ASU -STP 311 Mock Exam 4|Linear Algebra & Research Design |2026 actual Practice Mock Examination questions with answers solutions

ASU · STP 311 Mock Exam 4|Linear Algebra & Research Design |2026 actual Practice Mock Examination questions
with answers solutions


ASU · STP 311
Quantitative Statistics · Practice Mock Examination



Mock Exam 04
Linear Algebra & Research Design
Matrix structure behind covariance and least squares, plus open-ended research problems.



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] Binomial thinning of a Poisson count 5 pts
Q02 [Estimation] Coarsely rounded measurements 5 pts
Q03 [Estimation] Gamma shape and rate 5 pts
Q04 [Diagnostics] Positive overall alpha, negative in every regime 5 pts
Q05 [Diagnostics] Random k-fold CV on time series 5 pts
Q06 [Diagnostics] Overlapping-window correlation 5 pts
Q07 [Applied] Optimal reserve price 5 pts
Q08 [Applied] Recovering the spread from price bounces (Roll model) 5 pts
Q09 [Applied] Mean-reversion half-life 5 pts
Q10 [Applied] Valuing an unbounded-expectation payoff 5 pts
Q11 [Numerical] Conditional expectation, bivariate normal (numerical) 5 pts
Q12 [Numerical] Law of total variance (numerical) 5 pts
Q13 [Distribution Theory] Linear combination of independent normals 5 pts
Q14 [Market Making] The winner’s curse across n market makers 5 pts
Q15 [Market Making] A market with no fair value 5 pts
Q16 [Group Markets] Reading a Liar’s-poker bid 5 pts
Q17 [Group Markets] How many of us hold at least one 4 pts
Q18 [Linear Algebra] The expectation of a quadratic form 4 pts
Q19 [Linear Algebra] AB and BA share their nonzero eigenvalues 4 pts
Q20 [Linear Algebra] How many pairwise obtuse vectors fit in Rd? 4 pts
Q21 [Research Design] From R2 to position size 4 pts

,A7. Binomial thinning of a Poisson count

Setup.

In each period the number of underlying events N is Poisson with mean λ. Each event is detected




Derive an estimator of λ.

Solution.
Find the distribution of the observed count. We derive the law of K through its probability generat-
ing function GK(t) = E[tK ]. Condition on N and use the tower property:

GK(t) = E E[tK | N] .

Given N, K is a sum of N independent Bernoulli(p) detections, so its conditional generating function
is E[tK | N] = (1 − p + pt)N (each detection contributes a factor 1 − p + pt). Therefore

GK(t) = E (1 − p + pt) N .

Now quote the Poisson generating function E[sN ] = eλ(s−1) and evaluate it at s = 1 − p + pt:

GK(t) = exp λ(1 − p + pt − 1) = exp λp(t − 1) .

This is exactly the generating function of a Poisson with mean λp, so K~ Poisson(λp) and in partic-
ular
E[K] = λp.
Estimator. The MLE of a Poisson mean is the sample¯ ¯ log-likelihood ∑i(Ki log(λp) − λp)
mean (the
^
has derivative ∑i Ki/(λp) − n = 0, giving mean = K). So λp = K, and since p is known,

K¯
λ̂ = .
p

This is unbiased (E [ K̄ ] = λp) and consistent. Its variance is Var ( λ̂ ) = Var(K¯ )/p2 = λp/(np2) =
λ/(np), which blows up as p → 0: rare detection makes λ hard to pin down. (Simulation λ = 8, p =
0.3: λ̂ = 7.99.)




12

, A15. Coarsely rounded measurements

Setup.

Underlying values are normal with unknown mean µ and standard deviation σ, recorded only
after rounding to the nearest integer k. The rounding is not negligible: σ is comparable to the
step of 1.

Derive the likelihood-based estimator of (µ, σ).

Solution.
Likelihood of a rounded observation. Rounding X to the nearest integer records k exactly when the
true value lies in the bin (k − 12, k + 12]. So the probability of recording k is the normal mass over that
interval, which is a difference of CDF values:

k + 12 − µ k − 12 − µ
P(record = k) = P k — 21 <X ≤ k+ 2 1 =Φ —Φ .
σ σ

Maximize the log-likelihood. With observations k1, . . . , kn,
" #
n ki + 12 − µ ki − 12 − µ
ℓ(µ, σ) = ∑ log Φ —Φ ,
i=1
σ σ

maximized numerically over (µ, σ). This “interval-censored” likelihood is exact.
Why naive moments are biased, and Sheppard’s correction. Treating each ki as the exact value adds
a rounding error r = X − k that is approximately uniform on (− 1 , 1 ) and roughly independent of
2 2
∫ 1/2
X. A uniform on (− 1 ,21 )2 has variance −1/2r2 dr = 112. Since the recorded value is X − r with r
independent, its variance is inflated by that amount, so the naive sample variance overstates σ 2 by
1
12 :
ˆ2 2 1
σ ≈ snaive − 12 (Sheppard’s correction).
The mean is essentially unaffected (the rounding error is mean-zero). The interval MLE above is
preferred when σ is small relative to the step.
import numpy as np
from scipy. stats import norm
from scipy. optimize import minimize
def negll(p) :
mu, ls = p[0) , p[1) ; s = np. exp(ls)
hi = norm. cdf((k + 0. 5 - mu) /s) ; lo = norm. cdf((k - 0. 5 - mu) /s)
return -np. sum(np. log(hi - lo + 1e-12))
res = minimize(negll, [np. mean(k), np. log(np. std(k)) ) , method=' Nelder-Mead' )
mu_hat, sigma_hat = res. x[0) , np. exp(res. x[1) )




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