Hello:

Please help me figure this out.

I understand (please correct me if I'm wrong) that with the following statement, the model pool is of size 2, with or w/o the intercept, and there is no intercept in the selected model:

```
train_data |>
model(ARIMA(MW_Hourly ~ pdq(p = 1, d = 0, q = 2) + PDQ(P = 4, D = 1, Q = 1) + lag(HumidityLow, 24) + IsWeekend,
order_constraint = TRUE, stepwise = FALSE, greedy = FALSE, approximation = FALSE)) |>
report()
# Series: MW_Hourly
# Model: LM w/ ARIMA(1,0,2)(4,1,1)[24] errors
#
# Coefficients:
# ar1 ma1 ma2 sar1 sar2 sar3 sar4 sma1 lag(HumidityLow, 24) IsWeekendTRUE
# 0.9638 0.2247 -0.1032 -0.2097 -0.2230 -0.1849 -0.1512 -0.5088 -1.5179 -97.3029
# s.e. 0.0087 0.0320 0.0318 0.1012 0.0714 0.0571 0.0471 0.1030 1.6111 42.0355
#
# sigma^2 estimated as 32392: log likelihood=-7477.02
# AIC=14976.04 AICc=14976.27 BIC=15031.58
```

The intercept wasn't included in the final model. To get the same model explicitly, I exclude the intercept by hand, but it generates an error:

```
train_data |>
model(ARIMA(MW_Hourly ~ 0 + pdq(p = 1, d = 0, q = 2) + PDQ(P = 4, D = 1, Q = 1) + lag(HumidityLow, 24) + IsWeekend,
order_constraint = TRUE, stepwise = FALSE, greedy = FALSE, approximation = FALSE)) |>
report()
# Series: MW_Hourly
# Model: NULL model
# NULL modelWarning message:
# 1 error encountered for ARIMA(MW_Hourly ~ 0 + pdq(p = 1, d = 0, q = 2) + PDQ(P = 4, D = 1,
# Q = 1) + lag(HumidityLow, 24) + IsWeekend, order_constraint = TRUE,
# stepwise = FALSE, greedy = FALSE, approximation = FALSE)
# [1] non-finite value supplied by optim
```

## @robjhyndman

^{Referred here by Forecasting: Principles and Practice, by Rob J Hyndman and George Athanasopoulos}