Re: st: how to get marginal effect when dependent variable has ln function?

 From David Greenberg To statalist@hsphsun2.harvard.edu Subject Re: st: how to get marginal effect when dependent variable has ln function? Date Tue, 23 Jun 2009 14:46:10 -0400

There is an algebraic error in the algebraic manipulations below. The line of algebra ought to go like this:

y2/y1 = (y1 + Δy)/y1 = 1 + (Δy)/y1 and so
ln(y2/y1) = exp(bΔx1).
David Greenberg
Sociology Department
New York University

----- Original Message -----
From: gjhxmu@sina.com
Date: Tuesday, June 23, 2009 8:18 am
Subject: st: how to get marginal effect when dependent variable has ln function?
To: statalist <statalist@hsphsun2.harvard.edu>

> Thank you very much for all replies! I am sorry for my delayed reply.
>
>
> Austin mentioned that "Each observation may have a different marginal
> effect of x1 on y".
>
> I doubt that and I calculate below:
>
> Assuming one equation is lny=a0+a1x1+a2x2+a3x3+e
>
> and all other indenpendent variables are unchanged, then
>
> Δlny=bΔx1，Δlny=lny2－lny1 = ln(y2/y1)=ln(Δy+1)=bΔx1，Δy=exp(bΔx1)－1，
> %Δy=100*(y2-y1)/y1=100*( exp(bΔx1)－1).
>
> So as long as x1 changes one unit, the change percent of y is
> constant, not vary with various levels of x1.
>
>
> I found -glm- is helpful. Thank you!
>
> However, why are the results of -reg- and -glm- different?
>
>
> The result of -reg-:
>
>
> reg lny x1,eform("x1")
>
> Source | SS df MS Number of obs = 149
> -------------+------------------------------ F( 1, 147) = 13.02
> Model | 43.4816065 1 43.4816065 Prob > F = 0.0004
> Residual | 491.039393 147 3.34040403 R-squared = 0.0813
>
> -------------+------------------------------ Adj R-squared = 0.0751
> Total | 534.520999 148 3.61162837 Root MSE = 1.8277
>
> ------------------------------------------------------------------------------
> lny | x1 Std. Err. t P>|t| [95% Conf. Interval]
> -------------+----------------------------------------------------------------
> x1 | 1.287607 .0902158 3.61 0.000 1.121113 1.478828
>
>
> The result of -glm-with option link:
>
>
> . glm y x1, link(log)
>
> Iteration 0: log likelihood = -2298.4836
> Iteration 1: log likelihood = -2253.0398
> Iteration 2: log likelihood = -2227.9831
> Iteration 3: log likelihood = -2227.9678
> Iteration 4: log likelihood = -2227.9678
>
> Generalized linear models No. of obs = 149
> Optimization : ML Residual df = 147
> Scale parameter = 5.77e+11
> Deviance = 8.48306e+13 (1/df) Deviance = 5.77e+11
> Pearson = 8.48306e+13 (1/df) Pearson = 5.77e+11
>
> Variance function: V(u) = 1 [Gaussian]
> Link function : g(u) = ln(u) [Log]
>
> AIC = 29.93245
> Log likelihood = -2227.967803 BIC = 8.48e+13
>
> ------------------------------------------------------------------------------
> | OIM
> y | Coef. Std. Err. z P>|z| [95% Conf. Interval]
> -------------+----------------------------------------------------------------
> x1 | .1846949 .0662293 2.79 0.005 .0548878 .3145019
> _cons | 12.68207 .2616964 48.46 0.000 12.16916 13.19499
>
>
> The result of -glm- with option family:
>
>
> . glm y x1, family(poisson)
> note: y has noninteger values
>
> Iteration 0: log likelihood = -57268851
> Iteration 1: log likelihood = -55358804
> Iteration 2: log likelihood = -55358404
> Iteration 3: log likelihood = -55358404
>
> Generalized linear models No. of obs = 149
> Optimization : ML Residual df = 147
> Scale parameter = 1
> Deviance = 110714755.9 (1/df) Deviance = 753161.6
> Pearson = 159601721.9 (1/df) Pearson = 1085726
>
> Variance function: V(u) = u [Poisson]
> Link function : g(u) = ln(u) [Log]
>
> AIC = 743065.8
> Log likelihood = -55358403.61 BIC = 1.11e+08
>
> ------------------------------------------------------------------------------
> | OIM
> y | Coef. Std. Err. z P>|z| [95% Conf. Interval]
> -------------+----------------------------------------------------------------
> x1 | .1673077 .000054 3097.82 0.000 .1672018 .1674136
> _cons | 12.73727 .0001869 68143.38 0.000 12.7369 12.7376
>
>
>
> Please forgive my ignorance if any and thank you for any reply!
>
>
> sincerely,
>
> from ROSE
>
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