Let's solve each part step by step: --- ### (i) **Identify the model as MA(q) or AR(p), specify q or p respectively.** Given: \[ X_t = Z_t + 2Z_{t-1} - 8Z_{t-2} \] This is a linear combination of current and past white noise terms (\( Z_t \)), which fits the definition of a **Moving Average (MA)** model. The general form for MA(q): \[ X_t = Z_t + \theta_1 Z_{t-1} + \theta_2 Z_{t-2} + \cdots + \theta_q Z_{t-q} \] Comparing, we have nonzero coefficients up to \( Z_{t-2} \). So, **q = 2**. **Answer:** The model is **MA(2)**. --- ### (ii) **Is the model stationary and invertible? Explain fully and show calculations where needed.** #### **Stationarity** MA(q) models are **always stationary** because they are a finite linear combination of white noise terms. #### **Invertibility** For invertibility, the roots of the MA polynomial must lie **outside the unit circle**. The MA polynomial is: \[ \theta(B) = 1 + 2B - 8B^2 \] Set \(\theta(B) = 0\): \[ 1 + 2B - 8B^2 = 0 \implies 8B^2 - 2B - 1 = 0 \] Solve for \(B\): \[ B = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 8 \cdot (-1)}}{2 \cdot 8} = \frac{2 \pm \sqrt{4 + 32}}{16} = \frac{2 \pm 6}{16} \] So, - \(B_1 = \frac{2+6}{16} = \frac{8}{16} = 0.5\) - \(B_2 = \frac{2-6}{16} = \frac{-4}{16} = -0.25\) These roots are **inside** the unit circle (since \(|0.5| < 1\) and \(|-0.25| < 1\)), so the model is **not invertible**. **Answer:** - The model is **stationary** (all MA models are). - The model is **not invertible** because the roots of the MA polynomial are inside the unit circle. --- ### (iii) **Find \(\rho_X(2)\) (the autocorrelation at lag 2).** Recall for MA(2): \[ X_t = Z_t + \theta_1 Z_{t-1} + \theta_2 Z_{t-2} \] Here, \(\theta_1 = 2\), \(\theta_2 = -8\). The general autocovariance formula for MA(2) (assuming \(Z_t\) is white noise with variance \(\sigma^2\)): - \(\gamma_X(0) = (1^2 + \theta_1^2 + \theta_2^2)\sigma^2\) - \(\gamma_X(1) = (1 \cdot \theta_1 + \theta_1 \cdot \theta_2)\sigma^2\) - \(\gamma_X(2) = (1 \cdot \theta_2)\sigma^2\) - For \(h > 2\), \(\gamma_X(h) = 0\) So, \[ \gamma_X(0) = (1^2 + 2^2 + (-8)^2)\sigma^2 = (1+4+64)\sigma^2 = 69\sigma^2 \] \[ \gamma_X(2) = (1 \cdot -8)\sigma^2 = -8\sigma^2 \] \[ \rho_X(2) = \frac{\gamma_X(2)}{\gamma_X(0)} = \frac{-8\sigma^2}{69\sigma^2} = -\frac{8}{69} \] **Final Answer:** \[ \rho_X(2) = -\frac{8}{69} \approx -0.116 \] --- ### (iv) **Write an R script...** Below is an R script that simulates \(X_t\) for 300 and 10,000 values, computes sample acf, and compares \(\rho_X(2)\): ```r # (a) Simulate MA(2) process set.seed(123) # for reproducibility # Function to simulate and analyze analyze_MA2 <- function(n) { Z <- rnorm(n + 2) # extra 2 for lag X <- Z[3:(n+2)] + 2 * Z[2:(n+1)] - 8 * Z[1:n] # Plot sample acf acf(X, main = paste("Sample ACF (n =", n, ")")) # Sample autocorrelation at lag 2 acf_values <- acf(X, plot = FALSE)$acf sample_rho2 <- acf_values[3] # lag 2 is third value (lags: 0,1,2) cat("Sample rho_X(2) for n =", n, ":", sample_rho2, "\n") return(sample_rho2) } # (b) For n = 300 sample_rho2_300 <- analyze_MA2(300) cat("Theoretical rho_X(2):", -8/69, "\n") # (c) For n = 10,000 sample_rho2_10000 <- analyze_MA2(10000) # Compare cat("Sample rho_X(2) with n = 300:", sample_rho2_300, "\n") cat("Sample rho_X(2) with n = 10000:", sample_rho2_10000, "\n") cat("Theoretical rho_X(2):", -8/69, "\n") ``` #### **Comments on findings:** - With \(n = 300\), the sample autocorrelation at lag 2 (\(\rho_X(2)\)) may deviate from the theoretical value due to sampling variability. - With \(n = 10,000\), the sample autocorrelation should be much closer to the theoretical value (\(-8/69 \approx -0.116\)), illustrating the law of large numbers. --- ## **Summary of answers** **(i)** MA(2) model **(ii)** Stationary but not invertible **(iii)** \(\rho_X(2) = -\frac{8}{69} \approx -0.116\) **(iv)** See R code above; larger sample sizes yield sample autocorrelations closer to the theoretical value.