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I just read a paper where they used a log-log regression with log(1+x) as one of the independent variables and log(1+y) being the dependent variable.

The only inference from this regression in the paper was that, since the estimated coefficient is positive and significant, the impact of x on y is positive. This made me wonder whether the statement is true and if there is any possibility for a more detailed inference from the estimated coefficient.

Xdelta10
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    $f(x) = \log(x+1)$ is monotonically increasing, so the statement is true (with the caveats that regression in general entails). I don't understand the second question: what more about the coefficient would you like to know? – Firebug Feb 09 '23 at 11:56
  • In an ordinary linear regression, one can infer that a one unit change in x (holding the other independent variables fixed) would change y by the estimated coefficient. Is it possible to make any statement like this concerning the influence of x on y from the estimated coefficient in this regression setting? – Xdelta10 Feb 09 '23 at 14:06
  • Yes, of course, because the meaning of the coefficient does not change merely because of how you express your values. As always, a difference of one unit of $x$ is associated with a difference of $\hat\beta$ units of $\log(1+y).$ – whuber Feb 09 '23 at 14:51
  • Our thread at https://stats.stackexchange.com/questions/576504 perhaps offers the "more detailed inference" you seek. – whuber Feb 09 '23 at 15:51

1 Answers1

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The only inference from this regression in the paper was that, since the estimated coefficient is positive and significant, the impact of x on y is positive. This made me wonder whether the statement is true and if there is any possibility for a more detailed inference from the estimated coefficient.

$f(x)=log(x+1)$ is monotonically increasing, so the statement is true (with the caveats that regression in general entails).

The second part of the question, however, is no, we can't express changes in $y$ in terms of purely $\beta_1$, because the relation is non-linear, and changes for different values of $x$.

$$E[\log(y+1)|x] = \log(\hat y + 1) = \beta_0+\beta_1\log(x+1)$$

$$\frac{\partial{\log(\hat y + 1)}}{\partial x} = \frac{\partial{\log(\hat y + 1)}}{\partial \hat y}\frac{\partial{\hat y}}{\partial x}= \frac{\partial{\hat y}}{\partial x}\frac{1}{\hat y + 1} =\frac{\beta_1}{x+1}$$

So

$$\frac{\partial{\hat y}}{\partial x}={\beta_1}\frac{\hat y + 1}{x+1}= {\beta_1}\frac{\exp(\beta_0+\beta_1\log(x+1))}{x+1}$$

This can be further simplified, although no further insight ensues:

$$\frac{\partial{\hat y}}{\partial x}= {\beta_1}\frac{\exp(\beta_0+\beta_1\log(x+1))}{x+1} = {\beta_1}\frac{\exp(\beta_0)\exp(\beta_1\log(x+1))}{x+1} =\\ {\beta_1}\frac{\exp(\beta_0)(x+1)^{\beta_1}}{x+1} =\\ \beta_1\exp(\beta_0)(x+1)^{\beta_1-1} $$

Firebug
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