Modelli di raccordo in R in cui i coefficienti sono soggetti a restrizioni lineari

Come dovrei definire una formula modello in R, quando sono disponibili una (o più) restrizioni lineari esatte che legano i coefficienti. Ad esempio, supponiamo che tu sappia che b1 = 2 * b0 in un modello di regressione lineare semplice.

Grazie!

r regression modeling

— George Dontas
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Supponiamo che il tuo modello sia

$Y(t) = \beta_0 + \beta_1 \cdot X_1(t) + \beta_2 \cdot X_2(t) + \varepsilon(t)$

e stai pianificando di limitare i coefficienti, ad esempio come:

$\beta_1 = 2 \beta_2$

inserendo la restrizione, riscrivendo il modello di regressione originale che otterrai

$Y(t) = \beta_0 + 2 \beta_2 \cdot X_1(t) + \beta_2 \cdot X_2(t) + \varepsilon(t)$

$Y(t) = \beta_0 + \beta_2 (2 \cdot X_1(t) + X_2(t)) + \varepsilon(t)$

introduce a new variable $Z(t) = 2 \cdot X_1(t) + X_2(t)$ and your model with restriction will be

$Y(t) = \beta_0 + \beta_2 Z(t) + \varepsilon(t)$

In this way you can handle any exact restrictions, because the number of equal signs reduces the number of unknown parameters by the same number.

Playing with R formulas you can do directly by I() function

lm(formula = Y ~ I(1 + 2*X1) + X2 + X3 - 1, data = <your data>) 
lm(formula = Y ~ I(2*X1 + X2) + X3, data = <your data>)

— Dmitrij Celov
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This is pretty clear, but the question was suggesting a restriction between b0 and b1. Should I also create a new variable Z = 2X + 1 and fit a model without intercept?

— George Dontas

I think usualy I is used instead of eval in formulas, i.e. Y~I(1+2*X1)+X2+X3-1

— mpiktas

@gd047: I have updated with a code pieces, yes it is as you say. @mpiktas: will change this, yes it is shorter ;)

— Dmitrij Celov

This is a good answer for the general theoretical approach, but for an easier way to actually implement these hypotheses in R, which also has the advantage of not requiring one to estimate multiple models, see linearHypothesis() in the car package.

— Jake Westfall