ASU -STP 311 Quantitative Statistics ·2026 new update Practice Mock Examination Mock Exam 02 Diagnostics &
Model Failure|questions and answers solution
ASU · STP 311
Quantitative Statistics · Practice Mock Examination
Mock Exam 02
Diagnostics & Model Failure
Identifying what went wrong in a stated analysis, and what the corrected conclusion should be.
INSTRUCTIONS
• Time allowed: 3 hours. Questions: 21. Total: 100 points.
• Attempt every question. Show all algebraic steps; a correct final answer with no derivation earns partial
credit only.
• State any assumption you rely on, and note where an estimator is biased, inconsistent, or undefined.
• Worked solutions follow each question in this booklet. Cover them until you have committed to an answer.
• No calculator is required. Leave answers in closed form unless a number is explicitly requested.
QUESTION PAPER
Q01 [Estimation] Errors in variables (attenuation) 5 pts
Q02 [Estimation] Location under heavy tails 5 pts
Q03 [Estimation] Misclassified binary labels 5 pts
Q04 [Diagnostics] R2 always rises with more features 5 pts
Q05 [Diagnostics] Full-sample standardization then walk-forward 5 pts
Q06 [Diagnostics] Target leakage in a default model 5 pts
Q07 [Diagnostics] Controlling for a collider 5 pts
Q08 [Applied] Hidden-sign regression: sign redrawn each point 5 pts
Q09 [Applied] Optimal stopping for the best offer 5 pts
Q10 [Applied] Combining correlated signals 5 pts
Q11 [Applied] Optimal participation rate 5 pts
Q12 [Numerical] Attenuation bias (numerical) 5 pts
Q13 [Numerical] Sharpe annualization under autocorrelation (numerical) 5 pts
Q14 [Distribution Theory] Normal approximation to a binomial 5 pts
Q15 [Market Making] Quoting against someone who knows the answer 5 pts
Q16 [Market Making] Four markets that must agree 5 pts
Q17 [Group Markets] Wisdom of crowds with correlated errors 4 pts
Q18 [Group Markets] The jar of coins 4 pts
Q19 [Linear Algebra] Updating an inverse by a rank-one change 4 pts
Q20 [Linear Algebra] Minimum-variance weights 4 pts
Q21 [Research Design] Forecasting the closing auction imbalance 4 pts
,A3. Errors in variables (attenuation)
Setup.
2 2
ε
2
independent of X and ε. The error variance σ2 is known.
Show that regressing Y on X̃ is biased and derive the corrected estimator of β.
Solution.
Set up the naive regression. Fitting Y = b X̃ + error by OLS gives, in the limit, b⋆ = Cov( X̃ , Y)/ Var( X̃ ).
Evaluate the two moments.
Numerator. Substitute both definitions and expand by the bilinearity of covariance:
Cov( X̃ , Y) = Cov(X + u, βX + ε) = β Cov(X, X) + β Cov(u, X) + Cov(X, ε) + Cov(u, ε).
By independence, Cov(u, X) = Cov(X, ε) = Cov(u, ε) = 0, and Cov(X, X) = Var(X) = σx2, so
Cov( X̃ , Y) = βσx2.
Denominator. Since u is independent of X,
Var( X̃ ) = Var(X + u) = Var(X) + Var(u) = σ2 + σ2.
x u
The attenuation. Dividing,
βσ2 σ2
x x
b⋆ = 2 2 = λβ, λ= 2 2 < 1.
σx + σu σx + σu
The measurement noise adds to the denominator’s variance but contributes nothing to the numera-
tor’s covariance, so the slope is shrunk toward 0 by the reliability ratio λ — “attenuation.”
Correct it. The naive slope bˆ and the proxy variance Vˆa r ( X˜ ) are both estimable, and σu2 is known, so
recover the signal variance as σ̂ x = Vˆ
2 ˜ ) − σu and divide it out:
ar(X 2
˜ ˆ ˆ ˜
βˆ = C ô v ( X , Y) = b · V a r ( X ) .
Vâr( X˜ ) − σu2 ˜ ) − σu2
Vâr( X
Consistent because
2
each sample moment
2
converges
˜ to its population value and the denominator
ˆ
converges to σx > 0. Degenerate: if σu ≥ Var(X) (over-stated error) the denominator is nonpositive
and the estimate is meaningless. (Simulation β = 2, λ = 0.61: naive 1.22; corrected 2.00.)
8
, A11. Location under heavy tails
Setup.
Observations are i.i.d. from a symmetric distribution centered at θ, specifically a Student-t with
2 degrees of freedom shifted to center θ. This distribution has infinite variance.
Explain why the sample mean is poor and derive a consistent alternative.
Solution.
Why the mean fails. The t2 density about its center behaves like x| − θ|−3 far out. Test whether the
second moment exists by checking the integral’s tail:
∫ ∞ ∫ ∞
dx ∞
E[(X − θ)2] ∝ x2 · x−3 dx = = log x = ∞.
x
The variance is infinite. The central limit theorem requires a finite variance, so it does not apply: the
sample mean X̄ does not concentrate at the usual n−1/2 rate and remains volatile, dominated by rare
enormous draws. (The mean θ itself exists by symmetry, so X̄ is still consistent, but its sampling error
shrinks far too slowly to be useful.)
The median is consistent with finite variance. By symmetry of the density about θ, the population
median is θ: P(X ≤θ) = 1 . 2 For any distribution with positive density at its median, the sample
median m̂ is asymptotically normal (a quoted result),
1
˙ N θ,
m̂ ∼ ,
4n f (θ)2
which has finite variance even though the mean does not — the median depends on the bulk of the
data, not the tails, so heavy tails do not inflate it. Hence the sample median is a valid estimator of θ.
The efficient choice: the MLE. Maximizing the tν likelihood (density ∝ [1 + (x — θ)2/ν]−(ν+1)/2)
leads to the score equation
(ν + 1)(Xi − θ)
∑ ν + (X − θ)2 = 0 (ν = 2).
i i
Each summand is a bounded function of Xi (it → 0 as |Xi | → ∞), so a single outlier cannot dominate
— the MLE automatically down-weights extreme observations. Solve numerically. In practice use
the median or a trimmed mean when the tail index is uncertain, and the MLE when the tν form is
trusted.
17
Model Failure|questions and answers solution
ASU · STP 311
Quantitative Statistics · Practice Mock Examination
Mock Exam 02
Diagnostics & Model Failure
Identifying what went wrong in a stated analysis, and what the corrected conclusion should be.
INSTRUCTIONS
• Time allowed: 3 hours. Questions: 21. Total: 100 points.
• Attempt every question. Show all algebraic steps; a correct final answer with no derivation earns partial
credit only.
• State any assumption you rely on, and note where an estimator is biased, inconsistent, or undefined.
• Worked solutions follow each question in this booklet. Cover them until you have committed to an answer.
• No calculator is required. Leave answers in closed form unless a number is explicitly requested.
QUESTION PAPER
Q01 [Estimation] Errors in variables (attenuation) 5 pts
Q02 [Estimation] Location under heavy tails 5 pts
Q03 [Estimation] Misclassified binary labels 5 pts
Q04 [Diagnostics] R2 always rises with more features 5 pts
Q05 [Diagnostics] Full-sample standardization then walk-forward 5 pts
Q06 [Diagnostics] Target leakage in a default model 5 pts
Q07 [Diagnostics] Controlling for a collider 5 pts
Q08 [Applied] Hidden-sign regression: sign redrawn each point 5 pts
Q09 [Applied] Optimal stopping for the best offer 5 pts
Q10 [Applied] Combining correlated signals 5 pts
Q11 [Applied] Optimal participation rate 5 pts
Q12 [Numerical] Attenuation bias (numerical) 5 pts
Q13 [Numerical] Sharpe annualization under autocorrelation (numerical) 5 pts
Q14 [Distribution Theory] Normal approximation to a binomial 5 pts
Q15 [Market Making] Quoting against someone who knows the answer 5 pts
Q16 [Market Making] Four markets that must agree 5 pts
Q17 [Group Markets] Wisdom of crowds with correlated errors 4 pts
Q18 [Group Markets] The jar of coins 4 pts
Q19 [Linear Algebra] Updating an inverse by a rank-one change 4 pts
Q20 [Linear Algebra] Minimum-variance weights 4 pts
Q21 [Research Design] Forecasting the closing auction imbalance 4 pts
,A3. Errors in variables (attenuation)
Setup.
2 2
ε
2
independent of X and ε. The error variance σ2 is known.
Show that regressing Y on X̃ is biased and derive the corrected estimator of β.
Solution.
Set up the naive regression. Fitting Y = b X̃ + error by OLS gives, in the limit, b⋆ = Cov( X̃ , Y)/ Var( X̃ ).
Evaluate the two moments.
Numerator. Substitute both definitions and expand by the bilinearity of covariance:
Cov( X̃ , Y) = Cov(X + u, βX + ε) = β Cov(X, X) + β Cov(u, X) + Cov(X, ε) + Cov(u, ε).
By independence, Cov(u, X) = Cov(X, ε) = Cov(u, ε) = 0, and Cov(X, X) = Var(X) = σx2, so
Cov( X̃ , Y) = βσx2.
Denominator. Since u is independent of X,
Var( X̃ ) = Var(X + u) = Var(X) + Var(u) = σ2 + σ2.
x u
The attenuation. Dividing,
βσ2 σ2
x x
b⋆ = 2 2 = λβ, λ= 2 2 < 1.
σx + σu σx + σu
The measurement noise adds to the denominator’s variance but contributes nothing to the numera-
tor’s covariance, so the slope is shrunk toward 0 by the reliability ratio λ — “attenuation.”
Correct it. The naive slope bˆ and the proxy variance Vˆa r ( X˜ ) are both estimable, and σu2 is known, so
recover the signal variance as σ̂ x = Vˆ
2 ˜ ) − σu and divide it out:
ar(X 2
˜ ˆ ˆ ˜
βˆ = C ô v ( X , Y) = b · V a r ( X ) .
Vâr( X˜ ) − σu2 ˜ ) − σu2
Vâr( X
Consistent because
2
each sample moment
2
converges
˜ to its population value and the denominator
ˆ
converges to σx > 0. Degenerate: if σu ≥ Var(X) (over-stated error) the denominator is nonpositive
and the estimate is meaningless. (Simulation β = 2, λ = 0.61: naive 1.22; corrected 2.00.)
8
, A11. Location under heavy tails
Setup.
Observations are i.i.d. from a symmetric distribution centered at θ, specifically a Student-t with
2 degrees of freedom shifted to center θ. This distribution has infinite variance.
Explain why the sample mean is poor and derive a consistent alternative.
Solution.
Why the mean fails. The t2 density about its center behaves like x| − θ|−3 far out. Test whether the
second moment exists by checking the integral’s tail:
∫ ∞ ∫ ∞
dx ∞
E[(X − θ)2] ∝ x2 · x−3 dx = = log x = ∞.
x
The variance is infinite. The central limit theorem requires a finite variance, so it does not apply: the
sample mean X̄ does not concentrate at the usual n−1/2 rate and remains volatile, dominated by rare
enormous draws. (The mean θ itself exists by symmetry, so X̄ is still consistent, but its sampling error
shrinks far too slowly to be useful.)
The median is consistent with finite variance. By symmetry of the density about θ, the population
median is θ: P(X ≤θ) = 1 . 2 For any distribution with positive density at its median, the sample
median m̂ is asymptotically normal (a quoted result),
1
˙ N θ,
m̂ ∼ ,
4n f (θ)2
which has finite variance even though the mean does not — the median depends on the bulk of the
data, not the tails, so heavy tails do not inflate it. Hence the sample median is a valid estimator of θ.
The efficient choice: the MLE. Maximizing the tν likelihood (density ∝ [1 + (x — θ)2/ν]−(ν+1)/2)
leads to the score equation
(ν + 1)(Xi − θ)
∑ ν + (X − θ)2 = 0 (ν = 2).
i i
Each summand is a bounded function of Xi (it → 0 as |Xi | → ∞), so a single outlier cannot dominate
— the MLE automatically down-weights extreme observations. Solve numerically. In practice use
the median or a trimmed mean when the tail index is uncertain, and the MLE when the tν form is
trusted.
17
