Why a Coefficient Can Change When a Control Is Added
An OLS coefficient is defined relative to a particular regression. In a regression of an outcome on one regressor, its slope summarizes the best linear fit using all variation in that regressor. In a regression that also includes controls, the coefficient on the original regressor is determined by the variation left after accounting for the controls. Those are different comparisons, so the two coefficients need not match—even when both regressions use the same observations.
The distinction matters for interpreting empirical results. A coefficient change may reflect that the controlled regression uses a different portion of regressor variation, that the two regressions estimate different population projection targets, or both. It does not by itself establish that a control has removed bias or revealed a causal effect.
Assume the variables have finite second moments. The marginal slope is uniquely defined if \(\operatorname{Var}(X)>0\). The controlled slope is uniquely defined if \(X\) is not perfectly linearly explained by the intercept and \(Z\). Equivalently, if \(\widetilde X\) is the residual from the population linear projection of \(X\) on an intercept and \(Z\), then the identification condition is \(E[\widetilde X^2]>0\). Without this condition, there is no unique coefficient on \(X\) in the controlled projection.
Partialling Out: What the Controlled Coefficient Uses
The Frisch–Waugh–Lovell theorem gives a precise way to understand the controlled slope. First remove the linear association with the control from both \(X\) and \(Y\). Then regress the residualized outcome on the residualized regressor. The resulting slope is exactly the OLS coefficient on \(X\) in the regression that includes \(Z\).
Here \(M_WX\) and \(M_WY\) are the residuals after linearly projecting \(X\) and \(Y\), respectively, on the intercept and controls. The condition \(X'M_WX>0\) says that the sample regressor retains some variation after those projections.
The result follows directly from the OLS normal equations. Write the fitted regression as \(Y=Xb+W\gamma+e\), where \(W\) includes the intercept. The normal equations imply \(W'e=0\) and \(X'e=0\). The first condition gives \(M_We=e\), and the second implies \((M_WX)'e=X'M_We=X'e=0\). Applying \(M_W\) to the regression equation removes \(W\gamma\), leaving \(M_WY=bM_WX+e\). Taking the inner product with \(M_WX\) therefore gives \((M_WX)'M_WY=b(M_WX)'M_WX\). Dividing by the positive denominator yields the formula. This derivation also shows why the residualized regressor must have nonzero variation.
Worked Example: Residualizing Both Variables
Consider this hypothetical sample. Each row is one observation, and all variables are measured in arbitrary units.
| Observation | \(X\) | \(Z\) | \(Y\) |
|---|---|---|---|
| 1 | -3 | -1 | -9 |
| 2 | -1 | -1 | -5 |
| 3 | 1 | 1 | 5 |
| 4 | 3 | 1 | 9 |
All three sample means are zero, so the intercept does not change the slope calculations. The slope from regressing \(Z\) on \(X\) is
Thus the residuals from regressing \(Z\) on \(X\) are not the residuals needed for the coefficient on \(X\). For that coefficient, residualize \(X\) and \(Y\) on \(Z\). The slope from regressing \(X\) on \(Z\) is \(\sum_iX_iZ_i/\sum_iZ_i^2=8/4=2\), so the residualized \(X\) values are \((-1,1,-1,1)\). The slope from regressing \(Y\) on \(Z\) is \(\sum_iY_iZ_i/\sum_iZ_i^2=28/4=7\), giving residualized \(Y\) values \((-2,2,-2,2)\).
The partial-regression slope is therefore
The denominator is positive, so the controlled sample coefficient is identified. Its value is determined by the relationship between the parts of \(X\) and \(Y\) that remain after their linear associations with \(Z\) have been removed.
Shared Variation and the Population Projection Target
A control can share variation with \(X\). In a regression without that control, the slope uses this shared variation as part of the relationship between \(X\) and \(Y\). In the controlled regression, the coefficient on \(X\) instead uses only its residual variation. This is the partialling-out perspective: the controlled coefficient is based on \(X\) after its linear association with \(Z\) has been removed.
There is also a population-target perspective. The marginal slope minimizes squared prediction error over lines involving \(X\) alone. The controlled slope minimizes squared prediction error over planes involving both \(X\) and \(Z\). These are different optimization problems, so adding a regressor can change the target coefficient even when the population distribution has not changed. A coefficient change is not a comparison of two estimates of an automatically identical quantity.
For a single control, the relationship between the two population slopes can be stated exactly. Suppose the controlled projection is uniquely identified and write its residual as \(U=Y-a_c-b_cX-c_cZ\). The projection normal equations give \(\operatorname{Cov}(X,U)=0\). Taking covariance with \(X\) in the projection equation and dividing by \(\operatorname{Var}(X)\) gives
Thus the marginal slope and controlled slope coincide if, for example, \(c_c=0\) or \(\operatorname{Cov}(X,Z)=0\). Otherwise, their difference depends on the control’s projection coefficient and the amount of shared variation. This identity describes population linear projections; by itself, it does not say that either coefficient is causal.
Worked Example: A Population Target Changes
Suppose hypothetical centered population variables satisfy \(E[X^2]=4\), \(E[Z^2]=1\), and \(E[XZ]=1\). Let \(Y=X+2Z+U\), where \(U\) has zero covariance with both \(X\) and \(Z\).
The controlled projection has coefficients \(b_c=1\) and \(c_c=2\), with residual \(U\). These coefficients are uniquely identified: the residual from projecting \(X\) on \(Z\) is \(\widetilde X=X-Z\), and
The marginal slope, however, uses all variation in \(X\). Since \(Y=X+2Z+U\) and \(U\) is uncorrelated with \(X\),
The marginal slope is \(1.5\), while the controlled slope is \(1\). The difference is consistent with the population identity: \(1+2(1/4)=1.5\). These are coefficients for different linear projection targets. The calculation alone does not establish that the controlled coefficient is a causal effect.
Worked Example: A Control Need Not Change the Coefficient
Consider a second hypothetical sample with centered variables:
| Observation | \(X\) | \(Z\) | \(Y\) |
|---|---|---|---|
| 1 | -1 | -1 | -5 |
| 2 | -1 | 1 | 1 |
| 3 | 1 | -1 | -1 |
| 4 | 1 | 1 | 5 |
Here \(\sum_iX_iZ_i=0\), \(\sum_iX_i^2=4\), and \(\sum_iX_iY_i=8\). The regression of \(Y\) on an intercept and \(X\) therefore has slope \(8/4=2\). Because \(X\) and \(Z\) are orthogonal in this sample, residualizing \(X\) on \(Z\) leaves \(X\) unchanged. The slope from regressing \(Y\) on \(Z\) is \(\sum_iY_iZ_i/\sum_iZ_i^2=12/4=3\). Subtracting \(3Z\) from \(Y\) leaves the values \((-2, -2, 2, 2)\), which equal \(2X\). The partial-regression slope is consequently
In this example, adding the control does not change the slope on \(X\). Its residual variation is nonzero, so the controlled slope is identified; the equality occurs because the sample shared variation between \(X\) and \(Z\) is zero.
Identification and Interpretation
The partial-regression formula is useful only when the residualized \(X\) has positive variation. If \(X\) is perfectly explained by the intercept and controls, the denominator in the formula is zero. The data then do not distinguish a unique coefficient on \(X\) from coefficients on the controls. This is a lack of identification, not evidence that the coefficient is zero.
For example, if \(Z=X/2\) for every observation, then \(X\) is perfectly linearly explained by the control. A fitted value involving \(b_XX+b_ZZ\) depends only on \(b_X+b_Z/2\); many pairs \((b_X,b_Z)\) produce the same fitted values. The controlled coefficient on \(X\) is not unique. Always check that the regressor retains variation after partialling out the controls before interpreting its coefficient.
Even when the coefficient is uniquely identified, “holding \(Z\) fixed” is a description of the linear regression comparison, not a guarantee of a causal interpretation. A causal claim requires additional assumptions about how \(X\), \(Z\), and the outcome were generated. The regression coefficient change alone cannot verify those assumptions.
Common Mistakes
- Claiming that a coefficient changed because the estimate became less biased. A change shows that two fitted regressions produced different coefficients. It does not, by itself, establish which coefficient is closer to a causal effect.
- Assuming the controlled coefficient uses all variation in \(X\). By the partial-regression result, it uses the residual variation in \(X\) after its linear association with the controls is removed.
- Treating the two coefficients as the same population target. The marginal and controlled slopes minimize different prediction criteria. They need not be equal even in a fixed population.
- Interpreting a non-unique coefficient as an ordinary estimate. If \(E[\widetilde X^2]=0\) in the population, or \(X'M_WX=0\) in the sample, the controlled coefficient on \(X\) is not uniquely identified.
Check Your Understanding
Answer each question using the distinction between residualized variation and the population projection target.
- In a sample regression of \(Y\) on an intercept, \(X\), and controls \(W\), what two variables are residualized to obtain the coefficient on \(X\)?
- What condition ensures that the population controlled coefficient on \(X\) is uniquely identified?
- If \(X\) and \(Z\) have zero covariance, what does the single-control population slope identity imply about the marginal and controlled slopes?
- Why does a change in the coefficient on \(X\) after adding \(Z\) not, by itself, prove that the controlled coefficient is causal?
- If a control perfectly explains \(X\), what happens to the partial-regression denominator, and what does that imply about the coefficient on \(X\)?