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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions for Problem Set 6}
\centerline{October 22, 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}.
\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 $V(Y) = \frac{11}{20} - \left(\frac{17}{24}\right)^2 = \frac{139}{2880}$.
\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.
\end{document}
