Variance of model-averaged betas

[Morten Frederiksen]

I've been trying to figure out how to get correct variance estimates when you're averaging model betas, and the effect in question doesn't appear in all models. Getting the correct model-averaged beta is simple - you just set the estimate for models without this effect to zero (Burnham & Anderson 2004, Sociological Methods & Reseacrh 33, 261-304). However, things are not as simple for the model-averaged variance. If you set both estimated beat and variance to zero when the effect doesn't appear, and if the summed w(i) for that effect is low, the result is a model-averaged variance smaller than for models including the effect. This seems inappropriate. The alternative is averaging the variance only over models where the particular effect appears, i.e. has a non-zero estimate and variance, but I haven't been able to find any references recommending this - and it seems inconsistent with the way the model-averaged beta is derived.

Does anybody have any ideas?

Morten

[bmitchel]

Hello Morten,

My understanding is that if the parameter you are calculating variance for is not in the current model, you should use a variance of zero and a parameter estimate of zero. This will contribute to the unconditional variance an amount equal to the model weight times the square of the model averaged parameter estimate (if you use the formula in B&A 2004, which you should).

You are right that if the model weights of models that include the parameter of interest are low, then this will result in a model averaged variance that is lower than the variance you see for a model with the parameter that is included. I don't see this as a problem; if the parameter is not included in most of your models, the model-averaged parameter estimate should be close to zero, as should its variance.

When I asked this question of David Anderson last year, he confirmed to me that the correct approach is to treat the variance as zero when the parameter estimate is zero.

Brian

[Morten Frederiksen]

Thanks Brian! That is indeed what the equation in B&A 2004 seems to imply, and I suppose it makes sense when you look at it in the way you describe.

Morten