\magnification=\magstep1
\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions -- Problem Set 6}
\centerline{October 29, 2003}
\bigskip
\noindent
4.1/Since $Y$ takes only positive integer values, $p(y) =$
$$P(Y = y) = P(Y \le y) - P(Y \le y-1) = F(y) - F(y-1) = \cases{F(1)  & if $y = 1$\cr
                                           F(y) - F(y-1) & if $y > 1$\cr}$$
\bigskip
\noindent
4.2/a) The probability function for the (discrete) geometric
random variable is $p(y) = q^{y-1}p$.  Hence if $y = i \ge 1$,
$$\eqalign{
F(i) &= P(Y \le i) = P(Y = 1) + \cdots + P(Y = i)\cr
     &= p(1 + q + q^2 + \cdots + q^{i-1})\cr
     &= p\cdot {1 - q^i\over 1 - q}\cr
     &=1 - q^i\cr}$$
(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 $y$ is the largest integer $\le y$.  Hence 
$$F(y) = \cases{ 0  & if $y < 0$\cr
                 1-q^i & if $i \le y < i$ for $i = 0, 1, 2, \ldots$\cr}$$

b)  $\lim_{y\to -\infty}F(y) = 0$ 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
$$i_1 > i_2 \Rightarrow q^{i_1} < q^{i_2} \Rightarrow
-q^{i_1} > -q^{i_2} \Rightarrow 1-q^{i_1} > 1-q^{i_2}.$$

\bigskip
\noindent
4.4/a)  To be a legal density function we must have 
$$1 = \int_{-\infty}^\infty f(y)\ dy = \int_0^1 ky(1-y)\ dy = k 
\left({y^2\over 2} - {y^3\over 3}\right)|_0^1 = k\cdot {1\over 6}$$
So $k = 6$.

b) and c)  Since the distribution is continuous,
including the endpoints of the intervals or not makes no difference.
Both probabilities equal
$$\int_{.4}^1 6y(1-y)\ dy = .648.$$

d) and e)  Both conditional probabilities equal
$${\int_0^{.4} 6y(1-y)\ dy\over \int_0^{.8} 6y(1-y)\ dy}
\doteq {.352\over .896} \doteq .393.$$

\bigskip
\noindent
4.9/a)  We must show $f(y) \ge 0$ for all $y$ and 
$\int_{-\infty}^\infty f(y)\ dy = 1$.  The first condition
follows immediately since $b > 0$ and $y^2 > 0$ for all $y > 0$.  
For the second
$$\eqalign{\int_{-\infty}^\infty f(y)\ dy 
&= \int_b^\infty {b \over y^2}\ dy\cr
&= {-b\over y}|_b^\infty = 0 - (-1) = 1.\cr}$$

b)  The cumulative distribution function $F$ is
$$\eqalign{F(y) &= \int_{-\infty}^y f(t)\ dt\cr
                &= \cases{0 & if $y \le 0$\cr
                          \int_b^y {b\over t^2}\ dt & if $y > 0$\cr}\cr
                &= \cases{0 & if $y \le b$\cr
                          1 - {b\over y} & if $y > b$\cr}\cr}$$

c)  Using part b, 
$$P(Y > b+c) = 1 - F(b+c) = 1 - (1 - {b\over b+c}) = 
{b \over b + c}.$$

d)  With $d > c$, we have by the definition of conditional probability:
$$P(Y > b+d | Y > b + c) = {P(Y > b+d)\over P(Y > b+c)}.$$
Using part c, this equals
$$\left({b\over b + d}\right)/\left({b \over b + c}\right) = {b + c\over b+d}.$$

\bigskip
\noindent
4.11/a)  We must have $\int_{-\infty}^\infty f(y)\ dy = 1$.  Hence
$$1 = \int_0^1 cy^2 + y\ dy = {c\over 3} + {1\over 2}.$$
So $c = {3\over 2}$.

b)  The formula for the cdf $F(y)$ can be derived either 
by taking the indefinite integral of $f(y)$ ``piece by piece'' and 
adjusting the constants of integration to make the resulting function 
continuous and a legal cdf (as we did in an example from class),
or else by doing the definite integral $F(y) = \int_{-\infty}^y f(t)\ dt$
as in an example in the book.  The result either way is
$$F(y) = \cases{0 & if $y < 0$\cr
                y^3/2 + y^2/2 & if $0\le y < 1$\cr
                1 & if $y \ge 1$\cr}$$

c)  Plots omitted.  d) $F(-1) = F(0) = 0$.  $F(1) = 1$.

e)  $P(Y < 1/2) = \int_0^{1/2} f(y)\ dy = F(1/2) - F(0) = 3/16$.

f)  Using the definition of conditional probability:
$$P(Y > 1/2 | Y > 1/4) = P(Y > 1/2)/P(Y > 1/4) = {1 - F(1/2)\over 1 - F(1/4)}
= {13/16\over 123/128} = {104/123}.$$

\bigskip
\noindent
4.15/Using the formula for $f(y)$ from 4.11 above, we have
$$\eqalign{\mu = E(Y) &= \int_{-\infty}^{\infty} yf(y)\ dy\cr
&=\int_0^1 y({3\over 2}y^2 + y)\ dy\cr
&={17\over 24}\cr}$$
Then, to get $V(Y)$, we can proceed as follows:
$$E(Y^2) = \int_0^1 y^2({3\over 2}y^2 + y)\ dy = 11/20.$$
Finally, 
$$V(Y) = E(Y^2) - (E(Y))^2 = 11/20 - (17/24)^2 \doteq .0483$$

\bigskip
\noindent
4.18/The proof here is virtually the same as in the discrete case, 
using Theorem 4.5:  Let $\mu = E(Y)$.  Then 
$$\eqalign{V(Y) = E((Y-\mu)^2) &= E(Y^2 - 2\mu Y + \mu^2)\cr
                &= E(Y^2) - 2\mu E(Y) +\mu^2\cr
                &= E(Y^2) - 2\mu^2 + \mu^2\cr
                &= E(Y^2) - (E(Y))^2.\cr}$$

\bigskip
\noindent
4.19/To perform the desired calculations, we need the density
function $f(y) = F'(y)$.  We have
$$f(y) = \cases{1/8 & if $0 < y < 2$\cr
                y/8 & if $2 < y < 4$\cr
                 0 & otherwise\cr}$$
Then we integrate by splitting the integral up into two terms
based on the split in the definition of $f(y)$ at $y = 2$:
$$\eqalign{E(Y) &= \int_{-\infty}^{\infty} yf(y)\ dy\cr
&= \int_0^2 y/8\ dy + \int_2^4 y^2/8\ dy\cr
&=y^2/16|_0^2 + y^3/24|_2^4\cr
&\doteq 2.583\cr}$$
Similarly,
$$\eqalign{E(Y^2) &= \int_{-\infty}^{\infty} y^2f(y)\ dy\cr
&= \int_0^2 y^2/8\ dy + \int_2^4 y^3/8\ dy\cr
&=y^3/24|_0^2 + y^4/32|_2^4\cr
&\doteq 7.833\cr}$$
Then using the formula from 4.18,
$$V(Y) = E(Y^2) - (E(Y))^2 \doteq 7.833 - (2.583)^2 \doteq 1.16.$$

\bigskip
\noindent
4.20/a)  Using Theorem 4.5 (with $g_1(Y) = aY$ and $g_2(Y) = b$ (constant)), 
we have
$$E(aY+b) = aE(Y) + b = a\mu + b.$$

b) From the definition of the variance,
 $V(aY+b) = E( ((aY+b) - (a\mu + b))^2)$, and cancelling inside
the square and factoring out $a^2$ from what remains, we have:
$$\eqalign{&= E(a^2(Y-\mu)^2)\cr
           &= a^2E((Y-\mu)^2)\cr
           &= a^2V(Y).\cr}$$ 
           
\bigskip
\noindent
4.28/a)  The uniform cdf is this anti-derivative of the uniform
density ($f(y) = 1$ if $0 < y < 1$ and $0$ otherwise):
$$F(y) = \cases{0 & if $y < 0$\cr
                y & if $0 \le y < 1$\cr
                1 & if $y \ge 1$\cr}$$
          
b) Since $a \ge 0$ and $a + b\le 1$, we can just
use the ``middle case'' of $F$ to get:
$$P(a \le Y \le a+b) = \int_a^{a+b} 1\ dy = F(a+b) - F(a) = a+b-a = b.$$

\bigskip
\noindent
4.29/a) The parachutist is closer to $A$ than $B$ whenever
$y$ = the random location satisfies $A \le y \le {A+B\over 2}$.
We get 
$$P\left(A \le Y \le {A+B\over 2}\right) = \int_A^{(A+B)/2} {1\over B-A}\ dy = {1\over 2}$$

b)  The parachutist is at least 3 times farther from $A$ than from
$B$ whenever
$y$ = the random location satisfies ${A+3B\over 4} \le y \le B$.
(Find the cutoff point by solving for $y$ in $y - A = 3(b - y)$.)
We get 
$$P\left({A+3B\over 4} \le Y \le B\right) = \int_{(A+3B)/4}^B {1\over B-A}\ dy = {1\over 4}$$

\bigskip\noindent
4.35/a)  $Y =$ the delivery time is uniform on $[1,5]$. So its
density function is $f(y) = 1/4$ for $1 \le y \le 5$ and 0 otherwise.
Hence $P(Y > 2) = \int_2^5 1/4\ dy = 3/4$.

b)  The expected value of the cost is 
$$E(C) = \int_1^5 (c_0 + c_1y^2)\cdot {1\over 4}\ dy = 
(c_0 y/4 + c_1 y^3/12)|_1^5 = c_0 + 31c_1/3.$$

\bigskip
\noindent
4.46/Using the table of standard normal probabilities in 
the text,
\item{a)} $P(0 \le Z \le 1.2) = .5 - .1151 = .3849$.
\item{b)} By symmetry, $P(-.9 \le Z \le 0) = P(0 \le Z \le .9) = 
.5 - .1841 = .3159$.
\item{c)} $P(.3 \le Z \le 1.56) = .3821 - .0594 = .3227$.
\item{d)} By symmetry, $P(-1.56\le Z \le -.2) = P(.2 \le Z \le 1.56)
 = .4207-.0594 = .3613$.
 
 \bigskip
 \noindent
 4.57/a)
$$P(.12 \le Y \le .13) =
P\left({.12 - .13\over .005} \le {Y - .13\over .005} \le {.14 - .13\over .005}\right)
=P\left(-2 \le {Y - .13\over .005} \le 2\right)$$
From the table for the standard normal $Z$, this is $.9544$
(Note ${Y - .13\over .005} = {Y - \mu\over \sigma}$ is a standard
normal).

b)  The number that meet the specifications is a 
binomial (discrete) random variable with $n = 4$ and $p = .9544$ from 
part a(!).  Hence the probability that all four meet the specifications
is ${4\choose 4}(.9544)^4(.0456)\doteq .83$.

\bigskip
\noindent
4.58/
\item{a)}  Convert to a probability involving a standard normal:
$$P(Y > 11) = P\left({Y-7\over 2.6} > {11 - 7\over 2.6}\right) = P(Z > 1.538) = .0618$$
\item{b)}  Similarly,
$$P(Y < 9) = P\left({Y-7\over 2.6} < {9 - 7\over 2.6}\right) = P(Z < .77) = 1 - .2206 = .7794.$$

\bigskip
\noindent
4.61/Here we will use the standard normal table ``backwards''.
We want to pick $\mu$ (the mean fill amount) so that 
$P(Y > 8) = .01$.  Equivalently, so that 
$P({Y - \mu\over .3} > {8-\mu\over .3}) > .01$.
The entry in the standard normal table for $z = 2.33$ gives 
$P(Z > 2.33) = .0099$, so we will use that value.
${8 - \mu\over .3} = 2.33 \Rightarrow \mu \doteq 7.3$.

\bigskip
\noindent
4.65/The constant ${1\over \sqrt{2\pi\sigma^2}}$
in the formula for the normal density is irrelevant for the location
of the inflection points, so we will ignore it and consider the 
following function $g(y)$ and its derivatives:
$$\eqalign{g(y) &= e^{-(y-\mu)^2/(2\sigma^2)}\cr
\Rightarrow g'(y) &= -e^{-(y-\mu)^2/(2\sigma^2)}\cdot (y-\mu)/\sigma^2 \quad {\rm chain\ rule}\cr
\Rightarrow g''(y) &= -e^{-(y-\mu)^2/(2\sigma^2)}
\left({1\over \sigma^2} - {(y-\mu)^2\over\sigma^4}\right)\quad {\rm product\ rule}\cr}$$
We have $g''(y) = 0$ if and only if
$$\eqalign{ 0 &= {1\over \sigma^2} - {(y-\mu)^2\over\sigma^4}\cr
   \Rightarrow (y-\mu)^2 &= \sigma^2\cr
   \Rightarrow y -\mu &= \pm \sigma\cr
   \Rightarrow y &= \mu \pm \sigma\cr}$$
which is what we wanted to show.
\bye


\bye                 
    