## [1] 0.0725First-order auto correlation \(\rho_1 = 0.0725\)
## [1] 0.068875## [1] 0.95Second-order autocorrelation of \(y_t= 0.068875\), The ratio of the second-order to first-order autocorrelation equal to \(0.95\),which is \(\phi_1\) in this case.
We can prove this by:\[\begin{align} y_t = 0.95y_{t-1}-0.9\varepsilon_{t-1}+\varepsilon_t \tag{1.2} \end{align}\]
\[\begin{align*} E_t[y_{t+1}] &= 0.95E[y_t]+.9*E[\varepsilon_t]\\ &= 0.95y_t + 0.9\varepsilon_t \\ &= 0.95 \cdot 0.6 - 0.9 \cdot 0.1 \\ &= 0.48 \end{align*}\]
For \(y_{t+2}\), by iteration, we have, \[\begin{align*} y_{t+2} &= 0.95y_{t+1}-0.9\varepsilon_{t-1}+\varepsilon_t\\ &= 0.95(0.95y_{t}-0.9\varepsilon_{t}+\varepsilon_{t+1})-0.9\varepsilon_{t+1}+\varepsilon_{t+2}\\ &= 0.95^2 y_t -0.9*.95\varepsilon_{t} +0.95 \varepsilon_{t+1} -0.9\varepsilon_{t+1}+\varepsilon_{t+2}\\ &= 0.95^2 y_t -0.9*.95 \varepsilon_{t} +(0.95-0.9) \varepsilon_{t+1}\varepsilon_{t+2} \end{align*}\] Therefore, \[\begin{align*} E_t[y_{t+2}] &= 0.95 E_t[y_{t}] + (1-0.9*.95 )E_t[\varepsilon_{t}] +0.05 E_t[\varepsilon_{t+1}]-E_t[\varepsilon_{t+1}]\\ &= 0.456 \end{align*}\]
Taking expectation of equation (1.2) model: \[E[y_t]=0.95E[y_{t-1}]-0.9E[\varepsilon_{t-1}]+E[\varepsilon_t]\] \[E[y_t]-0.95E[y_{t-1}]=-0.9E[\varepsilon_{t-1}]+E[\varepsilon_t]\] By stationary and \(E[\varepsilon_i] = 0\) for all \(i\), \[E[y_t]=0\] Unconditional Mean:\[E[\hat{x_t}]=0.95E[y_t]=0\]
Calculating \(Var[y_t]\): Multiplying equation (1.2) both side by \(\varepsilon_t\) and take expectation,
\[\begin{align*} E[y_t\varepsilon_t] &= E[\varepsilon_t^2]-\theta_1E[\varepsilon_t\varepsilon_{t-1}]\\ &= E[\varepsilon_t^2]\\ &= \sigma_{\varepsilon}^2\\ &= 0.05^2 \end{align*}\]
Rewriting equation (1.2), \[y_t = \phi_1y_{t-1}+\varepsilon_t-\theta_1\varepsilon_{t-1}\] Taking Variance, we have \[Var(y_t) = \phi_1^2Var(y_{t-1})+\sigma_{\varepsilon}^2+\theta_1^2\sigma_{\varepsilon}^2 -2\phi_1\theta_1E[y_{t-1}\varepsilon_{t-1}]\] As \(\varepsilon_{t}\) and \(y_{t-1}\) are uncorrelated. We have,
\[\begin{align*} Var(y_t) &= \sigma^2(\frac{1+\theta^2 + 2\phi_1\theta_1}{1-\phi^2})\\ &= 0.05^2(1+0.9^2-2 \cdot .95 \cdot .9)/(1-.95^2)\\ &= 0.002564103 \end{align*}\]
\[\begin{align*} \sigma(\hat{x_t}) &= \sqrt{0.95^2Var(y_t)}\\ &= 0.04810512 \end{align*}\]
## [1] 0.04810512## [1] 0.002314103calculating ACF: As \(x_t = E_t(y_{t+1}) = 0.95y_t\) Multiply both side by \(x_{t-j}\) and take expectation,
\[\begin{align*} E[x_t x_{t-j}] &= E[(0.95y_t)(x_{t-j})]\\ &= E[(0.95y_t)(0.95y_{t-j})]\\ &= E[0.95 \gamma_j ] \end{align*}\]
As we only care about \(j=1\) in this case,
\[\begin{align*} E[x_t x_{t-1}] &= E[0.95 \gamma_1 ]\\ &= 0.95(\sigma_{\varepsilon}^2 \frac{(\phi_1+\theta_1)(1+\theta_1\phi_1)}{1-\phi_1^2})\\ &= 0.0001766026\\ E[(x_t)^2] &= Var(x_t)\\ &= 0.002314103\\ \rho_1 &= \frac{E[x_t x_{t-1}]}{E[(x_t)^2]}\\ &= .07631578 \end{align*}\]
## [1] 0.07631578\[\begin{align*} x_{t-1} &= e_{t-1}-e_{t-2}\\ x_{t-2} &= e_{t-2}-e_{t-3}\\ x_{t-3} &= e_{t-3}-e_{t-4} \end{align*}\]
So we can rewrite \(y_t\) as
\[\begin{align*} y_t &= e_t -e_{t-1}+e_{t-1}-e_{t-2}+e_{t-2}-e_{t-3}+e_{t-3}-e_{t-4}\\ &= x_t + x_{t-1} + x_{t-2} + x_{t-3} \end{align*}\]
As \(x_t\) is an AR(1) model, with \(\phi=0\) and covariance stationarity order 0 auto-covariances: \(\gamma_0 = 4E(x_{t-1}x_{t-1}) = E(\varepsilon_{t-1}^2)=4 \sigma_{\varepsilon}^2 = 4\)\ order 1 auto-covariances: \[\begin{align*} \gamma_1 &= E[(x_t + x_{t-1}+x_{t-2} + x_{t-3})(x_{t-1}+x_{t-2} + x_{t-3}+ x_{t-4})] \\ &= E[(\varepsilon_t + \varepsilon_{t-1}+\varepsilon_{t-2} + \varepsilon_{t-3})(\varepsilon_{t-1}+\varepsilon_{t-2} + \varepsilon_{t-3}+ \varepsilon_{t-4})]\\ &= Var(\varepsilon_{t-1})+Var(\varepsilon_{t-2})=Var(\varepsilon_{t-3})\\ &=3 \end{align*}\]
Therefore, when \(\phi = 0\),\(y_t\) is an ARMA (1,4) model. with AR coefficient \(=1\) and MA coefficient \(=-1\).
3)Optional when \(0<\phi<1\), for general model a. Autocovarainces: order-0: \(\gamma_0 = \frac{4\sigma_{\varepsilon}^2}{1-\phi}\)\ order-j (for \(0<j \leq 5\)): \(\gamma_j = (4\phi_1)^2\gamma_{j-1}\)\ We would expect a decreasing trend in the autocovariance graph as we are having a fractional coefficients.
b.With \(0<\phi<1\), we could modify our previous derivation as , \[\begin{align*} y_t &= e_t-e_{t-4}\\ &= e_{t-1} + x_t - e_{t-5} - x_{t-4}\\ &= e_{t-1} - e_{t-5} +(x_t - x_{t-4})\\ &= e_{t-1} - e_{t-5} +(\phi x_{t-1} + \varepsilon_t - (\phi x_{t-5} + \varepsilon_{t-4}))\\ &= e_{t-1} - e_{t-5} +\phi (e_{t-1}-e_{t-2}) -\phi (e_{t-5} - e_{t-6}) + \varepsilon_t - \varepsilon_{t-4}\\ &= (1+\phi) e_{t-1} -(1+\phi)e_{t-5} - \phi (e_{t-2} -e_{t-6} )+ \varepsilon_t - \varepsilon_{t-4}\\ &= (1+\phi) y_{t-1} - \phi y_{t-2} + \varepsilon_t - \varepsilon_{t-4} \end{align*}\]
Therefore, with \(0<\phi<1\), \(y-t\) is an ARMA(2,4) model. With AR coefficient \(1+\phi\) and \(-\phi\) and MA coefficient \(=-1\)