Statistics Graduate Level Quiz: Ar And Arma Models
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Ar And Arma ModelsQuestion 1 of 10

Consider the stationary ARMA model Xt=0.5Xt−1+εt−0.8εt−1,X_t=0.5X_{t-1}+\varepsilon_t-0.8\varepsilon_{t-1}, where the innovations have variance σ2.\sigma^2. Which pair gives the first two autocorrelations?

ρ1=−314\rho_1=-\dfrac{3}{14} and ρ2=−328\rho_2=-\dfrac{3}{28}
ρ1=−314\rho_1=-\dfrac{3}{14} and ρ2=328\rho_2=\dfrac{3}{28}
ρ1=−310\rho_1=-\dfrac{3}{10} and ρ2=−320\rho_2=-\dfrac{3}{20}
ρ1=12\rho_1=\dfrac{1}{2} and ρ2=14\rho_2=\dfrac{1}{4}
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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Ar And Arma Models

Practice Ar And Arma Models in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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Question 1

Consider the stationary ARMA model Xt=0.5Xt−1+εt−0.8εt−1,X_t=0.5X_{t-1}+\varepsilon_t-0.8\varepsilon_{t-1}, where the innovations have variance σ2.\sigma^2. Which pair gives the first two autocorrelations?

  1. ρ1=−314\rho_1=-\dfrac{3}{14} and ρ2=−328\rho_2=-\dfrac{3}{28} (correct answer)
  2. ρ1=−314\rho_1=-\dfrac{3}{14} and ρ2=328\rho_2=\dfrac{3}{28}
  3. ρ1=−310\rho_1=-\dfrac{3}{10} and ρ2=−320\rho_2=-\dfrac{3}{20}
  4. ρ1=12\rho_1=\dfrac{1}{2} and ρ2=14\rho_2=\dfrac{1}{4}
Explanation: When working with ARMA(1,1) models, your goal is to derive autocorrelations using the Yule-Walker framework combined with the autocovariance structure of the moving average component. For Xt=ϕXt−1+εt−θεt−1X_t = \phi X_{t-1} + \varepsilon_t - \theta\varepsilon_{t-1} with ϕ=0.5\phi = 0.5 and θ=0.8\theta = 0.8, first compute the variance γ0\gamma_0 and lag-1 autocovariance γ1\gamma_1 directly. Multiplying both sides by XtX_t and Xt−1X_{t-1} respectively and taking expectations: γ0=ϕγ1+σ2(1+θ2)−ϕθσ2⋅...\gamma_0 = \phi\gamma_1 + \sigma^2(1 + \theta^2) - \phi\theta\sigma^2\cdot... Using the known closed-form results: γ0=(1−2ϕθ+θ2)σ21−ϕ2=(1−2(0.5)(0.8)+0.64)σ20.75=0.84σ20.75=56σ250\gamma_0 = \frac{(1 - 2\phi\theta + \theta^2)\sigma^2}{1-\phi^2} = \frac{(1 - 2(0.5)(0.8) + 0.64)\sigma^2}{0.75} = \frac{0.84\sigma^2}{0.75} = \frac{56\sigma^2}{50} γ1=(ϕ−θ)(1−ϕθ)σ21−ϕ2=(−0.3)(0.6)σ20.75=−0.18σ20.75=−12σ250\gamma_1 = \frac{(\phi - \theta)(1 - \phi\theta)\sigma^2}{1 - \phi^2} = \frac{(-0.3)(0.6)\sigma^2}{0.75} = \frac{-0.18\sigma^2}{0.75} = \frac{-12\sigma^2}{50} So ρ1=γ1/γ0=−12/56=−3/14\rho_1 = \gamma_1/\gamma_0 = -12/56 = -3/14. For lags k≥2k \geq 2, the ACF satisfies the pure AR(1) recursion: ρk=ϕρk−1\rho_k = \phi\rho_{k-1}, giving ρ2=0.5×(−3/14)=−3/28\rho_2 = 0.5 \times (-3/14) = -3/28. This confirms answer A. Answer B gets ρ1\rho_1 right but incorrectly makes ρ2\rho_2 positive — a sign error in the recursion. Answer C applies the wrong formula, likely confusing ϕ−θ\phi - \theta terms. Answer D simply reports ϕ\phi and ϕ2\phi^2, ignoring the MA component entirely — a common trap when students forget ARMA autocorrelations differ from pure AR. Remember: for ARMA(1,1), only ρ1\rho_1 has a special formula; all higher lags follow ρk=ϕρk−1\rho_k = \phi\rho_{k-1}.

Question 2

A process is represented by Xt−0.7Xt−1=εt−0.7εt−1,X_t-0.7X_{t-1}=\varepsilon_t-0.7\varepsilon_{t-1}, where {εt}\{\varepsilon_t\} is white noise. Assuming the stationary causal solution is intended, which statement is correct?

  1. XtX_t is white noise because the common autoregressive and moving-average factors cancel (correct answer)
  2. XtX_t is an AR model with coefficient 0.70.7 because only the left side determines dependence
  3. XtX_t is an invertible MA model with coefficient −0.7-0.7 because the right side determines dependence
  4. XtX_t has autocorrelation 0.7k0.7^k because cancellation changes variance but not serial correlation
Explanation: When you encounter an ARMA model, your first instinct should be to factor the lag polynomials and check for common roots — this is the key concept being tested here. The process can be written using the backshift operator BB as (1−0.7B)Xt=(1−0.7B)εt(1 - 0.7B)X_t = (1 - 0.7B)\varepsilon_t. Notice that both sides share the identical factor (1−0.7B)(1 - 0.7B). Because the stationary causal solution is assumed, this common factor cancels legally, leaving simply Xt=εtX_t = \varepsilon_t. This means {Xt}\{X_t\} is white noise — serially uncorrelated, with constant variance and zero mean. So A is correct. B is wrong because focusing only on the left-hand side ignores the right-hand side entirely. The AR polynomial does not act in isolation; the full operator equation must be considered. C is wrong for the symmetric reason — focusing only on the MA side misses the cancellation. Even though −0.7-0.7 appears as the MA coefficient, the process after cancellation is not an MA(1). D is particularly tempting but incorrect: it imagines that cancellation affects variance while leaving autocorrelation structure intact. In reality, cancellation eliminates all serial correlation — after simplification, Xt=εtX_t = \varepsilon_t has zero autocorrelation at every lag k≥1k \geq 1, not 0.7k0.7^k. A useful rule of thumb: common factor = parameter redundancy = simplification. Whenever an ARMA model has matching AR and MA roots, the model is over-parameterized and reduces to a simpler process. Always factor both lag polynomials before interpreting any ARMA equation.

Question 3

A latent signal follows Xt=0.5Xt−1+εt,X_t=0.5X_{t-1}+\varepsilon_t, where the innovation variance is 3.3. Instead of observing the signal directly, an analyst observes Yt=Xt+ηt,Y_t=X_t+\eta_t, where ηt\eta_t is independent measurement noise with variance 4.4. All noise sequences are mutually independent over time.

Which statement correctly describes the second-order dependence of the observed process {Yt}\{Y_t\}?

  1. ρ1=0.25\rho_1=0.25 and ρ2=0.0625;\rho_2=0.0625; the observed process is AR(1) with effective coefficient 0.250.25
  2. ρ1=0.50\rho_1=0.50 and ρ2=0.25;\rho_2=0.25; measurement noise inflates the marginal variance but does not attenuate the autocorrelations
  3. ρ1=0.25\rho_1=0.25 and ρ2=0.125;\rho_2=0.125; these autocorrelations are consistent with an ARMA structure but not with a pure AR(1) model (correct answer)
  4. ρ1=0\rho_1=0 and ρ2=0;\rho_2=0; the independent measurement noise eliminates all serial dependence in the observed series
Explanation: When a latent AR(1) signal is observed through additive noise, the resulting process {Yt}\{Y_t\} is neither pure AR(1) nor white noise — it becomes an ARMA(1,1) process. Recognizing this structure is the key to the question. Start by computing the marginal variance of XtX_t. For a stationary AR(1) with coefficient ϕ=0.5\phi = 0.5 and innovation variance σε2=3\sigma_\varepsilon^2 = 3, you get Var(Xt)=31−0.25=4\text{Var}(X_t) = \frac{3}{1 - 0.25} = 4. The marginal variance of YtY_t is then Var(Yt)=4+4=8\text{Var}(Y_t) = 4 + 4 = 8. For lag-1, the autocovariance of YtY_t equals that of XtX_t (measurement noise terms are independent across time): γY(1)=γX(1)=ϕ⋅Var(Xt)=0.5×4=2\gamma_Y(1) = \gamma_X(1) = \phi \cdot \text{Var}(X_t) = 0.5 \times 4 = 2, giving ρ1=2/8=0.25\rho_1 = 2/8 = 0.25. At lag-2, γY(2)=γX(2)=ϕ2⋅Var(Xt)=0.25×4=1\gamma_Y(2) = \gamma_X(2) = \phi^2 \cdot \text{Var}(X_t) = 0.25 \times 4 = 1, giving ρ2=1/8=0.125\rho_2 = 1/8 = 0.125. Notice that ρ2/ρ1=0.5=ϕ\rho_2 / \rho_1 = 0.5 = \phi, which is consistent with the AR part of an ARMA(1,1) — but the ratio ρ1≠ϕ\rho_1 \neq \phi, ruling out a pure AR(1). This confirms C is correct. A is wrong because while ρ1=0.25\rho_1 = 0.25 is correct, ρ2=0.0625\rho_2 = 0.0625 would require ρ2=ρ12\rho_2 = \rho_1^2, the signature of a pure AR(1) — which this process is not. B is wrong because measurement noise does attenuate autocorrelations by inflating the denominator (marginal variance) without affecting the numerator. D is wrong because independent noise kills contemporaneous correlation but not lagged cross-signal dependence, which persists through XtX_t. Your study tip: whenever you add independent measurement noise to an AR(1), immediately check whether the resulting ACF pattern still satisfies the pure AR(1) Yule-Walker equations. If not, suspect ARMA.

Question 4

Suppose Xt=0.6Xt−1+εt,X_t=0.6X_{t-1}+\varepsilon_t, where {εt}\{\varepsilon_t\} is weak white noise with finite variance: distinct innovations are uncorrelated, but they are not assumed independent or Gaussian. For the causal stationary solution, which claim is justified by these assumptions alone?

  1. XtX_t is strictly stationary, Gaussian, and first-order Markov because its recursion has one lag
  2. XtX_t is weakly stationary with autocorrelation ρk=0.6∣k∣,\rho_k=0.6^{|k|}, but it need not be Gaussian or Markov (correct answer)
  3. XtX_t is weakly stationary with zero autocorrelation at every nonzero lag because the innovations are uncorrelated
  4. XtX_t has autocorrelation ρk=0.6∣k∣,\rho_k=0.6^{|k|}, which by itself guarantees a first-order Markov process
Explanation: When analyzing an AR(1) process under weak white noise assumptions, you need to carefully track exactly what the assumptions give you — and what they don't. For the causal stationary solution Xt=∑j=0∞0.6jεt−jX_t = \sum_{j=0}^{\infty} 0.6^j \varepsilon_{t-j}, weak stationarity follows directly: the mean is zero, variance is σ2/(1−0.36)\sigma^2/(1-0.36), and the autocovariance depends only on lag. The autocorrelation function is ρk=0.6∣k∣\rho_k = 0.6^{|k|}, derived purely from the uncorrelatedness of {εt}\{\varepsilon_t\} and the geometric decay of coefficients. This makes B correct — weak stationarity and ρk=0.6∣k∣\rho_k = 0.6^{|k|} are fully justified, but nothing forces Gaussianity or the Markov property. A fails on multiple grounds. Strict stationarity requires the joint distributions to be shift-invariant, which demands independence of innovations, not merely uncorrelatedness. Gaussianity requires either Gaussian inputs or a CLT argument with additional conditions. The Markov property requires that the conditional distribution of XtX_t given the entire past depends only on Xt−1X_{t-1} — true for Gaussian AR(1), but under weak white noise, higher-order conditional moments may still depend on further lags. C is wrong because the autocorrelation is not zero at nonzero lags. Zero autocorrelation would mean the process itself is white noise, which it isn't — the geometric decay 0.6∣k∣0.6^{|k|} reflects the persistence introduced by the autoregressive structure. D confuses a necessary condition with a sufficient one. Having ρk=0.6∣k∣\rho_k = 0.6^{|k|} characterizes the second-order structure, but the Markov property is a statement about conditional distributions — second-order properties alone cannot guarantee it without Gaussianity. Strategy tip: On time series questions, always ask: "Are my assumptions about distributions or just second moments?" Weak white noise only controls correlations, so properties requiring independence or distributional form (Markov, Gaussian, strict stationarity) need stronger assumptions.

Question 5

A stationary process has population autocorrelations ρ1=0.40,\rho_1=0.40, ρ2=0.28,\rho_2=0.28, and, for every k≥2,k\geq 2, satisfies ρk+1=0.70ρk.\rho_{k+1}=0.70\rho_k. Which conclusion best distinguishes an AR model from an ARMA model?

  1. The process must be MA of order one because the first autocorrelation differs from the geometric decay parameter
  2. The process must be AR of order one with coefficient 0.700.70 because its autocorrelations decay geometrically after lag one
  3. The process cannot be AR of order one because its lag-two partial autocorrelation is nonzero, but the pattern is compatible with ARMA of order one-one (correct answer)
  4. The process cannot be stationary because the lag-one autocorrelation differs from the ratio of later autocorrelations
Explanation: When identifying time series models, your key diagnostic tools are the autocorrelation function (ACF) and the partial autocorrelation function (PACF). A pure AR(1) process produces geometric decay in the ACF starting from lag one, and its PACF cuts off sharply after lag one — meaning the lag-two partial autocorrelation equals exactly zero. Here, the ACF does decay geometrically, but only after lag one: ρ2=0.28=0.70×0.40=0.70ρ1\rho_2 = 0.28 = 0.70 \times 0.40 = 0.70\rho_1, and ρk+1=0.70ρk\rho_{k+1} = 0.70\rho_k for k≥2k \geq 2. Notice that if this were a pure AR(1) with coefficient 0.700.70, you'd expect ρ1=0.70\rho_1 = 0.70 and ρ2=0.702=0.49\rho_2 = 0.70^2 = 0.49. The observed ρ1=0.40\rho_1 = 0.40 doesn't fit that mold. This "delayed" geometric start is the signature of an ARMA(1,1) process, where the moving-average component distorts the first autocorrelation before the AR component takes over. Crucially, an AR(1) would force the lag-two partial autocorrelation to zero, but because the lag-two ACF isn't consistent with a pure AR(1), the PACF will not cut off cleanly — confirming the process cannot be AR(1). Answer C captures this precisely. Answer A is wrong because the MA(1) ACF cuts off after lag one entirely — it cannot produce the persistent geometric decay seen here. Answer B incorrectly assumes the geometric decay parameter equals the AR(1) coefficient; it ignores that ρ1=0.40≠0.70\rho_1 = 0.40 \neq 0.70. Answer D is a red herring — nothing about these autocorrelations violates stationarity conditions. Study tip: When ACF decays geometrically but lag one is "off," immediately think ARMA(1,1). The MA component shifts the initial value without destroying the long-run AR pattern.

Question 6

A stationary process satisfies Xt=2+0.6Xt−1+εt,X_t=2+0.6X_{t-1}+\varepsilon_t, where the innovations are independent with mean zero and variance 4.4. If Xt=8,X_t=8, which pair gives the optimal linear forecast of Xt+2X_{t+2} and its forecast-error variance?

  1. Forecast 6.086.08; forecast-error variance 5.445.44 (correct answer)
  2. Forecast 6.806.80; forecast-error variance 5.445.44
  3. Forecast 6.086.08; forecast-error variance 4.004.00
  4. Forecast 6.806.80; forecast-error variance 6.406.40
Explanation: When forecasting from an AR(1) model, you apply the model recursively and track how forecast error variance accumulates with each additional step ahead. Start by finding the process mean. Setting E[Xt]=μE[X_t] = \mu gives μ=2+0.6μ\mu = 2 + 0.6\mu, so μ=5\mu = 5. The one-step-ahead forecast is straightforward: X^t+1=2+0.6(8)=6.8\hat{X}_{t+1} = 2 + 0.6(8) = 6.8, with error variance equal to the innovation variance, σ2=4\sigma^2 = 4. For the two-step forecast, substitute the one-step forecast into the model: X^t+2=2+0.6X^t+1=2+0.6(6.8)=2+4.08=6.08\hat{X}_{t+2} = 2 + 0.6\hat{X}_{t+1} = 2 + 0.6(6.8) = 2 + 4.08 = 6.08. The two-step forecast error variance compounds across both steps: Vart+2=σ2(1+ϕ2)=4(1+0.36)=4(1.36)=5.44\text{Var}_{t+2} = \sigma^2(1 + \phi^2) = 4(1 + 0.36) = 4(1.36) = 5.44. This confirms answer A is correct. Choice B reverses the forecasts — it reports the one-step forecast (6.80) but pairs it with the two-step error variance (5.44), mixing horizons incorrectly. Choice C gets the two-step point forecast right (6.08) but uses only σ2=4.00\sigma^2 = 4.00, forgetting that a second period of uncertainty from the first propagated error must be added. Choice D uses the one-step forecast again (6.80) and an incorrect variance of 6.40, which doesn't correspond to any standard formula. A useful strategy: always track the forecast horizon carefully. The hh-step forecast error variance for an AR(1) is σ2∑j=0h−1ϕ2j\sigma^2 \sum_{j=0}^{h-1} \phi^{2j}, not simply h⋅σ2h \cdot \sigma^2. Confusing these two is the most common trap on multi-step forecasting questions.

Question 7

An analyst fits the model Xt=1.2+0.8Xt−1+εt.X_t=1.2+0.8X_{t-1}+\varepsilon_t. Assuming stationarity, which pair gives the unconditional mean and the cumulative effect on the expected future path of a one-unit innovation occurring at time tt?

  1. Unconditional mean 1.21.2; cumulative expected effect 55
  2. Unconditional mean 66; cumulative expected effect 55 (correct answer)
  3. Unconditional mean 66; cumulative expected effect 44
  4. Unconditional mean 55; cumulative expected effect 66
Explanation: When you see an AR(1) model with a constant, two separate mechanics are being tested: the long-run equilibrium level and how a shock propagates through time. For an AR(1) model Xt=c+ϕXt−1+εtX_t = c + \phi X_{t-1} + \varepsilon_t, the unconditional mean is found by setting E[Xt]=E[Xt−1]=μE[X_t] = E[X_{t-1}] = \mu and solving: μ=c+ϕμ⇒μ=c1−ϕ.\mu = c + \phi\mu \Rightarrow \mu = \frac{c}{1-\phi}. Here, μ=1.21−0.8=1.20.2=6.\mu = \frac{1.2}{1-0.8} = \frac{1.2}{0.2} = 6. So the unconditional mean is 6. For the cumulative effect of a one-unit innovation, recognize that a shock at time tt decays geometrically. The impulse response at horizon hh is ϕh\phi^h, so the total cumulative effect across all future periods is the sum of a geometric series: ∑h=0∞ϕh=11−ϕ=11−0.8=5.\sum_{h=0}^{\infty} \phi^h = \frac{1}{1-\phi} = \frac{1}{1-0.8} = 5. This confirms answer B: unconditional mean 6, cumulative effect 5. Answer A gets the cumulative effect right (5) but incorrectly uses the raw constant 1.2 as the mean — confusing cc with μ\mu. Answer C correctly identifies the mean as 6 but reports a cumulative effect of 4, which would correspond to ϕ/(1−ϕ)\phi/(1-\phi), mistakenly starting the sum at h=1h=1 instead of h=0h=0 (omitting the contemporaneous impact). Answer D swaps the two values entirely, reversing mean and cumulative effect. A useful memory anchor: both the unconditional mean formula and the cumulative impulse response share the same denominator, 1−ϕ1-\phi. If you internalize 11−ϕ\frac{1}{1-\phi} as the "AR(1) multiplier," you can quickly compute both quantities without mixing them up.

Question 8

Let Xt=ϕXt−1+εtX_t=\phi X_{t-1}+\varepsilon_t with ∣ϕ∣<1,|\phi|<1, and define the differenced series Yt=Xt−Xt−1.Y_t=X_t-X_{t-1}. Which representation of YtY_t in terms of lag polynomials and the qualification of its moving-average component are correct?

  1. Yt=(1−ϕ)Xt−1+εt;Y_t=(1-\phi)X_{t-1}+\varepsilon_t; the differenced series reduces to white noise after centering
  2. (1−B)Yt=(1−ϕB)εt;(1-B)Y_t=(1-\phi B)\varepsilon_t; the MA factor (1−ϕB)(1-\phi B) is invertible since ∣ϕ∣<1|\phi|<1
  3. Yt=ϕYt−1+εt;Y_t=\phi Y_{t-1}+\varepsilon_t; a stationary AR model with the same AR coefficient and unchanged innovations
  4. (1−ϕB)Yt=(1−B)εt;(1-\phi B)Y_t=(1-B)\varepsilon_t; the MA factor (1−B)(1-B) has a unit root, so the representation is noninvertible (correct answer)
Explanation: When you apply the difference operator to an AR(1) process, you need to track what happens to both the autoregressive and moving-average structure simultaneously — this is a classic lag-polynomial manipulation problem. Start from Xt=ϕXt−1+εtX_t = \phi X_{t-1} + \varepsilon_t, which in lag-polynomial form is (1−ϕB)Xt=εt(1 - \phi B)X_t = \varepsilon_t. Now define Yt=(1−B)XtY_t = (1-B)X_t. You want to express YtY_t purely in terms of εt\varepsilon_t. Since (1−ϕB)Xt=εt(1-\phi B)X_t = \varepsilon_t, multiply both sides by (1−B)(1-B): (1−ϕB)(1−B)Xt=(1−B)εt(1-\phi B)(1-B)X_t = (1-B)\varepsilon_t, which gives (1−ϕB)Yt=(1−B)εt(1-\phi B)Y_t = (1-B)\varepsilon_t. The MA factor here is (1−B)(1-B), which has a root exactly on the unit circle (root at z=1z=1). A moving-average polynomial is invertible only when all its roots lie strictly outside the unit circle, so (1−B)(1-B) is noninvertible — making D correct. Choice A simply rearranges the AR(1) equation without applying the difference operator properly; it doesn't yield a valid ARMA representation for YtY_t. Choice B applies the difference operator to YtY_t a second time, producing (1−B)Yt(1-B)Y_t on the left — that's over-differencing — and incorrectly claims the MA factor (1−ϕB)(1-\phi B) is what appears. Choice C incorrectly asserts YtY_t satisfies the same AR(1) equation as XtX_t; differencing changes the model structure entirely and introduces a moving-average component. Your study tip: whenever you difference a stationary AR process, expect the result to have a noninvertible MA component with a unit root — this is the canonical signal that the series was already stationary and should not have been differenced.

Question 9

Suppose Xt=0.6Xt−1+εt+0.4εt−1,X_t=0.6X_{t-1}+\varepsilon_t+0.4\varepsilon_{t-1}, where the innovations are independent with variance 1.1. At time t,t, the observed values needed for forecasting are Xt=1X_t=1 and εt=0.5.\varepsilon_t=0.5. Which triple gives the one-step forecast, the two-step forecast, and the two-step forecast-error variance?

  1. 0.80, 0.48, 1.160.80,\ 0.48,\ 1.16
  2. 0.80, 0.48, 1.360.80,\ 0.48,\ 1.36
  3. 0.60, 0.36, 2.000.60,\ 0.36,\ 2.00
  4. 0.80, 0.48, 2.000.80,\ 0.48,\ 2.00 (correct answer)
Explanation: When you see an ARMA(1,1) model, the key is remembering that forecasts beyond one step ahead lose the moving-average correction term, since future innovations have conditional expectation zero. The model is Xt=0.6Xt−1+εt+0.4εt−1X_t = 0.6X_{t-1} + \varepsilon_t + 0.4\varepsilon_{t-1}. The one-step forecast from time tt is: X^t+1=0.6Xt+0.4εt=0.6(1)+0.4(0.5)=0.60+0.20=0.80.\hat{X}_{t+1} = 0.6X_t + 0.4\varepsilon_t = 0.6(1) + 0.4(0.5) = 0.60 + 0.20 = 0.80. For the two-step forecast, you take the conditional expectation of Xt+2=0.6Xt+1+εt+2+0.4εt+1X_{t+2} = 0.6X_{t+1} + \varepsilon_{t+2} + 0.4\varepsilon_{t+1}. Since E[εt+1]=E[εt+2]=0E[\varepsilon_{t+1}] = E[\varepsilon_{t+2}] = 0, only the AR part survives: X^t+2=0.6X^t+1=0.6(0.80)=0.48.\hat{X}_{t+2} = 0.6\hat{X}_{t+1} = 0.6(0.80) = 0.48. For the two-step forecast-error variance, write Xt+2−X^t+2=εt+2+(0.6+0.4)εt+1=εt+2+εt+1X_{t+2} - \hat{X}_{t+2} = \varepsilon_{t+2} + (0.6 + 0.4)\varepsilon_{t+1} = \varepsilon_{t+2} + \varepsilon_{t+1}. Each innovation has variance 1, so the total is 1+12=2.001 + 1^2 = 2.00. This gives the triple (0.80, 0.48, 2.00)(0.80,\ 0.48,\ 2.00), confirming D. Choice A uses 1.161.16, which incorrectly blends MA coefficients into a one-step error variance formula. Choice B gets 1.361.36, a common error from squaring only the MA(1) coefficient (1+0.42=1.16)(1 + 0.4^2 = 1.16) or using a misremembered recursion. Choice C uses 0.600.60 as the one-step forecast, forgetting to include the 0.4εt0.4\varepsilon_t correction available at time tt. Study tip: For ARMA forecast-error variance, build it step by step using the MA(∞\infty) representation — the hh-step variance is σ2∑j=0h−1ψj2\sigma^2\sum_{j=0}^{h-1}\psi_j^2, where ψj\psi_j are the impulse-response coefficients.

Question 10

The noninvertible moving-average process Xt=εt+2εt−1X_t=\varepsilon_t+2\varepsilon_{t-1} has innovation variance 1.1. Which invertible MA representation has the same autocovariance function?

  1. Xt=ut+0.5ut−1,X_t=u_t+0.5u_{t-1}, with innovation variance 44 (correct answer)
  2. Xt=ut+0.5ut−1,X_t=u_t+0.5u_{t-1}, with innovation variance 11
  3. Xt=ut−0.5ut−1,X_t=u_t-0.5u_{t-1}, with innovation variance 44
  4. Xt=ut+2ut−1,X_t=u_t+2u_{t-1}, with innovation variance 0.250.25
Explanation: When you encounter a noninvertible MA process, the key concept to recall is invertibility duality: any MA(1) process Xt=εt+θεt−1X_t = \varepsilon_t + \theta\varepsilon_{t-1} with ∣θ∣>1|\theta| > 1 shares its entire autocovariance function with an invertible process using coefficient 1/θ1/\theta — but the innovation variance must be rescaled to preserve those autocovariances. Here's the mechanics. For the given process Xt=εt+2εt−1X_t = \varepsilon_t + 2\varepsilon_{t-1} with σ2=1\sigma^2 = 1, compute the autocovariances: γ(0)=(1+4)⋅1=5\gamma(0) = (1 + 4)\cdot 1 = 5 and γ(1)=2⋅1=2\gamma(1) = 2 \cdot 1 = 2. The invertible partner replaces θ=2\theta = 2 with 1/θ=0.51/\theta = 0.5. For the new process Xt=ut+0.5ut−1X_t = u_t + 0.5u_{t-1} with variance σu2\sigma_u^2, you need γ(0)=(1+0.25)σu2=5\gamma(0) = (1 + 0.25)\sigma_u^2 = 5, giving σu2=4\sigma_u^2 = 4, and γ(1)=0.5σu2=2\gamma(1) = 0.5\sigma_u^2 = 2 ✓. This confirms answer A is correct. Answer B uses the right coefficient 0.50.5 but keeps the variance at 11, producing γ(0)=1.25≠5\gamma(0) = 1.25 \neq 5 — the autocovariance structure is completely wrong. Answer C introduces a sign flip to −0.5-0.5, which changes γ(1)\gamma(1) to a negative value, contradicting the original process. Answer D keeps θ=2\theta = 2 (still noninvertible) and tries to fix things with σu2=0.25\sigma_u^2 = 0.25, which produces γ(0)=1.25≠5\gamma(0) = 1.25 \neq 5 — wrong scaling entirely. Study tip: Always verify both autocovariances γ(0)\gamma(0) and γ(1)\gamma(1) match after the duality substitution. The rescaling rule is σu2=θ2σε2\sigma_u^2 = \theta^2 \sigma_\varepsilon^2, so swapping θ→1/θ\theta \to 1/\theta requires multiplying the variance by θ2\theta^2.