Solution 1

1) The process can be written as \[ (1 - 0.4L + 0.2L^2) X_t = 40 + \epsilon_t. \] The process is stationary if the roots of the polynomial lie outside the unit disk: \[ 1 - 0.4Z + 0.2Z^2 = 0 \quad \text{iff} \quad Z = \frac{0.4 \pm \sqrt{0.16}}{0.2} = \frac{0.4 \pm 0.4}{0.2}. \] That is, \[ Z = \begin{cases} Z = 1 + 2i \\ Z = 1 - 2i \end{cases} \] whose modulus is \(\sqrt{1 + 2^2} = \sqrt{5} > 1\): the process is stationary.

2) The process is an AR(2) with constant: it is not centered. \[ E(X_t) = 0.4 E(X_{t-1}) - 0.2 E(X_{t-2}) + 40 \] so \(\mu = E(X_t) = \dfrac{40}{0.8} = 50\).

3) Let \(h \geq p + 1\). We compute for any \(t\): \[ \begin{aligned} E(X_{t+h}\mid X_{t+1}, \dots ,X_{t+h-1}) &= E(\varepsilon_{t+h}+a_{1}X_{t+h-1} + \dots + a_{p}X_{t+h-p}\mid X_{t+1}, \dots ,X_{t+h-1})\\ &= a_{1}X_{t+h-1} + \dots + a_{p}X_{t+h-p}. \end{aligned} \] Hence \[ X_{t+h} - E(X_{t+h}\mid X_{t+1}, \dots ,X_{t+h-1}) = \varepsilon_{t+h}. \] Similarly, \[ \begin{aligned} E(X_{t}\mid X_{t+1}, \dots ,X_{t+h-1}) &=\frac{1}{a_{p}}(X_{t+p} - \varepsilon_{t+p}-a_{1}X_{t+p-1}-\dots-a_{p-1}X_{t+1}) -\frac{1}{a_{p}}E(\varepsilon_{t+p}\mid X_{t+1}, \dots ,X_{t+h-1}). \end{aligned} \] Therefore \(\operatorname{Cov}(X_{t+h} - E(X_{t+h}\mid \dots), X_{t} - E(X_{t}\mid \dots)) = 0\) because \(\varepsilon_{t+h}\) is independent of the past.

4) Let \(Y_t = X_t - 50\). Then \(Y_t = 0.4 Y_{t-1} - 0.2 Y_{t-2} + \epsilon_t\) with \(E(Y_t)=0\). The Yule–Walker equations give: \[ \gamma_h = 0.4 \gamma_{h-1} - 0.2 \gamma_{h-2}. \] Initial conditions: \[ \gamma_1 = \frac{0.4 \gamma_0}{1.2} = \frac{\gamma_0}{3},\qquad \gamma_2 = 0.4\gamma_1 - 0.2\gamma_0 = -\frac{0.2\gamma_0}{3}. \] Using \(\gamma_0 = 0.4\gamma_1 - 0.2\gamma_2 + \sigma^2\) we obtain \(\gamma_0 = 15\). Then \[ \gamma_1 = 5,\; \gamma_2 = -1,\; \gamma_3 = -1.4,\; \gamma_4 = 0.36,\; \gamma_5 = 0.136. \] Partial autocorrelations: \(\alpha(1)=0.33,\; \alpha(2)=-0.199,\; \alpha(3)\approx -0.00069\) (very small, confirming AR(2)).

Solution 2

a) Yes, because the roots of the polynomial are \(5\) and \(-10/7\).

b) \(Y_{t }+ \frac{1}{2}Y_{t-1} -\frac{7}{50}Y_{t-2} = \varepsilon_{t}\) and \[ Y_{t} = \frac{1}{(1 + \frac{7}{10} B)(1 - \frac{1}{5} B)}\varepsilon_{t}. \]

c) \(\psi_{k} =\frac{ \lambda^{k+1}_{1}-\lambda^{k+1}_{2}}{\lambda_{1}-\lambda_{2}}\) with \(\lambda_{1}=1/5,\; \lambda_{2}=-7/10\).

d) \(\rho_{1} = \frac{\varphi_{1}}{1-\varphi_{2}}= -\frac{25}{43}\).

e) \(\rho_{k} = -\frac{1}{2}\rho_{k-1}+ \frac{7}{50}\rho_{k-2}\) and \[ \rho_{k} = \frac{1}{129}\left[17\left(\frac{1}{5}\right)^{k} + 112\left(-\frac{7}{10}\right)^{k}\right]. \]

Solution 3

a) Roots \(2\) and \(-3\) ⇒ stationary, causal, invertible. (1 pt)

b) \(X_n = \sum a_i \varepsilon_{n-i}\) ⇒ \(\operatorname{cov}(\varepsilon_n, X_{n-1})=0\). (1.5 pt)

c) Autocovariance: \(\gamma(0)=\phi_1\gamma(1)+\dots+\phi_p\gamma(p)+\sigma^2\); for \(h>0\), \(\gamma(h)=\phi_1\gamma(h-1)+\dots+\phi_p\gamma(h-p)\). (1.5 pt)

d) For \(X_n = \frac16 X_{n-1}+\frac16 X_{n-2}+\varepsilon_n\): \[ \gamma(1)=\frac{\gamma(0)}{5},\quad 34\gamma(0)=2\gamma(1)+36\sigma^2 \] ⇒ \(\gamma(1)=\frac{3\sigma^2}{14},\; \gamma(0)=\frac{15}{14}\sigma^2\). Factorization \(P(z)=\bigl(1-\frac{z}{2}\bigr)\bigl(1+\frac{z}{3}\bigr)\) gives \(\gamma(h)=\frac{a}{2^h}+\frac{b}{(-3)^h}\) with \(a=\frac{24}{35}\sigma^2,\; b=\frac{27}{70}\sigma^2\). (2 pt)

e) MA(∞): \(a_i = \frac35\bigl(-\frac13\bigr)^i + \frac25\bigl(\frac12\bigr)^i,\; a_0=1\). (1 pt)