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Re: st: estimation of SE of correlation after mixlogit

From   Arne Risa Hole <[email protected]>
To   [email protected]
Subject   Re: st: estimation of SE of correlation after mixlogit
Date   Fri, 12 Aug 2011 12:26:56 +0100


Note that -mixlogit-  is a user-written command which is available
from the SSC and described in

Running the following command after -mixlogit- (on one line) will give
you the SE of the correlation:

nlcom (rho: [l21]_b[_cons]*[l11]_b[_cons] /
sqrt([l11]_b[_cons]*[l11]_b[_cons]*([l21]_b[_cons]*[l21]_b[_cons] +


On 12 August 2011 10:59, Nick Darson <[email protected]> wrote:
> Hi listers,
> After running mixlogit, I am wondering how I exactly calculate the
> standard error of the correlation.
> To my understanding, I have to do this manually after obtaining the
> covariance-variance components using the command "mixlcov" (please
> tell me if there is an easier automatic way that I have overlooked,
> such as using nlcom - I have tried this but always get error
> messages).
> For example, I have a multinomial model with three possible
> alternatives: low, mid, high (low as the base case; mid and high are
> random and allowed to be correlated).
> Hence, after estimation with mixlogit and using "mixlcov", I obtain
> the values for v21, v11, and v22.
> This way I can easily calculate the correlation of the two variables:
> rho= _b[v21] / sqrt (_b[v11] * _b[v22])
> However, how do I calculate the SE of this correlation?
> I found the following formula:  SE(rho) =  sqrt ( 1-rho^2 / n -2 )
> However, which value do I use for "n"? In my case, I have repeated
> measures (each of the 200 individuals makes 5 choices). Hence, should
> I use n=200 (the number of subjects), n= 1000 (total number of
> observations), or do I need to look how many cases actually have
> chosen mid and high (e.g. let's say from the 1000 total choices, 200
> are mid and 300 are high, hence n=500)?
> Thankful for any help on this apparently simple issue.
> Cheers,
> Nick
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