Solved – Linear regression with logarithmic dependent variable

interpretationlogarithmregression

I have the following regression:

$$\ln(1 + \text{hours}) = \beta_0 + \beta_1 \text{ educ} + \beta_{2-x} \text{ other independent variables} + \varepsilon$$

$\beta_2$ for educ is $0.0642$. How would one interpret this coefficient?

Best Answer

It's fairly straight forward. This is a log-linear model. A one-unit change in $x_i$ is associated with a $100 * \beta_i$ percent change in $y$.

$100\% * 0.0642 = 6.42\% $

We can now interpret it in percentages. For each one unit increase in educ, there is a 6.42% increase in your outcome variable.

ETA in light of whuber's comment:

To be more precise, the relationship between the percent change in $y$ and change in $x$ is:

$100\% * (e^{\beta_i} - 1)$

So,

$100\% * (e^{0.0642} -1) = 6.630564\%$

A whuber states, this is important as the absolute value of $\beta_i$ grows larger than 0.1. The simple $\beta_i * 100$ rule serves as an easy approximation when $|\beta_i| < 0.1$.

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Isn't the derivative that you want actually $\frac{dy}{dx_{1}}=\frac{300}{x_{1}}-\frac{30}{x_{1}}\ln x_{1}$? Your derivative is the change in $y$ for a small change in $\ln x_{1}$. I think it's probably easier to think about about changing $x_{1}$ on the original (non-logged) scale. That shows that the marginal effect of $x_{1}$ starts out positive for small $x_1$ and declines steeply until $x_{1}=e^{10}$ and is negative for $x_{1} \gt e^{10}$.

Here's some Stata code to make the graphs you asked about below. There might be a more elegant way to do the labels on the original scale, but that's all I got:

#delimit;
clear all;
set more off;

sysuse auto, clear;

gen lmpg=ln(mpg);

reg price c.lmpg##c.lmpg;

/* yhat at x of 18, 20 & 25 */
margins, at(lmpg=(`=ln(18)' `=ln(20)' `=ln(25)'));

sum lmpg;
local mean = r(mean);
local sd = r(sd);

/* yhat at mean of ln(x) and at mean plus 1 sd */
margins, at(lmpg=(`mean' `=`mean'+`sd''));
marginsplot, xlabels(`mean' "`=round(exp(`mean'),,01)'" `=`mean'+`sd'' "`=round(exp(`mean'+`sd'),.01)'");  


/* The MARGINAL effect of x on y at mean and mean of ln(x) plus 1sd */
margins, dydx(lmpg) at(lmpg=(`mean' `=`mean'+`sd''));
marginsplot, xlabels(`mean' "`=round(exp(`mean'),.01)'" `=`mean'+`sd'' "`=round(exp(`mean'+`sd'),.01)'");

Solved – Adding a quadratic variable to regression

Often the relationship between y and x is nonlinear. There are a variety of solutions. One solution is to add polynomial terms and the first one to look at is usually $x^2$. But you should first look at a scatterplot of x and y; you should also look at the residuals from the linear model without the quadratic term. But it turns out that many relationships are pretty well fit by $y \sim b_0 + b_1x + b_2x^2$ (plus any other x variables, of course).

It is also possible to add cubic, quartic and even higher order terms, but such models quickly become hard to interpret. Another possibility is to look at a spline regression.

I don't really understand your last bit about endogeneity. A variable can be endogenous or exogenous to a model, but that doesn't seem to relate to this ... at least, not in a way that's clear to me.

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