Sb estimates calc

[jincal]

I am wondering if someone could help me understand the idea behind the calc for Sb estimates.

I see it is taking the average of the Sb estimates for the 200 Halton draws (or whatever number/type of draws is specified). However, I also see the Sb mean t ratios are given as well.

If I take (1.96/(sb mean t ratio) )^2 I get a very different answer from the Sb mean estimates for betas that have a relatively high s.e.

Why not calculate the Sb mean esimates using (1.96/(sb mean t ratio) )^2 where the sb mean t comes from the average beta for the draws divided by the square root of the diagonals on the average covariance matrix instead of averaging the Sb estimates for the draws.

Any suggestions would be welcome.

Jim

[johnr]

Hi Jim

Ngene reports two values. The first are the Sp estimates. In the literature, p is sometimes used to designate fixed priors. In this case, Ngene is taking the mean of the prior distribution and calculating what the S-estimates would be if you used these fixed values as priors - that is, it ignores the Bayesian priors. This is not the mean of the Bayesian calculations - it is using the mean of the Baysian prior distributions only.

The second estimates, the Sb, estimates are using all the information provided for the Bayesian priors - mean + std dev. (or spread). Again, in the literature, b is sometimes used to designate Bayesian priors. In this case, it is calculating the S-estimate for each draw, and then averaging the estimate over the draws. Hence, we call it these the Sb mean estimates.

Hope this answers your question.

John

[jincal]

Thanks John.

Jim