Deriving Omitted-Confounder Bias In A Later Treatment Model

Why Derive the Coefficient Shift?

In a two-visit study, $L_1$ is a covariate measured before the later treatment $A_2$, and it may predict both $A_2$ and the outcome $Y$. If a linear outcome regression includes $A_2$ but leaves out $L_1$, the coefficient of $A_2$ can absorb part of the outcome association carried by $L_1$. The resulting coefficient shift is not mysterious: it is determined by how strongly $A_2$ and $L_1$ are associated, and how strongly $L_1$ predicts $Y$ after accounting for the other included variables.

The earlier material on treatment-confounder feedback motivates this calculation. Here the aim is narrower: derive the omitted-variable formula and make its assumptions and limits explicit. The formula concerns a linear regression coefficient. By itself, it does not establish that the coefficient is a causal effect.

Definition: The population omitted-variable coefficient shift is the difference between the coefficient of $A_2$ in a population linear regression that omits $L_1$ and the coefficient of $A_2$ in the specified outcome model that includes $L_1$. Under the assumptions below, this shift is the omitted-confounder bias formula. It is a population-projection discrepancy, not an assertion that finite-sample OLS bias is exactly the same quantity.

The Basic Formula

First consider one treatment variable and one omitted covariate, with no other regressors. Center the variables, and suppose the outcome follows the additive linear model

Y = \beta_A A_2 + \beta_L L_1 + \varepsilon,

where $E[\varepsilon\mid A_2,L_1]=0$, and $A_2$ has positive variance. Assume all variables have finite second moments. The coefficient from the population regression of $Y$ on $A_2$ alone is $\operatorname{Cov}(A_2,Y)/\operatorname{Var}(A_2)$. Substituting the outcome model gives

\frac{\operatorname{Cov}(A_2,Y)}{\operatorname{Var}(A_2)} = \beta_A + \beta_L\frac{\operatorname{Cov}(A_2,L_1)}{\operatorname{Var}(A_2)} + \frac{\operatorname{Cov}(A_2,\varepsilon)}{\operatorname{Var}(A_2)}.

The conditional-mean assumption implies $\operatorname{Cov}(A_2,\varepsilon)=0$: iterated expectation gives $E[A_2\varepsilon]=E[A_2E(\varepsilon\mid A_2,L_1)]=0$, and $E[\varepsilon]=0$. Thus the population coefficient is $\beta_A$ plus a bias component. Its sign depends on both $\beta_L$ and the association between treatment and the omitted covariate; it is not necessarily positive.

Result: Under the centered additive linear model above, with $E[\varepsilon\mid A_2,L_1]=0$ and $0<\operatorname{Var}(A_2)<\infty$, the population coefficient from regressing $Y$ on $A_2$ alone is $$ \beta_A^{\mathrm{omit}} = \beta_A + \beta_L\frac{\operatorname{Cov}(A_2,L_1)}{\operatorname{Var}(A_2)}. $$ The coefficient shift is the product of the outcome coefficient for $L_1$ and the slope from the population linear regression of $L_1$ on $A_2$.

Worked Example: Calculating Bias From a Treatment Model

Suppose $L_1$ and $V$ are independent, centered variables with variances $4$ and $1$, respectively. Let $A_2=0.6L_1+V$, and let $Y=1.5A_2+2L_1+\varepsilon$, where $E[\varepsilon\mid A_2,L_1]=0$. Find the population coefficient of $A_2$ when $L_1$ is omitted.

The treatment variance and treatment-covariate covariance are

\operatorname{Var}(A_2) =0.6^2\operatorname{Var}(L_1)+\operatorname{Var}(V) =0.36(4)+1=2.44,

\operatorname{Cov}(A_2,L_1) =0.6\operatorname{Var}(L_1)+\operatorname{Cov}(V,L_1) =0.6(4)+0=2.4.

Therefore the coefficient shift is

2\frac{2.4}{2.44} =\frac{4.8}{2.44} =\frac{120}{61} \approx 1.967.

The omitted-covariate coefficient is $1.5+1.967\approx 3.467$. The calculation checks: $L_1$ has a positive outcome coefficient and is positively associated with $A_2$, so omitting it shifts the treatment coefficient upward.

Keeping Other Covariates in the Regression

In practice, the regression may include baseline covariates or earlier treatment and covariate history, denoted collectively by $X$. The relevant association is then not the raw covariance between $A_2$ and $L_1$, but their association after linearly removing the part explained by $X$. This qualification matters: a covariate can be associated with treatment before adjustment but have little or no residual association after adjustment.

Assume the specified outcome model is

Y=\beta_0+\beta_A A_2+\beta_L L_1+\gamma^\mathsf{T}X+\varepsilon,

with $E[\varepsilon\mid A_2,L_1,X]=0$, finite second moments, and a nonsingular covariance matrix for the regressors needed in the regressions. Define $\widetilde A_2$ as the residual from the population linear regression of $A_2$ on an intercept and $X$; define $\widetilde L_1$ analogously. Assume $\operatorname{Var}(\widetilde A_2)>0$.

Result: The coefficient of $A_2$ in the population regression of $Y$ on an intercept, $X$, and $A_2$, but not $L_1$, is $$ \beta_A^{\mathrm{omit}} = \beta_A + \beta_L\frac{\operatorname{Cov}(\widetilde A_2,\widetilde L_1)} {\operatorname{Var}(\widetilde A_2)}. $$

To derive this expression, apply the Frisch–Waugh–Lovell theorem: the coefficient of $A_2$ after controlling linearly for $X$ is the slope from regressing the residualized outcome on $\widetilde A_2$. Residualizing the specified outcome model with respect to the intercept and $X$ gives

\widetilde Y = \beta_A\widetilde A_2+\beta_L\widetilde L_1+\widetilde\varepsilon.

Here residualizing is a linear projection, so it removes the intercept and the linear component in $X$. The projection residual $\widetilde A_2$ is orthogonal to $X$ and the intercept. Also, $E[\varepsilon\mid A_2,L_1,X]=0$ implies that $\varepsilon$ is orthogonal to every square-integrable function of $(A_2,L_1,X)$, including $\widetilde A_2$. Thus $\operatorname{Cov}(\widetilde A_2,\widetilde\varepsilon)=0$. Taking the covariance with $\widetilde A_2$ and dividing by its positive variance yields the stated formula. $\square$

Worked Example: Bias After Controlling for Prior History

Let $X$ be centered with variance $1$. Suppose $A_2=0.4X+U$ and $L_1=0.7X+V$, where $X$ is independent of the centered pair $(U,V)$. Let $\operatorname{Var}(U)=2$, $\operatorname{Var}(V)=3$, and $\operatorname{Cov}(U,V)=0.8$. The specified outcome model is $Y=0.9A_2-1.2L_1+0.5X+\varepsilon$, with conditional mean-zero error. Find the coefficient shift when the regression includes $X,A_2$ but omits $L_1$.

Because the components proportional to $X$ are removed when controlling for $X$, the residuals are $\widetilde A_2=U$ and $\widetilde L_1=V$. Hence

\frac{\operatorname{Cov}(\widetilde A_2,\widetilde L_1)} {\operatorname{Var}(\widetilde A_2)} = \frac{0.8}{2}=0.4.

The coefficient shift is $(-1.2)(0.4)=-0.48$, so the omitted-model coefficient is $0.9-0.48=0.42$. The covariance specification is feasible because $0.8^2=0.64<2(3)=6$, the positive-definiteness condition for this two-variable covariance matrix. In this example, the covariate’s negative outcome coefficient makes the shift negative despite the positive residual association.

How the Treatment Model Determines the Shift

A useful special case makes the treatment-association term easy to see. After controlling for $X$, suppose the residualized treatment satisfies $\widetilde A_2=\alpha\widetilde L_1+V$, where $V$ is uncorrelated with $\widetilde L_1$. Then

\operatorname{Cov}(\widetilde A_2,\widetilde L_1) =\alpha\operatorname{Var}(\widetilde L_1), \qquad \operatorname{Var}(\widetilde A_2) =\alpha^2\operatorname{Var}(\widetilde L_1)+\operatorname{Var}(V).

Substitution shows that the shift equals $\beta_L\alpha\operatorname{Var}(\widetilde L_1)/ [\alpha^2\operatorname{Var}(\widetilde L_1)+\operatorname{Var}(V)]$. This derivation requires uncorrelatedness, not independence. It also shows why the treatment-model slope alone is not the bias: the outcome coefficient and the relative variation in treatment due to $L_1$ and the residual $V$ both matter.

Worked Example: A Covariate That Reverses the Coefficient’s Sign

Suppose centered $L_1$ has variance $1$, and $A_2=0.4L_1+V$, where $V$ is uncorrelated with $L_1$ and has variance $0.84$. Let $Y=0.5A_2-3L_1+\varepsilon$, with $E[\varepsilon\mid A_2,L_1]=0$. Calculate the omitted-model coefficient.

First, $\operatorname{Cov}(A_2,L_1)=0.4(1)=0.4$, and $\operatorname{Var}(A_2)=0.4^2(1)+0.84=1$. Therefore the coefficient shift is $(-3)(0.4/1)=-1.2$. The coefficient from omitting $L_1$ is $0.5-1.2=-0.7$, which reverses the sign of the specified coefficient $0.5$. This is an algebraic feature of the model, not evidence that the treatment’s causal effect is negative.

Assumptions, Diagnostics, and Alternatives

The derivations require a meaningful additive linear outcome specification, a well-defined population projection, finite second moments, and positive residual treatment variance. The conditional-mean-zero error assumption rules out any remaining association of the outcome error with the included treatment, omitted covariate, and controls. If the outcome has interactions or nonlinear terms, the displayed formula need not describe the coefficient shift: a single linear coefficient may then be a projection summary rather than the coefficient in the proposed outcome model.

For independent and identically distributed observations, standard laws of large numbers imply that sample OLS coefficients converge to their population projection coefficients when the requisite moments exist and the regressor covariance matrix is nonsingular. Thus, under the stated model, a large sample tends toward the omitted-model coefficient in the formula, not automatically toward $\beta_A$. This is an asymptotic statement; the finite-sample expectation of an OLS ratio need not equal the population formula.

Useful diagnostics include checking whether $L_1$ predicts $A_2$ after the included controls, whether it predicts $Y$ conditional on treatment and controls, and whether the estimated coefficient changes when $L_1$ is added. These checks do not establish that all confounding has been measured or that the linear model is correct. A small change in one fitted model does not rule out omitted variables, nonlinearities, or sampling uncertainty.

The direct alternative for this algebraic problem is to include $L_1$ in the linear regression, provided the model is correctly specified and the needed variation and rank conditions hold. Another is to compare treatment slopes within levels or strata of relevant covariates; with continuous covariates, this requires a model or method for combining those comparisons. Both alternatives use measured covariate information, unlike simply omitting it. Neither automatically resolves every causal identification or model-specification problem. In particular, the omitted-variable formula quantifies one coefficient discrepancy; it does not determine which regression coefficient answers a broader causal question.

Common pitfall: “$L_1$ predicts $A_2$, so omitting $L_1$ must bias the treatment coefficient upward.” This is wrong. The shift is the product of the treatment–covariate association and the covariate’s outcome coefficient. If either factor is negative, the shift can be negative; if their product is zero, this particular shift is zero. Moreover, a zero shift for one omitted covariate does not certify that the regression is otherwise unbiased.

Check Your Understanding

For each question, identify the relevant formula before interpreting the result. Solutions are included for checking.

In the centered model with no other regressors, suppose $\beta_L=2$, $\operatorname{Cov}(A_2,L_1)=1.5$, and $\operatorname{Var}(A_2)=3$. What is the coefficient shift?
When a regression includes $X$, should the formula use the raw covariance of $A_2$ and $L_1$, or their residual covariance after linear adjustment for $X$?
If $\beta_L=0$ but $A_2$ and $L_1$ are strongly associated, what does the one-covariate formula imply about this particular coefficient shift?
Does a coefficient shift calculated from the formula establish that the omitted-model coefficient is a causal effect? Explain briefly.

Solutions. (1) The shift is $2(1.5/3)=1$. (2) Use the covariance of $\widetilde A_2$ and $\widetilde L_1$, after each has been linearly residualized on $X$ and an intercept. (3) The shift is zero because it is $\beta_L$ multiplied by the treatment–covariate slope; this says nothing about other sources of bias or misspecification. (4) No. The formula compares a specified linear-model coefficient with an omitted-model population projection. Causal interpretation requires additional causal assumptions and a clearly defined estimand.