4

suppose for a time series, the auto.arima output is:

ARIMA(1,0,1) with non-zero mean 

Coefficients:
         ar1     ma1  intercept
      0.7456  0.2775   767.7463
s.e.  0.0804  0.1197     0.1072

It is specified in help file of arima function that definition used for this function is

X[t] = a[1]X[t-1] + … + a[p]X[t-p] + e[t] + b[1]e[t-1] + … + b[q]e[t-q]

and if we include mean(m), this formula applies to X - m rather than X.

My questions are:

(1) For the model considered above, what will be the arima equation?

  (a) X[t] =  767.7463 + 0.7456*X[t-1] + 0.2775*e[t-1]

or (b) X[t] = 195.3258 + 0.7456*X[t-1] + 0.2775*e[t-1],

where constant term is achieved by 195.3258 = 767.7463*(1-0.7456).

(2) How to interpret this constant term if we have more than one AR coefficient?

e.g. ARIMA(2,0,2) with non-zero mean 

Coefficients:
          ar1      ar2     ma1     ma2  intercept
      -0.0124  -0.6945  0.3629  0.9414    -0.1288
s.e.   0.1172   0.1132  0.0598  0.1270     0.0636

(3) What should be the ARIMA equation in presence of drift term?

e.g. ARIMA(1,1,0) with drift         

Coefficients:
         ar1   drift
      0.3307  0.3981
s.e.  0.0946  0.1463
Richard Hardy
  • 67,272
na ja
  • 45

2 Answers2

5

The model with no differencing:

$$\phi(B)(X_t -c) = \theta(B)e_t$$ where $\phi(B)=1-\phi_1B - \cdots - \phi_pB^p$, $\theta(B)=1+\theta_1B + \cdots + \phi_pB^p$, and $c$ is the "intercept".

The model with differencing:

$$\phi(B)(1-B)(X_t -ct) = \theta(B)e_t$$ where $c$ is the "drift" parameter.

Example 1: $$(1-0.7456B)(X_t - 767.7463) = (1+0.2775B)e_t$$ So \begin{align*} X_t &= (1-0.7456)*(767.7463) + 0.7456*X_{t-1} + e_t + 0.2775 e_{t-1} \\ &= 195.3147 + 0.7456*X_{t-1} + e_t + 0.2775 e_{t-1} \end{align*}

Example 2: $$(1+0.0124B + 0.6945B^2)(X_t +0.1288) = (1+0.3629B + 0.9414B^2)e_t$$ So \begin{align*} X_t &= -(1+0.0124 + 0.6945)*(0.1288) - 0.0124 X_{t-1} -0.6945 X_{t-2} + 0.3629 e_{t-1} + 0.9414 e_{t-2} + e_t\\ &= -0.2198 - 0.0124 X_{t-1} -0.6945 X_{t-2} + 0.3629 e_{t-1} + 0.9414 e_{t-2} + e_t \end{align*}

Example 3: $$(1-0.3307B)(1-B)(X_t - 0.3981t) = e_t$$ So \begin{align*} X_t & = (1-0.3307)*0.3981 + 1.3307X_{t-1} - 0.3307 X_{t-2} + e_t \\ & = 0.2664 + 0.3307X_{t-1} - 0.3307 X_{t-2} + e_t \end{align*}

Rob Hyndman
  • 56,782
  • In the model with differencing, is the drift parameter ct multiplicative? That is, were I forecasting t + h periods in the future, would I get the appropriate drift value with c*(t + h)? – jpm_phd Feb 05 '20 at 18:05
3

Regarding your question 1, what ARIMA calls the intercept is in reality the estimated mean of the time series. So the obtained constant term

195.3258 = 767.7463*(1-0.7456)

is the estimated intercept. See issue 3 in http://www.stat.pitt.edu/stoffer/tsa4/Rissues.htm for an example:

Claude
  • 151