Suppose that I have the following model $$g(\mu)=\beta_0+\beta_1(x_1-\bar{x}_1)+\beta_2(x_2-\bar{x}_2)+\beta_3(x_2-\bar{x}_2)^2$$ where $g(\mu)$ is the complementary log-log function.

I calculated the increments in the mean for each unit change in the values of $x_1$ and $x_2$ fixing a value of $x_1$ then fixing the values of $x_2$.

Fixing the value of $x_2$ I can calculate the increment in $g(\mu)$ as

$$g_1(\mu)=\beta_0+\beta_1((x_1+1)-\bar{x}_1)+\beta_2(x_2-\bar{x}_2)+\beta_3(x_2-\bar{x}_2)^2$$ $$=g(\mu)+\beta_1$$

Fixing now the value of $x_1$ then

$$g_1(\mu)=\beta_0+\beta_1(x_1-\bar{x}_1)+\beta_2((x_2+1)-\bar{x}_2)+\beta_3((x_2+1)-\bar{x}_2)^2$$ $$=g(\mu)+\beta_2+\beta_3+2\beta_3(x_2-\bar{x}_2)$$

So these are the increments in the values of $g(\mu)$ to calculate the increments in $\mu$ I just calculate the inverse of link function?

Edit: The increments in the the complementary log-log in the first case are $$g_1(\mu)-g(\mu)=\beta_1$$ so the increment in $\mu$ is $$1-\exp(-\exp(\beta_1))$$

In the second case the increments in $g$ are $$g_1(\mu)-g(\mu)=\beta_2+\beta_3+2\beta_3(x_2-\bar{x_2})$$ so the increments in $\mu$ are $$1-\exp(-\exp(\beta_2+\beta_3+2\beta_3(x_2-\bar{x_2})))$$

Is it right?

EDIT2: In this model the parameters estimatives are $$\beta_0=-1.177,\quad\beta_1=-0.153,\quad\beta_2=0.153,\quad\beta_3=0.075$$

So in the first case the increment is $$1-\exp(-\exp(\beta_1))=0.57$$

It means that the increment in mean for each unit change in $x_1$ will be $0.57$? It doesn't make sense since $\mu\in [0,1]$.

EDIT3: I did a test and fixed a value for $x_2$ and calculated $\mu$ for the values $x_1=40$ and $x_1=41$ and the difference between those two values are $$0.3076309-0.309594=-0.0019$$ a small reduction in the response variable and it's a reasonable value (expecting something like it). I'm starting to think that this expression for increments is wrong.

If I fix a value for $x_2$ and start $x_1=30$, then start to calculate the values in $\mu$ for $x_1=40,50,60,70$ then I will have increments for $10,20,30,40$ right?

  • $\begingroup$ This matches my interpretation and understanding. I suppose the comment "remember these are multiplicative, not additive!" may be one to throw in as well. $\endgroup$
    – mflo-ByeSE
    Jun 16, 2017 at 0:25
  • $\begingroup$ @mfloren What you mean with "remember these are multiplicative, not additive!"? $\endgroup$
    – user72621
    Jun 16, 2017 at 2:15
  • $\begingroup$ Short answer: because all of these changes are within exponents, it multiplicatively affects the change in mean. Think: percentage change (and how that is different from additive change). $\endgroup$
    – mflo-ByeSE
    Jun 16, 2017 at 18:04

1 Answer 1


Again, I suggest looking at multiplicative change (as suggested in the comments). In other words, try comparing the changes from 0 to 1 to that of 40 to 41 (they should be different). Based on this: the value interpretation is that of multiplicative change. In other words: one uses the value gained from the parameter to multiply, not add, to the previous value. Here is an example.


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