Karsten W. Karsten W. - 4 days ago 4
R Question

Simulate a time series

In a recent online course on macroeconomic forecasting, there was an excercise to model

y(t) = 3.0 + 0.55 y(t-1) + e(t)


where
e(t)
is defined as

et <- c(-1.2138662, -0.2854597, 0.5902700, 0.8285463, -0.9954260, -0.3716332)


Now I tried to do this in R (the course used EViews), however I do not come to the solution given there: 5.648 for the 5th element.
I tried (similar to this blogpost):

y <- rep(NA,6)
y[1] <- 0
y[2] <- 3 + 0.55*y[1]+et[1]
y[3] <- 3 + 0.55*y[2]+et[2]
y[4] <- 3 + 0.55*y[3]+et[3]
y[5] <- 3 + 0.55*y[4]+et[4]
y[6] <- 3 + 0.55*y[5]+et[5]


and then

y <- rep(NA,6)
y[1] <- et[1]
y[2] <- 3 + 0.55*y[1]+et[2]
y[3] <- 3 + 0.55*y[2]+et[3]
y[4] <- 3 + 0.55*y[3]+et[4]
y[5] <- 3 + 0.55*y[4]+et[5]
y[6] <- 3 + 0.55*y[5]+et[6]


and then

arima.sim(list(order=c(1,0,0), ar=0.55), n=6, innov=head(et,6)+3)


However all three methods give different results. I wonder why this is, I am afraid I did not understand something basic.

Answer

arima.sim has a "burn-in" period in order to reach stationarity. Argument n.start sets the length of this period, and it must has length ar + ma. For an AR(1) process, n.start must be at least 1. Note, if you don't specify n.start, arima.sim will automatically compute a reasonable one for you.

Given n.start (either a specified one or auto-computed one), arima.sim then randomly sample n.start innovations for "burn-in". It is right because of this random procedure, you will get different result from one run to another:

## innovations of length 6
et <- c(-1.2138662, -0.2854597, 0.5902700, 0.8285463, -0.9954260, -0.3716332)

set.seed(0)
arima.sim(list(order = c(1,0,0), ar = 0.55), n = 6, innov = et, n.start = 1)
#[1] -0.5192413 -0.5710424  0.2761967  0.9804545 -0.4561760 -0.6225300

set.seed(1)
arima.sim(list(order = c(1,0,0), ar = 0.55), n = 6, innov = et, n.start = 1)
# [1] -1.55841580 -1.14258839 -0.03815361  0.80756181 -0.55126700 -0.67483005

We can eliminate such randomness, by providing a set of starting innovations using start.innov argument. Note, this vector must has length n.start:

## fixing starting innovation at 0; i.e., `y[0] = 0`, so `y[1] = et[1]`.
arima.sim(list(order = c(1,0,0), ar = 0.55), n = 6, innov = et, n.start = 1,
          start.innov = 0)
# [1] -1.21386620 -0.95308611  0.06607264  0.86488625 -0.51973856 -0.65748941

Now, compare this last arima.sim with the following:

y <- rep(NA, 6)
y[1] <- et[1]
y[2] <- 0.55 * y[1] + et[2]
y[3] <- 0.55 * y[2] + et[3]
y[4] <- 0.55 * y[3] + et[4]
y[5] <- 0.55 * y[4] + et[5]
y[6] <- 0.55 * y[5] + et[6]

# [1] -1.21386620 -0.95308611  0.06607264  0.86488625 -0.51973856 -0.65748941

We have everything completely reproducible.

The above generates zero-mean time series; if you want mean at 3, shift the result by +3 yourself:

y <- y + 3
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