Solved – Is every ARIMA(1,1,0) model equivalent to an AR(2) model

arimartime series

Assume I have a time series $ x_t $ that I want to fit using an ARIMA(1,1,0) model of the form:

$ \Delta x_t = \alpha \Delta x_{t-1} + w_t $

This could be rewritten as:

$ x_t – x_{t-1} = \alpha ( x_{t-1} – x_{t-2} )+ w_t $

$ x_t = ( 1 + \alpha)x_{t-1} – \alpha x_{t-2} + w_t $

The last equation describes an AR(2) model with coefficients $1+\alpha$ and $-\alpha$. I recognize that, depending on $\alpha$, this AR(2) model might be non-stationary. However, if I was taking a diff to begin with, then the series I am modeling shouldn't be stationary.

I know that if the model is non-stationary, a diff should be used. But how would the results differ if I used a AR(2) model vs an ARIMA(1,1,0) model? I assume (as hinted by R) that it has an issue with convergence. However, when I ask R to perform the fits, it will do both of them, and the coefficients are (mostly) consistent with my observations above. The forecasts are definitely different, though.

If anyone could shed some light on this, or point me to a good reference, I would appreciate it.

Here is the R code I used to generate both models. 

> set.seed(2)
> x <- arima.sim(n = 1000, model=list(order=c(1,1,0), ar=c(0.3)))
> plot(x)
> arima(x, order=c(1,1,0))

Call:
arima(x = x, order = c(1, 1, 0))

Coefficients:
         ar1
      0.3291
s.e.  0.0298

sigma^2 estimated as 1.03:  log likelihood = -1433.91,  aic = 2871.81
> arima(x, order=c(2,0,0))

Call:
arima(x = x, order = c(2, 0, 0))

Coefficients:
         ar1      ar2  intercept
      1.3290  -0.3294    50.9803
s.e.  0.0298   0.0299    35.9741

sigma^2 estimated as 1.03:  log likelihood = -1438.93,  aic = 2885.86
Warning messages:
1: In log(s2) : NaNs produced
2: In log(s2) : NaNs produced
3: In log(s2) : NaNs produced
4: In arima(x, order = c(2, 0, 0)) :
  possible convergence problem: optim gave code = 1

Best Answer

The forecast for the ARIMA(1,1,0) enforces the restriction that $d=1$.

It is maybe even easier to see in the AR(1) vs. ARIMA(0,1,0) case: The latter is just $$ \Delta y_t=\epsilon_t $$ whose optimal forecasts are 0 at all horizons (we expect $\epsilon_t$ to take the value zero). If we aim to forecast $y_t$ itself, we take the last in sample value and just accumulate the forecast changes of $y_t$. Basically, we expect the value tomorrow to be today's value plus the expected change from today to tomorrow.

So, as we do not expect any changes here, the optimal forecast for such a random walk is $y_T$ ($T$ being the last in sample observation) for all $h=T+1,\ldots$.

If, on the other hand, we fit an AR(1) model, we obtain an estimate $\hat\alpha$ and produce the optimal forecasts from an AR(1) model as $$ y_{T+h}=\hat\alpha^hy_T $$ If estimation errors (as they generally will in finite samples) are such that $\hat\alpha$ differs from the true value of 1, the forecasts will differ.