2
$\begingroup$

I have converted the ARIMA (1,0,0)(1,0,1)12 into the following equation,

$ (1-\phi_1B) (1-\zeta_1 B^{12}) Y_t = (1- \eta_1 B^{12})e_t$

where $\phi_1$ AR coefficient, $\zeta_1$ is SAR coeffiecient, and $\eta_1$ is SMA coefficient. When i expand this equation i get the following equation,

$ y_t- \phi_1 y_{t-1} + \zeta_1 \phi_1 y_{t-13} - \zeta_1 y_{t-12} = c + e_t - \eta_1 e_{t-12}$

My question is how do i get the $e_t$ and $e_{t-12}$ from R? My time series is univariate. R generates standard error when I run coef(df_arima) but does not give the previous error terms. Same goes with the $c$. I am not sure how to get this either. is c the intercept ?

My aim is to predict the value for the next month , taking into account, the previous data, by hand. I have all the values except $e_t$, $e_{t-12}$ and $c$.

$\endgroup$

1 Answer 1

1
$\begingroup$

Let's simulate some data so we are discussing the same thing:

set.seed(1) # for reproducibility
foo <- ts(rnorm(120),frequency=12)
library(forecast)
model <- Arima(foo,order=c(1,0,0),seasonal=c(1,0,1))
summary(model)

This gives us coefficients as follows (among other information I cut out):

Coefficients:
          ar1     sar1    sma1    mean
      -0.0099  -0.1702  0.1455  0.1094
s.e.   0.0845   0.1488  0.1301  0.0778

The mean column gives the estimate of the intercept $c$, so $\hat{c}=0.1094$. We also see that

$$ \hat{\phi}_1=-0.0099,\quad\hat{\zeta}_1=-0.1702,\quad\hat{\eta}_1=0.1455.$$

Now, don't confuse the standard errors of coefficients (the bottom row in that table) with the "errors" $e_t$ of the time series, which are often also called "innovations"! To calculate your forecast by hand, you will need the in-sample residuals, which you can get by residuals(model):

            Jan          Feb          Mar          Apr          May          Jun          Jul          Aug          Sep          Oct          Nov          Dec
1  -0.735553343  0.066955583 -0.943961915  1.476074374  0.234806723 -0.927359944  0.368693460  0.632512042  0.472470019 -0.409945615  1.397862585  0.294450877
2  -0.746077521 -2.329640213  0.969061751 -0.107573389 -0.121263921  0.810190805  0.729287954  0.507294672  0.826148959  0.670607748  0.006578943 -2.091766001
3   0.474295307 -0.218331743 -0.238847488 -1.591649628 -0.607182475  0.326703714  1.268850473 -0.189888368  0.294634012 -0.142091804 -1.493790693 -0.592080902
4  -0.494517692 -0.169220482  0.978319051  0.625789052 -0.281741566 -0.361444773  0.612571688  0.446748132 -0.789537517 -0.831412254  0.211147825  0.656072516
5  -0.229845305  0.764597566  0.322403594 -0.696611263  0.220069074 -1.246051099  1.321653144  1.896296747 -0.478259657 -1.177661559  0.460253138 -0.222673095
6   2.286638985 -0.106118277  0.582443469 -0.096566119 -0.847234754  0.041789417 -1.882604304  1.382027510  0.048955844  2.037863852  0.396060640 -0.824113175
7   0.550322552 -1.044477734 -1.359551587  0.169737483 -0.572862717 -0.107744203 -0.087922486 -0.672689316 -0.682331268 -0.196481153  1.074479552 -1.641277559
8   0.472271407  0.203579085  0.919979555 -0.400093431  0.246128733  0.156631224 -0.643675563  1.070879306  1.044609294  0.587086842  1.508505784  0.426411586
9  -1.370487392 -0.687123491 -1.311877423 -0.606575690 -0.727654495 -0.069989482 -1.038005727  0.068624068 -0.734687244  1.667274973  0.656806055  0.823737095
10  0.246994475  1.556966428 -0.766859421 -0.591604041  1.297917511 -0.749437420 -0.347055541 -0.508741319 -0.457409991 -0.354390605  0.391589413 -0.265387074

So, suppose we want to forecast for Jan 11. Your formula

$$ y_t- \phi_1 y_{t-1} + \zeta_1 \phi_1 y_{t-13} - \zeta_1 y_{t-12} = c + e_t - \eta_1 e_{t-12}$$

turns into

$$ y_t= \phi_1 y_{t-1} - \zeta_1 \phi_1 y_{t-13} + \zeta_1 y_{t-12} + c + e_t - \eta_1 e_{t-12}. $$

We replace the unknown parameters by their estimates as above. We take $y_{t-1}$, $y_{t-12}$ and $y_{t-13}$ from the series history, and take $e_{t-12}$ from the residuals(model) table - it's the entry for Jan 10, which is equal to $0.246994475$. Finally, we don't know $e_t$ yet, because it's our unknown new innovation, so we replace it by its expectation, which is zero. And there you are.

Note that I didn't calculate the actual predictions, because it's tedious, and because I am almost certain there is an error in the formula (but I still believe the description here is helpful). Please take a look at this earlier thread. Note in particular how Arima() with an intercept fits an ARIMA model to $Y_t-\hat{c}$, so your formula will need to be adapted.

$\endgroup$
5
  • $\begingroup$ Shouldn't $e_{t-12}$ be error term 12 months earlier? For example if i want to predict for January 2018 shouldn't i take the residual 12 month earlier(January 2017) ? $\endgroup$
    – CODE_DIY
    Jul 23, 2020 at 14:27
  • 1
    $\begingroup$ Yes, exactly. When we are forecasting for Jan 11 in my simulated example, then $e_{t-12}$ is the residual from one year earlier, Jan 10. $\endgroup$ Jul 23, 2020 at 14:29
  • $\begingroup$ Can you spot a mistake in the equation, according to generalized SARIMA formula, my equation is correct, $\varnothing_{P}\left(B^{s}\right) \varphi_{p}(B)(1-B)^{d}\left(1-B^{s}\right)^{D} Z_{t}=\theta_{q}(B) \vartheta_{Q}\left(B^{s}\right) e_{t}$ , I am getting wrong prediction after manual calculation. $\endgroup$
    – CODE_DIY
    Jul 23, 2020 at 15:07
  • $\begingroup$ Have you looked at the earlier thread I linked? $\endgroup$ Jul 23, 2020 at 15:17
  • $\begingroup$ Yes, But due to seasonal component the equation becomes something other than ARMA. Any resource you can refer for this problem? Thnak you for your answer. $\endgroup$
    – CODE_DIY
    Jul 23, 2020 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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