Hi Jia The easiest way to calculate this is using the formula for \( \rho_k \) of an \( ARMA(1,1) \) given on page 40 of the Tables, using \( \alpha=0.8 \) and \( \beta=0.2 \). However, the question asks us to derive this, which we can do as follows: For an \( ARMA(1,1) \) it is easiest to do two starter steps: \( Cov(Y_t, Z_t) = Cov(0.8Y_{t-1} + Z_t + 0.2Z_{t-1}, Z_t) = Cov(Z_t, Z_t) = \sigma^2 \) \( Cov(Y_t, Z_{t-1}) = Cov(0.8Y_{t-1} + Z_t + 0.2Z_{t-1}, Z_{t-1}) = 0.8Cov(Y_{t-1}, Z_{t-1}) + 0.2Cov(Z_{t-1}, Z_{t-1}) = 0.8\sigma^2 + 0.8\sigma^2 = \sigma^2 \) Then we can derive the autocovariance function, \( \gamma_k \): \( \gamma_0 = Cov(Y_t, Y_t) = Cov(0.8Y_{t-1} + Z_t + 0.2Z_{t-1}, Y_t) = 0.8\gamma_1 + \sigma^2 + 0.2\sigma^2 = 0.8\gamma_1 + 1.2\sigma^2 \) \( \gamma_1 = Cov(Y_t, Y_{t-1}) = Cov(0.8Y_{t-1} + Z_t + 0.2Z_{t-1}, Y_{t-1}) = 0.8\gamma_0 + 0.2\sigma^2 \) At this point, we can solve these equations simultaneously to give us \( \gamma_0 \) and \( \gamma_1 \) in terms of \( \sigma^2 \) only: \( \gamma_0 = 0.8(0.8\gamma_0 + 0.2\sigma^2) + 1.2\sigma^2 = 0.64\gamma_0 + 1.36\sigma^2 \ \Rightarrow \gamma_0 = \frac{34}{9}\sigma^2 \) \( \gamma_1 = 0.8\frac{34}{9}\sigma^2 + 0.2\sigma^2 = \frac{29}{9}\sigma^2 \) Continuing with higher values of \( k \), we get: \( \gamma_2 = Cov(Y_t, Y_{t-2}) = Cov(0.8Y_{t-1} + Z_t + 0.2Z_{t-1}, Y_{t-2}) = 0.8\gamma_1 \) \( \gamma_3 = Cov(Y_t, Y_{t-3}) = Cov(0.8Y_{t-1} + Z_t + 0.2Z_{t-1}, Y_{t-3}) = 0.8\gamma_2 \) etc. We can see generalises to \( \gamma_k = 0.8\gamma_{k-1} \) for \( k > 1 \). To get the autocorrelation function we apply the relationship \( \rho_k = \frac{\gamma_k}{\gamma_0} \). This gives: \( \rho_0 = 1 \) \( \rho_1 = \frac{29}{9} \times \frac{9}{34} = \frac{29}{34} \) \( \rho_2 = \frac{\gamma_2}{\gamma_0} = \frac{0.8\gamma_1}{\gamma_0} = 0.8\rho_1 \) \( \rho_3 = \frac{\gamma_3}{\gamma_0} = \frac{0.8\gamma_2}{\gamma_0} = \frac{0.8^2\gamma_1}{\gamma_0} = 0.8^2\rho_1 \) Again, we can see this generalises to \( \rho_k = 0.8^{k-1}\rho_1 \) for \( k > 1 \). This is a lengthy derivation for 6 marks, which was more common in written exams. However, I have shown more here than was required in the marking schedule. Hope this helps, Dave
