4
$\begingroup$

I am running a dlog-dlog (difference of logarithm*) regression and I want to convert the coefficients into marginal effects. I know that it's different from a log-log regression, in which the coefficients directly give us the elasticities.

How can we interpret the coefficients from a dlog-dlog regression?

* For example dlog (Y)= a + b dlog(X)+ error term.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
4
$\begingroup$

Since you have differences, this means that the data is time series and we can write

$$Y_t=Y_0+\sum_{s=1}^t\Delta Y_s$$

So if the true model is

$$\Delta Y_t=\alpha+\beta \Delta X_t$$

we have

$$Y_t=Y_0+\sum_{s=1}^t(\alpha+\beta \Delta X_t)=Y_0-\beta X_0+\alpha t+\beta X_t$$

So you can say that interpretation remains the same as in model with levels.

Improve this answer
$\endgroup$
3
  • $\begingroup$ So, is b equal to elasticity of Y with respect to X, as can be obtained from the regression "log (Y)= a + b log(X)+ error term"? $\endgroup$
    – Dlogger
    Commented May 22, 2011 at 13:20
  • $\begingroup$ @Dlogger, $\log(Y)=a_0+a_1t+b\log(X)+\varepsilon$ $\endgroup$
    – mpiktas
    Commented May 22, 2011 at 19:25
  • $\begingroup$ @ Mpiktas, Do you also know some interpretation on Dlog - log ... ? Is that similar to the interpretation of log-log? $\endgroup$
    – user36977
    Commented Jan 7, 2014 at 13:13
1
$\begingroup$

If this is indeed linear then I think your underlying model may be something like

$$Y_j \approx k \, \exp(aj) X_j^b $$

where your regression does not tell you about the value of the constant $k$, but you might perhaps be able to use it to pin the first and last points of your observed data.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.