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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 $MA(5)$ ?

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    $\begingroup$ It would be good to post the seed of your simulation so as to be able to replicate the series. $\endgroup$ – Christoph Hanck Sep 10 '18 at 15:59
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Essentially, it is randomness, yes, combined with the fact that the second process only has a small coefficient, so that its acf will decay quickly, just as that of an MA(1) process.

Comparing

acf(arima.sim(model = list(ar=0.05), n = 150))
acf(arima.sim(model = list(ma=0.27), n = 150))

indeed reveals similar patterns.

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