# Interpretation of Square of Logarithm-transformed term in linear regression

I am refreshing my understanding of econometrics and have a question regarding interpretation of square of logarithm-transformed term in linear regression. The interpretation of a square term or a logarithm-transformed term is standard. But what does it mean when we have a variable that is first logarithm-transformed and then taken square? For example, suppose we are interested in the relationship between income and household expenditure. We want to predict how income affects household expenditure decision. The LHS of the regression is

    log (yearly household expenditure)


measured by dollars and the RHS variable is

   log(income), (log(income))^2, and a vector of covariates.


Traditional double-log model's coefficient means how unit percentage change in independent variable is associated with percentage change in dependent variable. But do you have any ideas how to interpret regression results from a squared logarithm specification?

• Why would the interpretation be any different than that of the square of any variable in general? – whuber Jul 7 '17 at 19:55
• I might be wrong, but I think double-log model has a percentage interpretation, which means if we have a coefficient of 0.5, it means a 1 percentage increase in income is associated with 0.5 percentage increase in household expenditure. The interpretation is independent of the level of income. If you have any idea of interpretation, would you please let me know? Thanks – macintosh81 Jul 7 '17 at 20:08

It means the increase will vary depending on where the independent variable is. To evaluate the impact of a predictor in a linear regression you take the derivative with respect to the predictor variable, meaning (in the case of a squared predictor) :

$$\widehat{y_i} = \widehat{\beta_0} + \widehat{\beta_1} x_i + \widehat{\beta_2}x_i^2$$ $$\frac{\partial \widehat{y_i}}{\partial {x_i}} = \widehat{\beta_1} + 2\widehat{\beta_2}x_i$$

Which means that an increase in $x_i$ will have "stronger" impact in $\widehat{y_i}$ for more extreme values, since the impact it self depends on the level of $x_i$.

The rationale stays the same when using log variables , since the elasticity of a function, is defined in terms of the LHS of the last equation in the previous chunk, see here. In equation terms:

$$\widehat{\ln{y_i}} = \widehat{\beta_0} + \widehat{\beta_1} \ln{x_i} + \widehat{\beta_2}(\ln{x_i})^2$$ $$\frac{\partial \widehat{\ln{y_i}}}{\partial {\ln{x_i}}} = \widehat{\beta_1} + 2\widehat{\beta_2}\ln{x_i}$$

So, the percent change of $x_i$ impact on a $y_i$ will depend on the current value of $x_i$ (if $x_i$ is high, an increase will have a larger effect on $y_i$ (in % terms for the log case))

• Thanks for the clarification! It makes better sense now. – macintosh81 Jul 7 '17 at 20:57