Exam MAS-II · practice
Exam MAS-II — Time Series Practice Flashcards
Thirty exam-realistic multiple-choice problems on CAS Exam MAS-II time series — white noise and weak stationarity, AR(1)/AR(2) autocorrelations and variance via Yule–Walker, MA(1)/MA(2) autocorrelations and invertibility, ARMA/ARIMA differencing and random walks, ACF/PACF order identification, and one-step forecasting with error variance — each with a fully worked solution and distractors tied to common candidate errors.
8 free sample30 total · in appFree · fact-checked · LaTeX math
Tap card or press Space to flip
Answer
Unlock the full set
You're studying a free 8-problem sample. All 30 Time Series practice problems — plus every other Exam MAS-II subject and spaced-repetition scheduling — are built into the Willys AI Flashcards & Quizzes app. 14-day free trial, then $14.99.
Every deck is built into the Willys app
All of these decks — including the full practice problem banks — come built into Willys AI Flashcards & Quizzes for iPhone & iPad (Mac version coming soon), with FSRS + SM-2 spaced repetition, streaks, and exam-date cram mode. 14-day free trial, then $14.99. To load a deck in the app: Settings → Library → Browse, then pick your exam and deck.
More Exam MAS-II decks:
Bayesian Analysis Bayesian Analysis Practice Credibility Credibility Practice Generalized Linear Models Generalized Linear Models Practice
Browse all 30 problems as a list
- AR modelsAn AR(1) process $Y_t=\phi Y_{t-1}+\epsilon_t$ has $\phi=0.6$ and white-noise variance $\sigma^{2}=4$. Calculate the process variance $\gamma_0=\operatorname{Var}(Y_t)$. (A) $4.00$ (B) $5.00$ (C) $6.25$ (D) $10.00$ (E) $11.11$**Answer: (C).** For a stationary AR(1) the variance is $\gamma_0=\dfrac{\sigma^{2}}{1-\phi^{2}}$. $\gamma_0=\dfrac{4}{1-0.6^{2}}=\dfrac{4}{1-0.36}=\dfrac{4}{0.64}=6.25$. Using $1-\phi$ in the denominator instead of $1-\phi^{2}$ gives $\dfrac{4}{0.4}=10.0$ (distractor D); using $1-\phi^{2}$ but forgetting it gives $\sigma^{2}=4$ (distractor A); $\dfrac{4}{1-0.6^{3}}\approx 5.0$ is the wrong-power error.
- AR modelsAn AR(1) process $Y_t=\phi Y_{t-1}+\epsilon_t$ has $\phi=0.8$. Calculate the lag-3 autocorrelation $\rho_3$. (A) $0.240$ (B) $0.384$ (C) $0.512$ (D) $0.640$ (E) $0.800$**Answer: (C).** For an AR(1) the ACF decays geometrically: $\rho_k=\phi^{k}$. $\rho_3=0.8^{3}=0.512$. $\rho_1=0.8$ (E) and $\rho_2=0.64$ (D) are the earlier lags; $3\times 0.8\times 0.1=0.24$ (A) and $0.8^{2}\times 0.6$ (B) are arithmetic slips. The correct value is simply $\phi$ raised to the lag.
- AR modelsAn AR(1) process is $Y_t=-0.5\,Y_{t-1}+\epsilon_t$ with $\sigma^{2}=3$. Calculate $\rho_1$ and $\rho_2$. (A) $\rho_1=-0.50,\ \rho_2=-0.25$ (B) $\rho_1=-0.50,\ \rho_2=0.25$ (C) $\rho_1=0.50,\ \rho_2=0.25$ (D) $\rho_1=-0.25,\ \rho_2=0.125$ (E) $\rho_1=-0.50,\ \rho_2=0.50$**Answer: (B).** For an AR(1), $\rho_k=\phi^{k}$ with $\phi=-0.5$. $\rho_1=\phi=-0.5$ and $\rho_2=\phi^{2}=(-0.5)^{2}=0.25$. Because $\phi<0$ the ACF **alternates in sign**: negative at odd lags, positive at even lags. Keeping $\rho_2$ negative (A) forgets that squaring a negative gives a positive; the $\sigma^{2}$ value is not needed for the ACF.
- AR modelsAn AR(1) process is $Y_t=10+0.4\,Y_{t-1}+\epsilon_t$ with $\sigma^{2}=5$. Calculate the process mean $\mu$. (A) $4.00$ (B) $10.00$ (C) $14.00$ (D) $16.67$ (E) $25.00$**Answer: (D).** Taking expectations of the stationary AR(1) $\mu=c+\phi\mu$ gives $\mu=\dfrac{c}{1-\phi}$. $\mu=\dfrac{10}{1-0.4}=\dfrac{10}{0.6}\approx 16.67$. The intercept $c=10$ (B) is **not** the mean. Using $1-\phi^{2}=0.84$ in the denominator gives $\approx 11.9$ (not listed); treating $\mu=c/\phi$ gives $25$ (E). The variance $\sigma^{2}$ does not enter the mean.
- AR modelsAn AR(2) process is $Y_t=0.5\,Y_{t-1}+0.2\,Y_{t-2}+\epsilon_t$. Calculate the lag-1 autocorrelation $\rho_1$. (A) $0.500$ (B) $0.556$ (C) $0.625$ (D) $0.700$ (E) $0.800$**Answer: (C).** The first Yule–Walker equation for an AR(2) gives $\rho_1=\dfrac{\phi_1}{1-\phi_2}$. $\rho_1=\dfrac{0.5}{1-0.2}=\dfrac{0.5}{0.8}=0.625$. Using $\rho_1=\phi_1=0.5$ (A) ignores the $\phi_2$ feedback; $\dfrac{0.5}{1+0.2}\approx 0.417$ flips the sign of $\phi_2$; $\phi_1+\phi_2=0.7$ (D) is the wrong combination.
- AR modelsAn AR(2) process is $Y_t=0.5\,Y_{t-1}+0.2\,Y_{t-2}+\epsilon_t$, for which $\rho_1=0.625$. Calculate $\rho_2$. (A) $0.3125$ (B) $0.4025$ (C) $0.5000$ (D) $0.5125$ (E) $0.6250$**Answer: (D).** The lag-2 Yule–Walker equation is $\rho_2=\phi_1\rho_1+\phi_2$ (using $\rho_0=1$). $\rho_2=0.5(0.625)+0.2=0.3125+0.2=0.5125$. Forgetting the $\phi_2\rho_0=\phi_2$ term gives $0.3125$ (A); using $\phi_1\rho_1+\phi_2\rho_1=0.5(0.625)+0.2(0.625)=0.4375$ misapplies the recursion. The recursion $\rho_k=\phi_1\rho_{k-1}+\phi_2\rho_{k-2}$ uses $\rho_0=1$ at $k=2$.
- AR modelsSample autocorrelations $\hat\rho_1=0.7$ and $\hat\rho_2=0.5$ are observed for a series fit as AR(2). Solving the Yule–Walker equations, calculate $\hat\phi_1$. (A) $0.020$ (B) $0.350$ (C) $0.686$ (D) $0.700$ (E) $0.980$**Answer: (C).** The closed-form Yule–Walker solution is $\hat\phi_1=\dfrac{\rho_1(1-\rho_2)}{1-\rho_1^{2}}$. $\hat\phi_1=\dfrac{0.7(1-0.5)}{1-0.7^{2}}=\dfrac{0.7(0.5)}{1-0.49}=\dfrac{0.35}{0.51}\approx 0.686$. The value $0.020$ (A) is actually $\hat\phi_2=\dfrac{\rho_2-\rho_1^{2}}{1-\rho_1^{2}}=\dfrac{0.5-0.49}{0.51}$; using $\hat\phi_1=\rho_1=0.7$ (D) ignores the second equation; $0.35$ (B) is the unscaled numerator.
- AR modelsWhich set of parameters $(\phi_1,\phi_2)$ for an AR(2) model $Y_t=\phi_1 Y_{t-1}+\phi_2 Y_{t-2}+\epsilon_t$ FAILS the stationarity conditions? (A) $(0.5,\ 0.2)$ (B) $(1.0,\ -0.25)$ (C) $(0.3,\ 0.4)$ (D) $(0.6,\ 0.5)$ (E) $(-0.4,\ 0.3)$**Answer: (D).** AR(2) stationarity requires all three: $\phi_1+\phi_2<1$, $\phi_2-\phi_1<1$, and $|\phi_2|<1$. Check (D) $(0.6,0.5)$: $\phi_1+\phi_2=1.1\not<1$ — **fails** the first condition, so it is non-stationary. The others all pass: (A) $0.7,-0.3,0.2$; (B) $0.75,-1.25,0.25$; (C) $0.7,0.1,0.4$; (E) $-0.1,0.7,0.3$ — every inequality holds. The trap is checking only $|\phi_2|<1$ and missing the $\phi_1+\phi_2<1$ boundary.