Is coef() method return the best estimated beta?

[TigerZhao007]

First, thank you for the package! :)

In the Vignette, it is said that:
A formatted dataframe of the coefficient matrix of the final iteration of forecast evaluation can be obtained via the coef
method.

So it is NOT the best estimated beta under the best_lambda ?
To get the best estimated beta under the best_lambda, I have to run another BigVAR.fit() ? As code below:

library(BigVAR)
data(Y)
mod1<-constructModel(Y,p=4,"Basic",gran=c(150,10),h=1,cv="Rolling",verbose=FALSE,IC=TRUE,model.controls=list(intercept=TRUE))
results=cv.BigVAR(mod1)
best_lambda = results@OptimalLambda
coef1 = coef(results) # NOT the best

model2=BigVAR.fit(Y,p=4,"Basic",lambda=best_lambda,intercept=TRUE)
coef2 = model2[,,1] # best estimated beta under the best_lambda

[wbnicholson]

Thank you for the question. Both of these approaches are providing the coefficient matrix with the "best" lambda using all available data. The coefficients are slightly different because the result from cv.BigVAR uses the coefficient matrix from the previous iteration as a "warm start." However, the predictions generated are extremely similar:
abs(predict(results) - coef2%*%results@Zvals)

If you want to achieve the same results in both approaches or are more generally interested in support recovery. I would suggest decreasing the optimizer tolerance using the parameter tol:

mod1<- constructModel(Y,p=4,"Basic",gran=c(150,10),h=1,cv="Rolling",verbose=FALSE,IC=TRUE,model.controls=list(intercept=TRUE,tol=1e-6))
model2=BigVAR.fit(Y,p=4,"Basic",lambda=best_lambda,intercept=TRUE, tol=1e-6)