Consider a system process given by $$x_t = -0.9x_{t-2} + w_t, \hspace{2mm} t=1,\dots n$$ where $x_0 \sim N(0,\sigma^2_0), x_{-1} \sim N(0,\sigma_1^2)$, and $w_t$ is Gaussian white noise with variance $\sigma^2_w$.

My question is how can I simulate this series and include $x_0$ and $x_{-1}$? I know how to use the arima.sim function to simulate general AR processes, but I don't know how to include the initial conditions $x_0$ and $x_{-1}$.

  • 1
    $\begingroup$ if arima.sim doesn't let you do that, then you can just write your own R code. I don't have time right now but, if you're not familar with how to do that in R, I'm sure someone with more time than me at the moment, will write the code for you in an answer. $\endgroup$
    – mlofton
    Sep 22, 2022 at 18:03

1 Answer 1


Programming the simulation from scratch requires just a few lines of $\textsf{R}$ code:

n <- 500
sigma_sq_minus1 <- 3
sigma_sq_0 <- 2
sigma_sq_w <- 1
x_minus1 <- rnorm(1, sd = sqrt(sigma_sq_minus1))
x_0 <- rnorm(1, sd = sqrt(sigma_sq_0))
w <- rnorm(n, sd = sqrt(sigma_sq_w))
x <- c(x_minus1, x_0, rep(NA, n))

for (t in seq_len(n)+2) {
  x[t] <- -0.9 * x[t-2] + w[t-2]

plot(x, type = "l")
  • 1
    $\begingroup$ (+1) You can get arima.sim to do it but it's not really its purpose: it assumes that you always want the long-run distribution, so it forces an initial burn-in period that is discarded. You can put in extra zeros for $x_{-2}$ and $x_{-3}$ so that $x_{-1} = w_{-1}$ and $x_0 = w_0$, and then you can supply the initial values as initial innovations, like this: arima.sim(model = list(ar = c(0, -0.9)), innov = c(x_minus1, x_0, w), start.innov = c(0, 0), n = n+2, n.start = 2). I wouldn't recommend it; what you have here is a lot easier to read, and has similar performance. $\endgroup$
    – Chris Haug
    Sep 23, 2022 at 0:08
  • $\begingroup$ @ChrisHaug Thanks for showing how this can be achieved with arima.sim. Tbh, I'm not really familiar with $\textsf{R}$'s ARIMA functionality. Since the OP explicitly mentions the arima.sim function your comment (which I encourage you to repost as an answer) might actually be a better answer than mine. $\endgroup$
    – statmerkur
    Sep 23, 2022 at 10:10

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