使用来自 ARDL R ZEFE90A8E604A7C840E88D03A6B666Z 的 model object 进行预测

Question

I am trying to use forecast function on ardl model like this.

library(ARDL)
data(denmark)

models <- auto_ardl(LRM ~ LRY + IBO + IDE, data = denmark, max_order = 5)
ardl_3132 <- models$best_model

fabletools::forecast(object = ardl_3132, new_data = denmark[1:20, ])
forecast::forecast(object = ardl_3132, new_data = denmark[1:20, ])

However, it's throwing an error Error in L(LRM, 1): could not find function "L"

How can I make forecast work here?

Answer 1

I am seeting this error too when trying to use predict() on an ardl model.

L function is in tsDyn, I think, but loading that does not eliminate the error.

Answer 2

AFAIK, in the "core" of the ARDL model usually lies simple linear regression (unless it is specified as a dynamic model with dynlm), so... My best attempt would be to try building the ARDL model using a different package/function. It takes a bit more writing, but should give you a workable model object, which you could then use to build your prediction/forecast.

So what I would do, would be to estimate the ARDL model using the ARDL package, just like you did in your example above, extract the full formula from your model object, and rewrite it inside the lm() formula, using it's syntax, like so:

> ardl_3132$full_formula

LRM ~ L(LRM, 1) + L(LRM, 2) + L(LRM, 3) + LRY + L(LRY, 1) + IBO + L(IBO, 1) + L(IBO, 2) + L(IBO, 3) + IDE + L(IDE, 1) + L(IDE, 2)

We can specify the formula above inside an lm() model:

ardl_3132_lm <- lm(LRM ~ lag(LRM, 1) + lag(LRM, 2) + lag(LRM, 3) + LRY + lag(LRY, 1) + IBO + lag(IBO, 1) + lag(IBO, 2) + lag(IBO, 3) + IDE + lag(IDE, 1) + lag(IDE, 2), data = denmark)

You can use the predict.lm() function to create the forecast using the linearly specified ARDL model:

predict.lm(ardl_3132_lm)

The ARDL package itself is useful for estimating the model more conveniently than you would, using lm() or glm(). See summary() call on the model object to explore it's properties.

summary(ardl_3132)

Comparing it with the lm() specified model summary shows, however, that they are not identical.

summary(ardl_3132_lm)

The reason for this is that the function ARDL::auto_ardl() for this specific example has selected dynlm::dynlm(formula = full_formula, data = data, start = start, end = end) which specifies Dynamic Linear Models and Time Series Regression , and not a linear model.

To create a forecast from the dynlm model, you would need to use stats::predict() like so:

stats::predict(ardl_3132)

Comparing the dynlm forecasted values with the linear model predicted values,

stats::predict(ardl_3132_lm)

we can see, that the predictions are different.