07: Econometric Modeling
Various functional forms
Transcript
In a log-linear model, only the dependent variable is transformed. Read the first equation as log y sub i equals beta zero plus beta one times x sub i plus u sub i. Here y must be positive because its logarithm must exist, while x remains in its original units.
Look at the calculus line. Differentiate both sides with respect to x sub i. By the chain rule, the derivative of log y with respect to x is one over y times the derivative of y with respect to x. The derivative of beta zero is zero, the derivative of beta one x is beta one, and the usual marginal-effect calculation holds u fixed. That gives one over y times partial y over partial x equals beta one. For a small finite change, we write the proportional change delta y over y as approximately beta one times delta x.
Now translate that result into speech. If x rises by one unit, so delta x equals one, y changes by approximately one hundred times beta one percent. The unit of x still matters: one year of education and one month of education are different one-unit changes. Also notice that the slide’s delta expression is a local approximation. For a one-unit increase in x, the fitted conditional median of y is multiplied exactly by exponential of beta one, so the exact percentage change is one hundred times exponential beta one minus one. The approximation one hundred beta one is very close when beta one is small. This form is useful when percentage changes in y are more meaningful than level changes and when keeping fitted y positive matters. The next tab shows the shape that this equation creates in level space.
Model
\begin{align*} log(y_i)= \beta_0+\beta_1 x_i + u_i \notag \end{align*}Calculus
Differentiating the both sides wrt x_i,
\begin{align*} \frac{1}{y_i}\cdot\frac{\partial y_i}{\partial x_i} = \beta_1 \Rightarrow \frac{\Delta y_i}{y_i} = \beta_1 \Delta x_i \notag \end{align*}Interpretation
\beta_1 measures a percentage change in y_i (once multiplied by 100) when x_i is increased by one unit
Transcript
This figure translates the log-linear equation back into the original units of y. The horizontal axis is x, running from one to two, and the vertical axis is y. The legend labels each curve by the pair beta zero, beta one. The curve associated with two comma one uses y equals exponential of two plus x. The curve associated with four comma negative one uses y equals exponential of four minus x.
Why exponential? Starting from log y equals beta zero plus beta one x, exponentiate both sides. Ignoring the disturbance for the moment, y equals exponential of beta zero plus beta one x. That is why a model that is linear in log y becomes curved when we plot y itself. For the positive beta one curve, y rises as x rises, and the increase becomes steeper because the marginal effect in levels is beta one times y. A larger current y produces a larger absolute change. For the negative beta one curve, y falls as x rises and approaches zero from above.
The graph illustrates two restrictions. First, the sign of beta one determines whether the curve always rises or always falls; this simple form cannot turn around. Second, the fitted level is always positive because an exponential is positive. That can be substantively appealing for wages, prices, or yields, which should not have negative fitted values. A level-on-level straight line has no such protection when extrapolated. Do not confuse positivity with realism, however. The exponential shape is still a strong assumption. Flip to the example tab to see how the percentage interpretation is used for wages.
Transcript
Now apply the general log-linear derivation to a familiar labor-economics model. The first line says log wage equals beta zero plus beta one times education plus u. Wage is the hourly wage in positive level units, and education is measured in years. Beta zero is the fitted log wage at zero years of education, which may not be the quantity we care most about. Beta one is the slope of log wage with respect to education.
The calculus line makes that interpretation explicit. Differentiating log wage with respect to education gives one over wage times partial wage over partial education on the left and beta one on the right. For a small change, delta wage over wage is approximately beta one times delta education. If education rises by one year, delta education equals one, so the proportional wage change is approximately beta one. Multiplying by one hundred converts that proportion to a percentage. Thus we say that one additional year of education is associated with an approximately one hundred beta one percent change in hourly wage, holding the other included regressors fixed.
Use “associated with” unless the assumptions needed for a causal return are justified. Education may be correlated with ability, family background, or other determinants in u, so the functional form alone does not solve omitted-variable bias. The log specification changes the scale and interpretation, not identification. Its percentage effect is nevertheless useful because it is comparable across workers with different wage levels and across settings with different currencies or price levels. For a non-small beta one, remember the exact one-year percentage is one hundred times exponential beta one minus one. On the next tab we estimate the model and compare its curved level prediction with a straight-line wage regression.
Model
\begin{align*} log(wage)=\beta_0 + \beta_1 educ + u \notag \end{align*}Calculus
Differentiating both sides with respect to educ,
\begin{align*} \frac{1}{wage} \frac{\partial wage}{\partial educ} = \beta_1 \Rightarrow \frac{\Delta wage}{wage} = \beta_1\Delta educ\notag \end{align*}Interpretation
If education increases by 1 year (\Delta educ=1), then wage increases by \beta_1*100\% (\frac{\Delta wage}{wage}=\beta_1)
Transcript
The left side gives the estimated log-linear model. The first fitted equation says predicted log wage equals zero point five eight four plus zero point zero eight three times education. The slope means that one additional year of education is associated with approximately an eight point three percent higher wage. Using the exact calculation gives one hundred times exponential of zero point zero eight three minus one, which is about eight point seven percent. The second equation exponentiates the fitted log value, producing exponential of zero point five eight four plus zero point zero eight three education in wage units.
Now look at the figure. The horizontal axis is years of education and the vertical axis is hourly wage, limited on screen to zero through fifteen dollars. Each dot is an observation from the wage data. The red “OLS Regression Line” comes from regressing wage in levels directly on education, so it is straight. The blue “Log Fit” is the exponentiated log-linear prediction, so it curves upward. The two fits can be similar near the center of the education distribution but diverge near the ends because one imposes a constant dollar slope and the other imposes a constant proportional slope. The sparse endpoints also remind you to be cautious about extrapolating either fit.
The callout is essential. If log wage equals its fitted value plus u, exponentiating gives wage equals exponential of the fitted value times exponential u. Simply exponentiating the fitted line omits the conditional mean of exponential u. Under a symmetric log error, the fitted log line corresponds to the conditional median of wage, not generally its conditional mean. Jensen’s inequality gives expected exponential u greater than exponential of expected u, so recovering the mean requires a retransformation factor, such as a distributional correction or a smearing estimate. The slide deliberately labels the plotted blue curve as the exponentiated fit, not an automatically unbiased mean prediction.
When you estimate the following model using the wage dataset:
log(wage)=\beta_0 + \beta_1 educ + u \notag
Then, the estimated equation is the following:
\begin{align*} \widehat{log(wage)}=0.584+0.083 educ \notag \end{align*} \begin{align*} \widehat{wage}=e^{0.584+0.083 educ} \end{align*}
Transcript
In the linear-log model, y remains in its original units and only x is logged. The model line reads y sub i equals beta zero plus beta one times log x sub i plus u sub i. Because log x appears, x must be positive. Beta zero is the fitted value when log x is zero, which means x equals one, not when x equals zero.
Follow the calculus line carefully. The derivative of y with respect to x is the quantity we want. The derivative of beta one log x is beta one divided by x, so partial y over partial x equals beta one over x. For a small finite change, multiply by delta x: delta y is approximately beta one times delta x over x. The fraction delta x over x is a proportional change in x.
This produces a level response to a percentage change. If x rises by one percent, delta x over x is zero point zero one, and y changes by approximately zero point zero one times beta one units of y. Do not say y changes by beta one percent. Y is not logged, so its change is measured in y’s original units. Do not say a one-unit increase in x changes y by beta one either. The one-unit marginal effect is beta one over the current x and therefore varies with x.
That dependence creates diminishing absolute effects when beta one is positive: the same one-unit increase in x matters more when x is small than when x is large. Yet the effect of a fixed percentage increase is constant. This form is useful for variables such as income, acreage, or firm size when proportional changes in x are economically natural. The visualization tab shows the resulting shape and helps distinguish it from a quadratic.
Model
\begin{align*} y_i= \beta_0+\beta_1 log(x_i) +u_i \notag \end{align*}Calculus
Differentiating the both sides wrt x_i,
\begin{align*} \frac{\partial y_i}{\partial x_i} = \frac{\beta_1}{x_i} \Rightarrow \Delta y_i = \beta_1\frac{\Delta x_i}{x_i} \notag \end{align*}Interpretation
When x increases by 1\% (that is, \frac{\Delta x_i}{x_i} = 0.01), y increases by \beta_1 \times 0.01 units.
Transcript
The equation above the plot fixes beta zero at one and beta one at two, so the curve is y equals one plus two times log x. The horizontal axis is x, shown from one to twenty, and the vertical axis is y in level units. At x equal to one, log one is zero, so the curve begins at y equal to one. As x rises, y rises, but the line becomes progressively flatter.
The derivative explains that shape. Partial y over partial x equals two over x. At x equal to one, the slope is two. At x equal to ten, it is zero point two. At x equal to twenty, it is zero point one. Thus equal one-unit increases in x have smaller effects at larger starting values. By contrast, equal percentage increases have the same effect on y because beta one multiplies the proportional change in x.
This graph also shows what the linear-log form cannot do. With positive beta one it is always increasing over positive x, although at a diminishing rate. It never reaches a peak and turns downward. With negative beta one it would always decrease, with the absolute slope becoming smaller as x rises. A quadratic can turn because its marginal effect is linear in x and can cross zero. Choosing between these forms therefore encodes a substantive restriction. Use linear-log if theory suggests diminishing effects that retain the same sign; use a quadratic if a reversal is plausible. Also remember the domain restriction: the log is undefined at zero and for negative x, so this specification cannot be used mechanically for every regressor.
y = \beta_0 + \beta_1 log(x) = 1 + 2 \times log(x)
Transcript
In the log-log model, both the dependent variable and the regressor are transformed. The first equation reads log y sub i equals beta zero plus beta one times log x sub i plus u sub i. Both x and y must therefore be positive.
The calculus line is written as a ratio of proportional derivatives. Differentiating gives one over y times partial y over partial x equals beta one times one over x. Rearranging, the percentage change in y divided by the percentage change in x equals beta one. In finite-change language, delta y over y is approximately beta one times delta x over x. If x rises by one percent, y changes by approximately beta one percent. A negative beta one means the variables move in opposite directions, as in a downward-sloping demand curve; a positive beta one means they move together.
Beta one is therefore an elasticity. It is unit-free, so measuring x in dollars or cents and y in pounds or kilograms does not change it. That makes the specification especially common in demand, supply, and production models. The phrase “constant elasticity” on the slide is a restriction, not merely a convenient interpretation. The model requires the same percentage response at every positive value of x. In level form, ignoring u, y equals exponential beta zero times x raised to beta one, so the absolute slope can change even though the elasticity remains fixed.
Be precise about the approximation. For a small percentage change, beta one times that percentage is accurate. For a large discrete change, use the exact log-change relationship or compare fitted levels rather than treating the differential as exact. If theory says elasticity itself varies with price, income, or scale, this simple log-log model cannot capture that variation without additional terms. We now turn to a quadratic, where the marginal effect is explicitly allowed to change with x.
Model
\begin{align*} log(y_i)= \beta_0+\beta_1 log(x_i) +u_i \notag \end{align*}Calculus
Differentiating the both sides wrt x_i,
\begin{align*} \frac{\partial y_i}{y_i}/\frac{\partial x_i}{x_i} = \beta_1 \Rightarrow \frac{\Delta y_i}{y_i} = \beta_1 \frac{\Delta x_i}{x_i}\notag \end{align*}Interpretation
A percentage change in x would result in a \beta_1 percentage change in y_i (constant elasticity)
Transcript
The quadratic model adds x squared as a second regressor: y sub i equals beta zero plus beta one x sub i plus beta two x sub i squared plus u sub i. It is curved in x but still linear in the parameters because beta one and beta two simply multiply two known columns, x and x squared.
Look at the calculus line. Differentiate term by term with respect to x. Beta zero disappears, beta one x contributes beta one, and beta two x squared contributes two beta two x. The marginal effect is therefore beta one plus two times beta two times x. For a one-unit increase in x, the small-change approximation says y changes by that expression, evaluated at the starting value of x.
This dependence on x is the central feature. There is no single constant effect of x to report. Beta one alone is the marginal effect only at x equal to zero. If zero is outside the observed data or has no meaningful interpretation, treating beta one as “the effect of x” is misleading. Beta two controls how the slope changes: each one-unit increase in x changes the marginal effect by two beta two. A positive beta two makes the slope increase with x; a negative beta two makes it decrease.
Because both coefficients enter the marginal effect, inference at a particular x generally concerns a linear combination of beta one and beta two. The individual t-statistics do not answer whether that combination is zero. We will return to that point in the significance section. First, the next tabs show the range of shapes a quadratic can produce and apply it to returns to education.
Model
y_i= \beta_0 + \beta_1 x_i + \beta_2 x_i^2 + u_i
Calculus
Differentiating the both sides wrt x_i,
\frac{\partial y_i}{\partial x_i} = \beta_1 + 2*\beta_2 x_i\Rightarrow \Delta y_i = (\beta_1 + 2*\beta_2 x_i)\Delta x_i
Interpretation
When x increases by 1 unit (\Delta x_i=1), y increases by \beta_1 + 2*\beta_2 x_i
Transcript
These two panels plot quadratic functions over x from zero to two. In both panels, the horizontal axis is x and the vertical axis is y. The left equation is y equals x plus x squared, so beta one and beta two both equal one. Its marginal effect is one plus two x, which is positive and increasing throughout the plotted range. The curve therefore rises and becomes steeper as x increases.
The right equation is y equals three x minus two x squared. Here beta one is three and beta two is negative two. Its marginal effect is three minus four x. At x equal to zero the slope is three; it falls as x increases, reaches zero at x equal to three quarters, and becomes negative after that. The red curve consequently rises, peaks, and then falls. In general, a quadratic turning point occurs where beta one plus two beta two x equals zero, so x star is minus beta one divided by two beta two. When beta two is negative, that point is a maximum; when beta two is positive, it is a minimum.
This is why the slide calls the form flexible. It can represent increasing returns, diminishing returns, a peak, or a trough. But it always imposes a parabola, and its behavior can become extreme outside the data. Before discussing a turning point, calculate it and compare it with the observed support of x. If it lies outside that support, the estimated relationship may simply be increasing or decreasing over the relevant range. Do not claim that the outcome eventually reverses based on an extrapolated part of the parabola. The education example on the next tab shows why a changing slope can be economically plausible.
Quadratic functional form is quite flexible.
y = x + x^2 (\beta_1 = 1, \beta_2 = 1)
y = 3x-2x^2 (\beta_1 = 3, \beta_2 = -2)
Transcript
Use the three questions to see why a constant return to education may be too restrictive. The outcome of interest is income or wage, and the marginal impact of education means the change in expected income associated with a small increase in years of schooling, holding the other modeled factors fixed.
The first bullet asks about adding two years after elementary school. Those years might move a person toward completing a credential or acquiring foundational skills. The second asks about two years after college, perhaps a master’s degree or professional training. The third asks about adding two years after already spending six years in a doctoral program. Those increments all have the same numerical size, two years, but there is no strong reason to assume that the labor market values them identically. Credentials, field, opportunity cost, and the changing scarcity of skills can all make the return depend on the starting education level.
A model with only one education term forces all three cases to share the same slope. That equality comes from the specification, not from evidence. The observation at the bottom says the marginal impact does not appear to be linear, meaning more precisely that the level relationship may not have a constant marginal effect. Adding education squared is the simplest way to let the slope change smoothly with education. In that model the marginal effect is beta education plus two times beta education squared times education, so we can evaluate the return at elementary-school, college, or doctoral levels separately. This does not guarantee that a quadratic is correct, and it does not establish a causal effect. It is a transparent first relaxation of the constant-slope assumption. The next tab shows how to estimate it in R.
Education impacts on income
The marginal impact of education (the impact of a small change in education on income) may differ depending on the level of education you have had:
How much does it help to have two more years of education when you have had education until elementary school?
How much does it help to have two more years of education when you have graduated a college?
How much does it help to spend two more years as a Ph.D student if you have already spent six years in a Ph.D program
Observation
The marginal impact of education does not seem to be linear.
Transcript
The sentence above the code gives the formula rule to remember. Inside an R model formula, operators can have special formula meanings, so wrap literal arithmetic in capital I. Here I of education caret two tells R to calculate education squared observation by observation and include that numeric column. Without I, the caret is interpreted by the formula language rather than simply as the arithmetic square you intend.
The runnable regression call is fixest colon colon feols. Its first argument specifies wage as the dependent variable and includes female, education, and I of education squared on the right. Female controls for the average level difference associated with the binary indicator. The two education terms together create the quadratic. The vcov argument is set to “hetero,” asking fixest for heteroskedasticity-robust standard errors. The data argument points to wage1, which the preceding data call loads from the wooldridge package. Assigning the result to quad_reg lets later tabs reuse the same fitted model. Finally, broom colon colon tidy converts the model output into one row per coefficient with the estimate, standard error, test statistic, and p-value.
When you run the cell, the fitted education coefficient is about negative zero point four one six and the coefficient on education squared is about positive zero point zero three nine five. Those signs imply a U-shaped fitted parabola, and the marginal effect is negative zero point four one six plus two times zero point zero three nine five times education. Do not interpret the negative coefficient on education alone as education generally reducing wage. It is the slope at education equal to zero. The combination of both coefficients determines the slope at every education level. The setup code above prepares the WebR plotting theme and packages for the browser; it does not change the econometric specification. The next tab plots the fitted relationship so you can judge it against the observations.
When you want to include a variable that is a transformation of an existing variable, you can use I() function in which you write the mathematical expression of the desired transformation.
Transcript
The displayed fitted equation summarizes the quadratic regression: wage equals about five point six zero, minus two point one two times female, minus zero point four one six times education, plus zero point zero three nine times education squared. The intercept is the fitted wage for a man, because female equals zero, at zero years of education. The female coefficient shifts the fitted curve downward when female equals one. The final two terms determine how the curve changes with schooling.
Now read the plotting code and figure. The sequence x runs from the minimum to the maximum education observed in wage1. The new data table sets education equal to each value in that sequence and female equal to one, so the plotted line is specifically the fitted relationship for women. Predict with newdata calculates the model’s fitted wage at each of those combinations. In the graph, the horizontal axis is education in years and the vertical axis is hourly wage. The points are the observed wage and education pairs for everyone in the sample, while the line is the female-equals-one prediction. Because the points include both women and men but the line is for women, use the scatter mainly to see support and overall fit, not as a within-group residual display.
The line falls slightly at very low education and then rises increasingly steeply. Its turning point solves negative zero point four one six plus two times zero point zero three nine five times education equals zero, which is about five point three years. Very few observations lie at the lowest schooling levels, so the apparent decline rests on sparse support and the imposed parabolic shape. The safe conclusion is that the fitted marginal effect becomes more positive as education increases over the well-populated part of the sample. Always overlay or otherwise inspect the data before interpreting a quadratic’s turning point, especially near a boundary.
Estimated Model
wage = 5.60 - 2.12\times female -0.416\times educ + 0.039\times educ^2
Transcript
Start with the fitted equation from the previous tab and differentiate it with respect to education. The constant and female terms disappear. The derivative of negative zero point four one six times education is negative zero point four one six, and the derivative of zero point zero three nine times education squared is two times zero point zero three nine times education. That gives the displayed marginal effect: negative zero point four one six plus zero point zero seven eight times education, measured in dollars of hourly wage per additional year of schooling.
Evaluate the expression at the two values shown. At education equal to four, the fitted marginal effect is negative zero point four one six plus zero point zero seven eight times four, about negative zero point one zero four dollars per hour for an additional year. At education equal to ten, it is negative zero point four one six plus zero point zero seven eight times ten, about positive zero point three six four dollars per hour. The sign changes because the quadratic slope rises with education. These are local marginal effects, not the exact fitted change from four to five or ten to eleven; for a discrete one-year change, the squared term also contributes beta education squared times one squared. The derivative is the standard marginal-effect summary.
The next issue is uncertainty. The marginal effect at a chosen education level combines the estimated linear and squared coefficients. Its variance includes the variance of each estimate and their covariance. Neither individual t-statistic answers whether this combination differs from zero. The education t-test concerns the slope only at zero education, and the squared-term t-test concerns curvature. We need a test of one linear restriction at each education level. That is the purpose of the next section.
According to the estimated model, the marginal impact of educ is:
\frac{\partial wage}{\partial educ} = -0.416+0.039\times 2\times educ
When educ = 4, additional year of education is going to increase hourly wage by -0.104 on average
When educ = 10, additional year of education is going to increase hourly wage by 0.364 on average
Interaction terms
Transcript
An interaction term is a new regressor formed by multiplying two variables. The example is education times experience. Including this product allows the marginal effect of one variable to depend on the level of the other, which a purely additive model cannot do.
Read the displayed wage model carefully. It contains an intercept, experience by itself, and education times experience. Differentiate with respect to experience and you will obtain beta one plus beta two times education, so the return to another year of experience changes with schooling. Equivalently, differentiating with respect to education gives beta two times experience, so the association with schooling changes with experience. Interaction effects are symmetric in that algebraic sense even if the research question emphasizes one direction.
The important callout warns that this is a deliberately stripped-down example. Education does not enter on its own. That omission forces education to affect wage only through experience. When experience equals zero, the fitted marginal effect of education must be exactly zero. It also makes beta one the experience slope at education equal to zero. Those are strong restrictions chosen for a short derivation, not general modeling advice.
The rule of thumb in blue is the hierarchy principle: when you include x one times x two, ordinarily include x one and x two as separate main effects too. Then each main-effect coefficient describes a slope when the other variable is zero, and the interaction coefficient describes how that slope changes. If zero is not meaningful, centering a variable can make the main effects easier to interpret without changing fitted values. Interactions are not merely extra controls. They answer questions about heterogeneity, such as whether experience is more valuable at higher education levels. The next tab derives that marginal effect explicitly.
A variable that is a multiplication of two variables
Example
educ\times exper
A model with an interaction term
wage = \beta_0 + \beta_1 exper + \beta_2 educ \times exper + u
Transcript
Start from the stripped-down model on the previous tab: wage equals beta zero plus beta one times experience plus beta two times education times experience plus u. Differentiate with respect to experience. Beta zero drops out, beta one experience contributes beta one, and beta two education experience contributes beta two times education. The marginal impact is therefore beta one plus beta two education, exactly as displayed.
The first implication follows immediately: the experience slope is not constant. Evaluate it at a particular schooling level. Beta one is the marginal effect of experience when education equals zero. That is the reference value built into the parameterization, not an overall experience effect. If nobody in the relevant population has zero education, beta one alone is an extrapolated slope and may have little substantive meaning.
Beta two tells you how much the experience slope changes when education increases by one year. Its units are dollars of hourly wage per year of experience per year of education. If beta two is positive, another year of experience is associated with a larger wage increment for people with more schooling, so the two inputs reinforce each other in the fitted model. If beta two is negative, the experience return declines with schooling.
Remember the main-effect warning. Because this teaching model omits education by itself, it also says the marginal impact of education is beta two times experience and must be zero when experience is zero. A practical hierarchical specification would usually contain female, education, experience, and their product. The interaction coefficient would still measure how the two slopes change, while the separate education term would avoid that unintended zero restriction. On the R tab we deliberately estimate the stripped-down form so its coefficients match this algebra.
Marginal impact of experience:
\frac{\partial wage}{\partial exper} = \beta_1+\beta_2\times educ
Implications
The marginal impact of experience depends on education
\beta_1: the marginal impact of experience when educ=0
if \beta_2>0: additional year of experience is worth more when you have more years of education
Transcript
The code implements the stripped-down interaction model exactly. Fixest colon colon feols estimates wage on female, experience, and the literal product of experience and education. Female is included as a level control. Experience is the main effect shown in the algebra. Capital I around experience times education tells the formula parser to perform arithmetic multiplication and create only that product column. Without I, an asterisk in an R formula expands to both main effects plus their interaction.
The assignment stores the fitted model as reg_int so later tabs can extract coefficients, make predictions, and test restrictions. There is no vcov argument in this call, so fixest reports its default IID standard errors rather than the heteroskedasticity-robust standard errors used in the quadratic example. The surrounding parentheses print the model immediately after assigning it. Autorun evaluates the cell automatically.
When run on wage1, the estimates are approximately six point one two for the intercept, negative two point four one eight for female, negative zero point one eight eight for experience, and positive zero point zero two zero four for experience times education. The interaction estimate means each additional year of education raises the fitted marginal effect of a year of experience by about two cents per hour. The negative experience coefficient is the slope at education equal to zero, so it should not be described as the experience effect for a typical worker.
For your own work, the more standard fixest formula would use education star experience, which expands to education plus experience plus their interaction. Here we retain I of the product because the previous slide intentionally omitted the education main effect. The hidden knitr chunk fits the identical model for the static figures on the next tab; it does not add another specification.
Just like the quadratic case with educ^2, you can use I().
Transcript
The equation at the top reports the fitted model from the previous tab. Wage equals six point one two one, minus two point four one eight times female, minus zero point one eight eight times experience, plus zero point zero two zero times education times experience. Differentiating with respect to experience gives the displayed marginal effect, negative zero point one eight eight plus zero point zero two zero times education.
The code creates one thousand education values from the sample minimum to maximum and evaluates that expression at each value. In the left panel, the horizontal axis is education in years and the vertical axis is the marginal impact of experience, measured as dollars of hourly wage per additional year of experience. The horizontal reference line marks zero. The fitted effect begins negative at low education, increases linearly because the interaction coefficient is positive, and crosses zero at roughly zero point one eight eight divided by zero point zero two zero, about nine point two years of education. Above that value, the fitted experience slope is positive.
The right panel is a histogram of education. Its horizontal axis is again years of education and its vertical axis is the count of sample observations in each bin. Read the two panels together. The effect curve tells you what the model implies at each education value; the histogram tells you how much data support those evaluations. An effect near an education level with very few observations depends heavily on functional-form extrapolation.
The plot shows point estimates only. It does not include standard errors or a confidence band, so crossing the zero line does not by itself identify where the effect becomes statistically significant. For that we need the variance of the combination beta experience plus education times beta interaction. The next tabs construct a test at a particular education level.
Estimated Model
wage = 6.121 - 2.418 \times female - 0.188 \times exper + 0.020 \times educ \times exper
Marginal impact of experience
\frac{\partial wage}{\partial exper} = - 0.188 + 0.020 \times educ
Marginal impact of exper:
Histogram of education:
Transcript
This slide applies the quadratic-section logic to an interaction. Because the marginal impact of experience is negative zero point one eight eight plus zero point zero two zero times education, it varies with schooling. Substituting ten years gives about zero point zero one two dollars per hour for an additional year of experience using the rounded coefficients on screen. Substituting fifteen gives about zero point one one two dollars. These calculations are fitted marginal effects, not significance tests.
The research question below selects education equal to ten and asks whether the experience effect is positive or negative and statistically different from zero. In population notation, the null is beta experience plus ten times beta experience-education equals zero. The alternative says that combination is not zero. The interaction coefficient’s label stands for the coefficient on experience times education.
This is one linear restriction involving two population coefficients. A t-test of the combination is appropriate, or an equivalent one-restriction F-type test. Its numerator is beta hat experience plus ten beta hat interaction. Its variance is variance of beta hat experience, plus one hundred times variance of beta hat interaction, plus twenty times their covariance. Testing beta experience alone would ask about the experience slope at zero education. Testing the interaction alone would ask whether the experience slope changes with education. Neither is the question posed here.
The same framework can test fifteen years by replacing ten with fifteen. Because the estimated effect is larger there, the test may differ, but significance also depends on the combination’s standard error and the data available at that education level. The next tab asks R for the ten-year restriction.
Just like the case of the quadratic specification of education, marginal impact of experience is not constant
We can test if the marginal impact of experience is statistically significant for a given level of education
- When educ=10, \frac{\partial wage}{\partial exper} = - 0.188 + 0.020 \times 10=0.012
- When educ=15, \frac{\partial wage}{\partial exper} = - 0.188 + 0.020 \times 15=0.112
Question
Does an additional year of experience have a statistically significant impact (positive or negative) if your current education level is 10
Hypothesis
H_0: \beta_{exper} + \beta_{exper\_educ} \times 10=0
H_1: \beta_{exper} + \beta_{exper\_educ} \times 10\ne 0
Transcript
The code passes reg_int and one restriction string to car colon colon linearHypothesis. The string says the experience coefficient plus ten times the coefficient named I of experience times education equals zero. That is exactly the null from the previous tab at education equal to ten. Coefficient names inside the string must match the fitted model’s labels, including the capital I and the product term. If parsing fails, inspect names of coef reg_int and copy the stored label.
Using the unrounded estimates, the marginal effect at ten is about negative zero point one eight seven six plus ten times zero point zero two zero three seven, or positive zero point zero one six one dollars per hour. The test output gives a p-value of about zero point one one seven. At the five percent level, we therefore fail to reject the null that the experience effect is zero for a worker with ten years of education. The point estimate is positive, but it is not estimated precisely enough to distinguish from zero at that conventional level.
Do not turn “fail to reject” into proof of no effect. It means the sample and model do not provide sufficiently strong evidence against zero for this particular linear combination. It also does not settle the effect at fifteen years, where the combination and its standard error are different. The broader lesson is that whenever an interaction makes a marginal effect conditional on another variable, report both the conditioning value and the appropriate uncertainty for the full combination.
Including qualitative information
Transcript
So far our explanatory variables have naturally been numerical, such as education and experience. Many economically relevant characteristics are qualitative categories rather than measured amounts. The examples on screen include sex, marital status, and highest degree. Other examples would be region, crop type, treatment status, or occupation.
A regression still needs numerical columns, so we represent categories with indicator variables. For a binary characteristic such as married versus single, one zero-one variable contains all the information: code one for the category named by the variable and zero for the other category. Its coefficient compares those two groups, conditional on the other regressors.
For a characteristic with more than two states, such as high school, college, master’s, or doctoral degree, we need several indicators. With an intercept and c categories, include c minus one indicators. The omitted category becomes the baseline, and each included indicator coefficient measures a difference relative to it. Including all c indicators with an intercept creates exact redundancy because the indicators sum to one.
Coding does not create causality. A coefficient on a category describes a conditional mean difference after controlling for included variables, subject to the model assumptions. It should not automatically be labeled the effect of belonging to that category. We begin with binary variables, derive the comparison algebraically, and then extend the same idea to multiple categories and group-specific slopes.
Issue
How do we include qualitative information as an independent variable?
Examples
male or female (binary)
married or single (binary)
high-school, college, masters, or Ph.D (more than two states)
Transcript
The callout defines a dummy variable, also called an indicator or binary variable. It takes exactly two numerical values. One marks the category named by the variable and zero marks the other state. For example, in this dataset female equals one for women and zero for men; married equals one for married respondents and zero otherwise. The choice of which state receives one determines the sign and wording of the comparison, not the model’s fitted values.
The code beneath the definition uses dplyr colon colon select to keep six columns from wage1: wage, education, experience, female, and married, along with the listed variables, then pipes the result to head. Head displays the first six rows by default. Look across a row to see that continuous variables retain their measured values while female and married appear as zeros and ones. The pipe passes the selected table into head, making this a quick inspection rather than a transformation saved back to the data.
Why is zero-one coding convenient? In a regression, multiplying a coefficient by zero removes that category shift, while multiplying by one adds it. The conditional mean for the group coded one is therefore the baseline conditional mean plus the dummy coefficient. Subtracting the group-zero mean leaves exactly that coefficient. The next tab carries out this substitution for female and shows how “holding education fixed” enters the interpretation.
Example
Transcript
The model is wage equals beta zero plus sigma f times female plus beta two times education plus u. Under the zero conditional mean assumption, the conditional expectation drops u. To interpret sigma f, evaluate the equation for each value of the dummy while keeping education at the same value.
For female equal to one, the conditional expected wage is beta zero plus sigma f plus beta two education. For female equal to zero, the sigma term disappears and expected wage is beta zero plus beta two education. Because zero identifies men in this coding, men are the baseline group. Subtract the second expression from the first. Beta zero and beta two education cancel, leaving sigma f. That is the equation displayed in the middle of the slide.
Verbally, sigma f is the expected hourly-wage difference between women and men who have the same education in this model. If sigma f is negative, women have the lower fitted conditional mean by its absolute value; if positive, they have the higher one. The comparison is conditional because education is held fixed. It is not the raw difference between average female and male wages, which could also reflect different education distributions.
Notice another restriction. The education coefficient beta two is identical for both groups, so this model permits different intercepts but forces parallel wage-education lines. Also avoid automatically interpreting sigma f as discrimination or a causal effect. Unobserved experience, occupation, hours, and other wage determinants remain in u and may differ by group. The next tab estimates this conditional mean difference, and the visualization after it makes the parallel-line restriction visible.
Model
wage = \beta_0 +\sigma_f female +\beta_2 educ + u
Interpretation
female: E[wage|female=1,educ] = \beta_0 + \sigma_f +\beta_2 educmale: E[wage|female=0,educ] = \beta_0 + \beta_2 educ
This means that
\sigma_f = E[wage|female=1,educ]-E[wage|female=0,educ]
Verbally,
\sigma_f is the difference in the expected wage conditional on education between female and male
\sigma_f measures how much more (less) female workers make compared to male workers ( baseline ) if they were to have the same education level
Transcript
The code estimates the dummy-variable model with fixest colon colon feols. Wage is the dependent variable, and female plus education are the regressors. The data argument uses wage1. The result is stored as reg_df and then printed. Because no vcov option appears, the output uses fixest’s default IID standard errors.
The fitted female coefficient is about negative two point two seven three. Since female equals one for women and zero for men, read it as follows: among observations with the same education, the fitted hourly wage for women is about two dollars and twenty-seven cents lower than for men, on average. The inline R expression below the output pulls that exact coefficient from reg_df by its name, so the prose stays synchronized with the estimate rather than hardcoding a number.
The education coefficient is about zero point five zero six dollars per hour per additional year, and the intercept is about zero point six two three. Those terms define the baseline male line. Adding negative two point two seven three shifts that entire line downward for women. Because there is no female-times-education interaction, both groups share the same education slope.
Use conditional-association language. This regression controls only for education. It does not isolate discrimination, and it does not establish what the same person would earn under a different sex classification. Experience, occupation, tenure, work patterns, and many other factors are omitted and may be correlated with female. The coefficient is still a useful demonstration of dummy-variable interpretation, but the research design determines whether it has a causal meaning.
R implementation
Interpretation
Female workers make -2.2733619 ($/hour) less than male workers on average even though they have the same education level.
Transcript
This figure draws the two fitted regression lines implied by the dummy-variable model. The horizontal axis is education in years and the vertical axis is hourly wage in dollars. The legend distinguishes female and male. The male line is beta zero plus beta two times education because female equals zero. The female line is beta zero plus sigma f plus beta two times education because female equals one. The text labels on the plot write those two equations next to their lines.
Both lines have slope beta two, so they are parallel. Their vertical separation at every education value is sigma f. In this fitted model sigma f is negative, so the female line lies below the male line by about two point two seven dollars per hour everywhere. The code constructs education values from zero to twenty, evaluates each equation, combines the two data tables, and maps line color to group. The vertical and horizontal reference lines mark zero on the axes; they are coordinate guides, not estimated effects.
The picture reveals the restriction more clearly than the table. Adding a dummy by itself allows the intercept to differ across groups but does not allow the education slope to differ. If the lines actually converge, diverge, or cross, this model cannot represent that behavior. You would need a female-times-education interaction, which adds gamma to the female slope. Also notice that the plotted lines extend to zero education, even if that region has limited sample support. Interpret fitted differences primarily where the data exist. We first examine how changing the dummy coding affects the same parallel-line model.
Transcript
Now recode the same two categories with male equal to one and female equal to zero. The model is wage equals beta zero plus sigma m times male plus beta two education plus u. The algebra is identical, but the baseline changes.
For male equal to one, expected wage is beta zero plus sigma m plus beta two education. For male equal to zero, which now means female, expected wage is beta zero plus beta two education. Subtracting the female expression from the male expression leaves sigma m. Thus sigma m is the conditional expected wage difference, male minus female, at the same education level.
Compare this with the earlier female coding. Sigma f represented female minus male, so sigma m must be the negative of sigma f. The new intercept is the old intercept plus sigma f because the baseline line has switched from men to women. The education slope and every fitted wage remain unchanged. Coding changes the parameter labels and signs, not the substantive model.
The callout gives the general rule: whichever category receives zero is the baseline. State that baseline whenever you report a categorical coefficient. There is no universally correct choice. Pick a reference group that makes the comparisons meaningful or conventional for the application. The next tab verifies the sign and intercept changes in R.
Model
wage = \beta_0 +\sigma_m male +\beta_2 educ + u
Interpretation
male: E[wage|male = 1,educ] = \beta_0 + \sigma_m +\beta_2 educfemale: E[wage|male = 0,educ] = \beta_0 + \beta_2 educ
This means that
\sigma_m = E[wage|male=1,educ]-E[wage|male=0,educ]
Verbally,
\sigma_m is the difference in the expected wage conditional on education between male and female
\sigma_m measures how much more (less) male workers make compared to female workers (baseline) if they were to have the same education level
Transcript
The first code line creates male as one minus female. Because female is zero or one, this reverses the coding exactly: a female value of zero becomes male equal to one, and a female value of one becomes male equal to zero. Dplyr mutate returns the updated wage1 data with the new column.
The feols call then regresses wage on male and education. Wage1 is supplied through the data argument, and the fitted object is again named reg_df. The output shows a male coefficient of about positive two point two seven three, exactly the opposite sign of the earlier female coefficient. The intercept is about negative one point six five one rather than positive zero point six two three because it now describes the fitted female wage at zero education. The education coefficient remains about zero point five zero six.
The inline interpretation extracts the male coefficient by name. With women as the zero-coded baseline, men have a fitted hourly wage about two dollars and twenty-seven cents higher at the same education level. This says the same thing as women having a fitted wage two dollars and twenty-seven cents lower than men.
You can verify prediction invariance algebraically. For a man, the new equation adds the male coefficient to the new intercept, recovering the old male intercept. For a woman, it leaves the new intercept, which equals the old intercept plus the old female shift. Nothing about residuals, fitted values, R squared, or substantive conclusions changes. Only the reference category and coefficient parameterization change.
Regression results
Interpretation
Male workers make NA ($/hour) more than female workers on average even though they have the same education level.
Transcript
Before opening the answer, write the identity connecting these columns. For every observation, male plus female equals one. A regression intercept is itself a column of ones. Therefore the intercept column equals male plus female exactly.
If the model includes an intercept, male, and female, one regressor column is a perfect linear combination of the others. This is perfect multicollinearity. The design matrix does not have full column rank, so the normal equations cannot identify a unique coefficient on all three columns. Many coefficient triples produce the same fitted values because you can shift the intercept and offset both dummy coefficients without changing a prediction.
The answer panel lists the consequences. Male and female contain redundant information once the intercept is present. To estimate the model, drop either male or female and interpret the remaining indicator relative to the omitted group. This is the dummy-variable trap. More generally, with c mutually exclusive and exhaustive categories and an intercept, include c minus one category indicators.
An alternative is to omit the intercept and include all category indicators. Then each category coefficient can represent its own intercept, but the interpretation and other formula terms must be handled carefully. The standard approach in this course is to keep the intercept and omit one baseline category. The next tab lets you see how software handles the redundant specification.
What do you think will happen if we include both male and female dummy variables?
Answer
They contain redundant information
Indeed, including both of them along with the intercept would cause perfect collinearity problem
So, you need to drop either one of them
Transcript
The equation above the code states the exact dependence: intercept equals male plus female. The feols formula nevertheless requests wage on male, female, and education. Because those columns are perfectly collinear, all requested coefficients cannot be estimated uniquely.
When this cell runs, fixest detects the collinearity and removes female. It estimates the remaining model with an intercept, male, and education, then prints a message that one variable was removed because of collinearity. The reported coefficients match the male-baseline-choice regression from the previous tab because that is the estimable parameterization fixest retained.
The automatic removal prevents a matrix-inversion failure, but it should not be treated as permission to ignore the problem. The software’s choice of which variable to drop may depend on term order and implementation. If you overlook the message, you may interpret a baseline you did not intend or believe that a requested coefficient was estimated when it was not. Always inspect the coefficient table, removal notes, and stored collinearity information.
The proper remedy is to specify the reference category deliberately. Include either female or male, not both, when retaining the intercept. For a multi-category variable, use c minus one indicators or a factor-aware helper that explicitly names the reference. This preserves full rank and makes the comparison clear in the code and writeup. The next section returns to female coding and adds an interaction so the groups can differ in slope as well as intercept.
In the model, intercept = male + female, which causes perfect collinearity.
Here is what happens if you include both:
One of the variables that cause perfect collinearity is automatically dropped.
Interactions with a dummy variable
Transcript
The first bullet recalls the restriction in the dummy-only model: women and men could have different fitted wage levels, but education had exactly the same slope for both. The question here is whether we can allow the marginal association between education and wage to differ by group. We can, by interacting the female indicator with education while retaining both main effects.
Read the flexible model term by term. Beta zero is the male intercept because male is the female-equals-zero baseline. Sigma f times female allows the female intercept to differ. Beta two times education is the male education slope. Gamma times female times education allows that slope to change for women.
Substitute female equal to zero. Both terms containing female disappear, leaving expected wage for men equal to beta zero plus beta two education. Substitute female equal to one. The intercept becomes beta zero plus sigma f, and the two education terms combine into beta two plus gamma times education. Thus the model gives each group its own intercept and slope.
Gamma is the female education slope minus the male education slope. If gamma is positive, the fitted return to another year of schooling is larger for women; if negative, it is smaller. Sigma f now has a narrower interpretation than in the parallel-lines model: it is the female-minus-male wage difference at education equal to zero. At education level e, the fitted group gap is sigma f plus gamma times e, so it can widen, narrow, or cross zero.
Because zero education may not be a useful reference, you could center education at a meaningful value before interacting. Then sigma f would describe the group gap at that reference level while fitted values remain unchanged. The next tab estimates this full hierarchical specification.
In the previous example, the impact of education on wage was modeled to be exactly the same
Can we build a more flexible model that allows us to estimate the differential impacts of education on wage between male and female?
A more flexible model
wage = \beta_0 + \sigma_f female +\beta_2 educ + \gamma female\times educ + u
female: E[wage|female=1,educ] = \beta_0 + \sigma_f +(\beta_2+\gamma) educmale: E[wage|female=0,educ] = \beta_0 + \beta_2 educ
Interpretation
For female, education is more effective by \gamma than it is for male.
Transcript
The feols formula includes female, education, and I of female times education. Including all three respects the hierarchy rule: the two main effects remain alongside their product. Capital I tells the formula parser to multiply the two columns literally. The data argument uses wage1, and the surrounding parentheses cause the fitted model, stored as reg_di, to print immediately.
The output gives an intercept of about zero point two zero zero, a female coefficient of about negative one point one nine nine, an education coefficient of about zero point five three nine, and an interaction coefficient of about negative zero point zero eight six. No vcov argument is supplied, so these are paired with the default IID standard errors.
Interpret the slopes by substituting dummy values. For men, female equals zero, so the marginal effect of education is beta education, about fifty-four cents of hourly wage per additional year. For women, female equals one, so the marginal effect is beta education plus beta interaction, about zero point five three nine minus zero point zero eight six, or roughly forty-five cents. The interaction estimate is the difference between those slopes: the fitted education slope is about eight point six cents lower for women.
The female main-effect coefficient is not an overall group gap. In this interacted model it is the fitted female-minus-male difference at zero years of education. At education e, the gap is negative one point one nine nine minus zero point zero eight six times e. Also check uncertainty. The interaction t-statistic is about negative zero point eight three with p about zero point four zero seven, so this sample does not provide evidence that the slopes differ from zero at conventional levels. The next tab visualizes the point estimates and makes the changing group gap visible.
Transcript
The statement above the graph interprets gamma, the coefficient on female times education. Its estimate is negative zero point zero eight six, so the fitted marginal benefit of one more year of education is about eight point six cents per hour lower for women than for men. This is a difference in slopes, not the wage gap at a particular schooling level.
The graph has education in years on the horizontal axis and hourly wage in dollars on the vertical axis. The male line uses beta zero plus beta two education. The female line uses beta zero plus sigma f plus beta two plus gamma times education. The labels printed beside the lines show those equations, and the color legend identifies the groups. The vertical and horizontal lines mark zero on the axes.
Unlike the earlier dummy-only picture, these lines are not parallel. The male slope is about zero point five three nine, while the female slope is about zero point four five three. Because gamma is negative, the female line rises more slowly. Sigma f is also negative, so the female line starts below the male line at zero education and the fitted gap becomes more negative as education rises. At any education e, that gap is sigma f plus gamma e.
The figure displays fitted point estimates and does not show confidence bands. The interaction coefficient’s p-value is about zero point four zero seven, so we cannot statistically distinguish the estimated slope difference from zero at conventional levels. That uncertainty should accompany the point interpretation. Since gamma is one coefficient in this parameterization, its printed t-test directly addresses equality of the two education slopes. Testing whether the groups have identical regression lines altogether would require jointly testing both sigma f and gamma.
The marginal benefit of education is 0.086 ($/hour) less for females workers than for male workers on average.
