Why Put a Genetic Effect on a Standardized Scale?
A coefficient for a quantitative trait is easiest to interpret in the trait’s own units. For example, an additive dosage coefficient of 1.6 might mean that the fitted trait value increases by 1.6 units for each additional copy of an allele. But if two traits use different units—or have very different amounts of natural variation—their coefficients are not directly comparable just by their numerical sizes. Standardizing expresses a coefficient relative to one or more standard deviations. This changes its scale and interpretation, not the underlying fitted relationship.
The distinction between two standardized quantities matters. An outcome-standardized effect divides the original coefficient by the trait’s standard deviation, leaving the predictor in its original units. A fully standardized coefficient also expresses the predictor in standard-deviation units. For a genotype dosage coded 0, 1, or 2, these answer different questions: the first is the trait-standard-deviation change per additional allele; the second is the trait-standard-deviation change per one-standard-deviation increase in dosage.
Model and Definitions
Suppose a quantitative trait \(Y\) is modeled by a linear regression with an intercept, a genetic predictor \(X\), and possibly other covariates. For a simple model with only the genetic predictor, write
Here \(X_i\) is the chosen genotype coding for individual \(i\), \(\alpha\) is the intercept, \(\beta\) is the fitted change in the conditional mean of \(Y\) for a one-unit increase in \(X\), and \(\varepsilon_i\) is the residual. With additive dosage coding, \(X_i\) is the number of counted alleles, from 0 to 2, so \(\beta\) is the fitted trait change per additional counted allele. With covariates, \(\beta\) is the corresponding regression coefficient conditional on the included covariates.
Standardizing does not require a particular probability distribution for this calculation. To interpret a fitted coefficient as a linear change in the conditional mean, the model’s linear mean specification must be appropriate for the question and coding used. The observations must also refer to the same trait and predictor definitions. Standardizing alone does not establish causation, remove confounding, or validate a model. The standard deviations used below must be positive and must be identified as either sample or population quantities.
For dosage coding, the outcome-standardized effect retains the practical interpretation “trait standard deviations per additional allele.” The fully standardized coefficient instead describes a one-standard-deviation change in dosage. Since a dosage standard deviation depends on the population’s allele frequency and genotype distribution, it is not a fixed substitute for the per-allele effect.
Worked Example: Standardizing a Per-Allele Effect
Suppose a hypothetical analysis uses additive dosage \(X=0,1,2\) and estimates \(\beta=1.8\) mmHg per additional counted allele for a quantitative blood-pressure trait. The trait’s standard deviation in the analysis sample is \(s_Y=12\) mmHg. Find and interpret the outcome-standardized effect.
The formula is \(\beta/s_Y\). Substituting the coefficient and trait standard deviation gives
Thus, under the fitted model, one additional counted allele corresponds to a fitted increase of 0.150 standard deviations in the trait. The mmHg units cancel in the ratio. This is still a per-allele effect; it is not an effect per standard deviation of genotype dosage. The calculation describes the model coefficient and does not by itself imply that changing a person’s genotype would cause the trait to change.
How Rescaling Changes the Coefficient
To standardize both variables, define \(X_i^*=(X_i-\bar{x})/s_X\) and \(Y_i^*=(Y_i-\bar{y})/s_Y\), where the means and standard deviations are calculated in the analysis sample. A one-unit increase in \(X^*\) is an increase of \(s_X\) original predictor units; a one-unit increase in \(Y^*\) is an increase of \(s_Y\) original trait units. Substituting the inverse transformations, \(X_i=\bar{x}+s_X X_i^*\) and \(Y_i=\bar{y}+s_Y Y_i^*\), into the fitted line gives the standardized slope.
The coefficient of \(X_i^*\) is therefore \(\beta s_X/s_Y\), the fully standardized coefficient. If only the outcome is standardized, leaving \(X\) in its original units, the slope is \(\beta/s_Y\). Both formulas follow directly from changing measurement scales; neither changes which observations have higher or lower fitted values.
More precisely, using positive standard deviations preserves the ordering of fitted values: each fitted value is transformed by the same positive linear transformation. Residuals are divided by \(s_Y\) when the outcome is standardized. In a model with an intercept, centering and scaling the outcome and predictor in this way also leave the coefficient of determination unchanged: both the residual sum of squares and total sum of squares are divided by \(s_Y^2\). The intercept’s numerical value changes, as expected, because the origin and scale have changed.
Worked Example: Calculating a Fully Standardized Coefficient
In a hypothetical sample of 100 individuals, genotype dosages have counts 49 at dosage 0, 42 at dosage 1, and 9 at dosage 2. A fitted quantitative-trait model estimates \(\beta=1.6\) trait units per additional allele. The sample standard deviation of the trait is \(s_Y=10.5\) trait units. Calculate the dosage standard deviation and the fully standardized coefficient.
The dosage sum is \(42+2(9)=60\), and its squared-value sum is \(42+4(9)=78\). Thus the sample mean dosage is \(60/100=0.6\), and its sample variance is
Now apply the fully standardized formula:
The fitted trait value changes by about 0.0993 trait standard deviations for a one-standard-deviation increase in dosage. This is not the per-allele standardized effect. Dividing \(\beta\) by \(s_Y\) instead gives \(1.6/10.5\approx0.1524\) trait standard deviations per additional allele. The two answers differ because one dosage standard deviation is about 0.6513 dosage units, not one dosage unit.
Comparing Effects: What Standardization Does and Does Not Do
A coefficient in original units can be essential for practical interpretation: it tells a reader how much the fitted trait value differs for a specified change in the genetic predictor. Standardization can help describe the effect relative to the trait’s variability or compare effects whose traits use different units. It does not make coefficients universally comparable. In particular, an outcome-standardized effect retains the predictor’s scale, while a fully standardized coefficient depends on the predictor’s variability as well as the trait’s.
The following comparison illustrates the difference. Each example below is hypothetical; suppose both traits are analyzed using the same additive dosage definition, and each listed coefficient is a fitted per-allele coefficient.
Worked Example: Equal Standardized Effects, Different Raw Effects
Study A estimates a coefficient of 2.4 mg/dL per allele for a quantitative trait with standard deviation 12 mg/dL. Study B estimates a coefficient of 1.8 units per allele for a different trait with standard deviation 9 units. Calculate the outcome-standardized effects and state what the comparison means.
For Study A, \(2.4/12=0.200\) trait standard deviations per allele. For Study B, \(1.8/9=0.200\) trait standard deviations per allele. The standardized coefficients are equal, although the original-unit effects are 2.4 mg/dL and 1.8 units, respectively.
This equality says that the fitted per-allele change is the same fraction of each trait’s standard deviation. It does not say that the effects are equal in the traits’ original units, that the traits have the same biological meaning, or that the underlying genetic mechanisms are the same. Nor does it establish that either estimate is precise or unbiased. Those questions require information beyond rescaling the point estimates.
When studies differ in allele frequency, dosage variability can also differ. Consequently, their fully standardized coefficients can differ even if their per-allele coefficients and trait standard deviations are identical. Conversely, two fully standardized coefficients can agree while the corresponding per-allele effects differ. If the goal is to compare per-allele changes relative to trait variation, the outcome-standardized effect is the relevant scale. If the goal is a coefficient per standard deviation of the predictor, use the fully standardized coefficient and report how \(s_X\) was obtained.
Comparability also depends on using sufficiently similar definitions of the trait, the genotype predictor, and the fitted model. The standard deviation depends on the population or analysis sample: a more heterogeneous sample can have a larger \(s_Y\), producing a smaller outcome-standardized effect for the same raw coefficient. Standardization does not correct differences in covariate adjustment, measurement quality, or population composition.
A Practical Workflow
Before reporting a standardized genetic effect, specify what has been standardized and preserve the original coefficient alongside the rescaled result. A concise workflow is:
Identify the trait, its original units, the predictor coding, and the coefficient’s change per unit of that predictor.
Use \(\beta/s_Y\) for trait standard deviations per original predictor unit, or \(\beta s_X/s_Y\) for trait standard deviations per predictor standard deviation.
Calculate \(s_Y\), and if needed \(s_X\), from the specified analysis sample and state the basis for them.
Say whether the effect is per allele or per predictor standard deviation, retain the units or standard-deviation language, and avoid interpreting a rescaling as evidence of causation.
Common Mistake: Treating Standardized Coefficients as the Same Quantity
A common error is to divide every genetic coefficient by the trait standard deviation and label the result “the fully standardized effect.” This omits the predictor scale. Dividing \(\beta\) by \(s_Y\) standardizes only the outcome: for dosage coding, the result is trait standard deviations per additional allele. A fully standardized coefficient also multiplies by \(s_X\), giving trait standard deviations per dosage standard deviation. The correction is to name the scale explicitly and use the matching formula.
Check Your Understanding
Answer each question using the coefficient scale stated in the question.
- A dosage coefficient is 0.9 cm per allele, and the trait standard deviation is 6 cm. What is the outcome-standardized effect, and how should it be interpreted?
- In words, distinguish an effect measured in trait standard deviations per allele from a fully standardized coefficient.
- If both the predictor and outcome are centered and divided by their standard deviations, what happens to the fitted slope? State the formula.
- Why can two studies have different fully standardized coefficients even when their per-allele coefficients and trait standard deviations are the same?
- Does a standardized genetic coefficient establish a causal effect? Briefly explain.