# Why is auto.arima modeling an AR(1) process as an MA(1)?

Playing around with auto.arima to see how effective it is at model selection. I first simulated an $$AR(1)$$ process with $$X_{t+1} = 0.9 X_t + \epsilon_t$$

n <- 150
eps <- rnorm(n)
x0 <- rep(0, n)
for (i in seq.int(2, n))
x0[i] <- 0.9 * x0[i-1] + eps[i]
fit <- auto.arima(ts(x0))
fit
Series: ts(x0)
ARIMA(1,0,0) with non-zero mean

Coefficients:
ar1    mean
0.8265  0.7092
s.e.  0.0446  0.4405

sigma^2 estimated as 0.944:  log likelihood=-208.09
AIC=422.17   AICc=422.34   BIC=431.2


Ok, so far so good: auto.arima returned an $$AR(1)$$ model with $$a_1 = 0.8265$$, which is close to $$0.9$$.

The I tried the same thing with $$X_{t+1} = 0.05 X_t + \epsilon_t$$

n <- 150
eps <- rnorm(n)
x0 <- rep(0, n)
for (i in seq.int(2, n))
x0[i] <- 0.05 * x0[i-1] + eps[i]
fit <- auto.arima(ts(x0))
fit
Series: ts(x0)
ARIMA(0,0,1) with zero mean

Coefficients:
ma1
0.2738
s.e. 0.0801

sigma^2 estimated as 0.8831:  log likelihood=-203.05
AIC=410.1   AICc=410.18   BIC=416.12


Why is auto.arima returning an $$MA(1)$$ model? Why isn't it returning an $$AR(1)$$ model with an $$a_1$$ close to the coefficient I used in the simulation? Is there a theoretical reason for this or is just the randomness of the eps <- rnorm(n) that is giving something closer to an $$MA(1)$$ even though I'm trying to simulate an $$AR(1)$$?

And isn't an $$AR(1)$$ supposed to be equivalent to an $$MA(\infty)$$ (so that in practice I would get an $$MA(4)$$ or an $$MA(5)$$ ?

• It would be good to post the seed of your simulation so as to be able to replicate the series. – Christoph Hanck Sep 10 '18 at 15:59

acf(arima.sim(model = list(ar=0.05), n = 150))