Ash Ash - 1 month ago 15
R Question

Simulate `AR(1)` process with uniform innovations

I need to plot an

AR(1)
graph for the process

y[k] = 0.75 * y[k-1] + e[k] for y0 = 1.


Assume that
e[k]
is uniformly distributed on the interval
[-0.5, 0.5]
.

I am trying to use
arima.sim
:

library(tseries)
y.0 <- arima.sim(model=list(ar=.75), n=100)
plot(y.0)


It does not seem correct. Also, what parameters do I change if
y[0] = 10
?

Answer

We want to use R base function arima.sim for this task, and no extra libraries are required.

By default, arima.sim generates ARIMA with innovations ~ N(0,1). If we want to change this, we need to control the rand.gen or innov argument. For example, you want innovations from uniform distributions U[-0.5, 0.5], we can do either of the following:

arima.sim(model=list(ar=.75), n=100, rand.gen = runif, min = -0.5, max = 0.5)

arima.sim(model=list(ar=.75), n = 100, innov = runif(100, -0.5, 0.5))

Example

set.seed(0)
y <- arima.sim(model=list(ar=.75), n = 100, innov = runif(100, -0.5, 0.5))
ts.plot(y)

enter image description here


In case we want to have explicit control on y[0], we can just shift the above time series such that it starts from y[0]. Suppose y0 is our desired starting value, we can do

y <- y - y[1] + y0

For example, starting from y0 = 1:

y <- y - y[1] + 1
ts.plot(y)

enter image description here

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