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I want to estimate price of a product, lets say a boat, over time - panel data model.

I estimated a model where I control for number of characteristics denoted as $X$ ( e.g. engine size, weight etc.). However, I also included a trend and trend squared variable into the model to control for "other" market developments. Let's just assume that is correct specification for now.

$ln(price_i) = \alpha_i + \beta_i\sum X + \gamma_i~trend + \delta_i ~ trend^2$

My estimated coefficients for $\gamma$ is negative (-0.5) and $\delta$ positive (0.002). This indicates that the initially price is decreasing with time but after around 13 years the trend reverts (convex shape).

enter image description here

My question is how do I interpret the coefficients (ceteris paribus), meaning I know that per one unit increase in trend (year) there is approx. $-0.048 \approx (1- e^{-0.05})*1 + (1-e^{0.002})*1^2$ change in price in year 1, $-0.092$ in year 2, $...$, $-0.092$ in year 23, $...$, etc. However, is it change in price relative to "year 0" or year-to-year change? In short how I would label the y-axis on the plot above, would %$\Delta$ Price be correct?

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Personally, I don't find such coefficients to be very meaningful on their own. Here's how I would interpret these results.

The expected value for $\ln y$ is $$\alpha + \beta 'x + \gamma \cdot t + \delta \cdot t^2 $$

Taking the derivative of that with respect to $t$ and applying the chain rule, you get

$$\frac{\partial y}{\partial t}\cdot \frac{1}{y}= \gamma + 2 \cdot\delta \cdot t$$

That is the very definition of semi-elasticity, and it is a function of $t$ in your model. It is the relative change in price $y$ for a one unit change in $t$ since you can rearrange the LHS to $\frac{\Delta y/y}{\Delta t}$. I would multiply things by a 100 here to convert to a percentage change in $y$. This curve looks like:

enter image description here

You can see that the percent price change is negative when $t$ is small at -50% percent, and shrinks somewhat over 30 years to to -38%. It is always negative over this range. In short, time always leads to boat depreciation, but the price drop grows smaller over time.

You can also ask how that semi-elasticity changes with $t$ by taking the derivative again:

$$\frac{\partial \varepsilon}{\partial t} = 2 \cdot\delta $$

This makes $\delta$ a bit more interpretable: it tells you that the elasticity falls by constant $200 \cdot \delta$ for each additional year.

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  • $\begingroup$ Great answer, thank you! $\endgroup$ – An economist Jun 21 '19 at 20:32
  • $\begingroup$ +1 for sound application of discipline specific terminology and pointing out the relation between the linear model and the differential equation. You have interpreted the expected difference in price for a unit difference in time as being a linear function of time itself. Clever approach. $\endgroup$ – AdamO Jun 25 '19 at 20:14
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The question boils down to: 1) how do I interpret the coefficients in a log-linear model? And 2) how do I interpret the linear and quadratic terms in a linear model?

For 1. The (exponentiated) coefficient is a relative rate. Log-linear models are usually used model count data, but they amount to using a log-transform and a mean=variance relationship (using a quasilikelihood model is considerably more general). So a $e^\beta = 1.1$ coefficient for a predictor $X$ is interpreted as a 10% difference in (price) comparing groups differing by 1-unit of $X$.

For 2. The intercept has the usual interpretation. The linear term is the instantaneous relative rate when $X=0$. The quadratic term is like an interaction term between $X$ and itself. You can call it a relative rate ratio, i.e. it is the ratio of relative rates comparing groups differing by 1 unit in $X$.

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