Quantitative Investment Analysis  Chapter 5
library(tidyverse)
library(knitr)
library(kableExtra)
In this post we will solve the end of the chapter practice problems from chapter 5 of the book.
Problem 1
A European put option on stock conveys the right to sell the stock at a prespecified price, called the exercise price, at the maturity date of the option. The value of this put at maturity is (exercise price – stock price) or $0, whichever is greater. Suppose the exercise price is $100 and the underlying stock trades in ticks of $0.01. At any time before maturity, the terminal value of the put is a random variable.
 A Describe the distinct possible outcomes for terminal put value. (Think of the put’s maximum and minimum values and its minimum price increments.)
 B Is terminal put value, at a time before maturity, a discrete or continuous random variable?
 C Letting Y stand for terminal put value, express in standard notation the probability that terminal put value is less than or equal to $24. No calculations or formulas are necessary.
# Max value of the put at expiration
# When the stock price is 0
put_max_value = 100  0
# Min value of the put at expiration
# When the stock price is above 100
put_min_value = 0
# Since the put trades at 0.01 increments
# Distinct possible outcomes of put value are
distinct_values < seq(put_min_value, put_max_value, by = 0.01)
cat("The discrete possible outcomes are", distinct_values[1:25], "......", distinct_values[length(distinct_values)])
## The discrete possible outcomes are 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.2 0.21 0.22 0.23 0.24 ...... 100
A. The discrete possible outcomes are 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.2 0.21 0.22 0.23 0.24 …… 100
B. The terminal put value, at a time before maturity, is a discrete random variable.
C. P(Y <= 24)
Problem 2
Define the term “binomial random variable.” Describe the types of problems for which the binomial distribution is used.
A binomial random variable is defined as the number of successes in n Bernoulli trials. The binomial distribution is used to make probability statements about anything with binary outcomes (success or failure).
Problem 3
The value of the cumulative distribution function F(x), where x is a particular outcome, for a discrete uniform distribution:
 A sums to 1.
 B lies between 0 and 1.
 C decreases as x increases.
B. lies between 0 and 1
Problem 4
For a binomial random variable with five trials, and a probability of success on each trial of 0.50, the distribution will be:
 A skewed.
 B uniform.
 C symmetric
C. symmetric, since probability of success (and failure) is 0.50 each
Problem 5
In a discrete uniform distribution with 20 potential outcomes of integers 1 to 20, the probability that X is greater than or equal to 3 but less than 6, P(3 ≤ X < 6), is:
 A 0.10.
 B 0.15.
 C 0.20.
# Create our sample space
x < 1:20
x
## [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
# Lets create our event
y = x[x >= 3 & x < 6]
y
## [1] 3 4 5
# Get probability
p < length(y) / length(x)
cat("The probability that X is greater than or equal to 3 but less than 6, P(3 ≤ X < 6) is", p)
## The probability that X is greater than or equal to 3 but less than 6, P(3 ≤ X < 6) is 0.15
B. 0.15
Problem 6
Over the last 10 years, a company’s annual earnings increased year over year seven times and decreased year over year three times. You decide to model the number of earnings increases for the next decade as a binomial random variable.
 A What is your estimate of the probability of success, defined as an increase in annual earnings?
For Parts B, C, and D of this problem, assume the estimated probability is the actual probability for the next decade.
 B What is the probability that earnings will increase in exactly 5 of the next 10 years?
 C Calculate the expected number of yearly earnings increases during the next 10 years.
 D Calculate the variance and standard deviation of the number of yearly earnings increases during the next 10 years.
 E The expression for the probability function of a binomial random variable depends on two major assumptions. In the context of this problem, what must you assume about annual earnings increases to apply the binomial distribution in Part B? What reservations might you have about the validity of these assumptions?
event < 10
increase < 7
decrease < 3
p_success < increase / event
cat("The probability of success, defined as an increase in
annual earnings is", p_success)
## The probability of success, defined as an increase in
## annual earnings is 0.7
A. The probability of success, defined as an increase in annual earnings is 0.7
# http://www.rtutor.com/elementarystatistics/probabilitydistributions/binomialdistribution
p < dbinom(5,10,0.7)
cat("The probability that earnings will increase in exactly 5 of the next 10
years", p)
## The probability that earnings will increase in exactly 5 of the next 10
## years 0.1029193
B. The probability that earnings will increase in exactly 5 of the next 10 years 0.1029193
# lecture on expected value of bernoulli distribution
# https://www.khanacademy.org/math/statisticsprobability/randomvariablesstatslibrary/binomialmeanstandarddevformulas/v/expectedvalueofbinomialvariable
n = 10
p_success < 0.7
e < n * p_success
cat("The expected number of yearly earnings increases during the next
10 years", e)
## The expected number of yearly earnings increases during the next
## 10 years 7
C. The expected number of yearly earnings increases during the next 10 years 7
# Lecture on variance and standard deviation of benoulli distribution
# https://www.khanacademy.org/math/statisticsprobability/randomvariablesstatslibrary/binomialmeanstandarddevformulas/v/meanandvarianceofbernoullidistributionexample
# n * (1p) * p
n = 10
p = 0.7
# variane is
v < n * (1  p) * p
# sd is
s < sqrt(v)
cat("The variance and standard deviation of the number of yearly earnings
increases during the next 10 years is",v, "and", s, "respectively")
## The variance and standard deviation of the number of yearly earnings
## increases during the next 10 years is 2.1 and 1.449138 respectively
E. We assume the probability remains the same every year. We also assume that the trials are independent. This may not be the case as we know that probability will change every year and earnings are subject to economic cycles and thus may not be independent.
Problem 7
A portfolio manager annually outperforms her benchmark 60% of the time. Assuming independent annual trials, what is the probability that she will outperform her benchmark four or more times over the next five years?
 A 0.26
 B 0.34
 C 0.48
# There are two ways to solve this
p_success < 0.6
n < 5
# probability of exactly 4
p_4 < dbinom(4,5,0.6)
# probability of exactly 5
p_5 < dbinom(5,5,0.6)
p < p_4 + p_5
p
## [1] 0.33696
# probability of exactly 3 or less
p < pbinom(3, n, prob = p_success, lower.tail = FALSE)
p
## [1] 0.33696
cat("The probability that she will outperform
her benchmark four or more times over the next five years is", p)
## The probability that she will outperform
## her benchmark four or more times over the next five years is 0.33696
B. 0.34
Problem 8
You are examining the record of an investment newsletter writer who claims a 70 percent success rate in making investment recommendations that are profitable over a oneyear time horizon. You have the oneyear record of the newsletter’s seven most recent recommendations. Four of those recommendations were profitable. If all the recommendations are independent and the newsletter writer’s skill is as claimed, what is the probability of observing four or fewer profitable recommendations out of seven in total?
# Claimed success
p_success < 7/10
# Size
n < 7
# probability of observing four or fewer success
p < pbinom(4, n, prob = p_success)
cat("The probability of observing four or fewer
profitable recommendations out of seven in total is", p)
## The probability of observing four or fewer
## profitable recommendations out of seven in total is 0.3529305
The probability of observing four or fewer profitable recommendations out of seven in total is 0.3529305
Problem 9
You are forecasting sales for a company in the fourth quarter of its fiscal year. Your lowend estimate of sales is €14 million, and your highend estimate is €15 million. You decide to treat all outcomes for sales between these two values as equally likely, using a continuous uniform distribution.
 A What is the expected value of sales for the fourth quarter?
 B What is the probability that fourthquarter sales will be less than or equal to €14,125,000?
low_estimate < 14
high_estimate < 15
e < mean(c(low_estimate, high_estimate))
cat("The expected value of sales for the fourth quarter is", e)
## The expected value of sales for the fourth quarter is 14.5
# We can also do this using an experiment
# Let x be uniform distribution of 10000 values between 14 and 15
x < runif(10000, min = 14, max = 15)
# We can see that mean value of x is close to 14.5
mean(x)
## [1] 14.4971
A. The expected value of sales for the fourth quarter is 14.5
set.seed(123)
# Create our random uniform continuous variables
x < runif(100000, min = 14, max = 15)
# See how many values are less than 14.125
y = x[x <= 14.125]
# Count the values
p < length(y) / length(x)
cat("the probability that fourthquarter sales will be less than or equal to €14,125,000 is", p)
## the probability that fourthquarter sales will be less than or equal to €14,125,000 is 0.12493
The probability that fourthquarter sales will be less than or equal to €14,125,000 is 0.12493 ~12.50%
Problem 10
State the approximate probability that a normal random variable will fall within the following intervals:
 A Mean plus or minus one standard deviation.
 B Mean plus or minus two standard deviations.
 C Mean plus or minus three standard deviations
A Mean plus or minus one standard deviation will have approximately 68% of all outcomes. B Mean plus or minus two standard deviations will have approximately 95% of all outcomes. 95% C Mean plus or minus three standard deviations will have approximately 99% of all outcomes. 99%
Problem 11
Find the area under the normal curve up to z = 0.36; that is, find P(Z ≤ 0.36). Interpret this value.
cat("The area under the normal curve up to z = 0.36 is", pnorm(0.36))
## The area under the normal curve up to z = 0.36 is 0.6405764
When z = 0.36, we can interpret this as 64% of observations are smaller than 0.36.
Problem 12
If the probability that a portfolio outperforms its benchmark in any quarter is 0.75, the probability that the portfolio outperforms its benchmark in three or fewer quarters over the course of a year is closest to:
 A 0.26
 B 0.42
 C 0.68
# We know there are 4 quarters in a year
# The outcomes are outperform or not
# Binomial distribution
# Cumulative probability of having 3 out 4 quarters outperform is
pbinom(3,4,0.75)
## [1] 0.6835937
C. 0.68
Problem 13
In futures markets, profits or losses on contracts are settled at the end of each trading day. This procedure is called marking to market or daily resettlement. By preventing a trader’s losses from accumulating over many days, marking to market reduces the risk that traders will default on their obligations. A futures markets trader needs a liquidity pool to meet the daily mark to market. If liquidity is exhausted, the trader may be forced to unwind his position at an unfavorable time.
Suppose you are using financial futures contracts to hedge a risk in your portfolio. You have a liquidity pool (cash and cash equivalents) of λ dollars per contract and a time horizon of T trading days. For a given size liquidity pool, λ, Kolb, Gay, and Hunter (1985) developed an expression for the probability stating that you will exhaust your liquidity pool within a Tday horizon as a result of the daily mark to market. Kolb et al. assumed that the expected change in futures price is 0 and that futures price changes are normally distributed. With σ representing the standard deviation of daily futures price changes, the standard deviation of price changes over a time horizon to day T is σ T , given continuous compounding. With that background, the Kolb et al. expression is Probability of exhausting liquidity pool = 2[1 – N(x)] where x = λ/(σ√T). Here x is a standardized value of λ. N(x) is the standard normal cumulative distribution function. For some intuition about 1 − N(x) in the expression, note that the liquidity pool is exhausted if losses exceed the size of the liquidity pool at any time up to and including T; the probability of that event happening can be shown to be proportional to an area in the right tail of a standard normal distribution, 1 − N(x).
Using the Kolb et al. expression, answer the following questions:
 A Your hedging horizon is five days, and your liquidity pool is $2,000 per contract. You estimate that the standard deviation of daily price changes for the contract is $450. What is the probability that you will exhaust your liquidity pool in the fiveday period?
 B Suppose your hedging horizon is 20 days, but all the other facts given in Part A remain the same. What is the probability that you will exhaust your liquidity pool in the 20day period?
# x = λ/(σ√T)
# We know
# λ = 2000
# σ = 450
# T = 5
x < 2000 / (450 * sqrt(5))
x
## [1] 1.987616
# N(x) is the standard normal cumulative distribution function
N_x < pnorm(x)
# Probability of exhausting liquidity pool = 2[1 – N(x)]
p < 2 * (1  N_x)
p
## [1] 0.04685418
cat("The probability that you will exhaust your liquidity pool in the fiveday period", p)
## The probability that you will exhaust your liquidity pool in the fiveday period 0.04685418
The probability that you will exhaust your liquidity pool in the fiveday period 0.04685418
# the probability that you will exhaust your liquidity pool in the 20day
# We know
# λ = 2000
# σ = 450
# T = 20
x < 2000 / (450 * sqrt(20))
x
## [1] 0.993808
# N(x) is the standard normal cumulative distribution function
N_x < pnorm(x)
# Probability of exhausting liquidity pool = 2[1 – N(x)]
p < 2 * (1  N_x)
p
## [1] 0.3203164
cat("The probability that you will exhaust your liquidity pool in the twentyday period", p)
## The probability that you will exhaust your liquidity pool in the twentyday period 0.3203164
The probability that you will exhaust your liquidity pool in the twentyday period 0.3203164
Problem 14
Which of the following is characteristic of the normal distribution?
 A Asymmetry
 B Kurtosis of 3
 C Definitive limits or boundaries
B. Kurtosis of 3
Problem 15
Which of the following assets most likely requires the use of a multivariate distribution for modeling returns?
 A A call option on a bond
 B A portfolio of technology stocks
 C A stock in a market index
B. A portfolio of technology stocks. A multivariate distribution specify the probabilities of group of related random variables.
Problem 16
The total number of parameters that fully characterizes a multivariate normal distribution for the returns on two stocks is:
 A 3.
 B 4.
 C 5.
# multivariate normal distribution for the returns on two stocks will have
# 2 means
# 2 variance
# 1 correlation
C. 5
Problem 17
A client has a portfolio of common stocks and fixedincome instruments with a current value of £1,350,000. She intends to liquidate £50,000 from the portfolio at the end of the year to purchase a partnership share in a business. Furthermore, the client would like to be able to withdraw the £50,000 without reducing the initial capital of £1,350,000. The following table shows four alternative asset allocations
parameters  A  B  C  D 

Expected annual return  16  12  10  9 
Standard deviation of return  24  17  12  11 
Address the following questions (assume normality for Parts B and C):
 A Given the client’s desire not to invade the £1,350,000 principal, what is the shortfall level, RL? Use this shortfall level to answer Part B.
 B According to the safetyfirst criterion, which of the allocations is the best?
 C What is the probability that the return on the safetyfirst optimal portfolio will be less than the shortfall level, RL?
withdraw < 50000
portfolio < 1350000
shortfall_level < withdraw / portfolio
cat("The client's shortfall level is", shortfall_level)
## The client's shortfall level is 0.03703704
A. The client’s shortfall level is 0.03703704
# Safety first criteria
# allocates to assets with the highest ratio
# as expressed below
df %>%
gather(A:D, key = allocation, value = v) %>%
spread(parameters, value = v) %>%
mutate(ratio = (`Expected annual return`  shortfall_level * 100) / `Standard deviation of return`)
## # A tibble: 4 x 4
## allocation `Expected annual return` `Standard deviation of return` ratio
## <chr> <dbl> <dbl> <dbl>
## 1 A 16 24 0.512
## 2 B 12 17 0.488
## 3 C 10 12 0.525
## 4 D 9 11 0.481
B. Since allocation C has the highest ratio, it is the best allocation.
# Since allocation C has 0.5246 ratio
# round it up to 0.53
# We need to find the cumulative probability of z = 0.53
pnorm(0.53)
## [1] 0.298056
C. The safetyfirst optimal portfolio has a roughly 30 percent chance of not meeting a 3.7 percent return threshold
Problem 18
A portfolio has an expected mean return of 8 percent and standard deviation of 14 percent. The probability that its return falls between 8 and 11 percent is closest to:
 A 8.3%
 B 14.8%.
 C 58.3%.
z1 < (8  8) / 14
z1
## [1] 0
z2 < (11  8) / 14
z2
## [1] 0.2142857
p < pnorm(z2)  pnorm(z1)
cat("The probability that its return falls between 8 and 11 percent is
closest to is", p)
## The probability that its return falls between 8 and 11 percent is
## closest to is 0.08483787
A. 8.3%
Problem 19
A portfolio has an expected return of 7% with a standard deviation of 13%. For an investor with a minimum annual return target of 4%, the probability that the portfolio return will fail to meet the target is closest to:
 A 33%.
 B 41%.
 C 59%.
z < (4  7) / 13
p < pnorm(z)
cat("The probability that the
portfolio return will fail to meet the target is closest to is", p)
## The probability that the
## portfolio return will fail to meet the target is closest to is 0.408747
B. 41%
Problem 20
A Define Monte Carlo simulation and explain its use in finance. B Compared with analytical methods, what are the strengths and weaknesses of Monte Carlo simulation for use in valuing securities?
A. Monte Carlo simulation uses computer to program to find an approximate solution to a complex problem. Monte Carlo simulation generates a large number of random samples from a probability distribution.
 Strengths. It can be used to find approximate solutions to complex problems such as valuing exotic option.
 Weakness. It provides only extimates and not exact solutions.
Problem 21
A standard lookback call option on stock has a value at maturity equal to (Value of the stock at maturity – Minimum value of stock during the life of the option prior to maturity) or $0, whichever is greater. If the minimum value reached prior to maturity was $20.11 and the value of the stock at maturity is $23, for example, the call is worth $23 − $20.11 = $2.89. Briefly discuss how you might use Monte Carlo simulation in valuing a lookback call option.
Skip for now. We will cover Monte Carlo in a separate post.
Problem 22
Which of the following is a continuous random variable?
 A The value of a futures contract quoted in increments of $0.05
 B The total number of heads recorded in 1 million tosses of a coin
 C The rate of return on a diversified portfolio of stocks over a threemonth period
C. The rate of return on a diversified portfolio of stocks over a threemonth period
Problem 23
X is a discrete random variable with possible outcomes X = {1,2,3,4}. Three functions f(x), g(x), and h(x) are proposed to describe the probabilities of the outcomes in X.
X = x  f(x) = P(X = x)  g(x) = P(X = x) 


1  0.25  0.20  0.20 
2  0.25  0.25  0.25 
3  0.50  0.50  0.30 
4  0.25  0.05  0.35 
The conditions for a probability function are satisfied by: A f(x). B g(x). C h(x).
B. g(x). Since all values sum to 1
Problem 24
The cumulative distribution function for a discrete random variable is shown in the following table.
X = x  Cumulative Distribution Function F(x) = P(X ≤ x) 

1  0.15 
2  0.25 
3  0.50 
4  0.60 
5  0.95 
6  1.00 
The probability that X will take on a value of either 2 or 4 is closest to:
 A 0.20.
 B 0.35.
 C 0.85.
A. 0.20
Problem 25
Which of the following events can be represented as a Bernoulli trial?
 A The flip of a coin
 B The closing price of a stock
 C The picking of a random integer between 1 and 10
A. The flip of a coin
Problem 26
The weekly closing prices of Mordice Corporation shares are as follows:
Date  Closing Price (€) 

1 August  112 
8 August  160 
15 August  120 
The continuously compounded return of Mordice Corporation shares for the period August 1 to August 15 is closest to:
 A 6.90%
 B 7.14%
 C 8.95%
r < log(df$`Closing Price (€)`[length(df$Date)] / df$`Closing Price (€)`[1])
cat("The continuously compounded return of Mordice Corporation shares is", r)
## The continuously compounded return of Mordice Corporation shares is 0.06899287
A. 6.90%
Problem 27
A stock is priced at $100.00 and follows a oneperiod binomial process with an up move that equals 1.05 and a down move that equals 0.97. If 1 million Bernoulli trials are conducted, and the average terminal stock price is $102.00, the probability of an up move (p) is closest to:
 A 0.375.
 B 0.500.
 C 0.625
up_move < 100 * 1.05
down_move < 100 * 0.97
terminal_value < 102
# p(up_move) * up_move + p(down_move) * down_move = terminal_value
# p(up_move) * up_move + (1  p(up_move)) * down_move = terminal_value
# After doing some simple algebra
p < 5/8
cat("probability of an up move (p) is closest to", p)
## probability of an up move (p) is closest to 0.625
C. 0.625
Problem 28
A call option on a stock index is valued using a threestep binomial tree with an up move that equals 1.05 and a down move that equals 0.95. The current level of the index is $190, and the option exercise price is $200. If the option value is positive when the stock price exceeds the exercise price at expiration and $0 otherwise, the number of terminal nodes with a positive payoff is:
 A one.
 B two.
 C three.
# There are 4 nodes or paths.
# U=up, D=Down
# UUU
# UUD
# DUD
# DDD
node1 < 190 * 1.05 * 1.05 * 1.05
node2 < 190 * 1.05 * 1.05 * 0.95
node3 < 190 * 0.95 * 1.05 * 0.95
node4 < 190 * 0.95 * 0.95 * 0.95
cat("Node 1", node1,"\n",
"Node 2",node2,"\n",
"Node 3",node3,"\n",
"Node 4",node4)
## Node 1 219.9488
## Node 2 199.0012
## Node 3 180.0487
## Node 4 162.9012
# Only 1 value is above 200
# Node 1
A. one
Problem 29
A random number between zero and one is generated according to a continuous uniform distribution. What is the probability that the first number generated will have a value of exactly 0.30?
 A 0%
 B 30%
 C 70%
A. 0%. Getting an exact number on a continuous scale is not possible.
Problem 30
A Monte Carlo simulation can be used to:
 A directly provide precise valuations of call options.
 B simulate a process from historical records of returns.
 C test the sensitivity of a model to changes in assumptions.
C. test the sensitivity of a model to changes in assumptions.
Problem 31
A limitation of Monte Carlo simulation is:
 A its failure to do “what if” analysis.
 B that it requires historical records of returns
 C its inability to independently specify causeand effect relationships.
C. its inability to independently specify causeand effect relationships.
Problem 32
Which parameter equals zero in a normal distribution?
 A Kurtosis
 B Skewness
 C Standard deviation
B. Skewness
Problem 33
An analyst develops the following capital market projections.
## # A tibble: 2 x 3
## param Stocks Bonds
## <chr> <chr> <chr>
## 1 Mean Return 10% 2%
## 2 Standard Deviation 15% 5%
Assuming the returns of the asset classes are described by normal distributions, which of the following statements is correct?
 A Bonds have a higher probability of a negative return than stocks.
 B On average, 99% of stock returns will fall within two standard deviations of the mean.
 C The probability of a bond return less than or equal to 3% is determined using a Zscore of 0.25
A. Bonds have a higher probability of a negative return than stocks.
Problem 34
A client holding a £2,000,000 portfolio wants to withdraw £90,000 in one year without invading the principal. According to Roy’s safetyfirst criterion, which of the following portfolio allocations is optimal?
param  Allocation A  Allocation B  Allocation C 

Expected annual return  6.5%  7.5%  8.5% 
Standard deviation of returns  8.35%  10.21%  14.34% 
 A Allocation A
 B Allocation B
 C Allocation C
shortfall_level < 90000/2000000
df %>%
gather(`Allocation A`:`Allocation C`, key = `Allocation`, value = v) %>%
spread(param, value = v) %>%
mutate_at(c(2,3),.funs = parse_number) %>%
mutate(ratio = (`Expected annual return`  shortfall_level * 100) / `Standard deviation of returns`)
## # A tibble: 3 x 4
## Allocation `Expected annual return` `Standard deviation of retur… ratio
## <chr> <dbl> <dbl> <dbl>
## 1 Allocation A 6.5 8.35 0.240
## 2 Allocation B 7.5 10.2 0.294
## 3 Allocation C 8.5 14.3 0.279
B. Allocation B
Problem 35
In contrast to normal distributions, lognormal distributions:
 A are skewed to the left.
 B have outcomes that cannot be negative.
 C are more suitable for describing asset returns than asset prices.
B. have outcomes that cannot be negative.
Problem 36
The lognormal distribution is a more accurate model for the distribution of stock prices than the normal distribution because stock prices are:
 A symmetrical.
 B unbounded.
 C nonnegative.
C. nonnegative
Problem 37
The price of a stock at t = 0 is $208.25 and at t = 1 is $186.75. The continuously compounded rate of return for the stock from t = 0 to t = 1 is closest to:
 A –10.90%.
 B –10.32%.
 C 11.51%.
log(186.75/208.25)
## [1] 0.1089685
A. 10.90