Why Measurement Units Matter in Power Models
A coefficient in a regression equation is not a unit-free description of a relationship. Its numerical value depends on how the variables are measured, and raising a variable to a power changes that dependence. For example, changing a predictor from meters to centimeters multiplies its numerical values by 100; squaring those values multiplies them by 10,000. A coefficient attached to the squared predictor must therefore change if the fitted relationship is to stay the same.
The key distinction is between a numerical coefficient and the fitted relationship it represents. Changing units can change the coefficient substantially without changing any prediction for the same physical inputs and outputs, provided the coefficients are converted consistently.
The General Coefficient Conversion Rule
Consider a model with a power-transformed outcome and a power-transformed predictor:
Here \(X\) and \(Y\) are numerical measurements in specified units, \(q\) and \(r\) are fixed powers, \(\beta_0\) and \(\beta_1\) are coefficients, and \(\varepsilon\) is an error on the transformed-outcome scale. The powers must be defined for the observed values; in particular, the examples below use positive measurements. The model states that the conditional mean of \(Y^r\), given \(X\), is \(\beta_0+\beta_1X^q\), when \(E(\varepsilon\mid X)=0\). For ordinary least squares, the fitted equation minimizes the sum of squared residuals on the \(Y^r\) scale. Constant error variance is additionally needed for the usual standard-error calculations, but not for the unit-conversion algebra.
Under the new units, \(X_{\text{new}}=c_xX_{\text{old}}\) and \(Y_{\text{new}}=c_yY_{\text{old}}\). Raising these values to their respective powers gives \(X_{\text{new}}^q=c_x^qX_{\text{old}}^q\) and \(Y_{\text{new}}^r=c_y^rY_{\text{old}}^r\). Substituting into the model gives the converted equation and coefficients:
This follows by replacing \(Y_{\text{old}}^r\) with \(Y_{\text{new}}^r/c_y^r\) and \(X_{\text{old}}^q\) with \(X_{\text{new}}^q/c_x^q\), then multiplying the entire equation by \(c_y^r\). The converted error is \(c_y^r\varepsilon\), so the model’s error scale changes with the outcome units as well.
More generally, if the transformed-outcome model contains several power terms, \(Y^r=\beta_0+\sum_j\beta_jX^{q_j}+\varepsilon\), each coefficient changes according to its own predictor power: \(\beta_{j,\text{new}}=c_y^r\beta_{j,\text{old}}/c_x^{q_j}\). The intercept follows the same rule with predictor power zero, so its conversion is \(c_y^r\beta_0\).
Why Fitted Predictions Are Preserved
The conversion rule is not merely a way to adjust coefficient labels. It makes the numerical predictions agree after conversion. If the old fitted transformed value at \(x_{\text{old}}\) is \(\widehat{y_{\text{old}}^r}\), then the new fitted transformed value at \(x_{\text{new}}=c_xx_{\text{old}}\) is:
Thus, when the transformed prediction is converted back to the original outcome scale using the same inverse power, its numerical value changes by \(c_y\), exactly as the outcome units require. For example, a predicted mass of 2 kilograms and a predicted mass of 2,000 grams describe the same quantity. If the outcome transformation is nonlinear, this plug-in inverse prediction should not automatically be interpreted as the conditional arithmetic mean of \(Y\); the conclusion here is about consistent unit conversion, not about bias correction.
The same preservation holds for an ordinary least-squares fit to the transformed outcome. For any candidate set of coefficients, multiplying every transformed response and residual by \(c_y^r\) multiplies the residual sum of squares by \(c_y^{2r}\), a positive constant. The coefficient conversion also gives a one-to-one correspondence between the old and new equations because \(c_x\) and \(c_y\) are positive. Multiplication by a positive constant cannot change which equation minimizes the sum of squares. Therefore, refitting the same model after an exact unit conversion yields the converted coefficients and equivalent fitted predictions.
Worked Example: Squared Predictor in Meters and Centimeters
Suppose a hypothetical model predicts study time \(Y\), measured in hours, from the square of distance \(X\), measured in meters:
Assume this is the chosen mean model on the outcome scale, with residuals having conditional mean zero. Distance is nonnegative, the squared term is well-defined, and the unit change from meters to centimeters is exact. These conditions support the algebraic comparison; the example does not claim to assess whether the model is empirically appropriate.
Let \(X_{\text{cm}}=100X_{\text{m}}\). The outcome remains in hours, so \(c_y=1\); the predictor conversion factor is \(c_x=100\). Here \(r=1\) and \(q=2\). The intercept remains \(2.00\) hours, and the coefficient of the squared predictor becomes:
At a distance of 2.00 meters, the old equation gives \(2.00+0.0800(2.00)^2=2.32\) hours. The same distance is 200 centimeters, and the converted equation gives \(2.00+0.00000800(200)^2=2.32\) hours. The coefficient changed by a factor of 10,000, but the prediction did not.
When the Outcome Is Raised to a Power
If the outcome is transformed, its unit conversion affects every coefficient, including the intercept. With a square-root outcome, for instance, changing the outcome unit by a factor of \(c_y\) multiplies the transformed response and all fitted coefficients by \(\sqrt{c_y}\). Leaving the intercept unchanged would make the converted fitted equation inconsistent.
Worked Example: Square-Root Outcome in Kilograms and Grams
Consider a hypothetical model for a positive mass \(Y\), recorded in kilograms, with predictor \(X\) measured in hours:
Suppose the conditional mean of the square-root mass follows this equation and the transformed errors have conditional mean zero. The masses are positive, so the square root is defined. The predictor unit does not change, while converting kilograms to grams gives \(Y_{\text{g}}=1000Y_{\text{kg}}\).
For this model \(r=1/2\), \(q=1\), \(c_y=1000\), and \(c_x=1\). Both the intercept and slope must be multiplied by \(1000^{1/2}=\sqrt{1000}\):
Numerically, the converted equation is \(\sqrt{Y_{\text{g}}}=37.947+9.4868X+\sqrt{1000}\varepsilon\). At \(X=4.00\) hours, the old equation’s fitted square-root value is \(1.20+0.300(4.00)=2.40\sqrt{\text{kg}}\). The converted equation gives \(37.947+9.4868(4.00)=75.894\sqrt{\text{g}}\), equal to \(2.40\sqrt{1000}\) up to rounding.
Inverting these fitted values gives \(2.40^2=5.76\) kg and \(75.894^2\approx5{,}760\) g. Those are the same plug-in prediction expressed in different units. They are not necessarily estimates of the conditional arithmetic mean mass, since inverse-transforming a fitted transformed mean need not recover that mean.
Changing Units for Both Variables
When both variables change units, the predictor coefficient combines two factors: the outcome conversion multiplies it by \(c_y^r\), while the predictor conversion divides it by \(c_x^q\). A fractional predictor power can therefore produce a less familiar conversion factor, but the same rule applies.
Worked Example: Squared Outcome and Square-Root Predictor
Suppose a hypothetical model for positive mass \(Y\), measured in kilograms, uses positive distance \(X\), measured in kilometers:
Assume the conditional mean of the squared mass has this form and the transformed errors have conditional mean zero. Positivity makes both powers defined. Convert distance to meters and mass to grams, so \(c_x=1000\), \(c_y=1000\), \(q=1/2\), and \(r=2\). The converted intercept is \(1000^2(4.00)=4{,}000{,}000\), while the predictor coefficient is:
The converted equation is \(Y_{\text{g}}^2=4{,}000{,}000+15{,}811.4\sqrt{X_{\text{m}}}+\varepsilon_{\text{new}}\). At \(X=4.00\) km, the old fitted squared value is \(4.00+0.500\sqrt{4.00}=5.00\) kg\(^2\), giving the plug-in prediction \(\sqrt{5.00}\approx2.236\) kg. In the new units, \(X=4{,}000\) m, and the fitted squared value is approximately \(4{,}000{,}000+15{,}811.4\sqrt{4{,}000}=5{,}000{,}000\) g\(^2\). Its square root is approximately \(2{,}236\) g, the same predicted mass in grams.
Units, Interpretation, and a Common Pitfall
Dimensional units provide a useful check on the algebra. In a model \(Y^r=\beta_0+\beta_1X^q+\varepsilon\), the intercept and error have units of \(Y^r\), and \(\beta_1\) has units of \(Y^r/X^q\). For example, in the first worked example, \(0.0800\) has units of hours per square meter. Its converted value has units of hours per square centimeter. The numerical coefficient changes precisely because a square centimeter is a different unit of area from a square meter.
A frequent mistake is to convert the predictor values but leave the coefficient unchanged. In the first example, using \(0.0800\) with \(X=200\) centimeters would give \(2.00+0.0800(200)^2=3{,}202\) hours. That is not a different prediction from a valid model conversion; it is an inconsistent equation, since the coefficient still has units of hours per square meter while the input is now in centimeters.
The simple rule applies to multiplicative changes of units for unshifted powers. It does not mean every transformation changes by a constant factor. For example, adding an offset before a power changes the expression itself when the measurement origin changes, and logarithms do not obey the same power-scaling rule: \(\log(c_yY)=\log(Y)+\log(c_y)\). Always write the unit conversion first and substitute it into the model rather than guessing how the coefficients should change.
Check Your Understanding
Use the conversion factors in each question, and distinguish numerical coefficients from physical predictions.
- A model is \(\widehat{Y}=3+0.020X^2\), with \(X\) measured in meters and \(Y\) in hours. What should the coefficient of \(X^2\) be when \(X\) is measured in centimeters?
- In \(Y^3=\beta_0+\beta_1X+\varepsilon\), the outcome changes to units with numerical values \(Y_{\text{new}}=10Y_{\text{old}}\), while \(X\) is unchanged. By what factor do the intercept and slope change?
- A predictor changes from kilometers to meters in a model containing \(\sqrt{X}\). What factor converts the coefficient of that term if the outcome units are unchanged?
- Explain why an unchanged fitted physical prediction can have a different numerical value after converting the outcome from kilograms to grams.
- Why is multiplying every coefficient by the outcome conversion factor not generally correct when the outcome is squared or square-root transformed?