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Canvas Example MCQ — Q5

• A) Higher property income-growth expectations have no effect on cap rates.
• B) Higher property income-growth expectations result in lower cap rates.
• C) Higher Treasury / risk-free bond yields have no impact on cap rates.
• D) A higher property risk premium results in a lower cap rate.

Using the standard growth intuition, cap rate ≈ required return - expected growth. If expected growth rises while required return is unchanged, the cap rate falls.

Rule: Growth up -> cap rate down, ceteris paribus.

Canvas Example MCQ — Q8

• A) 0.10%
• B) 2.55%
• C) 10.00%
• D) 15%
• E) None of A-D is close; write the correct answer.

E[R] = 0.20(15%) + 0.60(10%) + 0.20(5%) = 3% + 6% + 1% = 10%.

Rule: Expected return is a probability-weighted average.

Old Exam — Dec 2022, Q2

• A) $2 750 000
• B) $1 214 100
• C) $ 1 514 828
• D) $1 412 556
• E) None of the above (A, B, C, D) is close to be correct.

PV = sum[250,000/(1.12)^t, t=1..5] + sum[300,000/(1.12)^t, t=6..10] + 3,000,000/(1.12)^10 = about $2,480,748. The listed A-D values miss it, so E.

Rule: Recognize cash-flow timing first. Discount each cash flow; terminal/sale proceeds are added in the sale year. IRR is the rate that sets NPV=0.

Old Exam — Dec 2022, Q3 · adapted

Given: Value at 12% ≈ $2 480 748; asking price $1 500 000.

• A) Yes, because the IRR is higher than the required rate of return.
• B) Yes, because the IRR is lower than the required rate of return.
• C) No, because the IRR is higher than the required rate of return.
• D) No, because the IRR is lower than the required rate of return.
• E) None of the above (A, B, C, D) is close to be correct.

The investment value at the 12% required return is about $2.481m, above the $1.5m asking price, so NPV>0. Equivalent IRR rule: IRR>required return -> accept.

Rule: Recognize cash-flow timing first. Discount each cash flow; terminal/sale proceeds are added in the sale year. IRR is the rate that sets NPV=0.

Old Exam — Dec 2022, Q4 · adapted

Given: Original required return 12%. New: nominal risk-free 8%, inflation 2%, risk premium 3%.

• A) The percentage change is 4.43%.
• B) The percentage change is ‒4.43%.
• C) The percentage change is 4.33%.
• D) The percentage change is ‒4.15%.
• E) None of the above (A, B, C, D) is close to be correct.

Nominal risk-free rate is already nominal, so new required return = 8%+3%=11% (do not subtract inflation). Revalue the same cash flows at 11%: about $2.639m. Change=(2.639/2.481)-1≈+6.36% -> E.

Rule: Recognize cash-flow timing first. Discount each cash flow; terminal/sale proceeds are added in the sale year. IRR is the rate that sets NPV=0.

Old Exam — Oct 2023, Q7 · adapted

Given: Severe recession: p = 10%, IRR = −4.75% Moderate recession: p = 30%, IRR = 0.55% Baseline: p = 35%, IRR = 6.53% Moderate expansion: p = 20%, IRR = 9.80% Strong boom: p = 5%, IRR = 18.94%

• A) 6.22%
• B) 4.53 %
• C) 4.88%
• D) 3.55 %
• E) None of the above (A, B, C, D) is close to be correct.

E[IRR] = 0.10(−4.75) + 0.30(0.55) + 0.35(6.53) + 0.20(9.80) + 0.05(18.94) = −0.475 + 0.165 + 2.286 + 1.960 + 0.947 ≈ 4.88% → C.

Rule: Probabilities sum to 1. Expected value is probability-weighted; variance uses squared deviations around the expected value.

Old Exam — Dec 2022, Q5

Given: Scenario probabilities: recession 28%, moderate growth 45%, strong growth X%. Probabilities must sum to 100%.

• A) 73 %
• B) 17 %
• C) 53 %
• D) 27 %
• E) None of the above (A, B, C, D) is close to be correct.

Probabilities must sum to 100%: X=100%-28%-45%=27%.

Rule: Probabilities sum to 1. Expected value is probability-weighted; variance uses squared deviations around the expected value.

Old Exam — Dec 2022, Q6

Given: Scenario setup: purchase price $45m; 15-year holding period; Year-1 NOI $2.5m; growth rates -0.5%, +1.5%, +2.5%; exit value at Year 15 uses Year-16 NOI / scenario exit cap (6%, 5.5%, 5%); probabilities 28%,45%,27%.

• A) 6.79 %
• B) 6.83 %
• C) 7.05 %
• D) 7.10 %
• E) None of the above (A, B, C, D) is close to be correct.

Method: compute the return/NPV separately in every scenario, then E[X]=sum p_s X_s. Risk: SD=sqrt(sum p_s(X_s-E[X])^2). Source correct answer: B) 6.83 %.

Rule: Probabilities sum to 1. Expected value is probability-weighted; variance uses squared deviations around the expected value.

Old Exam — Dec 2021 re-exam, Q24

Given: ln(y)=17.3-0.04x. A 1-unit fall in x changes ln(y) by +0.04; 2-unit fall -> about +8% in y.

• A) y decreases by 0.08 units.
• B) y increases by 8 percent.
• C) y increases by 4 units.
• D) y decreases by 8 percent.
• E) None of the above (A, B, C, D) is correct.

Delta x=-2. Delta ln(y)=(-0.04)(-2)=+0.08, so y rises by approximately 8% -> B.

Rule: Identify model units and form before calculating. Never confuse coefficient size, significance, R2 and prediction accuracy.

Old Exam — Dec 2021 re-exam, Q20

Given: Y is monthly residential lending in $m; X4=1 for Jun-Jul-Aug. Coefficient on X4 is +1.05.

• A) On an average, we would expect an additional 1.05 loans per month during June, July, and August.
• B) On an average, we would expect an additional $1.05 million in loans during June, July, and August.
• C) On an average, we would expect an additional 1.05% increase in loans during June, July, and August.
• D) There is no difference in the total residential loan amount between the months of June and September.
• E) None of the above (A, B, C, D) is correct.

X4 is a seasonal dummy and Y is measured in millions of dollars. Holding other regressors fixed, X4=1 adds 1.05 million dollars of expected loans -> B.

Rule: Interactions and quadratics mean the marginal effect is not constant. Log coefficients require percentage interpretation.

Old Exam — Dec 2022, Q14

Given: Hotel table (counts): Price 18/9/36; Location 23/15/42; Room 9/8/16; Other 12/14/25 across three hotels.

• A) There is a relationship between the primary reason for not returning and the hotel.
• B) There is no relationship between the primary reason for not returning and the hotel.
• C) There is a positive relationship between the primary reason for not returning and the hotel.
• D) There is a negative relationship between the primary reason for not returning and the hotel.
• E) None of the above (A, B, C, D) is close to be correct.

Chi-square workflow: H0=no association/same proportions; E=(row total x column total)/grand total; statistic=sum((O-E)^2/E); df=(r-1) (c-1); reject if p<alpha or statistic>critical. Source answer: B) There is no relationship between the primary reason for not returning and the hotel..

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Old Exam — Dec 2022, Q15 · adapted

Given: Hotel table (counts): Price 18/9/36; Location 23/15/42; Room 9/8/16; Other 12/14/25 across three hotels.

• A) 4.2962
• B) 3.6018
• C) 0.7304
• D) 27.4104
• E) None of the above (A, B, C, D) is close to be correct.

For a contingency table, chi-square=sum((O-E)^2/E). Using the supplied 4x3 hotel table gives about 3.6018.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Old Exam — Dec 2022, Q16 · adapted

Given: Hotel table (counts): Price 18/9/36; Location 23/15/42; Room 9/8/16; Other 12/14/25 across three hotels.

• A) 27.4104
• B) 12.5916
• C) 3.6018
• D) 16.8119
• E) None of the above (A, B, C, D) is close to be correct.

df=(rows-1)(columns-1)=(4-1)(3-1)=6. The 1% upper-tail chi-square critical value for df=6 is about 16.8119.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Old Exam — Dec 2022, Q17 · adapted

Given: Hotel table (counts): Price 18/9/36; Location 23/15/42; Room 9/8/16; Other 12/14/25 across three hotels.

• A) the null hypothesis will be rejected since the computed p-value is 0.0001.
• B) the null hypothesis will not be rejected since the computed p-value is 0.7304.
• C) the null hypothesis will be rejected since the computed p-value is 0.7304.
• D) the null hypothesis will be rejected since the computed p-value is 16.8119.
• E) None of the above (A, B, C, D) is close to be correct.

p≈0.7304>0.05, so fail to reject H0. Do not say that H0 is proved true.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Old Exam — Oct 2022, Q10

Given: Hotel table: 400 guests per hotel. Luxury yes counts 344,356,338; not-luxury 56,44,62.

• A) All hotel guests have experienced that the hotel visits were luxury.
• B) There are no differences among the three hotels in the proportion of guests who experienced the hotel visits as luxury.
• C) There are large differences among the three hotels in the proportion of guests who experienced the hotel visits as luxury.
• D) Fifty percent of the hotel guests in all hotels experienced the hotel visits as luxury.
• E) None of the above (A, B, C, D) is close to be correct.

For a chi-square homogeneity/independence test, H0 says the proportions/distributions are the same across groups - here, no hotel difference in the luxury proportion.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Old Exam — Oct 2022, Q11 · adapted

Given: Hotel table: 400 guests per hotel. Luxury yes counts 344,356,338; not-luxury 56,44,62.

• A) 3.5967
• B) 5.9915
• C) 159.2251
• D) 2.7165.
• E) None of the above (A, B, C, D) is close to be correct.

Compute expected counts from row total x column total / grand total, then sum (O-E)^2/E. Source result ≈3.5967.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Old Exam — Oct 2022, Q12 · adapted

Given: Hotel table: 400 guests per hotel. Luxury yes counts 344,356,338; not-luxury 56,44,62.

• A) The p-value must be lower than 5 %, implying that the null hypothesis will be rejected.
• B) The p-value must be lower than 5 %, implying that the null hypothesis will not be rejected.
• C) The p-value must be higher than 5 %, implying that the null hypothesis will not be rejected.
• D) The p-value must be higher than 5 %, implying that the null hypothesis will be rejected.
• E) None of the above (A, B, C, D) is close to be correct.

With df=2, chi-square≈3.5967 gives p>0.05. Therefore fail to reject H0.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Old Exam — Dec 2021 re-exam, Q13

Given: Restaurant table: satisfied counts 80,85,60; sample sizes 100,120,80. Total satisfied=225 of 300.

• A) 60
• B) 75
• C) 80
• D) 90
• E) None of the above (A, B, C, D) is close to be correct.

Expected satisfied in Los Angeles = row total 225 x LA column total 80 / grand total 300 = 60 -> A.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Old Exam — Dec 2021 re-exam, Q14 · adapted

Given: Restaurant table: satisfied counts 80,85,60; sample sizes 100,120,80. Total satisfied=225 of 300.

• A) 1.49
• B) 2.44
• C) 4.15
• D) 5.33
• E) None of the above (A, B, C, D) is close to be correct.

Compute all expected counts and sum (O-E)^2/E; source result≈2.44 -> B.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Old Exam — Dec 2021 re-exam, Q15 · adapted

Given: Restaurant table: satisfied counts 80,85,60; sample sizes 100,120,80. Total satisfied=225 of 300. Source question prints alpha=0.5, but the keyed 5.991 critical value is the 5% (0.05) critical value for df=2. Treat 0.5 as a source typo.

• A) 5.991
• B) 3.841
• C) 2.706
• D) 7.815
• E) None of the above (A, B, C, D) is close to be correct.

For df=(2-1)(3-1)=2 and alpha=0.05, chi-square critical≈5.991 -> A. The printed alpha=0.5 is inconsistent with the keyed answer and is treated as a typo.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Old Exam — Dec 2021 re-exam, Q16 · adapted

Given: Restaurant table: satisfied counts 80,85,60; sample sizes 100,120,80. Total satisfied=225 of 300. With chi-square≈2.44 and 5% critical≈5.991, do not reject H0.

• A) more than the critical value, we can reject the null hypothesis and conclude that there is a difference in proportion of satisfied customers between these three locations
• B) less than the critical value, we can reject the null hypothesis and conclude that there is a difference in proportion of satisfied customers between these three locations
• C) less than the critical value, we cannot reject the null hypothesis and cannot conclude that there is a difference in proportion of satisfied customers between these three locations
• D) more than the critical value, we cannot reject the null hypothesis and conclude that there is no difference in proportion of satisfied customers between these three locations. 6

2.44<5.991, so fail to reject H0; there is insufficient evidence of different satisfaction proportions -> C.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Old Exam — Oct 2021, Q11

Given: 2x2 table: Exclusive speakers: Like Dune=108, Not like=95; Not-exclusive: Like=33, Not like=88. df=1.

• A) 6.63.
• B) 5.99.
• C) 20.74.
• D) 2.71.
• E) None of the above (A, B, C, D) is close to be correct.

The table is 2x2, so df=(2-1)(2-1)=1. The 1% upper-tail chi-square critical value is about 6.63 -> A.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Old Exam — Oct 2021, Q12 · adapted

Given: 2x2 table: Exclusive speakers: Like Dune=108, Not like=95; Not-exclusive: Like=33, Not like=88

• A) 6.63.
• B) 5.99.
• C) 20.74.
• D) 2.71.
• E) None of the above (A, B, C, D) is close to be correct.

Using E=row total x column total / grand total and summing (O-E)^2/E gives chi-square≈20.74 -> C.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Old Exam — Oct 2021, Q13 · adapted

Given: 2x2 table: Exclusive speakers: Like Dune=108, Not like=95; Not-exclusive: Like=33, Not like=88

• A) The p-value must be higher than 10 %, implying that the null hypothesis will be rejected.
• B) The p-value must be higher than 10 %, implying that the null hypothesis will not be rejected.
• C) The p-value must be lower than 1 %, implying that the null hypothesis will not be rejected.
• D) The p-value must be lower than 1 %, implying that the null hypothesis will be rejected.
• E) None of the above (A, B, C, D) is close to be correct.

20.74 is far above the 1% critical value 6.63, so p<0.01 and H0 is rejected -> D.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Old Exam — Dec 2022, Q26

Given: Graph-based stationarity question. Exam-useful version: a series whose level wanders/persists with no stable mean is likely non- stationary.

• A) The variable is stationary for sure, and therefore can be used in the ARDL model without making any transformation.
• B) The variable is likely not stationary, and therefore can be used in the ARDL model without making any transformations.
• C) The variable is likely not stationary, and therefore cannot be used in the ARDL model. That is, it must first be made stationary before it can be used in the ARDL model.
• D) Since variables do not need to be stationary in ARDL models, it does not matter if the variable is stationary or not stationary.
• E) None of the above (A, B, C, D) is close to be correct.

A series with persistent level shifts/wandering mean is likely non-stationary. For dynamic time-series regression, the course's old exam expected it to be made stationary first.

Rule: Stationarity means stable distributional properties over time. Current relevance is secondary unless reinforced by current Canvas/labs.

Old Exam — Dec 2021 re-exam, Q27

• A) Only the expected mean remains constant over time.
• B) Only the expected variance remains constant over time.
• C) Both the expected mean and expected variance remain constant over time.
• D) The expected mean and expected variance are a linear function of time.
• E) None of the above (A, B, C, D) is correct.

Weak stationarity assumes constant mean and variance over time (and autocovariance depending on lag, not calendar time) -> C.

Rule: Stationarity means stable distributional properties over time. Current relevance is secondary unless reinforced by current Canvas/labs.

Computer Lab 1 — Exercise 1, Case B

• A) 8.57%, lower than Case A because higher growth raises property value
• B) 9.48%, same as Case A
• C) 10.48%, higher than Case A
• D) 9.00%, equal to the Base Case

Going-in cap rate = 100,000/1,166,992.19 = 8.57%. Since Case B has higher expected NOI growth, its property value is higher than Case A's, and with the same Year-1 NOI, a higher value implies a lower going-in cap rate.

Rule: With equal current income, higher expected growth → higher value → lower going-in cap rate.

Computer Lab 1 — Exercise 1 concept

• A) Year 6 NOI (the NOI expected in the year after sale)
• B) Year 5 NOI (the NOI in the year of sale)
• C) Year 1 NOI (initial NOI)
• D) The average NOI over Years 1-5

Reversion value = NOI_(T+1)/terminal cap rate, because the buyer at the end of Year 5 is purchasing rights to future (Year 6 onward) income.

Rule: Terminal/reversion value always capitalizes the NEXT year's NOI (forward-looking), not the NOI in the sale year.

Computer Lab 1 — Exercise 2C

• A) ≈20.03%
• B) ≈19.86% (simple average)
• C) ≈25.95%
• D) ≈5.83%

E[IRR] = 0.05(5.89%)+0.20(13.74%)+0.50(20.25%)+0.20(25.95%)+0.05(33.50%) = 20.03%, a probability-weighted average, not a simple average.

Rule: Expected value under scenario analysis = Σ (probability × outcome), not a simple unweighted average.

Computer Lab 1 — Exercise 2C

• A) Leverage increases expected return but increases risk proportionally even more, lowering the risk-adjusted return ratio
• B) Leverage increases expected return with no effect on risk
• C) Leverage decreases both expected return and risk
• D) Leverage has no effect on return or risk, only on cash flow timing

Both mean IRR and σ rise with LTV, but σ rises faster (1.98%→10.21%, a ~5.2x increase vs. mean IRR's ~2.4x increase), so the Mean/Std.Dev ratio falls from 6.18x to 2.91x — risk grows faster than return.

Rule: Financial leverage magnifies both expected equity returns and the variability (risk) of those returns; risk generally grows faster than return as leverage increases.

Computer Lab 1 — Exercise 3 Problem 2

• A) Investment I (CV=0.316) is lower risk per unit of return than Investment II (CV=0.350)
• B) Investment II is lower risk per unit of return because it has a lower standard deviation
• C) Both investments have identical CV
• D) Investment II is better because its expected IRR (14%) is higher

E[IRR]_I=10%, σ_I=3.16%, CV_I=0.316. E[IRR]_II=14%, σ_II=4.90%, CV_II=0.350. Lower CV means less risk per unit of expected return, so Investment I has the better risk-adjusted profile by this measure, even though II has a higher expected IRR.

Rule: Coefficient of Variation (σ/E[R]) measures risk per unit of expected return; lower CV means better reward-to-risk (though this does not mean universally 'better').

Computer Lab 1 — Exercise 3 Problem 3

• A) E[IRR]≈10.01%, σ≈1.62%
• B) E[IRR]≈10.02%, σ≈2.19%
• C) E[IRR]≈10.68%, σ≈1.62%
• D) E[IRR]≈10.01%, σ≈4.90%

E[IRR]=0.30(7.93%)+0.40(10.00%)+0.30(12.12%)=10.01%. Variance=Σp(IRR−E[IRR])²≈0.00026296, σ=√0.00026296≈1.62%.

Rule: Standard deviation = √[Σ p_i × (R_i − E[R])²] using scenario probabilities.

Computer Lab 1 — Exercise 3 Problem 3(d)

• A) The alternative has higher expected return, but the lab property has much lower risk per unit of return (lower CV)
• B) The lab property dominates on every metric: higher return and lower risk
• C) The alternative dominates on every metric: higher return and lower risk
• D) They are financially identical because CV differs by less than 0.2

The alternative's E[R]=12% exceeds 10.01%, but its CV (0.333) is roughly double the lab property's CV (0.162), meaning it carries much more risk per unit of expected return.

Rule: When comparing investments with different expected returns AND different risks, use CV = σ/E[R] to assess risk per unit of return.

Computer Lab 1 — Exercise 3 Problem 4

• A) Yes — standard deviation rises from 1.62% to 5.52% while expected return rises only slightly, because fixed debt service makes equity a residual claim
• B) No — leverage has almost no effect since expected return barely changed
• C) No — standard deviation actually falls with leverage because debt diversifies risk
• D) Cannot be determined without knowing the loan's credit rating

Fixed debt service must be paid regardless of NOI outcome, so equity cash flows (the residual) swing proportionally more with property performance. Standard deviation more than triples (1.62%→5.52%) while expected return only rises slightly (10.01%→10.14%).

Rule: Financial leverage magnifies equity risk because debt service is a fixed prior claim; equity absorbs the full variability on a proportionally smaller base.

Computer Lab 2 — Exercise 3, Problem 12.14

• A) Expected return = 2.12%; No, it cannot equal 2.36% because default risk lowers expected return below the promised yield
• B) Expected return = 2.36%; Yes, YTM always equals expected return
• C) Expected return = 1.36%; equals the risk-free Treasury yield
• D) Expected return = 0.24%; the default-loss amount itself

r_d = YTM − p×L = 2.36% − 0.5%×48% = 2.36%−0.24% = 2.12%. The 2.36% YTM is the best-case (promised) return only if every payment is received; with any default risk, it cannot reflect the true expected return (which is lower).

Rule: Expected return on risky debt: r_d = YTM − p×L, always ≤ the promised YTM.

Computer Lab 2 — Exercise 3, Problem 12.15

• A) ≈21.25%
• B) ≈28.83%
• C) ≈4.55%
• D) ≈12.75%

Debt CAPM: r_d = 3.0%+0.31×5.0% = 4.55%. Default probability: p = (YTM−r_d)/L = (17.3%−4.55%)/60% = 12.75%/60% = 21.25%.

Rule: When YTM and the CAPM-implied expected return are known, solve for p via p = (YTM − r_d)/L.

Computer Lab 2 — Exercise 3, Problem 12.16

• A) ≈3.70% (default-adjustment) and ≈3.85% (debt CAPM) — both far below the 8.5% promised YTM
• B) 8.5% under both methods, since YTM is a market-determined fair rate
• C) ≈4.8% default-loss amount is itself the expected return
• D) ≈11.5%, since recession risk should raise expected return

Method 1: r_d = 8.5%−8.0%×60% = 8.5%−4.8% = 3.70%. Method 2 (CAPM): r_d = 3.0%+0.17×5.0% = 3.85%. Both are far below the promised 8.5% YTM, reflecting significant recession default risk.

Rule: Under recession-level default risk, promised YTM substantially overstates the investor's true expected return; two independent methods (default-adjustment, debt CAPM) should roughly agree.

Computer Lab 2 — Exercise 3, Problem 12.17

• A) YTM≈4.80%; spread≈0.80%
• B) YTM≈4.50%; spread≈0.50%
• C) YTM≈4.00%; spread≈0.00%
• D) YTM≈6.90%; spread≈2.90%

Required expected return via CAPM: r_d = 4.0%+0.10×5.0% = 4.50%. YTM = r_d + p×L = 4.50%+0.5%×60% = 4.50%+0.30% = 4.80%. Spread over AAA = 4.80%−4.00% = 0.80%.

Rule: Set promised YTM = required expected return (from debt CAPM) plus the expected default-loss compensation (p×L).

Computer Lab 2 — Exercise 3 concept

• A) Promised YTM is generally higher than expected return, because YTM assumes all payments are received with certainty
• B) Promised YTM is generally lower than expected return
• C) They are always identical regardless of default risk
• D) Expected return exceeds YTM only for investment-grade bonds

YTM is calculated assuming every promised payment is made in full and on time. If there's any default probability, the investor's actual expected return (accounting for potential losses) must be lower than the promised YTM.

Rule: Promised YTM ≥ expected return whenever there is positive default probability; the gap widens as credit quality worsens.

Computer Lab 3 — Problem 2

• A) 10.00%
• B) 10.00% is wrong, it should be a simple average of 10%
• C) 11.67%
• D) 15.00%

E[IRR]=0.20(15%)+0.60(10%)+0.20(5%)=3%+6%+1%=10.00%.

Rule: Expected value = Σ (probability × outcome) across all scenarios.

Computer Lab 3 — Problem 3(b)

• A) ≈10.01%
• B) ≈10.02%
• C) ≈10.68%
• D) ≈11.00%

E[IRR]=0.30(7.93%)+0.40(10.00%)+0.30(12.12%)=2.379%+4.00%+3.636%=10.0144%≈10.01%.

Rule: Probability-weighted expected value: Σ p_i × IRR_i.

Computer Lab 3 — Problem 3(d)

• A) The property is stronger by E[R]/σ (6.18 vs 3.00), but the alternative offers higher absolute expected return with more risk — the best choice depends on the investor's risk aversion
• B) The alternative is stronger on every metric, so it is unambiguously better
• C) The property is stronger on every metric, so it is unambiguously better
• D) They are equally good since both ratios round to about 3-6x

E[R]/σ favors the property (6.18 > 3.00), reflecting better return-per-unit-of-risk. However, the alternative has a higher absolute expected return (12% vs 10.01%), so a return-maximizing investor willing to bear more risk might prefer it. No single metric makes one universally dominant.

Rule: When neither investment dominates on both return and risk simultaneously, the final choice depends on the investor's degree of risk aversion.

Computer Lab 3 — Problem 4A-B

• A) ≈10.14%
• B) ≈10.01%
• C) ≈10.42%
• D) ≈13.44%

E[IRRe]=0.30(2.82%)+0.40(10.42%)+0.30(17.07%)=0.846%+4.168%+5.121%=10.1356%≈10.14%.

Rule: Probability-weighted expected value across all scenarios: Σ p_i × IRR_i.

Computer Lab 3 — Problem 4B

• A) Yes — σ rises from 1.62% to 5.52% because fixed debt service makes equity a residual claim, amplifying the effect of NOI/sale-price changes on equity returns
• B) No — expected return only rose slightly (10.01%→10.14%), so risk is essentially unchanged
• C) No — leverage reduces risk by diversifying the capital structure
• D) Risk is unchanged because E[IRR]/σ actually increased with leverage

σ more than triples (1.62%→5.52%). Because debt service is fixed, all variability in property performance (NOI, sale price) flows through to the smaller equity base, magnifying the percentage swings in equity returns.

Rule: Leverage magnifies equity-return volatility because debt holders are paid first (fixed claim) while equity absorbs all residual variability.

Computer Lab 3 — Problem 2 recap

• A) Direct scenario-by-scenario (state-by-state) comparison / first-order stochastic dominance — it requires no assumption about risk preference
• B) The coefficient of variation (σ/E[R]), since it always gives the definitive ranking
• C) Standard deviation alone, since lower risk always wins
• D) Expected return alone, since higher return always wins

If an investment's payoff is ≥ in every scenario (state-by-state), it dominates regardless of any risk-preference assumption — this is a stronger and more general conclusion than CV, σ, or E[R] comparisons alone, each of which implicitly assumes something about risk tolerance or ignores risk entirely.

Rule: State-by-state (first-order stochastic) dominance, when it exists, is the strongest possible ranking tool because it holds for all risk preferences.

Computer Lab 4 — Exercise 1, Task 5

• A) The property should offer a higher expected return/cap rate than the Treasury yield to compensate for added risk
• B) They should offer identical returns since both are low-risk
• C) The Treasury should offer a higher return because it is less liquid
• D) Property risk is irrelevant once a cap rate is quoted

Real estate carries vacancy, operating, regulatory, liquidity and asset-specific risks beyond those of government bonds, so investors typically require a risk premium—i.e., a cap rate/expected return above the Treasury yield.

Rule: Riskier assets require a return premium above the risk-free (Treasury) rate.

Computer Lab 4 — Exercise 1, Task 13

• A) Lower inflation expectations, weaker growth, expected Fed cuts, and a lower term premium
• B) Persistent inflation and heavy Treasury issuance
• C) Higher-for-longer policy rate expectations and strong private capital demand
• D) Stronger economic growth and rising inflation expectations

A major decline in the long yield generally requires a lower expected path of future short-term rates (driven by lower inflation/growth expectations and anticipated rate cuts) and/or a lower term premium.

Rule: Long yield = expected average future short rates + term premium; both components must fall for a substantial yield decline.

Computer Lab 4 — Exercise 2, Task 2

• A) False — it is a conditional association and is causal only under additional identification assumptions
• B) True — any regression coefficient is causal by construction
• C) True, but only when R² exceeds 0.70
• D) False — coefficients from regression are always meaningless

Holding the other included regressors constant removes their confounding influence only, not all possible confounding from omitted variables; causal interpretation requires further assumptions (e.g., no omitted confounders, correct functional form).

Rule: Partial association ≠ causal effect unless identification assumptions hold.

Computer Lab 4 — Exercise 2, Task 6 (Model 1)

• A) A one-unit increase in STR is associated with about a 2.28-point decrease in expected test score
• B) A one-unit increase in STR causes exactly a 2.28-point decrease in every school's test score
• C) A 2.28% increase in STR decreases test scores by one point
• D) STR explains 2.28% of the variation in test scores

The coefficient -2.280 means expected test score falls by about 2.28 points per one-unit increase in STR, as an average association, not a guaranteed effect for every unit, and not a causal claim without further assumptions.

Rule: Interpret OLS slopes as changes in expected/average outcome, not as a guarantee for each individual observation.

Computer Lab 4 — Exercise 2, Task 6 (Model 5)

• A) About -2.028 points
• B) About -1.014 points
• C) About -2.280 points
• D) About +2.028 points

ΔTestScore = -1.014 × 2 = -2.028 points, holding the other regressors constant.

Rule: Multiply the partial-effect coefficient by the size of the change in X, holding other regressors fixed.

Computer Lab 5 — Exercise 1, Task 1

• A) +4.56 points
• B) -4.56 points
• C) +2.28 points
• D) -2.28 points per unit, so the direction is indeterminate

ΔTestScore = (-2.28)×(-2) = +4.56. A decrease in STR of 2 units raises the expected test score by about 4.56 points on average.

Rule: Multiply the slope coefficient by the change in X; watch the signs carefully.

Formula Battle — method recognition (built from the Formula Book)

• A) (IRR1 + IRR2 + IRR3) / 3
• B) p1·IRR1 + p2·IRR2 + p3·IRR3
• C) √[Σ p(IRR − E)²]
• D) The most likely scenario's IRR

Expected value is probability-weighted. Correct method: p1·IRR1 + p2·IRR2 + p3·IRR3.

Rule: E = p1x1 + p2x2 + p3x3