Rescaling Predictors and Coefficient Units

Why Predictor Units Matter

A numerical predictor can be recorded in different units without changing the underlying cases. A distance might be recorded in meters or centimeters; a dose might be recorded in milligrams or grams. The design-matrix entries then change, and the coefficient attached to those entries must change with them. The central check is simple: after a consistent conversion, each case must have the same linear-predictor value.

This is a change of scale, not a change in the modeled relationship. The coefficient’s numerical value depends on the chosen units, so it cannot be interpreted without naming them. The coefficient’s sign and the contribution of a given physical predictor value, however, do not change under a consistent unit conversion.

Definition: Suppose a numerical predictor is recorded as \(x\), and a new unit makes its recorded value \(x^*=cx\), where \(c>0\) is the conversion factor. Rescaling the predictor means replacing the design-matrix column \(x\) by \(x^*\). To preserve the linear predictor, its coefficient must be replaced by \(\beta^*=\beta/c\). The coefficient is therefore expressed per unit of the rescaled predictor.

For example, when a distance in meters is converted to centimeters, \(x_{\text{cm}}=100x_{\text{m}}\). The coefficient per centimeter is the coefficient per meter divided by 100. The intercept is unchanged by this pure multiplication of the predictor; a conversion that also adds a constant is not pure rescaling.

One Predictor: The Coefficient Conversion

Consider a linear predictor with an intercept and one numerical predictor:

\eta_i=\beta_0+\beta x_i.

Here \(i\) indexes observations, \(\beta_0\) is the intercept, and \(\beta\) is the coefficient for the predictor recorded in its original units. If the new recorded value is \(x_i^*=cx_i\), we want coefficients \(\beta_0^*\) and \(\beta^*\) such that the linear predictor for every observation is unchanged.

Result: Under a pure rescaling \(x_i^*=cx_i\), with \(c>0\), the same linear predictor is obtained by taking \(\beta_0^*=\beta_0\) and \(\beta^*=\beta/c\).

Write the rescaled predictor.
Using the new design-matrix entry and coefficient, it is \(\eta_i^*=\beta_0^*+\beta^*x_i^*\).

Substitute the conversion and coefficient.
Set \(\beta_0^*=\beta_0\), \(x_i^*=cx_i\), and \(\beta^*=\beta/c\).

Simplify.
The product \((\beta/c)(cx_i)\) is \(\beta x_i\), so the original predictor is recovered.

The calculation proves the result for every observation, not just for a particular data set:

\eta_i^*=\beta_0^*+\beta^*x_i^* =\beta_0+\frac{\beta}{c}(cx_i) =\beta_0+\beta x_i =\eta_i.

Thus the rescaled coefficient is not an arbitrary adjustment: it is exactly the value required to offset the changed design-matrix entry. Conversely, if the intercept is held fixed and the predictor is rescaled by \(c\), equality for all \(x_i\) requires \(c\beta^*=\beta\), hence \(\beta^*=\beta/c\).

Worked Example: Meters to Centimeters

Suppose a linear predictor is \(\eta=1.2+0.8x_{\text{m}}\), where distance \(x_{\text{m}}\) is measured in meters. Convert the predictor to centimeters and check the result for a distance of 1.7 meters.

Since \(x_{\text{cm}}=100x_{\text{m}}\), the conversion factor is \(c=100\). The coefficient per centimeter is

\beta_{\text{cm}}=\frac{0.8}{100}=0.008.

The intercept remains 1.2, so the converted expression is \(\eta=1.2+0.008x_{\text{cm}}\). At 1.7 meters, the original calculation is

\eta=1.2+0.8(1.7)=1.2+1.36=2.56.

The same distance is \(100(1.7)=170\) centimeters. Using the converted coefficient gives

\eta=1.2+0.008(170)=1.2+1.36=2.56.

Both calculations agree. The predictor column’s entries became 100 times larger, while the coefficient became 100 times smaller.

Coefficient Units, Signs, and Interpretation

A coefficient describes the change in the linear predictor per one unit of the predictor, holding the other terms fixed. If \(x\) is in meters, \(\beta\) is per meter on the linear-predictor scale. If the predictor is instead recorded in centimeters, \(\beta^*=\beta/100\) is per centimeter on that same scale. The coefficient’s value changes because “one meter” and “one centimeter” represent different increments.

It is important to distinguish the magnitude of a coefficient from its signed numerical value. When \(c>1\), dividing by \(c\) makes the magnitude smaller: \(|\beta^*|=|\beta|/c\). But the signed value is not always smaller as a real number. If the coefficient is negative, division by \(c>1\) moves it closer to zero, so it becomes greater as a signed number.

Worked Example: A Negative Coefficient and Its Magnitude

Suppose \(\eta=4-2x_{\text{m}}\), with distance measured in meters. Convert the predictor to centimeters and compare the signed coefficients.

Here \(\beta=-2\) and \(x_{\text{cm}}=100x_{\text{m}}\). Therefore,

\beta_{\text{cm}}=\frac{-2}{100}=-0.02.

The coefficient’s magnitude has decreased from \(|-2|=2\) to \(|-0.02|=0.02\), a factor of 100. But \(-0.02\) is greater than \(-2\) as a signed number. The correct general statement is that the magnitude is divided by 100; it is not generally correct to say the signed coefficient becomes smaller.

For a distance of 0.6 meters, the original calculation is \(\eta=4-2(0.6)=4-1.2=2.8\). In centimeters the distance is 60, giving \(\eta=4-0.02(60)=4-1.2=2.8\). The sign and contribution are preserved.

More generally, if the predictor increases by a physical amount \(\Delta x\), the linear predictor changes by \(\beta\Delta x\), with other terms fixed. In the rescaled units, the same physical increment is \(\Delta x^*=c\Delta x\). Since \(\beta^*=\beta/c\),

\beta^*\Delta x^* =\frac{\beta}{c}(c\Delta x) =\beta\Delta x.

This identity shows why the coefficient’s units matter: a coefficient per centimeter is numerically different from one per meter, but the predicted change associated with the same physical increment is identical.

Result: For any fixed change in the underlying predictor, the change in the linear predictor is invariant to a positive unit conversion, provided the coefficient is converted inversely. In fact, \(\beta^*\Delta x^*=\beta\Delta x\).

To verify this result, write \(\Delta x^*=c\Delta x\) and \(\beta^*=\beta/c\). Multiplying gives \(\beta^*\Delta x^*=(\beta/c)(c\Delta x)=\beta\Delta x\). No approximation is involved, and the identity holds for positive or negative coefficients and for positive or negative increments. Proof. \(\square\)

Rescaling Several Predictors

The same rule applies separately to each numerical predictor. Suppose there are \(p\) predictor columns and the linear predictor for observation \(i\) is

\eta_i=\beta_0+\sum_{j=1}^{p}\beta_jx_{ij}.

For predictor \(j\), let the new unit give \(x_{ij}^*=c_jx_{ij}\), with \(c_j>0\). Then set \(\beta_j^*=\beta_j/c_j\), leaving the intercept unchanged. Each product in the sum is preserved, so the full \(\eta_i\) is preserved.

In matrix notation, include the intercept column and write \(\eta=X\beta\). Define \(D\) as the diagonal matrix whose intercept entry is 1 and whose other diagonal entries are \(c_1,\ldots,c_p\). The rescaled design matrix is \(X^*=XD\). Taking \(\beta^*=D^{-1}\beta\) gives

X^*\beta^*=(XD)(D^{-1}\beta)=X\beta=\eta.

This matrix identity is a complete derivation: every rescaled column is multiplied by its conversion factor, and the corresponding coefficient is multiplied by the reciprocal factor. It applies to each row of the design matrix at once.

Worked Example: Rescaling Two Predictors

Consider the linear predictor

\eta=-1+0.4a_{\text{yr}}+0.03w_{\text{kg}},

where \(a_{\text{yr}}\) is age in years and \(w_{\text{kg}}\) is weight in kilograms. Change age to months and weight to grams. Since \(a_{\text{mo}}=12a_{\text{yr}}\) and \(w_{\text{g}}=1000w_{\text{kg}}\), the new coefficients are

\beta_{a,\text{mo}}=\frac{0.4}{12}=\frac{1}{30}, \qquad \beta_{w,\text{g}}=\frac{0.03}{1000}=0.00003.

The intercept remains \(-1\), so the rescaled expression is \(\eta=-1+\frac{1}{30}a_{\text{mo}}+0.00003w_{\text{g}}\). Check the case \(a_{\text{yr}}=4\), \(w_{\text{kg}}=20\). In the original units,

\eta=-1+0.4(4)+0.03(20) =-1+1.6+0.6 =1.2.

The same case has \(a_{\text{mo}}=12(4)=48\) and \(w_{\text{g}}=1000(20)=20000\). In the new units,

\eta=-1+\frac{1}{30}(48)+0.00003(20000) =-1+1.6+0.6 =1.2.

Each predictor’s coefficient was converted using its own factor. Rescaling one column does not require rescaling the coefficient of another column.

Common Mistakes and a Practical Check

A frequent error is to multiply the coefficient by the same conversion factor as the predictor. If \(x^*=100x\), using \(\beta^*=100\beta\) would make the product \(\beta^*x^*=(100\beta)(100x)=10000\beta x\), not \(\beta x\). The coefficient must instead be divided by 100.

Another error is to change the predictor values but leave the coefficient untouched. For instance, replacing meters by centimeters while retaining a coefficient per meter multiplies that predictor’s contribution by 100. Always check the product of the coefficient and the corresponding design-matrix entry.

Write the conversion explicitly, such as \(x_{\text{cm}}=100x_{\text{m}}\).
Divide the coefficient by the factor multiplying the predictor column.
Keep the intercept unchanged for a pure rescaling.
Substitute at least one case in both unit systems and compare the resulting \(\eta\).

Rescaling is useful when choosing units that make predictor values easier to read or coefficients easier to communicate. The coefficient itself is unit-dependent, so reports should state the predictor’s units. What remains fixed is the linear predictor for each case, provided the design column and coefficient are transformed consistently.

Check Your Understanding

Use the conversion rule and verify each answer by tracking the coefficient–predictor product.

A predictor is converted from kilograms to grams. What factor multiplies its design-matrix column, and by what factor must its coefficient be divided?
A coefficient is \(-3\) per meter. Write its value per centimeter, and state separately how its magnitude and its signed numerical value compare with the original.
For \(\eta=2+5x\), suppose \(x^*=10x\). Give the coefficient and intercept in the rescaled expression and show why \(\eta\) is unchanged.
Two predictor columns are rescaled by factors 12 and 1000. Describe how their two coefficients must change.
Explain why multiplying both a predictor column and its coefficient by the same conversion factor does not preserve the linear predictor.