Below is the summary of my SARIMA model:
Series: diff(log(data_final_ts))
ARIMA(0,0,2)(0,0,2)[4] with non-zero mean
Coefficients:
ma1 ma2 sma1 sma2 intercept exogeneous_var1 exogeneous_var2
-0.7756 0.2152 -0.5155 -0.4023 0.0413 -1e-04 0.0028
s.e. 0.1358 0.1782 0.2985 0.2311 0.0068 0e+00 0.0347
exogeneous_var3 exogeneous_var4
-0.0054 0.0012
s.e. 0.0035 0.0049
sigma^2 estimated as 0.0001283: log likelihood=182.13
AIC=-344.26 AICc=-339.68 BIC=-323.49
Training set error measures:
ME RMSE MAE MPE MAPE MASE ACF1
Training set -0.0002763522 0.01042586 0.008296043 83.07064 116.4007 0.4182354 0.006248187
I want to know if the following model equation is correct:
$$(1-B)(1-B^4)Y_t=(1+\phi_1 B^4+\phi_2 B^8)(1+\theta_1 B+\theta_2 B^2)e_t+\text{const}+\beta X_t$$
with $\text{const}$ being the mean of $Y_t$ and with $X_t$ including an intercept and a few exogenous variables.
Best Answer
Your model is a regression with SARIMA errors (rather than a SARIMAX model) and is given by the following equations:
$$ \begin{aligned} y_t &= 0.0413 - 0.0001 x_{1,t} + 0.0028 x_{2,t} - 0.0054 x_{3,t} + 0.0012 x_{4,t} + u_t, \\ u_t &= (1 - 0.7756\text{B} + 0.2152\text{B}^2)(1 - 0.5155\text{B}^4 - 0.4023\text{B}^{4\cdot2}) \varepsilon_t. \end{aligned} $$
What is left to do is to multiply the terms in brackets in the second equation carefully and to get the time indices right due to the backshift operator $\text{B}$.
(Note that the constant enters the model in a simple way because there are no autoregressive terms in the model. If there were some, the expression would be a bit more complicated. See Rob J. Hyndman's blog post "Constants and ARIMA models in R".)
Edit: It looks like the variable supplied to the
arima
function that produced the output in the OP was a first difference of some other variable. Therefore, the model for the original (nondifferenced) variable would contain $(1-\text{B})$ in front of $y_t$.