Could someone, please, write me, how to compute the spar-value for the smooth.spline-routine to get the same HP-filtered time-series with a parameter lambda for a function (see mail "Hodrick-Prescott-Filter example" from ggrothendieck at yifan.net): hpf <- function(y,lambda{ eye <- diag(length(y)) d <- diff(eye,d=2) z <- solve(eye+lambda*crossprod(d),y)} ? Second, is there a connection between the parameter lambda in this function and the value $lambda of the smooth.spline routine? For annual time series with smooth.spline(y,spar=0.71205) I get almost the same $y-values like with the program above and lambda=100, but I would like to know the reason for this. With many thanks in advance Albrecht Kauffmann -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._