[SOLVED] CS F70TS Time Series Computer Lab 3

F70TS Time Series Computer Lab 3
In this computer lab we use R to make forecasts for an ARIMA process. We will first consider the case when the coefficient values are known, before moving to the case where the coefficients must be estimated from observed data, and examine the accuracy of the forecasts in each case.
Key R Commands The key R packages used in this lab are forecast and ggplot2. A brief introduction to some relevant commands is provided here, but for more information see the R documentation.
As in the first computer practical, we must simulate data using the arima.sim() function. As a simple example, the following generates a series of length 100 from an ARIMA(2,1,2) process:
n <- 100model <- list(order = c(2,1,2), ar = c(0.1, 0.2), ma = c(0.3, 0.4))generated_data <- arima.sim(model, n)The autoplot() function from the ggplot2 package is an easy way to visualise this data autoplot(generated_data)Before forecasting using an ARIMA model, it is first necessary to fit it to the data, using the function arima().arima_model <- arima(generated_data, order=c(2,1,2))(N.B. The data provided to arima() must be a Time-Series object. A vector or matrix can be converted to a Time-Series using the function ts().)By default, arima() estimates the values of the coefficients from the data. The coefficients can alternatively be fixed by supplying them as an argument to the original functionarima_model <- arima(generated_data, order=c(2,1,2), fixed = c(phi1, phi2, psi1, psi2))The following command simulates a future m steps from the time series, according to the ARIMA model fitted on the generated data.simulate(arima_model, nsim=m)A forecast from the time series model can then be produced using the function forecast from theforecast package. The following forecasts the next 10 timesteps of arima model.arima_forecast <- forecast(arima_model, h = 10)The output of this forecast is a list including the forecasted mean, and the lower and upper lim- its of 80% and 95% prediction intervals. These can be accessed using arima forecast$mean, arima forecast$lower and arima forecast$upper respectively. The forecast may be plotted using autoplot:autoplot(arima_forecast)1ExerciseIn this exercise we consider an ARIMA(2,1,2) process with coefficients 1 = 0.7, 2 = 0.2, 1 = 0.5 and 2 = 0.8, and t N(0,1).1. Simulate a time series of length n = 1000 from this process. Plot this data.2. Fit an ARIMA(2,1,2) model to the first k = 100 datapoints of the series. Fix the coefficients of the model to take their true values. Provide the model with the true value of Var(t) using the command your_model_name$sigma2 = 13. Use the function forecast() to obtain a 95% prediction interval for the value of Xk+50. Test whether the true value of Xk+50 is contained in this interval. Produce a plot illustrating the forecasted prediction intervals.4. Investigate the accuracy of these forecasts. Using the command simulate(your model name, nsim = 100) simulate another 100 steps of the time series. Repeating 104 times, calculate the proportion for which the simulated value of Xk+50 lies within the 95% predicted interval.(If this takes a long time, try 103 simulations.)5. Repeat (2)-(4) for a model where the coefficients and variance must be estimated from the data. For part (4) remember that the simulated data should still be generated according to the true model, rather than the model with estimated coefficients. How does the accuracy compare to the case when the coefficients and variance are known?6. How does the accuracy change if you alter the number of datapoints, k, used to estimate the parameters? How does the accuracy for k = 10 compare to k = 900?2