Solved – Regression with Quadratic Term – Understanding Marginal Effect

marginal-effectregression

I am working with a regression with an x and x squared predictors. The equation is:

-0.0104x + (-0.00002)x^2.

I understand the marginal effect is calculated by differentiating to:

-0.0104 + 2(-0.00002)x, and that the ME is calculated generally at the mean of x.

However, I was also taught that given an x such as 10, one can simply insert 10 in the equation, so that y hat when x is 10 is -0.0104(10) + (-0.00002)(100).

Doing so, I get -0.106. But the marginal effect at 10 is -0.0108.

So there are two values: -0.106, and -0.0108. What is the interpretation of the first value, given the second value is the marginal effect. Is it correct that the marginal effect represents the slope of the inverse parabola at point x=10?

Best Answer

I suspect that this is just rounding error: The computer program just used more digits to compute the marginal effect than you did.

Alternatively, if your model is non-linear (e.g. a logit or Poisson) then it could be that the marginal effect you got was actualy an average marginal effect instead of a marginal effect at the average. For an average marginal effect, the marginal effect is computed for each individual and than those marginal effects are averaged. For linear models there is no difference between the two, but for non-linear models these marginal effects are (somewhat) different.

Related Solutions

Solved – Interpreting a quadratic logarithmic term

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 – Delta Method for Marginal Effects of Generalized Linear model

I was struggling with this as well and wanted to understand the mathematics behind averaging the Jacobian. Some searching led to this link: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3976195/

What i gather from this is that we want the variance of the Average Marginal Effect (AME) and hence our transformed function is the AME: $$g(\hat{\beta}) = AME = \frac{1}{N} \sum_{i=1}^N g_i(x_i,\hat{\beta}) $$ Then using the delta method we have $$Var \left( g(\hat{\beta}) \right) = J_g' \Omega_{\hat{\beta}} J_g \quad \quad where \ \ \Omega_{\hat{\beta}} = Var(\hat{\beta})$$ and $$J_g' = \frac{\partial\left[\frac{1}{N}\sum_{i=1}^N g (x_i,\hat{\beta})\right]}{\partial \hat\beta} = \frac{1}{N}{\left[\sum_{i=1}^N \frac{\partial \left (g (x_i,\hat{\beta})\right)}{\partial \hat\beta}\right]} $$

Which justifies using the "average Jacobian" in the delta method to calculate variance of the AME.

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Related Solutions

Solved – Interpreting a quadratic logarithmic term

Solved – Delta Method for Marginal Effects of Generalized Linear model

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