Rescaling the Outcome and Its Coefficients

Why Outcome Units Matter

A regression equation is written using particular units. A model for length in meters and a model for the same length in centimeters may describe the same relationship, but their numerical coefficients will not look alike. To interpret or compare a model correctly, we need to know how the outcome’s unit conversion changes each part of the equation.

This tutorial considers a change of outcome units by multiplication: the new numerical outcome equals the old numerical outcome multiplied by a positive conversion factor \(c\). For example, converting meters to centimeters uses \(c=100\), while converting minutes to seconds uses \(c=60\). The predictors remain unchanged. The result applies to the intercept, slopes, fitted values, and residuals, which are all quantities on the outcome’s linear scale. It does not mean that every quantity involving the outcome is multiplied by \(c\): a variance, for example, is multiplied by \(c^2\).

The Model and the Scaling Rule

Consider a multiple linear regression model with \(n\) observations and \(p\) predictors. For observation \(i\), let \(y_i\) be the outcome and let \(x_{ij}\) be the value of predictor \(j\). Write the model as

y_i=\beta_0+\beta_1x_{i1}+\cdots+\beta_px_{ip}+\varepsilon_i.

Here, \(\beta_0\) is the intercept, \(\beta_j\) is the slope for predictor \(j\), and \(\varepsilon_i\) is the error for observation \(i\). For the usual regression interpretation, assume the conditional mean has this additive linear form, errors have mean zero and constant variance at the predictor values, and errors from different observations are independent. Normality is an additional condition used for some exact small-sample inference procedures; it is not needed for the unit-conversion arithmetic below.

Let \(\hat{\beta}_0,\hat{\beta}_1,\ldots,\hat{\beta}_p\) be the fitted coefficients from ordinary least squares using the original outcome units and the given predictor columns. Define \(\hat{y}_i\) as the fitted value and \(e_i=y_i-\hat{y}_i\) as the residual. If the outcome is converted according to \(y_i^{\text{new}}=c y_i\), with \(c>0\), then the corresponding fitted coefficients and residuals change as follows.

Scaling rule: When the outcome is multiplied by \(c\), with the same observations, predictor columns, and ordinary least-squares model, the intercept and every slope are multiplied by \(c\). The fitted values and residuals are also multiplied by \(c\). Thus, \(\hat{\beta}_0^{\text{new}}=c\hat{\beta}_0\), \(\hat{\beta}_j^{\text{new}}=c\hat{\beta}_j\), \(\hat{y}_i^{\text{new}}=c\hat{y}_i\), and \(e_i^{\text{new}}=c e_i\).

The units provide a useful check. The intercept has the outcome’s units. A slope has outcome units per unit of its predictor. So, if the outcome changes from meters to centimeters while a predictor remains in kilograms, the intercept is in centimeters and the slope is in centimeters per kilogram. The numerical value of each is multiplied by 100; the predictor’s unit has not changed.

Why the Rule Holds

The rule follows from the least-squares criterion. For any proposed coefficients \(b_0,b_1,\ldots,b_p\), the residual sum of squares in the original units is

\operatorname{SSE}(b_0,\ldots,b_p) =\sum_{i=1}^{n}\left(y_i-b_0-\sum_{j=1}^{p}b_jx_{ij}\right)^2.

Now use the converted outcome \(y_i^{\text{new}}=c y_i\) and try coefficients \(b_j^{\text{new}}=c b_j\) for \(j=0,\ldots,p\). The residual at observation \(i\) becomes \(c\) times its original value. Therefore the new residual sum of squares for these converted coefficients is

\begin{aligned} \operatorname{SSE}_{\text{new}}(c b_0,\ldots,c b_p) &=\sum_{i=1}^{n}\left(cy_i-cb_0-\sum_{j=1}^{p}cb_jx_{ij}\right)^2\\ &=c^2\sum_{i=1}^{n}\left(y_i-b_0-\sum_{j=1}^{p}b_jx_{ij}\right)^2\\ &=c^2\operatorname{SSE}(b_0,\ldots,b_p). \end{aligned}

Because \(c^2\) is positive and does not depend on the proposed coefficients, multiplying every candidate sum of squares by \(c^2\) does not change which candidate minimizes it. Also, every possible set of new coefficients corresponds to exactly one original set, since \(c\ne 0\). Consequently, the least-squares minimizer in the new units is the original minimizer multiplied by \(c\). Substituting those fitted coefficients into the model shows that each fitted value is multiplied by \(c\); subtracting the fitted value from the converted observed outcome shows the same for each residual.

Worked Example: Meters to Millimeters in a Multiple Regression

Suppose a hypothetical model predicts a component’s deflection \(y\), measured in meters, from load \(x_1\) in kilonewtons and temperature \(x_2\) in degrees Celsius:

\hat{y}=0.012+0.0045x_1-0.0008x_2.

Convert the outcome to millimeters. Since \(1\) meter is \(1000\) millimeters, \(c=1000\). Multiply the intercept and both slopes by 1000:

\hat{y}_{\text{mm}} =1000(0.012)+1000(0.0045)x_1-1000(0.0008)x_2 =12+4.5x_1-0.8x_2.

The new intercept is \(12\) mm. The load slope is \(4.5\) mm per kilonewton, and the temperature slope is \(-0.8\) mm per degree Celsius. The signs and relative directions have not changed; only the outcome scale has changed.

For a hypothetical component with \(x_1=6\) kN and \(x_2=10^\circ\)C, the original fitted value is \(0.012+0.0045(6)-0.0008(10)=0.031\) m. In millimeters it is \(1000(0.031)=31\) mm; the converted equation gives \(12+4.5(6)-0.8(10)=31\) mm as well.

If the observed deflection for this component is \(0.034\) m, its residual is \(0.034-0.031=0.003\) m. In millimeters, the observed value is \(34\) mm and the residual is \(34-31=3\) mm, which also equals \(1000(0.003)\) mm. This checks that the residual uses the same outcome conversion as the fitted value.

Worked Example: Minutes to Seconds With a Binary Predictor

A hypothetical model predicts appointment duration \(y\), in minutes, using travel distance \(x_1\) in kilometers and an indicator \(x_2\) that equals 1 for an appointment at a particular clinic type and 0 otherwise:

\hat{y}=8.4+1.25x_1+2.1x_2.

Convert the outcome to seconds, using \(c=60\). The intercept becomes \(60(8.4)=504\) seconds. The distance slope becomes \(60(1.25)=75\) seconds per kilometer. The indicator coefficient becomes \(60(2.1)=126\) seconds. Thus,

\hat{y}_{\text{seconds}}=504+75x_1+126x_2.

For \(x_1=4\) km and \(x_2=1\), the original fitted duration is \(8.4+1.25(4)+2.1(1)=15.5\) minutes. Converting gives \(15.5(60)=930\) seconds. The new equation gives \(504+75(4)+126(1)=930\) seconds. The indicator coefficient represents the fitted difference between the two clinic types at the same distance: it is \(2.1\) minutes in the original outcome units and \(126\) seconds in the new units. This is a change of units, not a change in the fitted comparison.

Worked Example: Milligrams per Liter to Micrograms per Liter

Suppose a hypothetical regression predicts a concentration \(y\), in milligrams per liter, from a numeric predictor \(x_1\) and a second predictor \(x_2\):

\hat{y}=0.30+0.08x_1-0.05x_2.

Since one milligram equals 1000 micrograms, the concentration in micrograms per liter is \(1000\) times its value in milligrams per liter. The converted equation is

\hat{y}_{\text{micrograms/L}} =300+80x_1-50x_2.

For a hypothetical record with \(x_1=2\) and \(x_2=1\), the fitted value is \(0.30+0.08(2)-0.05(1)=0.41\) mg/L, or \(410\) micrograms/L. If the observed concentration is \(0.46\) mg/L, then the residual is \(0.46-0.41=0.05\) mg/L. In the converted units, it is \(460-410=50\) micrograms/L, equal to \(1000(0.05)\) micrograms/L.

What Does Not Scale by the Same Factor?

The scaling rule applies to the listed linear-scale quantities, not to every statistic computed from the outcome. For instance, multiplying each residual by \(c\) multiplies its square by \(c^2\). The residual sum of squares therefore changes by a factor of \(c^2\), as the derivation showed. A residual variance estimate, when computed from that sum of squares using the same degrees of freedom, also scales by \(c^2\). A residual standard deviation, which is the square root of a variance, scales by \(c\) for positive \(c\).

This distinction is important when reporting a model. A residual of \(3\) mm is \(0.003\) m, but a residual variance of \(9\) mm\(^2\) is \(0.000009\) m\(^2\), not \(0.000009\) mm\(^2\). The squared outcome units must be converted using the square of the factor. The present coefficient rule does not imply that all model summaries or inferential quantities can be converted by simply multiplying by \(c\).

A Reliable Conversion Check

Before using converted coefficients, record the direction of the conversion. If the new unit is smaller, such as millimeters instead of meters, the numerical outcome and coefficients become larger. If the new unit is larger, the numerical values become smaller. Then check that the new coefficient units are sensible and that a fitted value or residual converts as expected.

Write the conversion equation.
Express the new numerical outcome as \(y^{\text{new}}=c y\), making clear which unit is old and which is new.

Scale every coefficient.
Multiply the intercept and every slope by \(c\). Leave the predictor values and their units unchanged.

Check one record.
Convert its original fitted value and residual by \(c\), then compare them with the values calculated in the new units.

Common Mistakes

Scaling only the slopes. The intercept is also measured on the outcome scale, so it must be multiplied by \(c\).
Using the reciprocal conversion factor. For meters to centimeters, \(c=100\), because the numerical value in centimeters is 100 times the numerical value in meters. State the conversion equation before changing coefficients.
Changing predictor units at the same time without accounting for them. This rule holds predictors fixed. A predictor-unit conversion changes slope values for a separate reason.
Multiplying a variance by \(c\). Residuals scale by \(c\), but their squares and variance scale by \(c^2\).
Treating rescaled coefficients as a changed relationship. The numerical coefficients differ because the outcome unit differs. Their units and the fitted relationship on the physical scale remain consistent.

Key takeaway: For a positive multiplicative conversion of the outcome, multiply the intercept, every slope, each fitted value, and each residual by the same factor \(c\). Quantities on a squared scale, such as residual variance, instead scale by \(c^2\).

Check Your Understanding

Work through these questions by writing the outcome conversion factor first.

A model predicts length in meters. What factor converts its intercept and slopes to a model whose outcome is in centimeters?
A model for time in hours has an intercept of \(0.6\) and a slope of \(0.2\) hours per kilometer. Give both coefficients in minutes and state the slope’s units.
If an observed outcome and fitted value are both multiplied by 100, how does their residual change? Explain using the residual definition.
A residual standard deviation is reported in the original outcome units. What factor converts it to the new units, and what factor converts a residual variance?
A learner converts a response from meters to millimeters but leaves the intercept unchanged. Identify the error and state the correct factor for the intercept.