smars smars - 25 days ago 12
R Question

How can I fit the model “Y(t) = αX + βY(t-1) - βY(t-2)" in R?

I have to make a one-step ahead forecast for a time series Y(t) using R. Theory suggests the ideal model should be:

Y(t) = αX + βY(t-1) - βY(t-2)

However, I don't know how to deal with the following issues:


  • I have to take βY(t-1) minus βY(t-2).

  • There are both autoregressive (Y(t-1),Y(t-2)) and exogenous variables (X).

  • I have to test whether or not "βY(t-1) - βY(t-2)" is the best way to express the autoregression, instead of other ARIMA models.



The time series Y(t) in question is:

Y <- c(57.4, 51.6, 36.1, 34.8, 41.2, 59.1, 62.5, 55.0, 53.8, 52.4, 44.5, 42.2, 50.1, 61.3, 49.6, 38.2, 51.1, 44.7, 40.8, 46.1, 53.5, 54.7, 50.3, 48.8, 53.7, 52.0)


The exogenous variable X used is:

X <- c(-12.1, 30.0, 13.5, 30.0, -3.8, -24.3, 30.0, 30.0, 30.0, 30.0, -21.6, 30.0, 0.0, 26.5, -30.0, 20.5, -4.8, -9.2, 22.2, -7.3, 15.9, 16.0, 13.7, 5.6, 5.7, 1.8)


As you may notice, this actual X does not provide much help in predicting Y. Nevertheless, I reported it as an example since I am currently looking for the right values of X.

If anything was wrong or not clear, let me know and I will give the necessary explanations.

Thanks in advance.

Answer

You can use the lag function for the transformation and lm or glm for regression:

Y <- c(57.4, 51.6, 36.1, 34.8, 41.2, 59.1, 62.5, 55.0, 53.8, 52.4, 44.5, 42.2, 50.1, 61.3, 49.6, 38.2, 51.1, 44.7, 40.8, 46.1, 53.5, 54.7, 50.3, 48.8, 53.7, 52.0)
X <- c(-12.1, 30.0, 13.5, 30.0, -3.8, -24.3, 30.0, 30.0, 30.0, 30.0, -21.6, 30.0, 0.0, 26.5, -30.0, 20.5, -4.8, -9.2, 22.2, -7.3, 15.9, 16.0, 13.7, 5.6, 5.7, 1.8)

y_1 <- lag(Y)
y_2 <- lag(Y,2)

lm(Y~X+y_1+y_2)

You could also do the lag transformations directly in the regression equation:

lm(Y ~ X + I(lag(Y)) + I(lag(Y, 2)))

Finally, the difference is just a change of the operator to this:

lm(Y ~ X + I(lag(Y)) - I(lag(Y, 2)))
Call:
lm(formula = Y ~ X + I(lag(Y)) - I(lag(Y, 2)))

Coefficients:
(Intercept)            X    I(lag(Y))  
  3.906e-14   -4.693e-18    1.000e+00
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