Including NaN in function for HP-Filter

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Philipp
Philipp am 24 Mai 2013
Hi there, my question is the following: I have a function file to calculate HP-Filtered series of some series y. The problem is the following: If the series y contains some NaN, then the HP-filter function returns a series only with NaN, i.e. up till now it can only produce HP-filtered series when there are no NaN in y. Can somebody adjust my function file such that it works with NaN?
Below the code:
function [cycle,trend]=hpfilter(y,lambda)
long = size(y,1);
HP=[1+lambda -2*lambda lambda zeros(1,long-3);...
-2*lambda 1+5*lambda -4*lambda lambda zeros(1,long-4);...
zeros(long-4,long);...
zeros(1,long-4) lambda -4*lambda 1+5*lambda -2*lambda;...
zeros(1,long-3) lambda -2*lambda 1+lambda];
for i=3:long-2;
HP(i,i-2)=lambda;
HP(i,i-1)=-4*lambda;
HP(i,i)=1+6*lambda;
HP(i,i+1)=-4*lambda;
HP(i,i+2)=lambda;
end;
trend = HP\y;
cycle = y-HP\y;
Thanks very much! Philipp
  1 Kommentar
David Sanchez
David Sanchez am 24 Mai 2013
I tried with
y=str2array('[1 2 e 3]')
y =
1
2
NaN
3
lambda = .2;
and it seems to work, what's the problem?

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Philipp
Philipp am 24 Mai 2013
Bearbeitet: Philipp am 24 Mai 2013
Hello David, maybe I have to be more precise: I run this line of command in order to check for my results:
lambda=1600;
h=log(countrydata.GR.WPU./countrydata.GR.GDPDEF);
x=hpfilter(h,lambda);
later I have everything in a loop which looks like this:
for jj=1:lc;
eval(['hpw.' Countries{1,jj} '.WPU=hpfilter(log(countrydata.' Countries{1,jj} '.' Var{1,2} './countrydata.' Countries{1,jj} '.' Var{1,9} '),lambda);']);
eval(['hpw.' Countries{1,jj} '.WPR=hpfilter(log(countrydata.' Countries{1,jj} '.' Var{1,3} './countrydata.' Countries{1,jj} '.' Var{1,9} '),lambda);']);
end
Now h contains some real values and some NaN, but when I execute hpfilter(h,lambda), then x is all NaN?!?
  2 Kommentare
David Sanchez
David Sanchez am 24 Mai 2013
What's the error message you get?
Philipp
Philipp am 24 Mai 2013
I dont get an error but each observation is NaN. But what I would like to have is that only the ones that are NaN in h are also NaN in x. Therefor, as mentioned in another forum, the interpolation is not an appropriate solution.

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