Solution 7

a) ARMA(2,2) because AR roots 4 and 3, MA roots 2 and 4. (1 pt)

b) Yes: stationary, causal, invertible (roots >1). (1 pt)

c) Simplified ARMA(1,1): \((1-\frac13 L)X_t = (1-\frac14 L)\varepsilon_t\). (1 pt)

d) MA(∞): \(\psi_i = (\frac13)^i - \frac14(\frac13)^{i-1},\; \psi_0=1\).
AR(∞): \(\phi_i = (\frac14)^i - \frac13(\frac14)^{i-1},\; \phi_0=1\). (2 pt)

e) Autocovariance satisfies \(\gamma(h)=\sum\phi_i\gamma(h-i)\) for \(h\ge q+1\). (2 pt)

Solution 8

a) MA(4): \(x_t = 10 + u_t + 0.9u_{t-4},\; \sigma_u^2=4\).

b) Forecast: \(\hat{x}_{t_0+1}=10.9,\; \hat{x}_{t_0+2}=5.5\); variance \(\sigma_u^2=4\).
95% CI: \(10.9\pm 4\) and \(5.5\pm 4\).

c) As \(h\to\infty\), conditional expectation → unconditional mean, conditional variance → unconditional variance.

d) Unconditional mean=10, unconditional variance=3.24 ⇒ 95% CI \(10\pm 3.6\).

Solution 9

a) Common root between AR and MA ⇒ simplified MA(1): \(x_t = 2.857 + u_t - 0.4u_{t-1}\).

b) \(\operatorname{cor}(x_t,x_{t-1})\approx -0.34\), zero for lags \(>1\).

c) Forecasts: \(\hat{x}_{T+1}\approx 3.0,\; \hat{x}_{T+2}=2.86\). Variances: 2.0 and 2.32 ⇒ CIs \([0.23,5.77]\) and \([-0.13,5.85]\).

d) Use AR(∞) when \(u_t\) unobserved: \(u_t = \sum_{i\ge0}0.4^i x_{t-i} - 2.857/0.6\).

e) \(\mathbb{P}[x_t<0] = \mathbb{P}[z<-1.88]\approx 2.5\%\).