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))

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

s.e.  0.0298

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

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

         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
  • $\begingroup$ Should $ \triangledown x_t = \alpha \triangledown x_{t-1} + w_t $ read $\Delta x_t = \alpha \Delta x_{t-1} + w_t $ or is this some notation I haven't seen before? $\endgroup$
    – Silverfish
    Commented Apr 21, 2016 at 13:02
  • $\begingroup$ Oops. You're right, @Silverfish. Not sure why I wrote those upside down. Thanks. $\endgroup$
    – Beane
    Commented Apr 24, 2016 at 18:10

2 Answers 2


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.


The equivalence depends on definitions. General ARMA(p,q) process can be defined as a stochastic process which is the solution to the following equation:

$$ X_t-\phi_1X_{t-1}-...-\phi_pX_{t-p}=Z_t+\theta_1 Z_{t-1}+...+\theta_qZ_{t-q},$$

where $Z_t$ is a white noise process. We must require that polynomials $\phi(z)=1-\phi_1 z-...-\phi_pz^p$ and $\theta(z)= 1+\theta_1z+...\theta_pz^p$ should not have common roots, in order for the equation to be uniquely defined.

Now the question arises, when this equation has a solution. The answer relies on the properties of polynomials $\phi(z)$ and $\theta(z)$. The equation has a stationary solution, when polynomials do not have roots on the unit circle.

So in this sense ARIMA(1,1,0) is not AR(2) process, because it is not stationary. It can be written as satisfying the AR(2) equation, but since the polynomials have a root on a unit circle, you cannot solve the equation. However if polynomial $\phi(z)$ has a unit root, then $\Delta X_t$ satisfies the ARMA(p-1,q) equation (with different polynomials). So it is possible to solve for $\Delta X_t$ and get back to $X_t$. To mark this difference ARIMA(p,d,q) notation is used.

So to sum up, if we strictly define a ARMA(p,q) process as a stationary solution to ARMA(p,q) equation, then ARIMA(1,1,0) and AR(2) are not equivalent.

The fact that R manages to find the correct coefficients is an interesting property of estimation, i.e. it is possible to show that in the case of unit roots, the OLS[1] will give consistent estimates of the coefficients, however the inference would be incorrect, as the limiting distributions are not normal. The ADF tests are based on such estimates. However the actual mathematics to show that the estimates are ok is quite complicated and relies on certain assumptions. These assumptions do not generalize well, hence it is not advisable to use usual estimation methods for unit root processes.

[1] The MLE and OLS are equivalent for AR(p) type specifications.

  • $\begingroup$ MLE and OLS are asymptotically equivalent, I suppose. (But conditional MLE and OLS could be plainly equivalent; are they?) $\endgroup$ Commented Apr 21, 2016 at 9:12
  • $\begingroup$ If you condition on first $p$ observations, then yes they should be plainly equivalent. $\endgroup$
    – mpiktas
    Commented Apr 21, 2016 at 9:18

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