I'm trying to understand how fitted values are calculated for ARMA(p,q) models. I've already found a question on here concerning fitted values of ARMA processes but haven't been able to make sense of it.

If I have a ARMA(1,1) model, i. e.

$$X_t = \alpha_1X_{t-1}+\epsilon_t - \beta_1 \epsilon_{t-1}$$

and am given a (stationary) time series I can estimate the parameters. How would I calculate the fitted values using those estimates. For an AR(1) model the fitted values are given by

$$\hat{X_t} = \hat{\alpha_1}X_{t-1} . $$

Since the innovations in an ARMA model are unobservable how would I use the estimate of the MA parameter? Would I just ignore the MA part and calculate the fitted values of the AR part?


1 Answer 1


To answer your questions, you basically need to know how the residuals i.e. $e_t$ are calculated in an arma model. Because then $\hat{X_{t}}=X_{t}-e_{t}$. Let's first generate a fake data ($X_t$) from arima(.5,.6) and fit the arma model (without mean):

ts_AR <- arima.sim(n = n, list(ar = 0.5,ma=0.6))
    Series: ts_AR 
    ARIMA(1,0,1) with zero mean     

             ar1     ma1
          0.4879  0.5595
    s.e.  0.0335  0.0317

    sigma^2 estimated as 1.014:  log likelihood=-1426.7
    AIC=2859.4   AICc=2859.42   BIC=2874.12

    Training set error measures:
                         ME    RMSE       MAE      MPE     MAPE      MASE
    Training set 0.02102758 1.00722 0.8057205 40.05802 160.1078 0.6313145

Now I create the residuals as follows: $e_1=0$ (since there is no residual at 1) and for $t=2,...,n$ we have: $e_t=X_t-Ar*X_{t-1}-Ma*e_{t-1}$, where $Ar$ and $Ma$ are the estimated auto-regressive and moving average part in above fitted model. Here is the code:

e = rep(1,n)
e[1] = 0 ##since there is no residual at 1, e1 = 0
for (t in (2 : n)){
  e[t] = ts_AR[t]-coef(f)[1]*ts_AR[t-1]-coef(f)[2]*e[t-1]

Once you find the residuals $e_{t}$, the fitted values are just $\hat{X_{t}}=X_{t}-e_{t}$. So in the following, I compared the first 10 fitted values obtained from R and the ones I can calculate from $e_{t}$ I created above (i.e. manually).

      fitted.from.package fitted.calculated.manually
 [1,]          -0.4193068                 -1.1653515
 [2,]          -0.8395447                 -0.5685977
 [3,]          -0.4386956                 -0.6051324
 [4,]           0.3594109                  0.4403898
 [5,]           2.9358336                  2.9013738
 [6,]           1.3489537                  1.3682191
 [7,]           0.5329436                  0.5219576
 [8,]           1.0221220                  1.0283511
 [9,]           0.6083310                  0.6048668
[10,]          -0.5371484                 -0.5352324

As you see there are close but not exactly the same. The reason is that when I created the residuals I set $e_{1}=0$. There are other choices though. For example based on the help file to arima, the residuals and their variance found by a Kalman filter and therefore their calculation of $e_t$ will be slightly different from me. But as time goes on they are converging.
Now for the Ar(1) model. I fitted the model (without mean) and directly show you how to calculate the fitted values using the coefficients. This time I didn't calculate the residuals. Note that I reported the first 10 fitted values removing the first one (as again it would be different depending on how you define it). As you can see, they are completely the same.

      fitted.from.package fitted.calculated.manually
 [1,]          -0.8356307                 -0.8356307
 [2,]          -0.6320580                 -0.6320580
 [3,]           0.0696877                  0.0696877
 [4,]           2.1549019                  2.1549019
 [5,]           2.0480074                  2.0480074
 [6,]           0.8814094                  0.8814094
 [7,]           0.9039184                  0.9039184
 [8,]           0.8079823                  0.8079823
 [9,]          -0.1347165                 -0.1347165
  • $\begingroup$ In the help file to arima they say: "(...) the innovations and their variance found by a Kalman filter." So the function apparently somehow uses the Kalman filter for the initial values. $\endgroup$
    – DatamineR
    Commented May 16, 2014 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.