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My problem is interpreting coefficients of such time series model:

\begin{equation} \ln Y_t - \ln Y_{t-1} =b_1 \cdot \left(X_{t}-X_{t-1}\right)+b_2 \cdot Z_t.\end{equation}

I don't know how to interpret coefficients $b_1$ and $b_2$.

Hope that someone can help.

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  • $\begingroup$ Thanx ! i learned a lot from your answer $\endgroup$ – user78949 Jun 4 '15 at 13:45
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For $small$ changes, you can interpret logged differences as percentage changes after multiplying by 100.

For example, $y_t=9$ and $y_{t-1}=8$. Then $\ln 9 - \ln 8=.118$ or 11.8%, which is the logarithmic approximation to the actual 12.5% increase. Note that I had to multiply by 100 here. For $y_t=9$ and $y_{t-1}=8.5$ the approximation will be much better ($5.9\% \approx 5.7\%$).

Usually, a coefficient tells you the effect on $y$ of a one unit change in that explanatory variable, holding other variables constant. A one unit change in $\Delta \ln x$ corresponds to a 100% change. This means that $b_1$ tells you the percentage change in $y$ associated with a 1% increase in x.

But your $x$ is not logged, so the coefficient needs to be interpreted differently. When $x$ grows by one unit, you get $100 \cdot b_1\%$ more $y$.

Moreover, $100 \cdot b_2$ tells you the percentage change in $y$ associated with a 1 unit increase in $z$.

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  • $\begingroup$ Thanks for your answer! Now I know how to interpret Z, but the problem with x still remains, because there is Δx and not Δlnx. Hope you can help. $\endgroup$ – Testeris Nov 8 '13 at 6:48
  • $\begingroup$ Thank you! Hope that one day I would be able to help others! $\endgroup$ – Testeris Nov 8 '13 at 12:08
  • $\begingroup$ Dimitrly, yes this is super old but hope you get this notice. Does this answer mean that differencing your series (y and x) does not change the interpretation of the coefficient at all? Does this hold for any differencing (e.g. season)? $\endgroup$ – B_Miner Sep 2 '15 at 3:40
  • $\begingroup$ @B_Miner I think the interpretation depends on the DGP and the coefficient. For example, the constant in the differenced specification is actually the linear trend, not the intercept from the original levels specification. With that caveat, I think seasonal differencing should still work. $\endgroup$ – Dimitriy V. Masterov Sep 2 '15 at 17:04
  • $\begingroup$ Im going to post a new question if you get a chance to look. $\endgroup$ – B_Miner Sep 2 '15 at 17:35

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