{3(\theta_2 - \theta_1)}\\
&= \frac{\theta_2^2 + \theta_2 \theta_1 + \theta_1^2}{3}.
\end{align*}
Then 
\begin{align*}
V(Y) &= E(Y^2) - E(Y)^2\\
&= \frac{\theta_2^2 + \theta_2 \theta_1 + \theta_1^2}{3} -  \left(\frac{\theta_1 + \theta_2}{2}\right)^2\\
&=\frac{4(\theta_2^2 + \theta_2 \theta_1 + \theta_1^2) - 3(\theta_2^2 + 2\theta_2 \theta_1 + \theta_1^2)}{12}\\
&=\frac{(\theta_2 - \theta_1)^2}{12},
\end{align*}
which is what we wanted to show.
\vskip 10pt
\noindent
4.42.  We have 
$$
.5 = P(Y \le \phi_{.5}) = \int_{\theta_1}^{\phi_{.5}} \frac{1}{\theta_2 - \theta_1}\ dy
= \frac{\phi_{.5} - \theta_1}{\theta_2 - \theta_1}.
$$
Solving this for $\phi_{.5}$, we get $\displaystyle{\phi_{.5} = \frac{\theta_1 + \theta_2}{2}}$.  This is the 
median, by definition.
\vskip 10pt
\noindent
4.43.  The mean of the area is 
$\overline{A} = E(A) =  \int_0^1 \pi r^2\ dr = \frac{\pi}{3}$, 
and the variance is 
\begin{align*}
V(A) &= E(A^2) - E(A)^2\\
&= \int_0^1 (\pi r^2)^2\ dr - \left(\frac{\pi}{3}\right)^2\\
&= \frac{\pi^2}{5} - \frac{\pi^2}{9}\\
&= \frac{4\pi^2}{45}.
\end{align*}
\vskip 10pt
\noindent
4.47.  The density function for $Y$ is $f(y) = \frac{1}{4}$ for 
$1 \le y \le 5$, and zero otherwise.  Hence

(a)  $P(Y \ge 2) = \int_2^5 \frac{1}{4}\ dy = \frac{3}{4}$.

(b)  The expected cost is $E(C) = E(c_0 + c_1 y^2) = c_0 + c_1 E(Y^2)$.
We can compute $E(Y^2)$ directly, or else use the computations from 
4.41 above:  $E(Y^2) = \frac{25 + 5 + 1}{3} = \frac{31}{3}$.  Therefore,
$E(C) = c_0 + \frac{31 c_1}{3}$.  
\vskip 10pt
\noindent
4.51.  This is a conditional probability:  
$$
P(Y > 65|Y > 55) = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}.
$$
\end{document}
