Units of Measurement and Functional Form

A regression coefficient means nothing without its units and functional form. Learn how rescaling changes the numbers, and how log-level and log-log models turn slopes into percentages and elasticities.

A regression coefficient never means anything on its own; its meaning is fixed entirely by how the variables are measured and what shape the relationship is assumed to take. Two things follow from this. First, rescaling a variable, switching miles to kilometres or pounds to dollars, changes the number you read off the output without changing the underlying relationship at all. Second, by applying logarithms you can change a coefficient from a level effect into a percentage effect or an elasticity, which is often a far more natural way to describe an economic relationship. This article covers both: the mechanical rules for rescaling, and the family of logarithmic functional forms.

Why Units Matter

Consider a regression of travel cost on distance. The slope is the extra cost per extra unit of distance, so it depends jointly on how distance and cost are measured. Change distance from miles to kilometres and the slope changes; change cost from pounds to euros and it changes again. The economic relationship is identical throughout, and only the numeric value of the coefficient moves. Understanding exactly how rescaling propagates through the estimates is what prevents misreading results and lets you compare studies that happen to use different units.

Rescaling the Independent Variable

Suppose you divide the regressor by a positive constant, replacing x with x-star equal to x over c-two. Then the intercept is unchanged and the slope is multiplied by c-two.

x=xc2β~0=β^0,β~1=c2β^1x^* = \frac{x}{c_2} \implies \tilde{\beta}_0 = \hat{\beta}_0, \quad \tilde{\beta}_1 = c_2\hat{\beta}_1

The intuition is direct: a one-unit change in the rescaled variable corresponds to a c-two-unit change in the original, so to produce the same predicted change in y the coefficient must be c-two times larger. The R-squared is untouched, because the proportion of variation explained does not depend on scale.

Take the regression of CEO salary, measured in thousands of dollars, on return on equity measured as a percentage, which comes out as a constant of 963.191 plus a slope of 18.501 on roe. Now express return on equity in decimal form by dividing by 100. The intercept stays at 963.191, because roe of zero and roe-in-decimals of zero are the same point, while the slope becomes 1,850.1, exactly 100 times larger, and the R-squared of 0.0132 does not move. The check confirms it: a one percentage-point rise in roe is the same as a 0.01 rise in the decimal version, and both predict a salary change of 18.501, that is 18,501 dollars.

Rescaling the Dependent Variable

Now suppose you multiply the dependent variable by a positive constant, replacing y with c-one times y. Then both the intercept and the slope are multiplied by c-one, and again R-squared is unchanged.

y=c1yβ~0=c1β^0,β~1=c1β^1y^* = c_1 y \implies \tilde{\beta}_0 = c_1\hat{\beta}_0, \quad \tilde{\beta}_1 = c_1\hat{\beta}_1

Converting the CEO salary from thousands of dollars into dollars, by multiplying by 1,000, turns the constant of 963.191 into 963,191 and the slope of 18.501 into 18,501, both scaled by 1,000, while the R-squared of 0.0132 is unaffected. Both coefficients scale with y because the whole left-hand side has been stretched, but the fit of the line is unchanged by a proportional stretch, which is why R-squared holds steady.

Rescaling Both at Once

Combining the two cases, if y is multiplied by c-one and x by c-two, the slope is multiplied by the ratio c-one over c-two and the intercept only by c-one.

β~1=c1c2β^1β~0=c1β^0\tilde{\beta}_1 = \frac{c_1}{c_2}\hat{\beta}_1 \qquad \tilde{\beta}_0 = c_1\hat{\beta}_0

The slope result drops out of the OLS formula directly. Substituting the rescaled variables, the numerator picks up a factor of c-one times c-two while the denominator picks up c-two squared, because the deviations of x are squared there. One factor of c-two cancels, leaving c-one over c-two times the original slope.

β~1=(c2xic2x)(c1yic1y)(c2xic2x)2=c1c2c22β^1=c1c2β^1\tilde{\beta}_1 = \frac{\sum (c_2 x_i – c_2\bar{x})(c_1 y_i – c_1\bar{y})}{\sum (c_2 x_i – c_2\bar{x})^2} = \frac{c_1 c_2}{c_2^2}\hat{\beta}_1 = \frac{c_1}{c_2}\hat{\beta}_1

The intercept then follows from the usual through-the-means relationship, and the c-two factors cancel cleanly, so only the rescaling of y survives.

β~0=c1yc1c2β^1c2x=c1(yβ^1x)=c1β^0 \tilde{\beta}_0 = c_1\bar{y} – \frac{c_1}{c_2}\hat{\beta}_1 \cdot c_2\bar{x} = c_1(\bar{y} – \hat{\beta}_1\bar{x}) = c_1\hat{\beta}_0

In words, the intercept responds only to rescaling the dependent variable, not the regressor.

Logarithmic Functional Forms

Rescaling keeps the relationship a straight line; logarithms change its shape and, with it, the interpretation of the slope. Taking logs of one or both variables captures proportional rather than absolute relationships, and it converts the slope from a level change into a percentage change or an elasticity. There are four standard combinations. In the level-level model, the slope is a unit-for-unit effect: a one-unit change in x changes y by the slope in its own units. In the log-level model, where only y is logged, one hundred times the slope is approximately the percentage change in y per one-unit change in x. In the level-log model, where only x is logged, the slope over one hundred is approximately the change in y for a one-percent change in x. And in the log-log model, where both are logged, the slope is an elasticity, the percentage change in y for a one-percent change in x.

The Log-Level Model

When y is expected to grow by a roughly constant percentage for each additional unit of x, logging y and regressing on x is the natural choice.

log(wage)=β0+β1,educ+u\log(\text{wage}) = \beta_0 + \beta_1,\text{educ} + u

Here one hundred times the slope is approximately the percentage change in wage per extra year of education. The reason rests on a property of logs: holding the error fixed, the change in log wage equals the slope times the change in education, and because the change in a log is approximately the proportional change in the variable for small movements, one hundred times the slope reads as a percentage change per unit of x.

Estimated on wage data, this gives a constant of 0.584 and a slope of 0.083.

log(wage)^=0.584+0.083,educ\widehat{\log(\text{wage})} = 0.584 + 0.083,\text{educ}

Each additional year of education raises predicted wage by roughly 8.3 percent, the familiar “return to education.” Contrast this with the level-level version, which puts the effect at a flat 0.54 dollars per year regardless of the starting wage. The log-level form is the more realistic of the two if higher-paid workers also gain more in absolute terms from another year of schooling, because a constant percentage of a larger wage is a larger dollar amount.

The Log-Log Model

When both variables are logged, the slope becomes a constant elasticity, the percentage change in y for a one-percent change in x.

log(salary)=β0+β1log(sales)+u\log(\text{salary}) = \beta_0 + \beta_1\log(\text{sales}) + u

Estimated for CEO salary against firm sales, the slope is 0.257.

log(salary)^=4.822+0.257log(sales)\widehat{\log(\text{salary})} = 4.822 + 0.257\log(\text{sales})

So a one-percent increase in sales raises predicted CEO salary by about 0.257 percent, an elasticity of salary with respect to sales. The defining virtue of the log-log model is that, because the slope is a ratio of percentage changes, it is completely unaffected by the units in which either variable is measured. Rescaling salary from dollars to thousands of dollars shifts the intercept but leaves the elasticity exactly where it was.

Why Rescaling Leaves the Log Slope Alone

This unit-invariance can be shown directly. If y is multiplied by c-one and then logged, the log of the product splits into a constant plus the log of y.

log(c1yi)=log(c1)+log(yi)\log(c_1 y_i) = \log(c_1) + \log(y_i)

Regressing this on x, the slope is unchanged and only the intercept moves, by the amount log of c-one.

β~1=β^1β~0=log(c1)+β^0 \tilde{\beta}_1 = \hat{\beta}_1 \qquad \tilde{\beta}_0 = \log(c_1) + \hat{\beta}_0

The slope is unchanged because the added constant log of c-one shifts every observation and the mean by the same amount, so it cancels out of the deviations-from-the-mean that the slope formula is built from. The intercept then absorbs the shift through the through-the-means relationship. This is the formal confirmation that percentage-change and elasticity interpretations are immune to the choice of units, which is a large part of why logs are so widely used.

A workbook with a few exercises here: https://datalad.co.uk/units-of-measurement-and-functional-form-10-exercises-with-full-solutions/

Conclusion

Rescaling a variable changes the numbers in your regression output in entirely predictable ways without altering the underlying relationship. Dividing the regressor by a constant multiplies the slope by that constant and leaves the intercept alone; multiplying the dependent variable by a constant scales both coefficients; doing both scales the slope by the ratio of the two constants. In every case R-squared is untouched, because fit does not depend on scale. Logarithms do something deeper: they reshape the relationship and reinterpret the slope, turning it into a percentage effect in the log-level model and into a unit-free elasticity in the log-log model. That unit-free quality is the practical payoff, because an elasticity means the same thing no matter how anyone chooses to measure the variables, which makes logged specifications both more comparable across studies and often more economically sensible.

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