if p1, p2, etc.. just represent the same quantities at different times, I'm
not sure why you don't reshape long?
reshape long p, i(idvar) j(time)
then all your statements could be done in 2 steps:
replace p=invlogit(`lgtp1' ) in 1
replace p=p[_n-1]*`ka'+(1-p[_n-1])*`mu' in 2/l
Michael Blasnik
[email protected]
----- Original Message -----
From: "Ben Pelzer" <[email protected]>
To: <[email protected]>
Sent: Wednesday, February 04, 2004 9:10 AM
Subject: RE: st: assignment by indexing
> Nick,
>
> Your answer yesterday was clear. I was looking for a neat way to solve the
> problem, but finally decided to use the following set of commands for a
> similar, yet smaller, problem of only 13 timepoints:
>
> quietly replace `p1' = invlogit(`lgtp1');
> quietly replace `p2' = `p1' * `ka' + (1-`p1' ) * `mu'
> quietly replace `p3' = `p2' * `ka' + (1-`p2' ) * `mu'
> quietly replace `p4' = `p3' * `ka' + (1-`p3' ) * `mu'
> quietly replace `p5' = `p4' * `ka' + (1-`p4' ) * `mu'
> quietly replace `p6' = `p5' * `ka' + (1-`p5' ) * `mu'
> quietly replace `p7' = `p6' * `ka' + (1-`p6' ) * `mu'
> quietly replace `p8' = `p7' * `ka' + (1-`p7' ) * `mu'
> quietly replace `p9' = `p8' * `ka' + (1-`p8' ) * `mu'
> quietly replace `p10' = `p9' * `ka' + (1-`p9' ) * `mu'
> quietly replace `p11' = `p10' * `ka' + (1-`p10' ) * `mu'
> quietly replace `p12' = `p11' * `ka' + (1-`p11' ) * `mu'
> quietly replace `p13' = `p12' * `ka' + (1-`p12' ) * `mu'
>
> Here, ka and mu are the transition probs. Of course, following your
> matrix-suggestion would solve all problems but also cause some extra work,
> while copying statements is very fast... I didn't know the 'egen'
statement,
> so, I'll have a look at it right away. Thanks again, Ben.
>
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