Solved – Interpreting a quadratic logarithmic term

interpretationlogarithmmultiple regressionregressionstata

Given the regression model

$y=\beta_0 + 300*ln(x_1) – 15 * (ln(x_1))^2 + \beta_3*x_2 + … + u $

$x$ ranges from 1 000 to 30 000 (6.9 to 10.3 on a logarithmic scale). How to interpret the diminishing effect that $x_1$ has on $y$? Is it possible to interpret it for small changes in $x_1$ analog to a "normal" quadratic term, i.e. the partial effect can be calculated as

$\Delta y = [\beta_1 – 2*\beta_2 * ln(x1)]$

and the maximum (turning point) as $x_{1} = – \beta_1 / (2*\beta_2)$, which would yield 300 / 30 = 10 with e^10=22026? How to give a meaningful and illustrative interpretation of the effect? For example, the effect of a 1% increase of $x_1$ or going from 20 000 to 21 000?

Edit

Thank you for the code and sorry for the delayed response. I managed to plot
the marginal effect of x1 but decided to use the plot of the actual and predicted values of y against x1 while holding all other variables at their mean using the following Stata code:

reg y ln(x1) (ln(x1))^2 x2 x3 x4 x5
adjust x2 x3 x4 x5 if e(sample), gen(predict)
twoway line predict x1, sort || scatter y x1

I think this illustrates the effect of x1 on y really well and the plot is
easy to understand despite the quadratic logarithmic term in the regression model.

Best Answer

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)'");

Related Solutions

Logistic Regression – Interpreting Logistic Regression with Interaction and Quadratic Terms

The only model that would really make sense is

$$Y = \beta_{0} + \beta_{1}X + \beta_{2}X^{2} + \beta_{3}(Z=b) + \beta_{4}(Z=c) + \beta_{5}X(Z=b) + \beta_{6}X(Z=c) + \beta_{7}X^{2}(Z=b) + \beta_{8}X^{2}(Z=c)$$

where $(Z=k)$ denotes 1 if $Z=k$ and 0 otherwise.

In this model the test for interaction is $H_{0}: \beta_{5}\dots \beta_{8} = 0$.

The $X$-effect depends on $Z$ and on the starting point for $X$ since $X$ is nonlinear. To get the effect of $X$ going from $u$ to $v$ when $Z=k$ write down the special case of the above formula when $X=v, Z=k$ then evaluate it when $X=u, Z=k$ and subtract term by term. What is left is the formula for that $X$ effect at $Z=k$ which you estimate by plugging in the the $\beta$ estimates from the model fit. Anti-log and you have the odds ratio. When $k=a$ the result is $\beta_{2}(v^{2} - u^{2}) + \beta_{1}(v - u)$.

Note that it is not common that $v-u = 1$, i.e., to get a meaningful $X$ effect you might use the quartiles of $X$ and not assume that a 1-unit change is that meaningful for the scale of $X$.

Note also that there is no such thing as the $X$ effect when $Z$ is not set.

Using the R rms package one can get any contrast of interest easily. To get the above use this example for comparing 30 year olds with 20 year olds for the first race group, where race has 3 categories a, b, c:

require(rms)
f <- lrm(y ~ pol(age, 2) * race) # quadratic with interaction
contrast(f, list(age=30, race='a'),
            list(age=20, race='a'))

Solved – Linear regression with logarithmic dependent variable

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 $|\beta_i| < 0.1$.

Edit

Best Answer

Related Solutions

Logistic Regression – Interpreting Logistic Regression with Interaction and Quadratic Terms

Solved – Linear regression with logarithmic dependent variable

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