ASU · STP 311 Quantitative Statistics · Practice Mock Examination Mock Exam 01 Estimation & Inference|Original
questions with solutions 2026
ASU · STP 311
Quantitative Statistics · Practice Mock Examination
Mock Exam 01
Estimation & Inference
Deriving estimators from first principles, then stating bias, consistency, and failure modes.
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] Sign-scrambled linear model 5 pts
Q02 [Estimation] Right-censored lifetimes 5 pts
Q03 [Estimation] Overdispersed counts 5 pts
Q04 [Estimation] Observation probability depends on the value 5 pts
Q05 [Diagnostics] Regression to the mean 5 pts
Q06 [Diagnostics] k = 1 nearest neighbors “fits perfectly” 5 pts
Q07 [Diagnostics] Arithmetic mean overstates compounding 5 pts
Q08 [Applied] Asymmetric information and a price floor 5 pts
Q09 [Applied] Direction of the next tick 5 pts
Q10 [Applied] Expected value of a barrier-stopped bet 5 pts
Q11 [Applied] Aggregating independent alphas 5 pts
Q12 [Numerical] Omitted-variable bias (numerical) 5 pts
Q13 [Numerical] Shrinkage / regression to the mean (numerical) 5 pts
Q14 [Distribution Theory] The coin majority-matching game 5 pts
Q15 [Market Making] Price impact in a two-sided market 5 pts
Q16 [Market Making] A market on the high-water mark 5 pts
Q17 [Group Markets] Dollar bill serial numbers: three rounds 4 pts
Q18 [Group Markets] The one liar 4 pts
Q19 [Linear Algebra] The equicorrelation matrix 4 pts
Q20 [Linear Algebra] Stochastic matrices and how fast a chain forgets 4 pts
Q21 [Research Design] Nowcasting the weekly natural gas storage report 4 pts
,A1. Sign-scrambled linear model
Setup.
2 2
probability 1 obability 21 , drawn independently of X and . With a constant
Derive an estimator of β.
Solution.
Try the linear estimator first. Consider fitting Y = bX + error by ordinary least squares. The OLS
setup minimizes ∑i(Yi — bXi)2; its probability limit is b⋆ = Cov(X, Y)/ Var(X). To evaluate this, first
record the moments of the sign from its definition:
E[Z] = (+1) 12 + (−1) 12 = 0, E[Z2] = (+1)2 12 + (−1)2 12 = 1.
Now expand the covariance. By Cov(X, Y) = E[XY] — E[X] E[Y] and E[X] = 0, only E[XY] matters.
Substitute Y = XZβ + ε and use linearity:
E[XY] = E[X · (XZβ + ε)] = β E[X2Z] + E[Xε].
Because X, Z, ε are mutually independent, each expectation of a product factorizes:
E[X2Z] = E[X2] E[Z] = σ2 x· 0 = 0, E[Xε] = E[X] E[ε] = 0.
Hence E[XY] = 0, so Cov(X, Y) = 0 and b⋆ = 0 for every β: the random sign averages the linear
signal to nothing. A linear fit is useless.
Find a statistic the sign cannot cancel. The obstruction was the odd power of Z; since Z2 = 1
identically, an even power is immune. Square the model and take the expectation conditional on X:
E[Y2 | X] = E (XZβ + ε)2 | X = E X2Z2β2 + 2XZβε + ε2 X .
Treat X as fixed and apply linearity, then independence of Z, ε term by term:
E[X2Z2β2 | X] = β2X2 E[Z2] = β2X2, E[2XZβε | X] = 2βX E[Z] E[ε] = 0, E[ε2 | X] = σε2.
Summing the three pieces,
E[Y2 | X] = β2X2 + σε2.
Estimator. This says the conditional mean of Y2 is a straight line in the variable X2, with slope β2 and
intercept σε2. So regress Y2 on X2 (with an intercept). By the OLS setup applied to (X2, Y2), the slope
i i i i
is q
∑ (X2 − X22)(Y2i − Y2) Côv(X ,2 Y 2) βˆ = max(β^2, 0).
β^2 = i i 2 2 = 2 ,
∑i(Xi − X ) Vâr(X )
ˆ ˆ
Properties. By the law of large numbers each sample moment converges to its population value, so
β^2 → β2 and β√→ β: co√nsistent. In finite samples β is biased low, for two reasons: the square root is
concave so E[ W] ≤ E[W] by Jensen, and the truncation at 0 raises the smallest estimates. Only
|β| is identified — the sign is absorbed into Z. If σx2 = 0 then X2 is constant, Vâr(X 2) = 0, and the
estimator is undefined. (Simulation β = 3, σε = 1.5: |βˆ| = 3.00, σ̂ ε = 1.50.)
6
, A9. Right-censored lifetimes
Setup.
or 0 if the unit was still alive at c (censored).
Derive the MLE of λ.
Solution.
Build the likelihood contribution of each unit. ∫ The exponential has density f (t) = λe−λt and
∞
survival function S(t) = P(T > t) = e− λt (from t λe− λu du = e− λt). An observed lifetime (di = 1)
contributes its density f (yi) = λe−λyi : we know the exact value. A censored unit (di = 0) contributes
only the information “it survived past c,” i.e. the survival probability S(c) = e−λc = e−λyi (with
yi = c). Writing both cases with the indicator,
n
di 1−di
L(λ) = ∏
i=1
λe−λyi e−λyi = ∏
i
λd e−λy .
i i
Maximize the log-likelihood. The exponential factor is common to both cases, so
ℓ(λ) = log L = ∑ di log λ − λyi = ∑ di log λ − λ ∑ yi.
i i i
Differentiate and set to zero:
∑ d ∑ d
ℓ′(λ) = i i − ∑ yi = 0 =⇒ λ̂ = i i = #events observed .
λ i ∑i yi total time on test
The second derivative ℓ′′(λ) = —∑i di/λ2 < 0 confirms a maximum. Each censored unit correctly
contributes its survived time yi = c to the denominator (exposure) without adding to the numerator
(event count) — the “occurrence over exposure” estimator at the foundation of survival analysis.
14
questions with solutions 2026
ASU · STP 311
Quantitative Statistics · Practice Mock Examination
Mock Exam 01
Estimation & Inference
Deriving estimators from first principles, then stating bias, consistency, and failure modes.
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] Sign-scrambled linear model 5 pts
Q02 [Estimation] Right-censored lifetimes 5 pts
Q03 [Estimation] Overdispersed counts 5 pts
Q04 [Estimation] Observation probability depends on the value 5 pts
Q05 [Diagnostics] Regression to the mean 5 pts
Q06 [Diagnostics] k = 1 nearest neighbors “fits perfectly” 5 pts
Q07 [Diagnostics] Arithmetic mean overstates compounding 5 pts
Q08 [Applied] Asymmetric information and a price floor 5 pts
Q09 [Applied] Direction of the next tick 5 pts
Q10 [Applied] Expected value of a barrier-stopped bet 5 pts
Q11 [Applied] Aggregating independent alphas 5 pts
Q12 [Numerical] Omitted-variable bias (numerical) 5 pts
Q13 [Numerical] Shrinkage / regression to the mean (numerical) 5 pts
Q14 [Distribution Theory] The coin majority-matching game 5 pts
Q15 [Market Making] Price impact in a two-sided market 5 pts
Q16 [Market Making] A market on the high-water mark 5 pts
Q17 [Group Markets] Dollar bill serial numbers: three rounds 4 pts
Q18 [Group Markets] The one liar 4 pts
Q19 [Linear Algebra] The equicorrelation matrix 4 pts
Q20 [Linear Algebra] Stochastic matrices and how fast a chain forgets 4 pts
Q21 [Research Design] Nowcasting the weekly natural gas storage report 4 pts
,A1. Sign-scrambled linear model
Setup.
2 2
probability 1 obability 21 , drawn independently of X and . With a constant
Derive an estimator of β.
Solution.
Try the linear estimator first. Consider fitting Y = bX + error by ordinary least squares. The OLS
setup minimizes ∑i(Yi — bXi)2; its probability limit is b⋆ = Cov(X, Y)/ Var(X). To evaluate this, first
record the moments of the sign from its definition:
E[Z] = (+1) 12 + (−1) 12 = 0, E[Z2] = (+1)2 12 + (−1)2 12 = 1.
Now expand the covariance. By Cov(X, Y) = E[XY] — E[X] E[Y] and E[X] = 0, only E[XY] matters.
Substitute Y = XZβ + ε and use linearity:
E[XY] = E[X · (XZβ + ε)] = β E[X2Z] + E[Xε].
Because X, Z, ε are mutually independent, each expectation of a product factorizes:
E[X2Z] = E[X2] E[Z] = σ2 x· 0 = 0, E[Xε] = E[X] E[ε] = 0.
Hence E[XY] = 0, so Cov(X, Y) = 0 and b⋆ = 0 for every β: the random sign averages the linear
signal to nothing. A linear fit is useless.
Find a statistic the sign cannot cancel. The obstruction was the odd power of Z; since Z2 = 1
identically, an even power is immune. Square the model and take the expectation conditional on X:
E[Y2 | X] = E (XZβ + ε)2 | X = E X2Z2β2 + 2XZβε + ε2 X .
Treat X as fixed and apply linearity, then independence of Z, ε term by term:
E[X2Z2β2 | X] = β2X2 E[Z2] = β2X2, E[2XZβε | X] = 2βX E[Z] E[ε] = 0, E[ε2 | X] = σε2.
Summing the three pieces,
E[Y2 | X] = β2X2 + σε2.
Estimator. This says the conditional mean of Y2 is a straight line in the variable X2, with slope β2 and
intercept σε2. So regress Y2 on X2 (with an intercept). By the OLS setup applied to (X2, Y2), the slope
i i i i
is q
∑ (X2 − X22)(Y2i − Y2) Côv(X ,2 Y 2) βˆ = max(β^2, 0).
β^2 = i i 2 2 = 2 ,
∑i(Xi − X ) Vâr(X )
ˆ ˆ
Properties. By the law of large numbers each sample moment converges to its population value, so
β^2 → β2 and β√→ β: co√nsistent. In finite samples β is biased low, for two reasons: the square root is
concave so E[ W] ≤ E[W] by Jensen, and the truncation at 0 raises the smallest estimates. Only
|β| is identified — the sign is absorbed into Z. If σx2 = 0 then X2 is constant, Vâr(X 2) = 0, and the
estimator is undefined. (Simulation β = 3, σε = 1.5: |βˆ| = 3.00, σ̂ ε = 1.50.)
6
, A9. Right-censored lifetimes
Setup.
or 0 if the unit was still alive at c (censored).
Derive the MLE of λ.
Solution.
Build the likelihood contribution of each unit. ∫ The exponential has density f (t) = λe−λt and
∞
survival function S(t) = P(T > t) = e− λt (from t λe− λu du = e− λt). An observed lifetime (di = 1)
contributes its density f (yi) = λe−λyi : we know the exact value. A censored unit (di = 0) contributes
only the information “it survived past c,” i.e. the survival probability S(c) = e−λc = e−λyi (with
yi = c). Writing both cases with the indicator,
n
di 1−di
L(λ) = ∏
i=1
λe−λyi e−λyi = ∏
i
λd e−λy .
i i
Maximize the log-likelihood. The exponential factor is common to both cases, so
ℓ(λ) = log L = ∑ di log λ − λyi = ∑ di log λ − λ ∑ yi.
i i i
Differentiate and set to zero:
∑ d ∑ d
ℓ′(λ) = i i − ∑ yi = 0 =⇒ λ̂ = i i = #events observed .
λ i ∑i yi total time on test
The second derivative ℓ′′(λ) = —∑i di/λ2 < 0 confirms a maximum. Each censored unit correctly
contributes its survived time yi = c to the denominator (exposure) without adding to the numerator
(event count) — the “occurrence over exposure” estimator at the foundation of survival analysis.
14
