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\centerline{Mathematics 375 -- Probability and Statistics I}
\centerline{Solutions -- Problem Set 6}
\centerline{October 21, 2005}
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\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
$$i_1 > i_2 \Rightarrow q^{i_1} < q^{i_2} \Rightarrow
-q^{i_1} > -q^{i_2} \Rightarrow 1-q^{i_1} > 1-q^{i_2}.$$

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\noindent
4.3/ a) From the table, $P(2 < Y < 5) = .967 -.678 = .289$.
$P(2\le Y < 5) = .967 - .376 = .591$
b) $P(2 < Y \le 5) = .994 - .678 = .316$.
$P(2\le Y \le 5) = .994 - .376 = .618$.
c)  No -- $Y$ is not a continuous random variable (it's
discrete, and the claim does not apply).  

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\noindent
4.5/  a) To be a pdf, we must have 
$1 = c \int_0^2 y\ dy = 2c \Rightarrow c = 1/2$.
b)  The cdf is $F(y) = \int_{-\infty}^y f(t)\ dt = \cases{
0 & if $y \le 0$\cr y^2/4 & if $y > 0$\cr}$
c) graphs omitted.
d) $F(2) - F(1) = 1 - 1/4 = 3/4$.
e) In geometric terms, $P(1 \le Y \le 2)$ is represented
by the area under $z = y/2$ for $1 \le y \le 2$.  This 
is the area of a trapezoid, so by the usual formula
$ = (1/2 + 1)/2\cdot 1 = 3/4$.

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\noindent
4.6/ a,b) $F(y) \ge 0$ for all $y$.  $\lim_{y\to -\infty}F(y) = 0$
and  $\lim_{y\to +\infty}F(y) = 1$ are clear.  Moreover, 
$f(y) = F'(y) = \cases{0 & if $y \le 0$\cr 2y e^{-y^2}  & if $y > 0$\cr}$,
which is always $\ge 0$.  Hence $F$ is nondecreasing.
Therefore, $F$ has the properties of a distribution function
according to Theorem 4.1.  c) (Note that times are
given in 100's of hours. So, for instance, 200 hours means $y = 2$.)
$P(Y \ge 2) = 1 - F(2) = e^{-4} \doteq .018$.
d) 
$$P(Y > 1 | Y \le 2) = {P(1 < Y \le 2)\over P(Y \le 2)}
= {F(2) - F(1) \over F(2)} = {e^{-1} - e^{-4}\over 1 - e^{-4}} 
\doteq .356.$$

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

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:
$$P(Y > 1/2 | Y > 1/4) = P(Y > 1/2)/P(Y > 1/4) = {1 - F(1/2)\over 1 - F(1/4)}
= {13/16\over 123/128} = {104/123}.$$

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

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

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\noindent
4.21/ Easiest method here is to use general facts
about expected values of linear combinations:
$W = 5 - .5Y$ where $Y$ is as in problem 4.15.
Hence $E(W) = 5 - .5E(Y) = 5 - .5(17/24) \doteq 4.65$. 
$V(W) = (.5)^2 V(Y) \doteq .012$. 

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\noindent
4.26/ a)  $E(Y) = {3\over 64}\int_0^4 y^3(4 - y)\ dy =
{3\over 64}\int_0^4 4y^3  - y^4\ dy = 2.4$.
Then $E(Y^2) = {3\over 64}\int_0^4 4y^4  - y^5\ dy = 6.4$
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:
$E(C) = 200(2.4) = \$480$.  $V(C) = 200^2(.64) = 25600$
(units of variance are dollars${}^2$), so $\sigma = 160$.

c) $P(C > 600) = P(Y > 3) = {3\over 64}\int_3^4 y^2(4-y)\ dy \doteq .26$. 

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\noindent
4.33/  a) With uniform distribution on $[20,25]$, 
$P(Y < 22)  =\int_{20}^{22} {1\over 5}\ dy = {2\over 5}$.
b)  Similarly, $P(Y > 24) = 1/5$.

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\noindent
4.36/  a) Uniform distribution on $[0,500]$ so
$P(Y \ge 475) = \int_{475}^{500} {1\over 500}\ dy = {25\over 500} = .05$.
 
b) is also .05.  c)  $P(Y < 250) = .5$.

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\noindent
4.47/  Using the normal curve area table:
a) $z_0 = 0$ by symmetry of the standard normal pdf. b) $z_0 = 1.10$.
c) By the symmetry of the standard normal pdf,
to get $P(-z_0 < Z < z_0) = .90$, we must have 
$P(Z \ge z_0) = .05$.  From the table, $z_0 \doteq 1.645$.  d)  
Similar to c.  Answer:  $z_0 \doteq 2.575$.

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\noindent
4.49/  $\mu = 16$ and $\sigma = 1$.  Hence
$P(Y > 17) = P(Y - 16 > 17 - 16) = P(Z > 1)$ where 
$Z$ is standard normal.  $P(Z > 1) = .1587$ from the table

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\noindent
4.58/ a)  $P(Y > 11) = P\left({Y-7\over 2.6} > {11 - 7\over 2.6}\right)
=P(Z > 1.54) = .0618$.  b)  $P(Y < 9) = 1 - P(Y \ge 9) 
= 1 - P\left(Z \ge  {9 - 7\over 2.6}\right) = 1 - .2206 = .7794$.

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\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