[R] Hodrick-Prescott-Filter as smooth.spline

[Albrecht Kauffmann]

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

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