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st: RE: Re: RE: Re: Generating predicted values for OLS with transformed dependent variables
You are correct. The results are not identical.
If they were, there would be hardly any need
for -glm- as the results could be obtained simply
by transforming the response first.
The difference between E(ln response) and ln(E (response))
Rodrigo A. Alfaro
> Dear Nick:
> I didn't explore glm command before, then I tried the following:
> sysuse auto
> g lnp=ln(price)
> glm price, family(normal) link(log) nohead nolog
> reg lnp, nohead
> and the coefficient (let's say mu) is different. Is there
> somethig than
> I am missing? a normalization issue?
"Nick Cox" <firstname.lastname@example.org>
> As posted earlier, -glm- offers the back-to-basics
> Jacobin solution as an alternative to this use
> of Jacobians.
Rodrigo A. Alfaro
> > Let me simplify the problem. Considere u~normal(0,s_u^2) and
> > g(y)=u. You
> > want E(y)... right? Sometimes you can find the distribution
> > of y using the
> > jacobian transformation. Suppose that h() is the inverse of g() then
> > y=h(u)... then you need to find the distribution of y. This
> > is a change of
> > variable, you have to evaluate the normal with h() and
> > multiply by the
> > jacobian.
> > Confuse? take g() =ln()... then ln(y)=u which is a
> > simplification of the
> > regression with log in the dependent variable. Note that the
> > inverse of g()
> > is known then h()=exp() and finally y=exp(u). You need to know the
> > distribution of y, which is lognormal!!!
> > (http://www.xycoon.com/logn_relationships1.htm). Using this
> > distribution we
> > can get the expected value of y E(y) = exp(0+0.5*s_u^2)
> > (http://www.xycoon.com/logn_expectedvalue.htm). If
> > ln(y)=bx+u, you can find
> > that for nonstochastic x
> > E(y)=exp(bx+0.5*s_u^2)=exp(bx)*exp(0.5*s_u^2), the
> > second term is the "adjustment".
> > In your problem you have to find the distribution of y for y=u^4 and
> > u~normal. I understand that you cannot use the jacobian
> > transformation, but
> > the proof of "the square of a standard normal is a chi-square
> > 1" is a useful
> > source to solve your problem.
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