Interpreting variables in Log-Linear models

[Jaweriah Abdullah]

I have a question regarding interpreting coefficients of independent variables in a Log-Linear model.
I am running a fixed effects regression where my dependent variable is log transformed total consumption expenditure and my main independent variable is the absolute deviation of rainfall below it’s long-term annual mean. I want to interpret my result in percentage terms, however looking at the literature and at other existing posts I am confused.

For example I look at a similar paper (How do droughts impact household food consumption and nutritional intake? A study of rural India).
The author runs regression with log dependent variables and take absolute deviation of rainfall below it’s long-term annual mean as the main independent variable. The results show a coefficient of -0.060 for the rainfall deviations variable (Table 4 of the paper). The author explains this by stating Considering a median dry shock of 0.15 m, total expenditure decreases by around 1 percent.
How is the coefficient converted to percentage terms?

[Dimitriy V. Masterov]

In the log-linear model, the literal interpretation of the estimated coefficient b is that a one-unit increase in X will produce an expected increase in log Y of b units.

In terms of Y itself, this means that the expected value of Y is multiplied by exp(b). So in terms of the effects of changes in X on Y (unlogged), each 1-unit increase in X multiplies the expected value of Y by exp(b).

To compute the effects on Y of another change in X than an increase of one unit, call this change d, we need to include d in the exponent. The effect of a d-unit increase in X is to multiply the expected value of Y by exp(d*b), so exp(.15*0.060) = 1.0090406.

[Dimitriy V. Masterov]

The above calculation is exact. For small values of b, exp(b) ≈ 1+b. We can use this for the following approximation for a quick interpretation of the coefficients: 100 * b is the expected percentage change in Y for a unit increase in X. So, exp(d*b) ≈ d*b, and .060*.15*100 = 0.9%, which is about 1%. The bigger the b, the worse the approximation.

[Jaweriah Abdullah]

Thank you for your reply. I get the concept now.