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I have a two stage model where coefficients of model 1 become the $y$-vector for model 2. I have standard errors for those coefficients, and I want to weigh the observations in model 2 according to the standard error that the corresponding coefficient had in model 1. I believe the correct way to do this is weigh the observations in 2 by 1/SE, where SE is the standard error of the corresponding coefficient in model 1.

Now, the data for model 1 actually gets artificially created by me. I therefore know that coefficients with a SE of $\geq X$ are completely useless (I know this given my setup). Now, in model 2, giving those useless observations a weight of $1/\text{SE}$, with $\text{SE}\geq X$, seems like giving them too much weight, since their actual weight should be 0

What is the best method to weigh observations in such a case?

Glen_b
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Jeff G.
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    This looks to be quite an odd experiment I have to say. I've had a go at answering because I think there is useful information in my answer on using inverse of variance to weight. But to give more meaningful advice we'd need to know exactly what you're trying to do - for example, why do you know that coefficients with a SE >=X are completely useless whereas those just below that threshold presumeably aren't. But most importantly what is the point of this whole thing? – Peter Ellis Apr 07 '13 at 00:07
  • I know that coefficients > X are useless because of the way I artificially create that dataset: \n – Jeff G. Apr 07 '13 at 08:32
  • I know that coefficients > X are useless because of the way I artificially create that dataset: I create "fake coefficients" using random.gauss with mean Z, stdv X. I then create the y-vector using the fake coefficients while also adding some noise. If the finally computed SE for that coefficient is >= the stdv I created it with, I'm assuming it's useless for model 2 – Jeff G. Apr 07 '13 at 08:55

1 Answers1

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First, you should probably use the inverse of the variance as your weights; that is, the square of what you are proposing. See for example the discussion on this question or any general discussion of weighted least squares regression.

Second, it's not clear to me how you can be so sure that the coefficients where SE > X are completely useless - in fact, I would have thought simply weighting these like others by the inverse of the estimated variance was ample. However, if you really are sure that this hard cut-off point is meaningful then you are saying that the real variance of those points is infinite (rather than the actual number you estimated in stage 1), so go ahead and give them zero weight.

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Peter Ellis
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