Search

23 questions · 1 formulas

Standardization

z = (x − μ) / σ; P(X < x) = Φ(z); P(X > x) = 1 − Φ(z)

Trap: N(μ, σ²): second parameter is variance — use σ = √σ².

Normal DistributionD–A level★ High priority — official Canvas example

In a regression with ln(price) as dependent variable, pool dummy coefficient = 0.0774. Best interpretation?

→ A) A house with a pool is on average about 7.74% more expensive than a house without a pool, holding the other regressors fixed.

Canvas Example MCQ — Q53

A) A house with a pool is on average about 7.74% more expensive than a house without a pool, holding the other regressors fixed.
B) A house with a pool is on average $7,740 more expensive than a house without a pool.
C) A house with a pool is on average 7.74% less expensive than a house without a pool.
D) A house with a pool is on average $77,740 more expensive than a house without a pool.
E) None of A-D is close; write the correct answer.

For a dummy in log-Y, the standard approximation is 100b% = 7.74%. Exact effect = 100(e^0.0774-1) ≈ 8.05%. The source MCQ is testing the approximation.

Rule: Know both approximate and exact dummy-in-log effects.

Normal DistributionD–A level★ High priority — official Canvas example

In ln(price) on ln(sqft), coefficient on ln(sqft) = 0.7244. Interpret.

→ A) A 1% increase in house size is associated with about a 0.7244% increase in sales price.

Canvas Example MCQ — Q54

A) A 1% increase in house size is associated with about a 0.7244% increase in sales price.
B) A 1,000 sqft increase in size increases sales price by about $7,244.
C) A 1% increase in size increases sales price by about $72.44.
D) A 10% increase in size increases sales price by about 0.8925%.
E) None of A-D is close; write the correct answer.

This is a log-log specification. The slope is elasticity directly.

Rule: ln(Y) on ln(X): coefficient = elasticity.

Normal DistributionD–A level★ High priority — official Canvas example

Interpret b1 in ln(Y)=b0+b1X+u.

→ B) A one-unit change in X is associated with approximately a 100*b1% change in Y.

Canvas Example MCQ — Q57

A) A 1% change in X is associated with a b1% change in Y.
B) A one-unit change in X is associated with approximately a 100*b1% change in Y.
C) A 1% change in X is associated with a change in Y of 0.01*b1.
D) A one-unit change in X is associated with a b1-unit change in Y.

This is a log-level model. For small b, multiply the coefficient by 100 to get the approximate percent change in Y for one unit of X.

Rule: ln(Y) on level X -> 100b% per unit X.

Normal DistributionD–A level★ High priority — official Canvas example

Interpret b1 in ln(Y)=b0+b1 ln(X)+u.

→ A) A 1% change in X is associated with a b1% change in Y.

Canvas Example MCQ — Q58

A) A 1% change in X is associated with a b1% change in Y.
B) A one-unit change in X is associated with a b1-unit change in Y.
C) A one-unit change in X is associated with a 100*b1% change in Y.
D) A 1% change in X is associated with a change in Y of 0.01*b1.

This is log-log. b1 is the elasticity.

Rule: ln-ln coefficient = elasticity.

Omitted Variable BiasD–A level★ High priority — official Canvas example

Which two conditions are required for OVB?

→ B) The omitted variable must affect Y and be correlated with an included regressor.

Canvas Example MCQ — Q81

A) The omitted variable must have a large variance and a non-zero mean.
B) The omitted variable must affect Y and be correlated with an included regressor.
C) The omitted variable must be normally distributed and correlated with Y.
D) The omitted variable must be statistically significant and correlated with the error term.
E) None of the above.

Exactly the two-condition rule.

Rule: Memorize verbatim.

ProbitD–A level★ High priority — official Canvas example

Probit: z = -4.60 + .40 FIXRATE - .20 MARGIN + .10 NETWORTH. At FIXRATE=12, MARGIN=2, NETWORTH=7, predicted probability?

→ C) approximately 69.1%

Canvas Example MCQ — Q89

A) approximately 30.9%
B) approximately 50.0%
C) approximately 69.1%
D) approximately 84.1%
E) None of A-D is close; write the correct answer.

z = -4.60 + .40(12) - .20(2) + .10(7) = 0.50. P = Φ(0.50) ≈ 0.6915 = 69.15%.

Rule: Compute the index first; then apply the standard normal CDF.

ProbitD–A level★ High priority — official Canvas example

Same Probit point; MARGIN coefficient = -0.20 and φ(0.50)=0.3521. Marginal effect of MARGIN?

→ C) approximately -7.0 percentage points

Canvas Example MCQ — Q90

A) approximately -20.0 percentage points
B) approximately -15.2 percentage points
C) approximately -7.0 percentage points
D) approximately +7.0 percentage points
E) None of A-D is close; write the correct answer.

ME = -0.20 x 0.3521 = -0.07042, i.e. about -7.04 percentage points for a one-unit local increase in MARGIN.

Rule: Probit ME = coefficient x normal PDF at the current index.

Normal DistributionD–A levelVery high priority

The 99% confidence interval for β-coefficient for the lsqrft variable is closest to__________.

→ A) (0.5027, 0.9026)

Old Exam — Dec 2022, Q21

Given: For ln(sqft): coefficient=0.7026, SE=0.0759. Use a 99% critical value about 2.63.

A) (0.5027, 0.9026)
B) (0.5517, 0.8535)
C) (0.5887, 0.8174)
D) Not possible to compute given the information in the summary output.
E) None of the above (A, B, C, D) is close to be correct.

99% CI ≈ b ± 2.63 SE = 0.7026 ± 2.63(0.0759) ≈ (0.503,0.902), matching A.

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

Normal DistributionD–A levelVery high priority

Which of the following interpretations of the coefficient of TRADITIONAL is most correct?

→ A) A traditional house in on average about 15.5 % less expensive than non-traditional houses.

Old Exam — Oct 2022, Q16

Given: TRADITIONAL coefficient=-0.1547 in a log-price model.

A) A traditional house in on average about 15.5 % less expensive than non-traditional houses.
B) A waterfront house in on average about 1.55 % less expensive than non-traditional houses.
C) A waterfront house in on average about 1.55 times less expensive than non-traditional houses.
D) A waterfront house in on average about 0.15% less expensive than non-traditional houses.
E) None of the above (A, B, C, D) is close to be correct.

TRADITIONAL is a dummy in ln(PRICE), b=-0.1547. Approximate effect=-15.47%. Exact effect=100(e^-0.1547-1)≈-14.3%. Among the choices, A is the intended approximation.

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

Normal DistributionD–A levelVery high priority

You have estimated a model of two variables related such that: ln(y) = 17.3 - 0.04x. If x decreases by 2 units, what is the expected change in y?

→ B) y increases by 8 percent.

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.

Normal DistributionD–A levelVery high priority

What is the income elasticity of your firm’s product?

→ A) 1.212

Old Exam — Dec 2021 re-exam, Q25

Given: ln(Q)=3.418+1.212 ln(I): the ln-ln slope is elasticity.

A) 1.212
B) 2.206
C) 3.418
D) 4.630
E) None of the above (A, B, C, D) is correct.

In a log-log model, the slope is elasticity. Income elasticity=1.212 -> A.

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

Normal DistributionD–A levelVery high priority

You want to test the null hypothesis that parameter (coefficient) of the ln(I) variable is equal to 1. Then the value of the test statistic (t- statistic) is

→ D) 0.54

Old Exam — Dec 2021 re-exam, Q26

Given: Test H0: beta=1. t=(1.212-1)/0.392≈0.54.

A) 3.09
B) 0.392
C) 3.418
D) 0.54
E) None of the above (A, B, C, D) is correct.

t=(1.212-1)/0.392=0.212/0.392≈0.54 -> D.

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

Normal DistributionD–A levelVery high priority

The t-statistic for the ln(SQFT) (see ln_sqft) is closest to_________.

→ C) 56.99

Old Exam — Oct 2021, Q19

Given: ln(SQFT) coefficient=.892479, SE=.015659.

A) 5.69
B) 0.00
C) 56.99
D) Not possible to compute given the information in the summary output.

t=b/SE=0.892479/0.015659≈56.99 -> C.

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

Normal DistributionD–A levelVery high priority

Which of the following interpretations of the coefficient of WATERFRONT is most correct?

→ B) A waterfront house in on average 13.5 % more expensive than not waterfront houses.

Old Exam — Oct 2021, Q21

Given: WATERFRONT is a dummy in a log-price model with coefficient≈0.135.

A) A waterfront house in on average 1.35 times more expensive than not waterfront houses.
B) A waterfront house in on average 13.5 % more expensive than not waterfront houses.
C) A waterfront house in on average 13.5 % more expensive than traditional houses.
D) A waterfront house in on average 135 % more expensive than not waterfront houses.
E) None of the above (A, B, C, D) is close to be correct.

Dummy in ln(price): b≈0.135. Approx effect≈13.5% higher; exact=100(e^0.135-1)≈14.5%. Intended answer B.

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

Bonds & Cap RatesE-levelVery high priority

NY CBD office cap rate is 5.75% and the comparable 10-year Treasury yield is 5.00%. What is the cap rate spread, and how should it be interpreted?

→ A) 75 bps; a relatively small premium given office property's additional risks

Computer Lab 4 — Exercise 1, Task 8

A) 75 bps; a relatively small premium given office property's additional risks
B) 675 bps; an unusually large premium
C) 75 bps; abnormally large for office property
D) 0.75 bps; essentially no premium

Spread = 5.75% − 5.00% = 0.75 percentage points = 75 basis points, which is a comparatively modest premium relative to the vacancy, leasing, operating and liquidity risks of office property, making Treasuries relatively attractive by comparison.

Rule: Compare the computed spread's economic magnitude to the asset's risk profile, not just its sign.

Normal DistributionE-levelVery high priority

Monthly return R ~ N(μ=1%, σ=4%). What is the probability of a negative monthly return, P(R<0)?

→ A) ≈40.13%

Computer Lab 6 — Exercise 2, (a)

A) ≈40.13%
B) ≈50.00%
C) ≈25.00%
D) ≈59.87%

z = (0−0.01)/0.04 = −0.25. P(R<0) = Φ(−0.25) ≈ 0.4013 = 40.13%.

Rule: Standardize with z=(x−μ)/σ, then use Φ(z) for a left-tail probability.

Normal DistributionE-levelVery high priority

With R ~ N(1%, 4%²), what is P(R > 5%)?

→ A) ≈15.87%

Computer Lab 6 — Exercise 2, (b)

A) ≈15.87%
B) ≈84.13%
C) ≈100%
D) ≈1.00%

z = (0.05−0.01)/0.04 = 1.00. P(R>0.05) = 1 − Φ(1.00) = 1 − 0.8413 = 0.1587 = 15.87%.

Rule: For 'greater than', use the complement: P(X>x) = 1 − Φ(z).

Normal DistributionE-levelVery high priority

With R ~ N(1%, 4%²), what is P(−2% < R < 4%)?

→ A) ≈54.68%

Computer Lab 6 — Exercise 2, (c)

A) ≈54.68%
B) ≈100.00%
C) ≈77.34%
D) ≈22.66%

z_L = (−0.02−0.01)/0.04 = −0.75; z_U = (0.04−0.01)/0.04 = 0.75. P = Φ(0.75) − Φ(−0.75) = 0.7734 − 0.2266 = 54.68%.

Rule: For an interval probability, compute Φ(upper z) − Φ(lower z).

Normal DistributionD–A levelVery high priority

With R ~ N(1%, 4%²), find the return value r such that P(R < r) = 5% (the 5th percentile), using z₀.₀₅ ≈ -1.645 and r = μ + zσ.

→ A) ≈-5.58%

Computer Lab 6 — Exercise 2, (d)

A) ≈-5.58%
B) ≈6.58%
C) ≈-1.645%
D) ≈-4.00%

r = 0.01 + (−1.645)(0.04) = 0.01 − 0.0658 = −0.0558 = −5.58%.

Rule: To go from a given probability to a value, first find the z-quantile, then compute x = μ + zσ.

ProbitD–A levelVery high priority

Using Model 1, P(ADJUST=1)=Φ[-7.0166+0.5301·FIXRATE-0.2675·MARGIN+0.0864·NETWORTH], compute the predicted probability for FIXRATE=12, MARGIN=2, NETWORTH=3. The linear index works out to z=-0.9312.

→ A) ≈17.59%

Computer Lab 6 — Exercise 3, Task c

A) ≈17.59%
B) ≈82.41%
C) ≈37.21%
D) ≈93.12%

P = Φ(-0.9312) ≈ 0.1759 = 17.59%, reading the standard normal CDF at z = -0.9312.

Rule: Compute z=Xβ first, then look up Φ(z) for the predicted probability.

LogitE-levelVery high priority

The Logit marginal effect formula is ME_j = β_j × P(1−P), while the Probit marginal effect formula is ME_j = β_j × φ(Xβ). A workbook table mistakenly uses a normal-distribution PDF for the Logit marginal effect. Why is this an error?

→ A) Logit's marginal effect uses the logistic density P(1−P), not the normal PDF φ(·), which belongs to Probit

Computer Lab 6 — Exercise 4, Logit vs Probit marginal effect

A) Logit's marginal effect uses the logistic density P(1−P), not the normal PDF φ(·), which belongs to Probit
B) Logit and Probit always use the exact same marginal effect formula
C) Logit does not have a marginal effect concept
D) P(1-P) and φ(Xβ) always give numerically identical results

Logit's link function is the logistic CDF Λ(z), whose derivative is the logistic density P(1-P); the normal PDF φ(·) is specific to Probit's standard normal CDF link, so using φ for Logit's marginal effect is a template/model-mismatch error.

Rule: Match the marginal-effect density to the model: Probit uses φ(Xβ); Logit uses P(1−P).

LogitE-levelVery high priority

Probit and Logit models for DENY give similar predicted probabilities (e.g., Black at PIRAT=0.35: 27.73% Probit vs 27.28% Logit), but their raw coefficients differ substantially in scale (e.g., PIRAT coefficient ≈2.74 in Probit vs ≈5.37 in Logit). What is the correct conclusion?

→ A) Compare predicted probabilities or marginal effects across models rather than comparing raw coefficients directly, since the two models' coefficient scales differ

Computer Lab 6 — Exercise 4, Probit vs Logit comparison

A) Compare predicted probabilities or marginal effects across models rather than comparing raw coefficients directly, since the two models' coefficient scales differ
B) Since the coefficients differ so much, one of the two models must be wrong
C) The Logit model is always exactly twice the Probit coefficient
D) Raw coefficients from Probit and Logit can be directly compared and interpreted the same way

Probit and Logit use different link functions (normal vs. logistic CDF) with different inherent scales, so their raw coefficients are not directly comparable; however, their implied predicted probabilities and marginal effects typically tell a similar story, as seen here.

Rule: Compare nonlinear-model results via predicted probabilities or marginal effects, not raw coefficient magnitudes, across different link functions.

Normal DistributionE-levelVery high priority

X ~ N(10, 16). To find P(X < 14) you standardize with:

→ B) z = (14 − 10)/4

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

A) z = (14 − 10)/16
B) z = (14 − 10)/4
C) z = (14 − 10)/√4
D) z = 14/16

Second parameter is variance. Correct method: z = (14 − 10)/4.

Rule: z = (x − μ)/σ, σ = √16 = 4