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ISyE 6402 Midterm Prep With Complete Solutions 2022/2023 | Time Series Analysis | Georgia Tech

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This comprehensive study guide provides ISyE 6402 Midterm Prep with complete solutions for the 2022/2023 academic year at Georgia Tech. Covering essential time series analysis topics including ARIMA modeling, forecasting techniques, seasonality analysis, stationarity, and statistical methods for time-dependent data analysis in industrial and systems engineering.

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Uploaded on
November 28, 2025
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Written in
2025/2026
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ISyE 6402 Midterm Prep With Complete
Solutions 2022/2023


Introduction

This practice mid-term simulates the structure, depth and analytical expectations of
Georgia Tech ISyE 6402 (Time-Series Analysis).
Domains covered include stationarity testing, classical decomposition, ARIMA/SARIMA
identification, forecasting theory, spectral/frequency domain tools, multivariate
extensions and non-stationary modelling.
All questions are original and aligned with the 2022/2023 course specification to
support mastery-level performance.



General Instructions

• Answer all questions.

• Total: 50 points (equal weight).

• Closed-book; one handwritten formula sheet permitted.

• Time: 120 minutes.

• Where coding is referenced, pseudo-code is acceptable.



Questions

Question 1
Let {yₜ} be a zero-mean discrete-time series. Consider
Δyₜ = yₜ − yₜ₋₁.
Which condition guarantees that {Δyₜ} is weakly stationary?
A. E[yₜ] is finite for every t
B. Var(yₜ) is constant but autocovariance depends on t
C. {yₜ} is I(1)
D. {yₜ} is I(0)

Answer: C. {yₜ} is I(1)
Solution: If {yₜ} is integrated of order 1, its first difference Δyₜ is I(0) ⇔ weakly
stationary with constant mean, finite variance and time-invariant autocovariance.




pg. 1

, Question 2
Given the AR(1):
yₜ = 0.8 yₜ₋₁ + εₜ, εₜ ~ WN(0, σ²).
Compute the 2-step ahead forecast E[yₜ₊₂ | yₜ].
A. 0.8 yₜ
B. 0.64 yₜ
C. 0.5 yₜ
D. 0.36 yₜ

Answer: B. 0.64 yₜ
Solution: yₜ₊₁|ₜ = 0.8 yₜ; yₜ₊₂|ₜ = 0.8 yₜ₊₁|ₜ = 0.8² yₜ = 0.64 yₜ.



Question 3
For an MA(1): yₜ = εₜ + θ εₜ₋₁, |θ|<1, which statement about the PACF is correct?
A. Cuts off after lag 1
B. Cuts off after lag 2
C. Tails off exponentially
D. Zero for odd lags only

Answer: C. Tails off exponentially
Solution: MA(1) PACF tails off → θᵏ/(1+θ²+...+θ²ᵏ⁻²) decays exponentially.



Question 4
Match the ACF plot:
lag 1: 0.65, lag 2: 0.62, lag 3: 0.59 ... (slow almost linear decay)
to the most likely model:
A. AR(1) with φ=0.7
B. MA(1) with θ=0.7
C. Random walk without drift
D. White noise

Answer: C. Random walk without drift
Solution: Linearly declining ACF ≈1−k/n is signature of I(1) process; random walk is
simplest I(1) example.



Question 5
Simulated series n = 100. Dickey-Fuller test statistic = −2.85 (no constant). 5 % critical =
−2.89. Decision:
A. Reject unit root; series is stationary
B. Fail to reject; treat as non-stationary




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