[Gretl-users] replace with
Hi,
I have the following problem. Let x(t) a time series with T=200 observations.
I must create a new time series y(t), with 250 observations, resampling with
replacement the elements of x. I have used the command
genr y = replace(x)
but gretl creates y(t) with 200 observations. Can I create y with 250
observations?
Thanks.
Best regard
Alessandro Attanasio