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# Solutions to practice problems {#solutions}
```{r setup14, include=FALSE}
knitr::opts_chunk$set(echo = TRUE,
prompt = FALSE,
tidy = TRUE,
collapse = TRUE)
library("tidyverse")
EmpData <- read_csv("sampledata/EmploymentData.csv")
# Make permanent changes to EmpData
EmpData <- EmpData %>%
mutate(MonthYr = as.Date(MonthYr, "%m/%d/%Y")) %>%
mutate(UnempPct = 100*UnempRate) %>%
mutate(LFPPct = 100*LFPRate)
```
This appendix provides solutions to all end-of-chapter practice problems.
## **2** Basic data cleaning with Excel {-#answers-basic-data-cleaning-with-excel}
[Click here to see the problems](#problems-basic-data-cleaning-with-excel)
1. Table (b) shows a tidy data set.
2. Master, working, and archive versions have the following features:
a. The master version represents the project at its most recent completed
stage. There should be only one, and it should not be edited directly.
b. A working copy represents current work in progress. There should be one,
and it can be edited directly.
c. An archive copy represents an earlier master version or working copy that
you are keeping for future reference. There can be many, and they should
not be edited directly.
3. Characteristic (a) is necessary for a variable to function as an ID variable.
4. The relevant tools are:
a. Fill (copy/paste is also OK here).
b. Copy/paste (fill only works on adjacent cells).
c. Series.
d. Insert.
5. The formulas are:
a. `=SQRT(A2)`
b. `=MIN(A2:A100)`
c. `=ABS(A2-B2)`
6. Cell E15 is 2 columns and 3 rows away from cell C12. So we change all
relative references from column B to column D, and all relative references
from row 2 to row 5, and all relative references from row 10 to row 13.
Absolute references are not changed.
a. `=D5`
b. `=$B$2`
c. `=$B5`
d. `=D$2`
e. `=SUM(D5:D13)`
f. `=SUM($B$2:$B$10)`
g. `=SUM($B5,$B13)`
h. `=SUM(D$2,D$10)`
7. The formulas are:
a. `=IF(A2<0.05,"Reject","Fail to reject")`
b. `=CONCAT("A2 = ",A2)`
c. `=LEFT(A2,2)`
8. The formulas are:
a. `=MONTH(TODAY())`
b. `=TODAY()+100`
c. `=TODAY()-DATE(1969,11,20)` - using my birthday; yours will obviously be
different.
9. If you ran into trouble doing this, review the section on
[saving and exporting data](saving-and-exporting-data).
a. Note that there are actually multiple options for exporting CSV files,
including "CSV UTF-8", "CSV (comma delimited)", "CSV (Macintosh)" and
"CSV (MS-DOS)". These all have tiny differences in how the data is
stored, but any of them will work for our purposes.
b. Note that it is important to close Excel here. Saving the worksheet as a
CSV file involves the loss of information, but Excel will hold onto that
information until you close the file.
c. Note that Excel usually makes itself the default application for CSV
files, so you can open them in Excel directly or by double-clicking on
them.
d. The worksheets differ in many ways:
- The *Raw data* worksheet is not in the CSV file.
- The formatting (e.g. the boldface in the top row) is gone.
- Columns E, F, G, and H contain values rather than formulas.
## **3** Probability and random events {-#answers-probability}
[Click here to see the problems](#problems-probability)
1. The sample space is the set of all possible outcomes for $(r,w)$, and its
cardinality is just the number of elements it has.
a. The sample space is:
\begin{align}
\Omega &= \left\{ \begin{aligned}
& (1,1),(1,2),(1,3),(1,4),(1,5),(1,6), \\
& (2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\
& (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \\
& (4,1),(4,2),(4,3),(4,4),(4,5),(4,6), \\
& (5,1),(5,2),(5,3),(5,4),(5,5),(5,6), \\
& (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \\
\end{aligned}
\right\}
\end{align}
b. Counting up the number of elements in the set, we get $|\Omega| = 36$.
2. Each event is just a set listing the $(r,w)$ outcomes that satisfy the
relevant conditions.
a. Yo wins whenever $r+w = 11$.
\begin{align}
YoWins &= \{(5,6),(6,5)\}
\end{align}
b. Snake eyes wins when $r+w = 2$:
\begin{align}
SnakeEyesWins = \{(1,1)\}
\end{align}
c. Boxcars wins when $r+w = 12$:
\begin{align}
BoxcarsWins = \{(6,6)\}
\end{align}
d. Field wins when $r+w \in \{2,3,4,9,10,11,12\}$:
\begin{align}
FieldWins &= \left\{\begin{aligned}
& (1,1),(1,2),(1,3), \\
& (2,1),(2,2), \\
& (3,1),(3,6), \\
& (4,5),(4,6), \\
& (5,4),(5,5),(5,6), \\
& (6,3),(6,4),(6,5),(6,6) \\
\end{aligned}\right\}
\end{align}
3. Statements (b) and (c) are true.
4. Statements (a) and (d) are true.
5. Statements (b) and (c) are true.
6. Event (c) is an elementary event.
7. Statements (a), (c), (e), (g), and (i) are true.
8. Statements (b) and (c) are true.
9. Statements (a), (d), (e), (f), and (h) are true.
10. Statements (a) and (b) are true.
11. All three elementary events have probability $1/36 \approx 0.028$
12. The probability of each event can be calculated by adding up the
probabilities of its elementary events.
a. The probability a bet on Yo wins is:
\begin{align}
\Pr(YoWins) &= \Pr(\{(5,6),(6,5)\}) \\
&= \Pr(\{(5,6)\}) + \Pr(\{(6,5)\}) \\
&= 2/36 \\
&\approx 0.056
\end{align}
b. The probability a bet on Snake eyes wins is:
\begin{align}
\Pr(SnakeEyesWins) &= \Pr(\{(1,1)\}) \\
&= 1/36 \\
&\approx 0.028
\end{align}
c. The probability a bet on Boxcars wins is:
\begin{align}
\Pr(BoxcarsWins) &= \Pr(\{(6,6)\}) \\
&= 1/36 \\
&\approx 0.028
\end{align}
d. The probability a bet on Field wins is:
\begin{align}
\Pr(FieldWins) &= \Pr\left(\left\{\begin{aligned}
& (1,1),(1,2),(1,3), \\
& (2,1),(2,2), \\
& (3,1),(3,6), \\
& (4,5),(4,6), \\
& (5,4),(5,5),(5,6), \\
& (6,3),(6,4),(6,5),(6,6) \\
\end{aligned}\right\}\right) \\
&= \Pr(\{(1,1)\}) + \Pr(\{(1,2)\}) + \cdots + \Pr(\{(6,6)\}) \\
&= 16/36 \\
&\approx 0.444
\end{align}
13. To calculate the joint probability, start by calculating the intersection
of the two events:
a. The joint probability is:
\begin{align}
\Pr(YoWins \cap BoxcarsWins)
&= \Pr(\{(5,6),(6,5)\} \cap \{(6,6)\}) \\
&= \Pr(\emptyset) \\
&= 0
\end{align}
b. The joint probability is:
\begin{align}
\Pr(YoWins \cap FieldWins)
&= \Pr\left(\{(5,6),(6,5)\} \cap \left\{\begin{aligned}
& (1,1),(1,2),(1,3), \\
& (2,1),(2,2), \\
& (3,1),(3,6), \\
& (4,5),(4,6), \\
& (5,4),(5,5),(5,6), \\
& (6,3),(6,4),(6,5),(6,6) \\
\end{aligned}\right\}\right) \\
&= \Pr(\{(5,6),(6,5)\}) \\
&= 2/36 \\
&\approx 0.056
\end{align}
c. The joint probability is:
\begin{align}
\Pr(YoWins \cap BoxcarsWins^C)
&= \Pr(\{(5,6),(6,5)\} \cap \{(6,6)\}^C) \\
&= \Pr(\{(5,6),(6,5)\}) \\
&= 2/36 \\
&\approx 0.056
\end{align}
14. The conditional probability is just the ratio of the joint probability to
the probability of the event we are conditioning on:
a. The conditional probability is:
\begin{align}
\Pr(YoWins | BoxcarsWins)
&= \frac{Pr(YoWins \cap BoxcarsWins)}{\Pr(BoxcarsWins)} \\
&= \frac{0}{1/36} \\
&= 0
\end{align}
b. The conditional probability is:
\begin{align}
\Pr(YoWins | FieldWins)
&= \frac{Pr(YoWins \cap FieldWins)}{\Pr(FieldWins)} \\
&= \frac{2/36}{16/36} \\
&= 0.125
\end{align}
c. The conditional probability is:
\begin{align}
\Pr(YoWins | BoxcarsWins^C)
&= \frac{Pr(YoWins \cap BoxcarsWins^C)}{\Pr(BoxcarsWins^C)} \\
&= \frac{2/36}{1-1/36} \\
&\approx 0.057
\end{align}
d. The conditional probability is:
\begin{align}
\Pr(FieldWins | YoWins)
&= \frac{Pr(FieldWins \cap YoWins)}{\Pr(YoWins)} \\
&= \frac{2/36}{2/36} \\
&= 1
\end{align}
e. The conditional probability is:
\begin{align}
\Pr(BoxcarsWins | YoWins)
&= \frac{Pr(BoxcarsWins \cap Yo)}{\Pr(Yo)} \\
&= \frac{0}{2/36} \\
&= 0
\end{align}
15. The events in (e) are independent.
16. Statements (b) and (c) are true.
17. The probability that Pass wins is:
\begin{align*}
\Pr(PassWins) &= \Pr(PassWins|c = 2)\Pr(c = 2) + \cdots + \Pr(PassWins|c = 12)\Pr(c = 12) \nonumber \\
&= \begin{aligned}[t]
& (0 \times 1/36) + (0 \times 2/36) + (3/9 \times 3/36) \\
& + (4/10 \times 4/36) + (5/11 \times 5/36) + (1 \times 6/36) \\
& + (5/11 \times 5/36) + (4/10 \times 4/36) + (3/9 \times 3/36) \\
& + (1 \times 2/36) + (0 \times 1/36) \\
\end{aligned} \\
&= 244/495 \\
&\approx 0.4929
\end{align*}
Notice that this is less than 50\% (the house always has an advantage) but
it is greater than the probability of winning a bet on red or black in
roulette, which we earlier calculated to be about 0.486. Craps is typically
the most favorable casino game to players (lowest house advantage), if you
play Pass.
18. The conditional return probabilities are:
\begin{align*}
\Pr(Return|BodyHit) &= \frac{\Pr(BodyHit|Return)\Pr(Return)}{\Pr(BodyHit)} \\
&= \frac{0.7 \times 0.8}{0.6} \\
&\approx 0.93 \\
\Pr(Return|EngineHit) &= \frac{\Pr(EngineHit|Return)\Pr(Return)}{\Pr(EngineHit)} \\
&= \frac{0.1 \times 0.8}{0.6} \\
&\approx 0.13
\end{align*}
## **4** Introduction to random variables {-#answers-random-variables}
[Click here to see the problems](#problems-random-variables)
1. We can define $t = r+w$.
2. We can define $y = I(r+w = 11)$.
3. The support of a random variable is the set of all values with positive
probability.
a. The support of $r$ is $S_r = \{1,2,3,4,5,6\}$.
b. The support of $t$ is $S_t = \{2,3,4,5,6,7,8,9,10,11,12\}$.
c. The support of $y$ is $S_y = \{0,1\}$.
4. The range of a random variable is just the interval defined by the support's
minimum and maximum values.
a. The range of $r$ is $[1,6]$.
b. The range of $t$ is $[2,12]$.
c. The range of $y$ is $[0,1]$.
5. The PDF can be derived from the probability distribution of the underlying
outcome $(r,w)$, which was calculated in the previous chapter.
a. The PDF of $r$ is:
\begin{align}
f_r(a) &= \begin{cases}
1/6 & a \in \{1,2,3,4,5,6\} \\
0 & \textrm{otherwise} \\
\end{cases}
\end{align}
b. The PDF of $t$ is:
\begin{align}
f_t(a) &= \begin{cases}
1/36 & a \in \{2,12\} \\
2/36 \textrm{ (or $1/18$)} & a \in \{3,11\} \\
3/36 \textrm{ (or $1/12$)} & a \in \{4,10\} \\
4/36 \textrm{ (or $1/9$)} & a \in \{5,9\} \\
5/36 & a \in \{6,8\} \\
6/36 \textrm{ (or $1/6$)} & a = 7 \\
0 & \textrm{otherwise} \\
\end{cases}
\end{align}
c. The PDF of $y$ is:
\begin{align}
f_y(a) &= \begin{cases}
17/18 & a = 0 \\
1/18 & a = 1 \\
0 & \textrm{otherwise} \\
\end{cases}
\end{align}
6. We can construct the CDF by cumulatively summing up the PDF:
a. The CDF of $r$ is:
\begin{align}
F_r(a) &= \begin{cases}
0 & a < 1 \\
1/6 & 1 \leq a < 2 \\
1/3 & 2 \leq a < 3 \\
1/2 & 3 \leq a < 4 \\
2/3 & 4 \leq a < 5 \\
5/6 & 5 \leq a < 6 \\
1 & a \geq 6 \\
\end{cases}
\end{align}
b. The CDF of $y$ is:
\begin{align}
F_y(a) &= \begin{cases}
0 & a < 0 \\
17/18 & 0 \leq a < 1 \\
1 & a \geq 1 \\
\end{cases}
\end{align}
7. The interval probabilities are:
a. $\Pr(x \leq 5) = F_x(5) = 0.8$
b. $\Pr(x < 5) = F_x(5) - f_x(5) = 0.7$
c. $\Pr(x > 5) = 1 - F_x(5) = 0.2$
d. $\Pr(x \geq 5) 1 - F_x(5) + f_x(5) = 0.3$
e. $\Pr(0 < x \leq 5) = F_x(5) - F_x(0) = 0.5$
f. $\Pr(0 \leq x \leq 5) = F_x(5) - F_x(0) + f_x(0) = 0.6$
g. $\Pr(0 < x < 5) = F_x(5) - F_x(0) - f_x(5) = 0.4$
h. $\Pr(0 \leq x < 5) = F_x(5) - F_x(0) + f_x(0) - f_x(5) = 0.5$
8. We find the expected value by applying the definition:
a. The expected value is
\begin{align}
E(r) &= \sum_{a \in S_r} a\Pr(r=a) \\
&= 1*\frac{1}{6} + 2*\frac{1}{6} + 3*\frac{1}{6} + 4*\frac{1}{6} + 5*\frac{1}{6} + 6*\frac{1}{6} \\
&= 21/6 \\
&= 3.5
\end{align}
b. The expected value is
\begin{align}
E(r^2) &= \sum_{a \in S_r} a^2\Pr(r=a) \\
&= 1^2*\frac{1}{6} + 2^2*\frac{1}{6} + 3^2*\frac{1}{6} + 4^2*\frac{1}{6} + 5^2*\frac{1}{6} + 6^2*\frac{1}{6} \\
&= 91/6 \\
&= 15.17
\end{align}
9. The key here is to write down the definition of the specific quantile you are
looking for, then substitute the CDF you derived earlier.
a. The median of $r$ is:
\begin{align}
F_r^{-1}(0.5) &= \min \{a: \Pr(r \leq a) \geq 0.5\} \\
&= \min \{a: F_r(a) \geq 0.5\} \\
&= \min \{a: a \geq 3\} \\
&= 3
\end{align}
b. The 0.25 quantile of $r$ is
\begin{align}
F_r^{-1}(0.25) &= \min \{a: \Pr(r \leq a) \geq 0.25\} \\
&= \min \{a: F_r(a) \geq 0.25\} \\
&= \min \{a: a \geq 2\} \\
&= 2
\end{align}
c. The 75th percentile of $r$ is just its 0.75 quantile:
\begin{align}
F_r^{-1}(0.75) &= \min \{a: \Pr(r \leq a) \geq 0.75\} \\
&= \min \{a: F_r(a) \geq 0.75\} \\
&= \min \{a: a \geq 5\} \\
&= 5
\end{align}
10. Let $d = (y - E(y))^2$.
a. We can derive the PDF of $d$ from the PDF of $y$:
\begin{align}
f_{d}(a) &= \begin{cases}
17/18 & a = (0-1/18)^2 \\
1/18 & a = (1-1/18)^2 \\
0 & \textrm{otherwise} \\
\end{cases}
\end{align}
b. The expected value is:
\begin{align}
E(d) &= (0-1/18)^2 * \frac{17}{18} + (1-1/18)^2 *\frac{1}{18} \\
&\approx 0.0525
\end{align}
c. The variance is:
\begin{align}
var(y) &= E(d) \\
&\approx 0.0525
\end{align}
11. Earlier, you found $E(r) = 3.5$ and $E(r^2) = 15.17$. So we can apply
our result that $var(x) = E(x^2) - E(x)^2$ for a simpler way of calculating
the variance:
\begin{align}
var(r) &= E(r^2) - E(r)^2 \\
&= 15.17 - 3.5^2 \\
&= 2.92
\end{align}
12. The standard deviation is just the square root of the variance.
a. The standard deviation is:
\begin{align}
sd(y) = \sqrt{var(y)} \approx \sqrt{0.0525} \approx 0.229
\end{align}
b. The standard deviation is:
\begin{align}
sd(r) &= \sqrt{var(r)} = \sqrt{2.92} \approx 1.71
\end{align}
13. The key here is to apply the formulas for the expected value and variance
of a linear function of a random variable.
a. The expected value is:
\begin{align}
E(W) &= E(160*y - 10) \\
&= 160*E(y) - 10 \\
&= 160*1/18 - 10 \\
&\approx -1.11
\end{align}
b. The variance is:
\begin{align}
var(W) &= var(160*y - 10) \\
&= 160^2*var(y) \\
&= 160^2*1/18*17/18 \\
&\approx 1343.2
\end{align}
c. The event probability is
\begin{align}
\Pr(W > 0) &= \Pr(y=1) = 1/18 \approx 0.056
\end{align}
14. Applying the formulas for a linear function:
a. The expected value is:
\begin{align}
E(W_{10}) &= E(16*Y_{10} - 10) \\
&= 16*E(Y_{10}) - 10 \\
&= 16*10*1/18 - 10 \\
&\approx -1.11
\end{align}
b. The variance is:
\begin{align}
var(W_{10}) &= var(16*Y_{10} - 10) \\
&= 16^2var(Y_10) \\
&= 16^2 * 10*1/18*17/18 \\
&\approx 134.32
\end{align}
c. The event probability is
\begin{align}
\Pr(W_{10} > 0) &= \Pr(Y_{10} > 10/16) \approx 0.435
\end{align}
15. The standardized form is constructed by subtracting the mean and dividing
by the standard deviation.
a. Since $E(y) = 1/18 \approx 0.056$ and $sd(y) \approx 0.229$, the
standardized variable will be:
\begin{equation}
z = \frac{y - E(y)}{sd(y)} = \frac{y - 0.056}{0.229}
\end{equation}
b. The support of $y$ is $S_y = \{0,1\}$, so the support of $z$ is:
\begin{align}
S_z &= \left\{\frac{0 - 0.056}{0.229}, \frac{1 - 0.056}{0.229} \right\} \\
&= \{-0.245, 4.122\}
\end{align}
c. The PDF of $z$ is:
\begin{align}
f_z(a) &= \begin{cases}
17/18 & a = -0.245 \\
1/18 & a = 4.122 \\
0 & \textrm{otherwise} \\
\end{cases}
\end{align}
16. The Bernoulli distribution describes any random variable (like $y$) that
has a binary $\{0,1\}$ support.
a. $y$ has the $Bernouili(p)$ distribution with $p=1/18$, or
\begin{align}
y \sim Bernoulli(1/18)
\end{align}
b. In the Bernoulli distribution $E(y)=p =1/18$.
c. In the Bernoulli distribution, $var(y) = p*(1-p) = 1/18 * 17/18 \approx 0.525$.
17. The binomial distribution describes any random variable that describes the
number of times an event happens in a fixed number of independent trials.
a. $Y_{10}$ has the binomial distribution with $n=10$ and $p=1/18$. We can
also write that as:
\begin{align}
Y_{10} \sim Binomial(10,1/18)
\end{align}
b. For the Binomial distribution we have:
\begin{align}
E(Y_{10}) = np = 10*1/18 \approx 0.556
\end{align}
c. For the binomial distribution we have:
\begin{align}
var(Y_{10}) = np(1-p) = 10*1/18*17/18 \approx 0.525
\end{align}
d. The Excel formula for $\Pr(Y_{10} = 0)$ would be
`=BINOM.DIST(0,10,1/18,FALSE)` which produces the result
\begin{align}
\Pr(Y_{10} = 0) \approx 0.565
\end{align}
e. The Excel formula for $\Pr(Y_{10} \leq 10/16)$ would be
`=BINOM.DIST(10/16,10,1/18,TRUE)` which produces the result:
\begin{align}
\Pr(Y_{10} \leq 10/16) \approx 0.565
\end{align}
f. The Excel formula for $\Pr(Y_{10} > 10/16)$ would be
`=1-BINOM.DIST(10/16,10,1/18,TRUE)` which produces the result:
\begin{align}
\Pr(Y_{10} > 10/16) \approx 0.435
\end{align}
## **5** Basic data analysis with Excel {-#answers-basic-data-analysis}
[Click here to see the problems](#problems-basic-data-analysis)
1. The table should look like this:
2. The answers are:
a. Sample size is 3
b. Sample average of age is 45.
c. Sample median of age is 32.
d. Sample variance of age is 829.
e. Sample standard deviation of age is 28.8.
3. The numerical answers and Excel formulas (assuming the table starts in cell
A1) are:
a. 3 `=COUNT(B2:B4)`
b. 45 `=AVERAGE(B2:B4)`
c. 32 `=MEDIAN(B2:B4)`
d. 28.5 `=PERCENTILE.INC(B2:B4,0.25)`
e. 829 `=VAR.S(B2:B4)`
f. 28.8 `=STDEV.S(B2:B4)`
4. The binned frequency table should look something like this:
| Range | Frequency | % Freq |
|:------|:---------:|:------:|
| 0-10 | 0 | 0 |
| 11-20 | 0 | 0 |
| 21-30 | 1 | 33 |
| 31-40 | 1 | 33 |
| 41-50 | 0 | 0 |
| 51-60 | 0 | 0 |
| 61-70 | 0 | 0 |
| 71-80 | 1 | 33 |
5. The time series graph should look something like this:
6. The histogram should look something like this:
## **6** More on random variables {-#answers-more-on-random-variables}
[Click here to see the problems](#problems-more-on-random-variables)
1. Based on these figures:
a. $x$ is continuous.
b. The top graph shows the PDF and the bottom graph shows the CDF.
c. The median of $x$ is about 3.
d. $x$ is more likely to be between 0 and 10.
2. This question uses various formulas for the uniform distribution.
a. The PDF is:
\begin{align}
f_x(a) &= \begin{cases}
0 & a < -1 \\
0.5 & -1 \leq a \leq 1 \\
0 & a > 1 \\
\end{cases}
\end{align}
b. The CDF is:
\begin{align}
f_x(a) &= \begin{cases}
0 & a < -1 \\
\frac{a+1}{2} & -1 \leq a \leq 1 \\
1 & a > 1 \\
\end{cases}
\end{align}
c. $x$ is a continuous random variable so $\Pr(x = 0) = 0$
d. $x$ is a continuous random variable so:
\begin{align}
\Pr(0 < x < 0.5) &= F_x(0.5) - F_x(0) \\
&= \frac{0.5+1}{2} - \frac{1}{2} \\
&= 0.25
\end{align}
e. $x$ is a continuous random variable so:
\begin{align}
\Pr(0 \leq x \leq 0.5) &= F_x(0.5) - F_x(0) \\
&= \frac{0.5+1}{2} - \frac{1}{2} \\
&= 0.25
\end{align}
f. The median is:
\begin{align}
F_x^{-1}(0.5) &= \min \{a: \Pr(x \leq a) \geq 0.5\} \\
&= \min \{a: F_x(a) \geq 0.5\} \\
&= \min \{a: \frac{a+1}{2} \geq 0.5\} \\
&= \textrm{the $a$ that solves the equation $\frac{a+1}{2} = 0.5$} \\
&= 0
\end{align}
g. The 75th percentile is
\begin{align}
F_x^{-1}(0.75) &= \min \{a: \Pr(x \leq a) \geq 0.75\} \\
&= \min \{a: F_x(a) \geq 0.75\} \\
&= \min \{a: \frac{a+1}{2} \geq 0.75\} \\
&= \textrm{the $a$ that solves the equation $\frac{a+1}{2} = 0.75$} \\
&= 0.5
\end{align}
h. The mean of a $Uniform(a,b)$ random variable is $(a+b)/2$, so
\begin{align}
E(x) &= \frac{1 + (-1)}{2} \\
&= 0
\end{align}
i. The variance of a $Uniform(a,b)$ random variable is $(b-a)^2/12$, so
\begin{align}
var(x) &= (b-a)^2/12 \\
&= \frac{(1 - (-1))^2}{12} \\
&\approx 0.33
\end{align}
3. A linear function of a uniform random variable is also uniform.
a. When $x = -1$, we have $y = 3*(-1)+5 = 2$. When $x = 1$, we have
$y = 3*(1)+5 = 8$. Therefore $y \sim U(2,8)$
b. The expected value is $E(y) = (2+8)/2 = 5$.
4. This question applies various formulas for the normal distribution.
a. The expected value of a $N(\mu,\sigma^2)$ random variable is $\mu$,
so $E(x) = 10$
b. The median of a $N(\mu,\sigma^2)$ random variable is $\mu$,
so $Median(x) = 10$.
c. The variance of a $N(\mu,\sigma^2)$ random variable is $\sigma^2$,
so $var(x) = 4$.
d. The standard deviation is $sd(x) = \sqrt{var(x)} = \sqrt{4} = 2$
b. The Excel formula would be `=NORM.DIST(11,10,2,TRUE)` which
yields the result $\Pr(x \leq 11) \approx 0.691$
5. A linear function of a normal random variable is also normal.
a. The distribution of $y$ will be normal with mean $3E(x) + 5 = 35$ and
variance $3^2 var(x) = 36$, so $y \sim N(35,36)$
b. If $x \sim N(10,4)$ then $z = \frac{x - 10}{2} \sim N(0,1)$.
c. Solving for $x$ in terms of $z$ we get $x = 2z + 10$
\begin{align}
\Pr(x \leq 11) &= \Pr(2z + 10 \leq 11) \\
&= \Pr(z \leq (11-10)/2) \\
&= \Pr(z \leq 0.5) \\
&= \Phi(0.5)
\end{align}
d. The correct Excel formula would be `=NORM.S.DIST(0.5,TRUE)` which
yields the result $\Pr(x \leq 11) \approx 0.691$.
6. The key here is to redefine joint events for $y$ and $b$ as events for the
single random variable $t$.
a. The joint PDF is:
\begin{align}
f_{y,b}(1,1) &= \Pr(y = 1 \cap b = 1) \\
&= \Pr(t = 11 \cap t = 12) \\
&= \Pr(\emptyset) \\
&= 0
\end{align}
b. The joint PDF is:
\begin{align}
f_{y,b}(0,1) &= \Pr(y \neq 1 \cap b = 1) \\
&= \Pr(t \neq 11 \cap t = 12) \\
&= \Pr(t = 12) \\
&= f_t(12) \\
&= 1/36 \\
&\approx 0.028
\end{align}
c. The joint PDF is:
\begin{align}
f_{y,b}(1,0) &= \Pr(y = 1 \cap b = 0) \\
&= \Pr(t = 11 \cap t \neq 12) \\
&= \Pr(t = 11) \\
&= f_t(11) \\
&= 1/18 \\
&\approx 0.056
\end{align}
d. The joint PDF is:
\begin{align}
f_{y,b}(0,0) &= \Pr(y = 0 \cap b = 0) \\
&= \Pr(t \neq 11 \cap t \neq 12) \\
&= \Pr(t \notin \{11,12\}) \\
&= 1 - \Pr(t \in \{11,12\}) \\
&= 1 - 1/36 - 1/18 \\
&= 11/12 \\
&\approx 0.917
\end{align}
7. The marginal PDF is constructed by adding together the joint PDFs.
a. The marginal PDF is:
\begin{align}
f_b(0) &= f_{y,b}(0,0) + f_{y,b}(1,0) \\
&= 11/12 + 1/18 \\
&= 35/36 \\
&\approx 0.972
\end{align}
b. The marginal PDF is:
\begin{align}
f_b(1) &= f_{y,b}(0,1) + f_{y,b}(1,1) \\
&= 1/36 + 0 \\
&= 1/36 \\
&\approx 0.028
\end{align}
c. The expected value is:
\begin{align}
E(b) &= 0*f_b(0) + 1*f_b(1) \\
&= f_b(1) \\
&= 1/36 \\
&\approx 0.028
\end{align}
8. The conditional PDF is the ratio of the joint PDF to the marginal PDF:
a. The conditional PDF is:
\begin{align}
f_{y|b}(1,1) &= \Pr(y=1|b=1) \\
&= \frac{f_{y,b}(1,1)}{f_b(1)} \\
&= \frac{0}{1/36} \\
&= 0
\end{align}
b. The conditional PDF is:
\begin{align}
f_{y|b}(0,1) &= \Pr(y=0|b=1) \\
&= \frac{f_{y,b}(0,1)}{f_b(1)} \\
&= \frac{1/36}{1/36} \\
&= 1
\end{align}
c. The conditional PDF is:
\begin{align}
f_{y|b}(1,0) &= \Pr(y=1|b=0) \\
&= \frac{f_{y,b}(1,0)}{f_b(0)} \\
&= \frac{1/18}{35/36} \\
&\approx 0.057
\end{align}
d. The conditional PDF is:
\begin{align}
f_{y|b}(0,0) &= \Pr(y=0|b=0) \\
&= \frac{f_{y,b}(0,0)}{f_b(0)} \\
&= \frac{11/12}{35/36} \\
&\approx 0.943
\end{align}
9. Based on your previous calculations:
a. The probability you and Betty both win is $f_{y,b}(1,1) = 0$.
b. The probability you and Betty both lose is $f_{y,b}(0,0) \approx 0.917$.
c. The probability Betty wins is $f_b(1) \approx 0.028$.
d. The probability that you win when Betty loses is $f_{y|b}(1,0) \approx 0.057$.
10. $r$ and $w$ (item c) are independent.
11. Since $E(y) = 1/18$ and $E(b) = 1/36$ we have
\begin{align}
cov(y,b) &= E((y-E(y))(b-E(b))) \\
&= \begin{aligned}[t]
& (0-E(y))*(0-E(b))*f_{y,b}(0,0) \\
&+ (1-E(y))*(0-E(b))*f_{y,b}(1,0) \\
&+ (0-E(y))*(1-E(b))*f_{y,b}(0,1) \\
&+ (1-E(y))*(1-E(b))*f_{y,b}(1,1) \\
\end{aligned} \\
&= \begin{aligned}[t]
& (0-1/18)*(0-1/36)*(11/12) \\
&+ (1-1/18)*(0-1/36)*(1/18) \\
&+ (0-1/18)*(1-1/36)*(1/36) \\
&+ (1-1/18)*(1-1/36)*(0) \\
\end{aligned} \\
&\approx -0.00154
\end{align}
12. The alternate formula for the covariance is $cov(y,b) = E(yb) - E(y)E(b)$.
a. The expected value is:
\begin{align}
E(yb) &= \begin{aligned}
& 0*0*f_{y,b}(0,0) \\
&+ 1*0*f_{y,b}(1,0) \\
&+ 0*1*f_{y,b}(0,1) \\
&+ 1*1*f_{y,b}(1,1) \\
\end{aligned} \\
&= f_{y,b}(1,1) \\
&= 0
\end{align}
b. The covariance is:
\begin{align}
cov(y,b) &= E(yb) - E(y)E(b) \\
&= 0 - (1/18)*(1/36) \\
&\approx -0.00154
\end{align}
c. Yes, if you have done it right you should get the same answer.
13. The correlation is:
\begin{align}
corr(y,b) &= \frac{cov(y,b)}{\sqrt{var(y)var(b)}} \\
&\approx \frac{-0.00154}{\sqrt{ 0.0525*0.027}} \\
&\approx -0.04
\end{align}
14. In this question we know the correlation and several values of the formula
defining it, and we use this to solve for the missing value.
a. We know that $corr(b,t) = \frac{cov(b,t)}{\sqrt{var(b)var(t)}}$, so
we can substitute known values to get:
\begin{align}
0.35 &= \frac{cov(b,t)}{\sqrt{0.027*5.83}}
\end{align}
Solving for $cov(b,t)$ we get
\begin{align}
cov(b,t) &= 0.35*\sqrt{0.027*5.83} \\
&= 0.139
\end{align}
b. We know that $cov(b,t) = E(bt) - E(b)E(t)$, so we can substitute
known values to get:
\begin{align}
0.139 &= E(bt) - (1/36)*7
\end{align}
Solving for $E(bt)$ we get:
\begin{align}
E(bt) &= 0.139 + (1/36)*7 \\
&= 0.333
\end{align}
15. Remember that independent events are also uncorrelated. So:
a. The covariance is $cov(r,w) = 0$.
b. The correlation is $corr(r,w) = 0$.
16. This question uses the formulas for the mean of a linear function of two
random variables.
a. The expected value is:
\begin{align}
E(y+b) &= E(y) + E(b) \\
&= 1/18 + 1/36 \\
&= 3/36 \\
&= 1/12
\end{align}
b. The expected value is:
\begin{align}
E(16y + 31b - 2) &= 16 E(y) + 31 E(b) - 2 \\
&= 16/18 + 31/36 - 2 \\
&= -9/36
\end{align}
17. This question uses the formulas for the variance and covariance of a linear
function of two random variables.
a. The covariance is:
\begin{align}
cov(16y,31b) &= 16 * 31 * cov(y,b) \\
&\approx 16 * 31 * (-0.00154) \\
&\approx -0.7638
\end{align}
b. The variance is:
\begin{align}
var(y+b) &= var(y) + var(b) + 2 cov(y,b) \\
&\approx 0.0525 + 0.027 + 2*(-0.00154) \\
&\approx 0.07642
\end{align}
18. Statement (b) is correct: the result of a bet on Boxcars is *negatively*
related to the result of a bet on Yo.
## **7** Statistics {-#answers-statistics}
[Click here to see the problems](#problems-statistics)
1. Since we have a random sample, the joint PDF is just a product of the
marginal PDFs.
a. The support is just the set of all possible combinations:
\begin{align}
S_{D_n} = \{(1,1),(1,2),(2,1),(2,2)\}
\end{align}
b. Since we have a random sample, the joint PDF is the product of the
marginal PDFs:
\begin{align}
f_{D_n}(1,1) &= f_x(1)f_x(1) \\
&= 0.4*0.4 \\
&= 0.16
\end{align}
c. Since we have a random sample, the joint PDF is the product of the
marginal PDFs:
\begin{align}
f_{D_n}(2,1) &= f_x(2)f_x(1) \\
&= 0.6*0.4 \\
&= 0.24
\end{align}
d. Since we have a random sample, the joint PDF is the product of the
marginal PDFs:
\begin{align}
f_{D_n}(1,2) &= f_x(1)f_x(2) \\
&= 0.4*0.6 \\
&= 0.24
\end{align}
e. Since we have a random sample, the joint PDF is the product of the
marginal PDFs:
\begin{align}
f_{D_n}(2,2) &= f_x(2)f_x(2) \\
&= 0.6*0.6 \\
&= 0.36
\end{align}
2. A random sample needs to be independent (not just uncorrelated) and
identically distributed (not just the same mean and variance).
a. Possibly a random sample.
b. Possibly a random sample.
c. Not a random sample.
d. Definitely a random sample.
e. Possibly a random sample.
f. Not a random sample.
3. Identify the sampling type (random sample, time series, stratified sample.
cluster sample, census, convenience sample) for each of the following data
sets.
a. This is a convenience sample.
b. This is a random sample.
c. This is a stratified sample.
d. This is a time series.
e. This is a census.
4. This is a long question but we can make it easier by preparing a simple table
mapping each of the four possible values of $D_n$ into a value for each
statistic:
| $x_1$ | $x_2$ |$f_{D_n}(\cdot)$|$\hat{f}_1$ |$\bar{x}$ | $\hat{\sigma}_x^2$ | $\hat{\sigma}_x$ | $xmin$ | $xmax$ |
|-------|-------|--------------|-------------|----------|--------------------|------------------|--------|--------|
| 1 | 1 | 0.16 | 1 | 1 | 0 | 0 | 1 | 1 |
| 2 | 1 | 0.24 | 0.5 | 1.5 | 0.5 | 0.71 | 1 | 2 |
| 1 | 2 | 0.24 | 0.5 | 1.5 | 0.5 | 0.71 | 1 | 2 |
| 2 | 2 | 0.36 | 0 | 2 | 0 | 0 | 2 | 2 |
This will make our lives much easier.
a. The support is $S_{\hat{f}_1} = \{0,0.5,1\}$ and the sampling distribution
is:
\begin{align}
f_{\hat{f}_1}(0) &= \Pr(\hat{f}_1 = 0) \\
&= f_{D_n}(2,2) \\
&= 0.36 \\
f_{\hat{f}_1}(0.5) &= \Pr(\hat{f}_1 = 0.5) \\
&= f_{D_n}(2,1) + f_{D_n}(1,2) \\
&= 0.24 + 0.24 \\
&= 0.48 \\
f_{\hat{f}_1}(1) &= \Pr(\hat{f}_1 = 1) \\
&= f_{D_n}(1,1) \\
&= 0.16
\end{align}
b. The support is $S_{\bar{x}} = \{1,1.5,2\}$ and the sampling distribution
is:
\begin{align}
f_{\bar{x}}(1) &= \Pr(\bar{x} = 1) \\
&= f_{D_n}(1,1) \\
&= 0.16 \\
f_{\bar{x}}(1.5) &= \Pr(\bar{x} = 1.5) \\
&= f_{D_n}(2,1) + f_{D_n}(1,2) \\
&= 0.24 + 0.24 \\
&= 0.48 \\
f_{\bar{x}}(2) &= \Pr(\bar{x} = 2) \\
&= f_{D_n}(2,2) \\
&= 0.36
\end{align}
c. The support is $S_{\hat{\sigma}_x^2} = \{0,0.5\}$ and the sampling
distribution is:
\begin{align}
f_{\hat{\sigma}_x^2}(0) &= \Pr(\hat{\sigma}_x^2 = 0) \\
&= f_{D_n}(1,1) + f_{D_n}(2,2)\\
&= 0.16 + 0.36 \\
&= 0.52 \\
f_{\hat{\sigma}_x^2}(0.5) &= \Pr(\hat{\sigma}_x^2 = 0.5) \\
&= f_{D_n}(2,1) + f_{D_n}(1,2) \\
&= 0.24 + 0.24 \\
&= 0.48
\end{align}
d. The support is $S_{\hat{\sigma}_x} = \{0,0.71\}$ and the sampling
distribution is:
\begin{align}
f_{\hat{\sigma}_x^2}(0) &= \Pr(\hat{\sigma}_x = 0) \\
&= f_{D_n}(1,1) + f_{D_n}(2,2)\\
&= 0.16 + 0.36 \\
&= 0.52 \\
f_{\hat{\sigma}_x^2}(0.71) &= \Pr(\hat{\sigma}_x = 0.71) \\
&= f_{D_n}(2,1) + f_{D_n}(1,2) \\
&= 0.24 + 0.24 \\
&= 0.48
\end{align}
e. The support is $S_{xmin} = \{1,2\}$ and the sampling distribution is:
\begin{align}
f_{xmin}(1) &= \Pr(xmin = 1) \\
&= f_{D_n}(1,1) + f_{D_n}(2,1) + f_{D_n}(1,2) \\
&= 0.16 + 0.24 + 0.24 \\
&= 0.64 \\
f_{xmin}(2) &= \Pr(xmin = 2) \\
&= f_{D_n}(2,2) \\
&= 0.36
\end{align}
f. The support is $S_{xmax} = \{1,2\}$ and the sampling distribution is:
\begin{align}
f_{xmax}(1) &= \Pr(xmax = 1) \\
&= f_{D_n}(1,1) \\
&= 0.16 \\
f_{xmax}(2) &= \Pr(xmax = 2) \\
&= f_{D_n}(2,2) + f_{D_n}(2,1) + f_{D_n}(1,2) \\
&= 0.36 + 0.24 + 0.24 \\
&= 0.84
\end{align}
5. The mean can be calculated from each statistic's PDF.
a. The mean is:
\begin{align}
E(\hat{f}_1) &= 0*f_{\hat{f}_1}(0) + 0.5*f_{\hat{f}_1}(0) + 1*f_{\hat{f}_1}(1) \\
&= 0*0.36 + 0.5*0.48 + 1*0.16 \\
&= 0.4
\end{align}
b. The mean is:
\begin{align}
E(\bar{x}) &= 1*f_{\bar{x}}(1) +1.5*f_{\bar{x}}(1.5) + 2*f_{\bar{x}}(2) \\
&= 1*0.16 +1.5*0.48 + 2*0.36 \\
&= 1.6
\end{align}
c. The mean is:
\begin{align}
E(\hat{\sigma}_x^2) &= 0*f_{\hat{\sigma}_x^2}(0) + 0.5*f_{\hat{\sigma}_x^2}(0.5) \\
&= 0*0.52 + 0.5*0.48 \\
&= 0.24
\end{align}
d. The mean is:
\begin{align}
E(\hat{\sigma}_x) &= 0*f_{\hat{\sigma}_x}(0) + 0.71*f_{\hat{\sigma}_x}(0.71) \\
&= 0*0.52 + 0.71*0.48 \\
&= 0.34
\end{align}
e. The mean is: