Elasticity from differenced log-level regression

Question

I have the regression

$ \Delta \ln Y_i = \alpha + \beta \Delta X_i + \varepsilon_i $

where $\Delta \ln Y_i = \ln Y_{t,i} - \ln Y_{t-1,i}$ and $\Delta X_i = ((X_{t,i} - X_{t-1,i}) / T_{t-1,i} ) \cdot 100$. T is the total population and X is a subset, i are regions.

Suppose $\beta$ is -0.073. I interpret the semi-elasticity that a 1 percentage point increase in X relativ to baseline T decreases Y by 7.3%. (Is this correct?)

Question: How can I back out the elasticity? Maybe divide $\beta$ by the mean of $\Delta X_i$?

Answer 1

1. Your interpretation of $\beta$ as a semi-elasticity is correct, though I would say "1 percent increase in X as a proportion of the lagged total" rather than "percentage point."

2. It helps to write down the equation for what you have and what you want and then try to figure out how to transform one into the other. You know that here the semi-elasticity is $$\beta \approx \frac{\frac{Y_t-Y_{t-1}}{Y_{t-1}}}{100 \cdot \frac{X_t-X_{t-1}}{T_{t-1}}}$$ An elasticity is $$\epsilon \approx \frac{\frac{Y_t-Y_{t-1}}{Y_{t-1}}}{\frac{X_t-X_{t-1}}{X_{t-1}}} = \frac{\frac{Y_t-Y_{t-1}}{Y_{t-1}}}{100 \cdot \frac{X_t-X_{t-1}}{T_{t-1}} \cdot \frac{T_{t-1}}{X_{t-1}} \cdot \frac{1}{100}}=\beta \cdot 100 \cdot \frac{X_{t-1}}{T_{t-1}}.$$ This is a function that depends on the ratio $R_{t-1}$ of $X_{t-1}$ to $T_{t-1}$ for each $i$, so really $\epsilon_{i,t}$. You can plot it for $R_{t-1}$ from $0$ to $1$. There are other aggregations that could make sense here. For instance, you can also take the mean over your data to get an average elasticity, which simplifies to $$\epsilon = \beta \cdot 100 \cdot \bar R.$$ Since you are effectively conditioning on $R_{t-1}$ in your model, you have a random variable multiplied by a non-stochastic constant, which makes the variance of the elasticity easy to calculate: $$\mathbf{Var}(\epsilon) = (100 \cdot \bar R)^2 \cdot \mathbf{Var}(\beta)$$ Take the square root to get the standard error of $\epsilon$.