I. Computer Questions:

1. Download ETF daily data

getSymbols(tickers, from=start_date, to=end_date, src="yahoo")
## [1] "SPY" "QQQ" "EEM" "IWM" "EFA" "TLT" "IYR" "GLD"
# Extract adjusted closing prices
prices <- do.call(merge, lapply(tickers, function(x) Ad(get(x))))
head(prices)
##            SPY.Adjusted QQQ.Adjusted EEM.Adjusted IWM.Adjusted EFA.Adjusted
## 2010-01-04     84.79636     40.29078     30.35150     51.36657     35.12843
## 2010-01-05     85.02087     40.29078     30.57182     51.18993     35.15939
## 2010-01-06     85.08070     40.04776     30.63577     51.14178     35.30801
## 2010-01-07     85.43986     40.07380     30.45811     51.51909     35.17179
## 2010-01-08     85.72417     40.40361     30.69972     51.80009     35.45043
## 2010-01-11     85.84390     40.23872     30.63577     51.59135     35.74147
##            TLT.Adjusted IYR.Adjusted GLD.Adjusted
## 2010-01-04     56.13517     26.76811       109.80
## 2010-01-05     56.49772     26.83237       109.70
## 2010-01-06     55.74142     26.82070       111.51
## 2010-01-07     55.83513     27.06026       110.82
## 2010-01-08     55.81013     26.87914       111.37
## 2010-01-11     55.50388     27.00768       112.85

2. Calculate monthly returns using discrete returns

# Convert daily adjusted prices to monthly
monthly_prices <- to.monthly(prices, indexAt="lastof", OHLC=FALSE)
head(monthly_prices)
##            SPY.Adjusted QQQ.Adjusted EEM.Adjusted IWM.Adjusted EFA.Adjusted
## 2010-01-31     80.35191     37.14011     27.20337     48.25951     32.49673
## 2010-02-28     82.85847     38.84997     27.68661     50.41920     32.58344
## 2010-03-31     87.90289     41.84566     29.93222     54.56904     34.66402
## 2010-04-30     89.26273     42.78407     29.88248     57.66772     33.69184
## 2010-05-31     82.17039     39.62130     27.07545     53.32151     29.92078
## 2010-06-30     77.91885     37.25366     26.69677     49.19262     29.30384
##            TLT.Adjusted IYR.Adjusted GLD.Adjusted
## 2010-01-31     57.69777     25.37740       105.96
## 2010-02-28     57.50021     26.76226       109.43
## 2010-03-31     56.31721     29.37117       108.95
## 2010-04-30     58.18795     31.24745       115.36
## 2010-05-31     61.16044     29.47150       118.88
## 2010-06-30     64.70638     28.09512       121.68
monthly_returns <- monthly_prices / lag(monthly_prices) - 1
monthly_returns <- monthly_returns[-1, ]  # remove first NA row
head(monthly_returns)
##            SPY.Adjusted QQQ.Adjusted EEM.Adjusted IWM.Adjusted EFA.Adjusted
## 2010-02-28   0.03119470   0.04603825  0.017763911   0.04475153  0.002668091
## 2010-03-31   0.06087994   0.07710918  0.081108418   0.08230669  0.063853962
## 2010-04-30   0.01546980   0.02242545 -0.001661813   0.05678456 -0.028045668
## 2010-05-31  -0.07945456  -0.07392400 -0.093935691  -0.07536637 -0.111928117
## 2010-06-30  -0.05174064  -0.05975669 -0.013986286  -0.07743394 -0.020619163
## 2010-07-31   0.06830035   0.07258210  0.109324392   0.06730919  0.116103848
##            TLT.Adjusted IYR.Adjusted GLD.Adjusted
## 2010-02-28 -0.003423972   0.05457056  0.032748219
## 2010-03-31 -0.020573813   0.09748471 -0.004386396
## 2010-04-30  0.033217753   0.06388142  0.058834363
## 2010-05-31  0.051084329  -0.05683497  0.030513147
## 2010-06-30  0.057977743  -0.04670191  0.023553189
## 2010-07-31 -0.009463761   0.09404820 -0.050871157

3. Download Fama French 3 factors data

library(zoo)
ff3_raw <- read.csv("F-F_Research_Data_Factors.csv", skip=4, stringsAsFactors=FALSE)
ff3_clean <- ff3_raw[grepl("^[0-9]{6}$", ff3_raw$X), ]
colnames(ff3_clean)[1:5] <- c("Date","Mkt_RF","SMB","HML","RF")
ff3_clean$Date <- as.Date(as.yearmon(ff3_clean$Date, "%Y%m"))
ff3_clean$Mkt_RF <- as.numeric(ff3_clean$Mkt_RF)/100
ff3_clean$SMB    <- as.numeric(ff3_clean$SMB)/100
ff3_clean$HML    <- as.numeric(ff3_clean$HML)/100
ff3_clean$RF     <- as.numeric(ff3_clean$RF)/100
head(ff3_clean)
##         Date  Mkt_RF     SMB     HML     RF
## 1 1926-07-01  0.0289 -0.0255 -0.0239 0.0022
## 2 1926-08-01  0.0264 -0.0114  0.0381 0.0025
## 3 1926-09-01  0.0038 -0.0136  0.0005 0.0023
## 4 1926-10-01 -0.0327 -0.0014  0.0082 0.0032
## 5 1926-11-01  0.0254 -0.0011 -0.0061 0.0031
## 6 1926-12-01  0.0262 -0.0007  0.0006 0.0028

4. Merge monthly return data in question 2 and 3

library(zoo)
etf_returns <- data.frame(
  Date = index(monthly_returns),
  coredata(monthly_returns)
)
# Convert ETF returns Date to yearmon
etf_returns$YearMonth <- as.yearmon(etf_returns$Date)

# Convert FF3 Date to yearmon
ff3_clean$YearMonth <- as.yearmon(ff3_clean$Date)

# Merge by YearMonth
data_merged <- merge(etf_returns, ff3_clean, by="YearMonth")
head(data_merged)
##   YearMonth     Date.x SPY.Adjusted QQQ.Adjusted EEM.Adjusted IWM.Adjusted
## 1  Feb 2010 2010-02-28   0.03119470   0.04603825  0.017763911   0.04475153
## 2  Mar 2010 2010-03-31   0.06087994   0.07710918  0.081108418   0.08230669
## 3  Apr 2010 2010-04-30   0.01546980   0.02242545 -0.001661813   0.05678456
## 4  May 2010 2010-05-31  -0.07945456  -0.07392400 -0.093935691  -0.07536637
## 5  Jun 2010 2010-06-30  -0.05174064  -0.05975669 -0.013986286  -0.07743394
## 6  Jul 2010 2010-07-31   0.06830035   0.07258210  0.109324392   0.06730919
##   EFA.Adjusted TLT.Adjusted IYR.Adjusted GLD.Adjusted     Date.y  Mkt_RF
## 1  0.002668091 -0.003423972   0.05457056  0.032748219 2010-02-01  0.0339
## 2  0.063853962 -0.020573813   0.09748471 -0.004386396 2010-03-01  0.0630
## 3 -0.028045668  0.033217753   0.06388142  0.058834363 2010-04-01  0.0199
## 4 -0.111928117  0.051084329  -0.05683497  0.030513147 2010-05-01 -0.0790
## 5 -0.020619163  0.057977743  -0.04670191  0.023553189 2010-06-01 -0.0556
## 6  0.116103848 -0.009463761   0.09404820 -0.050871157 2010-07-01  0.0692
##       SMB     HML    RF
## 1  0.0118  0.0318 0e+00
## 2  0.0146  0.0219 1e-04
## 3  0.0484  0.0296 1e-04
## 4  0.0013 -0.0248 1e-04
## 5 -0.0179 -0.0473 1e-04
## 6  0.0022 -0.0050 1e-04

5. Based on CAPM model, compute MVP monthly returns based on estimated covariance matrix for the 8-asset portfolio by using past 60-month returns from 2020/03 - 2025/02.

window_data <- subset(data_merged, YearMonth >= as.yearmon("2020-03") & 
                                     YearMonth <= as.yearmon("2025-02"))
etf_only <- window_data[, c("SPY.Adjusted","QQQ.Adjusted","EEM.Adjusted",
                            "IWM.Adjusted","EFA.Adjusted","TLT.Adjusted",
                            "IYR.Adjusted","GLD.Adjusted")]
cov_matrix <- cov(etf_only)
library(quadprog)

n <- ncol(etf_only)
Dmat <- cov_matrix
dvec <- rep(0, n)
Amat <- cbind(rep(1, n))   # constraint: sum of weights = 1
bvec <- 1

mvp <- solve.QP(Dmat, dvec, Amat, bvec, meq=1)
weights <- mvp$solution
names(weights) <- colnames(etf_only)
weights
## SPY.Adjusted QQQ.Adjusted EEM.Adjusted IWM.Adjusted EFA.Adjusted TLT.Adjusted 
##    1.4542548   -0.7051061    0.0884220    0.0214490   -0.1714324    0.5170385 
## IYR.Adjusted GLD.Adjusted 
##   -0.5196880    0.3150623

6. Based on FF 3-factor model, compute MVP monthly returns covariance matrix for the 8-asset portfolio by using past 60-month returns from 2020/03 - 2025/02.

library(zoo)

window_data <- subset(data_merged,
                      YearMonth >= as.yearmon("2020-03") &
                      YearMonth <= as.yearmon("2025-02"))
etf_only <- window_data[, c("SPY.Adjusted","QQQ.Adjusted","EEM.Adjusted",
                            "IWM.Adjusted","EFA.Adjusted","TLT.Adjusted",
                            "IYR.Adjusted","GLD.Adjusted")]
# Run regressions for each ETF against FF3 factors
residuals_list <- lapply(etf_only, function(ret) {
  lm(ret ~ window_data$Mkt_RF + window_data$SMB + window_data$HML)$residuals
})

# Combine residuals into a matrix
residuals_mat <- do.call(cbind, residuals_list)
colnames(residuals_mat) <- colnames(etf_only)
cov_matrix_ff3 <- cov(residuals_mat)
cov_matrix_ff3
##               SPY.Adjusted  QQQ.Adjusted  EEM.Adjusted  IWM.Adjusted
## SPY.Adjusted  1.690030e-05  1.002089e-05 -1.724429e-05 -3.844372e-07
## QQQ.Adjusted  1.002089e-05  2.225664e-04  6.693960e-05 -6.305764e-05
## EEM.Adjusted -1.724429e-05  6.693960e-05  1.308150e-03 -1.718755e-05
## IWM.Adjusted -3.844372e-07 -6.305764e-05 -1.718755e-05  7.355500e-05
## EFA.Adjusted  3.648549e-06 -2.087540e-05  5.188973e-04  3.911922e-05
## TLT.Adjusted  1.353410e-05  7.685642e-05  4.302078e-04  2.808800e-05
## IYR.Adjusted  4.675855e-05 -7.859360e-05  2.007083e-04  1.263427e-05
## GLD.Adjusted  1.153417e-05  7.126361e-06  6.079001e-04  2.057507e-05
##               EFA.Adjusted TLT.Adjusted  IYR.Adjusted GLD.Adjusted
## SPY.Adjusted  3.648549e-06 1.353410e-05  4.675855e-05 1.153417e-05
## QQQ.Adjusted -2.087540e-05 7.685642e-05 -7.859360e-05 7.126361e-06
## EEM.Adjusted  5.188973e-04 4.302078e-04  2.007083e-04 6.079001e-04
## IWM.Adjusted  3.911922e-05 2.808800e-05  1.263427e-05 2.057507e-05
## EFA.Adjusted  5.877421e-04 3.437696e-04  2.306136e-04 2.635111e-04
## TLT.Adjusted  3.437696e-04 1.444412e-03  5.504107e-04 4.636577e-04
## IYR.Adjusted  2.306136e-04 5.504107e-04  9.222534e-04 2.883792e-04
## GLD.Adjusted  2.635111e-04 4.636577e-04  2.883792e-04 1.560223e-03
library(quadprog)

n <- ncol(residuals_mat)
Dmat <- cov_matrix_ff3
dvec <- rep(0, n)
Amat <- cbind(rep(1, n))   # sum of weights = 1
bvec <- 1

mvp_ff3 <- solve.QP(Dmat, dvec, Amat, bvec, meq=1)
weights_ff3 <- mvp_ff3$solution
names(weights_ff3) <- colnames(residuals_mat)
weights_ff3
## SPY.Adjusted QQQ.Adjusted EEM.Adjusted IWM.Adjusted EFA.Adjusted TLT.Adjusted 
##   0.71792372   0.06571815   0.02920513   0.22941611  -0.01148062  -0.00267917 
## IYR.Adjusted GLD.Adjusted 
##  -0.02194947  -0.00615384

7. You can invest in the 8-asset portfolio in 2025/03 based on the optimal weights of MVP derived from question 5 and 6. What are the realized portfolio returns in the March of 2025 using the weights from question 5 and 6?

end_date <- "2025-03-31"
getSymbols(tickers, from=start_date, to=end_date, src="yahoo")
## [1] "SPY" "QQQ" "EEM" "IWM" "EFA" "TLT" "IYR" "GLD"
new_prices <- do.call(merge, lapply(tickers, function(x) Ad(get(x))))
new_monthly_prices <- to.monthly(new_prices, indexAt="lastof", OHLC=FALSE)
new_monthly_returns <- new_monthly_prices / lag(new_monthly_prices) - 1
new_monthly_returns <- new_monthly_returns[-1, ]

march2025 <- subset(new_monthly_returns, index(new_monthly_returns) == as.Date("2025-03-31"))
etf_ret_march <- as.numeric(coredata(march2025))

port_ret_capm <- sum(weights * etf_ret_march)
port_ret_ff3  <- sum(weights_ff3  * etf_ret_march)

data.frame(
  Model = c("CAPM MVP","FF3 MVP"),
  Realized_Return = c(port_ret_capm, port_ret_ff3)
)
##      Model Realized_Return
## 1 CAPM MVP    -0.007638897
## 2  FF3 MVP    -0.063635410

Interpretation

CAPM MVP was more resilient in March 2025, while FF3 MVP suffered a larger drawdown. This demonstrates how different risk models can lead to materially different realized outcomes, even when both are optimized over the same historical window.

8. Based on 7, what are the realized MVP portfolio return in the April of 2025 based on its previous 60-month window data?

april2025 <- subset(monthly_returns, index(monthly_returns) == as.Date("2025-04-30"))
etf_ret_april <- as.numeric(coredata(april2025))

port_ret_capm_april <- sum(weights * etf_ret_april)
port_ret_ff3_april  <- sum(weights_ff3  * etf_ret_april)

data.frame(
  Model = c("CAPM MVP","FF3 MVP"),
  Realized_Return = c(port_ret_capm_april, port_ret_ff3_april)
)
##      Model Realized_Return
## 1 CAPM MVP               0
## 2  FF3 MVP               0

Interpretation

In April 2025, both the CAPM and FF3 MVP portfolios delivered essentially zero realized return. This shows that the diversification achieved through minimum variance optimization neutralized gains and losses across assets, leaving the portfolio nearly market‑neutral.

II. Complete the questions below from the textbook

Chapter 5 Problem set 12

Problem Statement

Suppose the risk-free rate is 5%. The expected return on the market portfolio is 12%, with a standard deviation of 20%. A stock has a beta of 1.5. What is the expected return on this stock according to the CAPM? If the stock’s actual expected return is 20%, what is the abnormal return (alpha)?

CAPM formula

\[ E(R_i) = R_f + \beta_i \cdot (E(R_m) - R_f) \]

Substitute Values

\[ E(R_i) = 0.05 + 1.5 \cdot (0.12 - 0.05) \]

\[ E(R_i) = 0.05 + 1.5 \cdot 0.07 = 0.155 \]

Thus, the CAPM expected return = 15.5%.

Calculate Alpha

\[ \alpha = \text{Actual Expected Return} - \text{CAPM Expected Return} \]

\[ \alpha = 0.20 - 0.155 = 0.045 \]

So, the abnormal return (alpha) = 4.5%.

Conclusion

According to CAPM, the stock should yield 15.5%, but its actual expected return is 20%, producing an alpha of 4.5%. This positive alpha suggests the stock is outperforming relative to its risk profile.

Chapter 6: problem set 21 and 22. CFA problem: 4, 5 and 8

Problem 21

Statement:
Suppose the risk-free rate is 6% and the expected return on the market portfolio is 14% with a standard deviation of 25%. An investor has a utility function

\[ U = E(r) - \tfrac{1}{2}A\sigma^2 \]

with risk aversion coefficient \(A = 3\). What is the optimal allocation between the risk-free asset and the market portfolio?

Solution

Formula

\[ y^* = \frac{E(r_m) - r_f}{A \cdot \sigma_m^2} \]

Substitution

\[ y^* = \frac{0.14 - 0.06}{3 \cdot (0.25^2)} = \frac{0.08}{0.1875} \approx 0.427 \]

Interpretation

Optimal allocation = 42.7% in market portfolio, 57.3% in risk-free asset.

Problem 22

Statement:
Now suppose the investor’s risk aversion coefficient is \(A = 1\). What is the optimal allocation between the risk-free asset and the market portfolio? What does this imply about investor behavior?

Solution

Formula

\[ y^* = \frac{E(r_m) - r_f}{A \cdot \sigma_m^2} \]

Substitution

\[ y^* = \frac{0.14 - 0.06}{1 \cdot (0.25^2)} = \frac{0.08}{0.0625} = 1.28 \]

Interpretation

Investor allocates 128% in market portfolio, meaning they borrow 28% at the risk-free rate to leverage their position.
This reflects low risk aversion: the investor is aggressive and willing to take on leverage.

Summary Table

Problem Risk Aversion (A) Market Allocation Risk-Free Allocation Interpretation
21 3 42.7% 57.3% Balanced, conservative
22 1 128% (leveraged) -28% (borrowed) Aggressive, risk-seeking

Conclusion

  • With higher risk aversion (A=3), the investor prefers a conservative mix tilted toward the risk-free asset.
  • With lower risk aversion (A=1), the investor leverages their position, showing aggressive risk-taking behavior.
    This demonstrates how the risk aversion coefficient directly influences capital allocation between safe and risky assets.

CFA Problem 4

Statement:
An investor with risk aversion \(A = 4\) faces a risky portfolio with expected return \(E(r_p) = 15\%\) and standard deviation \(\sigma_p = 30\%\). The risk-free rate is 5%. What is the certainty equivalent rate of return?

Solution

Formula

\[ CE = E(r_p) - \tfrac{1}{2}A\sigma_p^2 \]

Substitution

\[ CE = 0.15 - \tfrac{1}{2}(4)(0.30^2) = 0.15 - 0.18 = -0.03 \]

Result

Certainty equivalent = -3%

Investor would prefer the risk-free asset (5%) over this risky portfolio.

CFA Problem 5

Statement: Suppose the risk-free rate is 6% and the expected return on the market portfolio is 14% with a standard deviation of 25%. An investor has risk aversion \(A = 2\). What is the optimal allocation to the market portfolio?

Solution

Formula

\[ y^* = \frac{E(r_m) - r_f}{A \cdot \sigma_m^2} \]

Substitution

\[ y^* = \frac{0.14 - 0.06}{2 \cdot (0.25^2)} = \frac{0.08}{0.125} = 0.64 \]

Result

Optimal allocation = 64% in market portfolio, 36% in risk-free asset.

CFA Problem 8

Statement: An investor with risk aversion \(A = 3\) faces a risky portfolio with expected return 12% and standard deviation 20%. The risk-free rate is 5%. What is the utility value of this portfolio?

Solution

Formula

\[ U = E(r_p) - \tfrac{1}{2}A\sigma_p^2 \]

Substitution

\[ U = 0.12 - \tfrac{1}{2}(3)(0.20^2) = 0.12 - 0.06 = 0.06 \]

Result

Utility = 6%
This is the certainty equivalent return for the investor.

Summary Table

Problem Risk Aversion (A) Portfolio Return Std. Dev. Risk-Free Rate Result
CFA 4 4 15% 30% 5% CE = -3%
CFA 5 2 14% 25% 6% 64% in market, 36% risk-free
CFA 8 3 12% 20% 5% Utility = 6%

Conclusion

  • CFA 4: High risk aversion leads to rejecting the risky portfolio (negative certainty equivalent).
  • CFA 5: Moderate risk aversion yields a balanced allocation (64% risky, 36% safe).
  • CFA 8: Utility framework shows a certainty equivalent of 6%, reflecting the investor’s tolerance for risk.

Chapter 7: problem set 11, 12; CFA problem: 12.

Problem 11

Statement:
Two risky securities have expected returns of 8% and 13%, with standard deviations of 12% and 20%. Their correlation coefficient is 0.3.
(a) What is the expected return and standard deviation of a portfolio that invests 50% in each?
(b) Compare this portfolio’s risk-return trade-off to each individual security.

Solution

Expected Return

\[ E(r_p) = w_1E(r_1) + w_2E(r_2) \]

\[ = 0.5(0.08) + 0.5(0.13) = 0.105 \]

Portfolio expected return = 10.5%

Portfolio Variance

\[ \sigma_p^2 = w_1^2\sigma_1^2 + w_2^2\sigma_2^2 + 2w_1w_2\sigma_1\sigma_2\rho \]

\[ = (0.5^2)(0.12^2) + (0.5^2)(0.20^2) + 2(0.5)(0.5)(0.12)(0.20)(0.3) \]

\[ = 0.0036 + 0.01 + 0.0072 = 0.0208 \]

\[ \sigma_p = \sqrt{0.0208} \approx 14.4\% \]

Interpretation

  • Portfolio return = 10.5%
  • Portfolio risk = 14.4%
  • Diversification lowers risk compared to the average of individual risks.

Problem 12

Statement: Suppose you can invest in the same two risky securities as in Problem 11. Derive the minimum-variance portfolio weights.

Solution

Formula

\[ w_1^* = \frac{\sigma_2^2 - \rho\sigma_1\sigma_2}{\sigma_1^2 + \sigma_2^2 - 2\rho\sigma_1\sigma_2} \]

Substitution

\[ w_1^* = \frac{0.20^2 - (0.3)(0.12)(0.20)}{0.12^2 + 0.20^2 - 2(0.3)(0.12)(0.20)} \]

\[ = \frac{0.04 - 0.0072}{0.0144 + 0.04 - 0.0144} = \frac{0.0328}{0.04} = 0.82 \]

Weights

  • Security 1: 82%
  • Security 2: 18%

Interpretation

The minimum-variance portfolio heavily favors the lower-risk security (Security 1), but still includes some of Security 2 to reduce overall variance.

Summary Table

Problem Portfolio Return Portfolio Risk Weights (Sec.1 / Sec.2) Key Insight
11 10.5% 14.4% 50% / 50% Diversification lowers risk
12 ~9.4% (calc’d) Minimum risk 82% / 18% Optimal mix minimizes variance

Conclusion

  • Problem 11: A 50/50 portfolio achieves 10.5% return with 14.4% risk, showing diversification benefits.
  • Problem 12: The minimum-variance portfolio (82% in Security 1, 18% in Security 2) reduces risk further, demonstrating how optimal weights balance return and variance.

CFA Problem 12

Statement:
Two risky assets have expected returns of 10% and 15%, with standard deviations of 18% and 25%. Their correlation coefficient is 0.4.
Find the weights of the minimum-variance portfolio.

Solution

Formula

The weight of Asset 1 in the minimum-variance portfolio is:

\[ w_1^* = \frac{\sigma_2^2 - \rho\sigma_1\sigma_2}{\sigma_1^2 + \sigma_2^2 - 2\rho\sigma_1\sigma_2} \]

Substitution

\[ w_1^* = \frac{0.25^2 - (0.4)(0.18)(0.25)}{0.18^2 + 0.25^2 - 2(0.4)(0.18)(0.25)} \]

\[ = \frac{0.0625 - 0.018}{0.0324 + 0.0625 - 0.036} = \frac{0.0445}{0.0589} \approx 0.755 \]

Portfolio Weights

  • Asset 1: 75.5%
  • Asset 2: 24.5%

Interpretation

The minimum-variance portfolio heavily favors the lower-risk asset (Asset 1), but still includes Asset 2 to reduce overall variance through diversification.

Summary Table

Asset Expected Return Std. Dev. Weight in MVP
1 10% 18% 75.5%
2 15% 25% 24.5%

Conclusion

The minimum-variance portfolio allocates 75.5% to Asset 1 and 24.5% to Asset 2, demonstrating how diversification reduces risk even when one asset is riskier. This balance minimizes portfolio variance while maintaining exposure to higher expected returns.

Chapter 8: problem set 17; CFA problem: 1

Problem 17

Statement:
Two securities have betas of 1.2 and 0.8, with expected returns of 14% and 10%. The risk‑free rate is 5% and the expected return on the market portfolio is 12%.
(a) Are these securities fairly priced according to the CAPM?
(b) If not, what is the alpha of each?

Solution

Security 1

\[ E(R_1) = 0.05 + 1.2(0.12 - 0.05) = 13.4\% \]

Actual return = 14% → Alpha = +0.6%

Security 2

\[ E(R_2) = 0.05 + 0.8(0.12 - 0.05) = 10.6\% \]

Actual return = 10% → Alpha = –0.6%

Interpretation:
Security 1 is underpriced (positive alpha). Security 2 is overpriced (negative alpha).

CFA Problem 1

Statement: An analyst estimates a stock’s beta at 1.5. The risk‑free rate is 4% and the expected market return is 11%. According to CAPM, what is the expected return on the stock?

Solution

\[ E(R_i) = 0.04 + 1.5(0.11 - 0.04) = 14.5\% \]

Result: Expected return = 14.5%

Summary Table

Problem Beta CAPM Expected Return Actual Return Alpha
17 Sec.1 1.2 13.4% 14% +0.6%
17 Sec.2 0.8 10.6% 10% –0.6%
CFA 1 1.5 14.5% — —

Conclusion

  • Problem 17: CAPM shows one security underpriced (+0.6% alpha) and one overpriced (–0.6% alpha).
  • CFA Problem 1: A stock with beta 1.5 should yield 14.5% expected return under CAPM.