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I have a question regarding parameter interpretation for a GLM with a gamma distributed dependent variable. This is what R returns for my GLM with a log-link:

Call:
glm(formula = income ~ height + age + educat + married + sex + language + highschool, 
    family = Gamma(link = log), data = fakesoep)

Deviance Residuals: 
       Min        1Q    Median        3Q       Max  
  -1.47399  -0.31490  -0.05961   0.18374   1.94176  

Coefficients:
              Estimate Std. Error t value Pr(>|t|)    
(Intercept)  6.2202325  0.2182771  28.497  < 2e-16 ***
height       0.0082530  0.0011930   6.918 5.58e-12 ***
age          0.0001786  0.0009345   0.191    0.848    
educat       0.0119425  0.0009816  12.166  < 2e-16 ***
married     -0.0178813  0.0173453  -1.031    0.303    
sex         -0.3179608  0.0216168 -14.709  < 2e-16 ***
language     0.0050755  0.0279452   0.182    0.856    
highschool   0.3466434  0.0167621  20.680  < 2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for Gamma family taken to be 0.1747557)

Null deviance: 757.46  on 2999  degrees of freedom
Residual deviance: 502.50  on 2992  degrees of freedom
AIC: 49184

How do I interpret the parameters? If I calculate exp(coef()) of my model, I get ~ 500 for the intercept. Now I believe that doesn't mean the expected income if all other variables are held constant, does it? Since the average or mean(age) lies at ~ 2000. I furthermore have no clue how to interpret the direction and value of the covariates' coefficients.

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    $\begingroup$ 500 would be close to the expected income if all other variables were exactly zero (not merely constant) --- just like in regression, really. $\endgroup$ – Glen_b -Reinstate Monica May 8 '14 at 22:32
  • $\begingroup$ @Glen_b why would it be expected income when exponential of coefficients is the multiplicative effect on the income when there is a change in explanatory variable? $\endgroup$ – tatami Oct 27 '17 at 15:12
  • $\begingroup$ The case under discussion is the conditional mean when all the explanatory variables are 0. $\endgroup$ – Glen_b -Reinstate Monica Oct 27 '17 at 16:12
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The log-linked gamma GLM specification is identical to exponential regression:

$$E[y \vert x,z] = \exp \left( \alpha + \beta \cdot x +\gamma \cdot z \right)=\hat y$$

This means that $E[y \vert x=0,z=0]=\exp(\alpha)$. That's not a very meaningful value (unless you centered your variables to be be mean zero beforehand).

There are at least three way to interpret your model. One is to take derivative of the expected value of $y$ given $x$ with respect to $x$:

$$\frac{\partial E[y \vert x,z]}{\partial x} = \exp \left( \alpha + \beta \cdot x +\gamma \cdot z\right)\cdot \beta=\hat y \cdot \beta$$

This quantity depends on $x$ and $z$, so you can either evaluate this at the mean/median/modal or representative values of $x$ and $z$, or take the average of $\hat y \cdot \beta$ over your sample. These are both called marginal effects. These derivatives only make sense for continuous variables (like height) and tell you an additive effect of a small change in $x$ on $y$.

If $x$ was binary (like sex), you might consider calculating finite differences instead: $$E[y \vert z,x=1]-E[y \vert z,x=0]=\exp \left( \alpha + \beta +\gamma \cdot z\right) - \exp \left( \alpha +\gamma \cdot z\right)= \exp \left( \alpha +\gamma \cdot z\right) \cdot\left( \exp(\beta)-1 \right)$$

This makes more sense since it's hard to imagine an infinitesimal change in sex. Of course, you can also do this with a continuous variable. These are additive effects from a one unit change in $x$, rather than a tiny one.

The third method is to exponentiate the coefficients. Note that:

$$ \begin{array} _E[y \vert z,x+1] &= \exp \left( \alpha + \beta \cdot (x+1) +\gamma \cdot z \right) \\ &=\exp \left( \alpha + \beta \cdot x+\beta +\gamma \cdot z \right)\\ &=\exp \left( \alpha + \beta \cdot x +\gamma \cdot z \right)\cdot \exp(\beta) \\ &= E[y \vert z,x]\cdot \exp(\beta) \end{array} $$

This means that you can interpret the exponentiated coefficients multiplicatively rather than additively. They give you the multiplier on the expected value when $x$ changes by 1.

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    $\begingroup$ Would you be able to illustrate the second interpretation? $\endgroup$ – tatami Oct 24 '17 at 16:57
  • $\begingroup$ @tatami I fixed a mistake in the binary case. Does it makes more sense now? $\endgroup$ – Dimitriy V. Masterov Oct 24 '17 at 17:10
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First I would look at the residuals to see how well the model fits. If it's OK, I would try using other link functions unless I had reason to believe it really came from a gamma distribution. If the gamma still looked convincing, I would conclude that the statistically significant terms are the intercept, height, education, sex, and high school (the ones marked with three stars). Among themselves one can't say more unless they are standardized (have the same range).

Response to comment: I understand your question better now. You absolutely can do that! A unit increase in the height causes a exp(0.0082530)-1 ~= 0.0082530 (using the exp x = 1 + x approximation for small x) relative change in the income. Very easy to interpret, no?

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    $\begingroup$ So I can't actually interpret the parameters e.g. the income increases by xy if the height increases by one? $\endgroup$ – Cajira Apr 29 '14 at 19:09
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    $\begingroup$ I believe now that I have to interpret it multiplicatively: exp(Intercept)*exp(height) would be the income with a 1 unit increase in height. Thank you nonetheless! :) $\endgroup$ – Cajira Apr 30 '14 at 8:51

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