If I have a log-log regression, like:

$\ln(\text {Price}) = b_0 + b_1 \times (\Delta \ln (\text{emp}))$

Where $\Delta(\ln (\text{emp})) = \ln(\text{employment growth_year2}) - \ln(\text{employment growth_year1})$

Is my interpretation: a $\,1\%$ increase in the change in employment growth, increases price by $b_1\,\%$.

Or does the 'change' aspect make for a different interpretation.

  • $\begingroup$ Are you referring to a change of a change? Like, say, the first year the $ln(emp) = 10$ and the second year it is $ln(emp) = 11$, so that the $\Delta ln(emp) = 10\%$? $\endgroup$ – Antoni Parellada Jan 18 '16 at 17:51
  • $\begingroup$ No I don't think so. In your example, my change in employment variable will be recorded as "1", the difference between the two logs, rather than being expressed as a % of year ones employment $\endgroup$ – Kelly Jan 18 '16 at 17:57
  • $\begingroup$ I interpreted it as $\Delta ln(emp) = 10\%$ increase from year one in $ln(empl)$. By the way, did I mess up in any way the OP - parentheses, logs...? $\endgroup$ – Antoni Parellada Jan 18 '16 at 18:03
  • $\begingroup$ I guess the key part is in the OP, when you write " $1\%$ increase in the change in employment growth." In your comment you don't seem to refer to an increase in the "change" (i.e. a change of a change - or a change of $\Delta$), but to the actual change in employment. $\endgroup$ – Antoni Parellada Jan 18 '16 at 18:09
  • $\begingroup$ I think I know now where we are confused. I should not have put the change symbol infront of Δln(emp)), it should just be ln(emp). I meant it as a variable name, my bad, I should be more careful. So in this case, my interpretation is okay? $\endgroup$ – Kelly Jan 18 '16 at 18:22

If $ln(p_t)=\beta_0 + \beta_1 (ln(E_t)-ln(E_{t-1}))$ then it holds that $ln(p_t)=\beta_0 + \beta_1 ln\left(\frac{E_t}{E_{t-1}}\right)$ or $ln(p_t)=\beta_0 + ln\left( \left(\frac{E_t}{E_{t-1}}\right)^{\beta_1}\right)$.

After 'exponentiation' this becomes $p_t=e^{\beta_0}\left(\frac{E_t}{E_{t-1}}\right)^{\beta_1}$.

So if $\frac{E_t}{E_{t-1}}$ is multiplied by a factor $f$, i.e. $\frac{E_t}{E_{t-1}} \to f \frac{E_t}{E_{t-1}}$, then the new $p_t$ will become $e^{\beta_0}\left(f\frac{E_t}{E_{t-1}}\right)^{\beta_1}=f^{\beta_1}e^{\beta_0}\left(\frac{E_t}{E_{t-1}}\right)^{\beta_1}=f^{\beta_1}p_t$.

So if $\frac{E_t}{E_{t-1}} \to f \frac{E_t}{E_{t-1}}$ then the price $p_t$ will become $p_t \to f^{\beta_1}p_t$.

For example, if $\frac{E_t}{E_{t-1}}$ increases by $1\%$ then $f=1.01$ and the price then increases by $(1.01^{\beta_1} - 1) \times 100 \%$.

Note that, for very small $x$ it holds that ( Taylor series expansion) $(1+x)^{\beta_1} \approx 1+ \beta_1 x$.

Applying this approximation to the above example we find that $(1.01)^{\beta_1}=(1+0.01)^{\beta_1} \approx 1+ \beta_1 \times 0.01$ so the increase in $p$ is $(1.01^{\beta_1} - 1) \times 100 \% \approx (1+\beta_1 \times 0.01 -1) \times 100\%=100 \times 0.01 \times \beta_1 \%=\beta_1 \%$.

So your interpretation is fine with the nuance that you talk about relative changes in growth rate, so if the growth rate is $\frac{E_t}{E_{t-1}}= 5\%$ then a 10% change brings this growth rate to 5.5% ($=1.1 \times 5\%$) , a 1% change brings it to 5.05% ($=1.01 \times 5\%$) (moreover, the approximation only holds for small percentages)

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