Why an Association Can Weaken After Adjustment
A predictor can have a noticeable association with an outcome when considered on its own, yet a smaller estimated association after another predictor is included in the regression. This happens when the added predictor is related to both the focal predictor and the outcome. Some of the outcome variation that accompanied the focal predictor in the simpler analysis is then accounted for by the added predictor.
The comparison is between two different regression summaries. A marginal slope comes from regressing the outcome on the focal predictor alone. An adjusted slope comes from a regression that includes the focal predictor and another predictor. The adjusted slope describes the fitted association with the focal predictor when the added predictor is held fixed. Neither slope, by itself, establishes a causal effect.
The change is about the size of the estimated slope, not necessarily about its sign. This tutorial focuses on a particular situation: the predictors are positively associated, and the added predictor has a positive adjusted slope for the outcome. In that situation, the marginal slope can include a positive contribution associated with \(Z\), making it larger than the adjusted slope for \(X\).
The Two-Predictor Calculation
Suppose both regression equations include an intercept and use the same observations. Let \(S_{XX}\), \(S_{ZZ}\), and \(S_{XZ}\) be centered sums of squares and cross-products for the predictors, and let \(S_{XY}\) and \(S_{ZY}\) be their centered cross-products with the outcome. For example, \(S_{XZ}=\sum_i (x_i-\bar{x})(z_i-\bar{z})\). The marginal slope is \(S_{XY}/S_{XX}\). If \(X\) and \(Z\) both vary and are not perfectly linearly related, the adjusted slopes are:
The denominator \(S_{XX}S_{ZZ}-S_{XZ}^{\,2}\) is positive when the predictors are not perfectly linearly related. If it is zero, the data do not contain separate information to estimate both adjusted slopes by this formula. These are least-squares calculations; no normality assumption is needed simply to compute the slopes. Any additional assumptions required for inference are separate from this comparison.
This identity gives a compact explanation of weakening. The term \(S_{XZ}/S_{XX}\) is the slope from regressing \(Z\) on \(X\). If this term is positive and \(b_Z\) is positive, their product is positive, so the marginal slope exceeds \(b_X\). In that case, adjustment removes that positive part from the marginal slope. This describes how the fitted coefficients relate in the observed data; it does not show that changing \(X\) would cause a change in \(Y\).
Why the Identity Holds
The adjusted regression uses an intercept and fitted slopes \(b_X\) and \(b_Z\). Least squares makes the fitted residuals have zero centered cross-product with each predictor. In particular, the normal equation for \(X\) is:
This equation says that the centered association between \(X\) and \(Y\) is accounted for by the adjusted contribution of \(X\) and by the part associated with \(Z\) that accompanies \(X\). Divide both sides by \(S_{XX}\), which is positive when \(X\) varies:
The left side is precisely the marginal slope. Thus the identity follows directly from the least-squares normal equation; it is not an approximation or a claim about a population relationship. The equation also shows why predictor correlation matters: if \(S_{XZ}=0\), the marginal slope equals \(b_X\) for this two-predictor comparison. When the cross-product is positive, \(Z\)'s adjusted slope helps determine whether the marginal slope is larger or smaller.
Worked Example: Study Time and Prior Scores
Consider a hypothetical data set of four students. Let \(X\) be study time in hours, \(Z\) be a prior assessment score in points, and \(Y\) be a later assessment score in points. The values below are constructed for illustration, not taken from a real study.
| Student | \(X\), study hours | \(Z\), prior score | \(Y\), later score |
|---|---|---|---|
| A | 4 | 70 | 79 |
| B | 4 | 68 | 75 |
| C | 6 | 70 | 81 |
| D | 6 | 72 | 85 |
The sample means are \(\bar{x}=5\) hours, \(\bar{z}=70\) points, and \(\bar{y}=80\) points. The centered values are \(X=(-1,-1,1,1)\), \(Z=(0,-2,0,2)\), and \(Y=(-1,-5,1,5)\). Therefore:
The marginal slope for study time is \(S_{XY}/S_{XX}=12/4=3\). In this sample, the separate regression of later score on study time estimates an association of 3 score points per additional study hour. For the adjusted slopes, first calculate the denominator:
Then calculate the adjusted slope for study time:
The adjusted slope for the prior score is:
So the estimated association with study time decreases from 3 to 1 point per hour when prior score is included. In the two-predictor fit, a one-hour difference in study time corresponds to a 1-point difference in fitted later score among students with the same prior score. This is an adjusted association, not a causal effect. The arithmetic identity checks the result: \(S_{XZ}/S_{XX}=4/4=1\), and \(b_X+b_Z(S_{XZ}/S_{XX})=1+2(1)=3\), the marginal slope.
Checking the Calculation With Summary Statistics
A weakening comparison can also be checked directly from centered summaries, without listing every observation. The summaries must come from the same set of observations and must use the same definitions of \(X\), \(Z\), and \(Y\). In particular, a cross-product such as \(S_{ZY}\) must be calculated from the centered \(Z\) and \(Y\) values, not inferred from a different quantity.
Worked Example: Comparing Slopes From Cross-Products
Suppose a hypothetical sample relating practice time \(X\) in hours, a baseline score \(Z\) in points, and a final score \(Y\) in points has centered summaries \(S_{XX}=10\), \(S_{ZZ}=10\), \(S_{XZ}=5\), \(S_{XY}=25\), and \(S_{ZY}=20\). These summaries indicate that both predictors vary and that their cross-product is positive. Calculate the marginal and adjusted slopes.
The marginal slope for practice time is:
The denominator for the adjusted slopes is:
The adjusted slope for practice time is:
The adjusted slope for baseline score is:
The practice-time slope weakens from 2.5 to 2 points per hour. As a check, the slope for \(Z\) on \(X\) is \(S_{XZ}/S_{XX}=5/10=0.5\), so the identity gives \(2+1(0.5)=2.5\). The weakening is consistent with the positive association between the predictors and the positive adjusted slope for baseline score.
Reading the Size of the Weakening
The difference between the marginal and adjusted slopes can be calculated as \(b_Z(S_{XZ}/S_{XX})\). Its units are the units of \(Y\) per unit of \(X\): \(b_Z\) has units of \(Y\) per unit of \(Z\), while \(S_{XZ}/S_{XX}\) has units of \(Z\) per unit of \(X\). Multiplying them gives units of \(Y\) per unit of \(X\), as required.
Worked Example: Explaining a Two-Point Difference
In another hypothetical data summary, let \(X\) be practice hours, \(Z\) a baseline score, and \(Y\) a final score. Suppose \(S_{XX}=4\), \(S_{ZZ}=10\), \(S_{XZ}=5\), \(S_{XY}=16\), and \(S_{ZY}=27.5\). The marginal slope is \(16/4=4\) final-score points per practice hour. The adjusted slope for \(X\) is:
The other adjusted slope is:
The slope of baseline score on practice hours is \(S_{XZ}/S_{XX}=5/4=1.25\) baseline-score points per hour. Its contribution to the marginal slope is \(2(1.25)=2.5\) final-score points per hour. Thus \(1.5+2.5=4\): the marginal slope exceeds the adjusted slope by 2.5 points per hour. The units and arithmetic agree, and the result describes regression associations in this hypothetical data set only.
Common Pitfall and Practical Meaning
A common mistake is to say, “The slope got smaller, so the added predictor removed part of the causal effect of \(X\).” That conclusion does not follow from a coefficient comparison. The arithmetic shows how including \(Z\) changes the fitted association with \(X\); it does not establish why the predictors are related, whether the model is appropriate for a causal question, or what would happen under an intervention.
A second mistake is to assume that every adjustment must weaken an association. It does not. The direction and size of the change depend on how \(X\) and \(Z\) are associated and on the adjusted slope for \(Z\). Here, both quantities made a positive contribution to the marginal slope. Always calculate or inspect those ingredients rather than treating “adjustment” as a synonym for “smaller.”
The comparison is useful because it makes clear what an adjusted coefficient does and does not summarize. A smaller adjusted slope means that, in the fitted model, the focal predictor’s association with the outcome is weaker at a fixed value of the added predictor than its overall one-predictor association. It should be reported in context, with units, and without presenting the difference as proof of causation.
Check Your Understanding
Use the definitions and calculations above to answer these questions.
- What is the difference between a marginal slope and an adjusted slope in a two-predictor regression?
- If \(S_{XX}=8\), \(S_{XZ}=4\), and \(S_{ZZ}=6\), calculate the slope from regressing \(Z\) on \(X\).
- Suppose \(b_X=1.2\), \(b_Z=2\), and \(S_{XZ}/S_{XX}=0.5\). Use the identity to calculate the marginal slope.
- Why must \(S_{ZY}\) be calculated from the centered \(Z\) and \(Y\) values when using the adjusted-slope formula?
- A smaller adjusted slope is observed after adding a predictor. What can be concluded about the fitted association, and what cannot be concluded about causation?