What Variation Can Identify a Fixed-Effects Slope?
A fixed-effects slope is identified by comparisons over time within the same unit. The comparison might be between a patient at two visits, a firm in two years, or a region in two periods. Differences between units can be substantial, but unit fixed effects absorb those differences rather than using them to estimate the slope.
This distinction is central to interpreting a fixed-effects estimate. The coefficient describes how the outcome changes with a regressor’s within-unit changes, under the model’s assumptions. It is not identified merely because units with different regressor averages also have different outcome averages.
The fixed-effects transformation subtracts each unit’s time average. Define \(\bar y_i=T_i^{-1}\sum_t y_{it}\), \(\bar x_i=T_i^{-1}\sum_t x_{it}\), and \(\bar u_i=T_i^{-1}\sum_t u_{it}\). The demeaned variables are \(y_{it}-\bar y_i\) and \(x_{it}-\bar x_i\). Subtracting the unit average of the model from each observation gives
The intercept disappears because it is constant within unit: \(\alpha_i-\alpha_i=0\). Every regressor is replaced by its deviation from that unit’s own average. Thus, observations contribute to the slope through their within-unit deviations, not through the unit averages.
This is an algebraic condition: it says whether the transformed data contain enough distinct regressor variation to determine the slope or slopes. It does not by itself make the estimate causal. A causal interpretation additionally requires an appropriate exogeneity condition, such as \(E[u_{it}\mid \alpha_i,x_{i1},\ldots,x_{iT_i}]=0\), together with the relevant sampling and model conditions. If time-varying omitted factors move with a regressor, within-unit variation alone does not remove that source of bias.
Why Between-Unit Differences Disappear
The removal of between-unit variation follows directly from the transformation. Write each regressor as its unit average plus a deviation: \(x_{it}=\bar x_i+(x_{it}-\bar x_i)\). Subtracting \(\bar x_i\) leaves only the deviation. Also, \(\sum_t(x_{it}-\bar x_i)=0\), so a unit’s average regressor value cannot create a nonzero within-transformed cross-product with the outcome.
The within-estimator minimizes the sum of squared transformed residuals. For any proposed \(\beta\), the fitted unit intercept in the original model is \(\hat\alpha_i=\bar y_i-\bar x_i'\beta\). Substituting this into the least-squares objective yields the within objective:
Differentiating this quadratic objective with respect to \(\beta\) gives the normal equations. When the within-regressor matrix has full rank, the solution is unique:
The inverse exists precisely when the matrix has rank \(K\). With one regressor, the formula is the sum of within-unit regressor–outcome cross-products divided by the sum of within-unit regressor squares. These expressions make the identifying information visible: all terms use deviations from unit means.
Worked Example: A Slope From Within-Unit Changes
Suppose three hypothetical units are observed twice. The table gives a regressor \(x\) and an outcome \(y\). Calculate the fixed-effects slope for this one-regressor model.
| Unit | Period | \(x_{it}\) | \(y_{it}\) |
|---|---|---|---|
| A | 1 | 1 | 5 |
| A | 2 | 3 | 9 |
| B | 1 | 2 | 8 |
| B | 2 | 4 | 10 |
| C | 1 | 5 | 12 |
| C | 2 | 5 | 14 |
First calculate the unit averages. For A, \(\bar x_A=2\) and \(\bar y_A=7\); for B, \(\bar x_B=3\) and \(\bar y_B=9\); for C, \(\bar x_C=5\) and \(\bar y_C=13\). The within deviations are:
| Unit | Period | \(x_{it}-\bar x_i\) | \(y_{it}-\bar y_i\) | Product | Squared \(x\) deviation |
|---|---|---|---|---|---|
| A | 1 | -1 | -2 | 2 | 1 |
| A | 2 | 1 | 2 | 2 | 1 |
| B | 1 | -1 | -1 | 1 | 1 |
| B | 2 | 1 | 1 | 1 | 1 |
| C | 1 | 0 | -1 | 0 | 0 |
| C | 2 | 0 | 1 | 0 | 0 |
The numerator is the sum of the products, \(2+2+1+1+0+0=6\). The denominator is the sum of squared \(x\) deviations, \(1+1+1+1+0+0=4\). Therefore,
The estimated within-unit association is 1.5 outcome units per one-unit increase in \(x\), under the fixed-effects model. Unit C has no change in \(x\), so its observations add nothing to either the numerator or denominator. Differences in average \(x\) across A, B, and C do not determine this estimate: they were removed when each unit’s average was subtracted.
When Within Variation Is Absent
A regressor can vary substantially across units but still provide no information for its fixed-effects slope. If it is constant over time within each unit, its demeaned value is zero everywhere. Its between-unit differences are absorbed by the unit intercepts.
Worked Example: A Regressor With No Within-Unit Change
Consider three hypothetical units observed in two periods. Suppose \(x\) is constant within each unit: unit A has \(x=1\), unit B has \(x=2\), and unit C has \(x=3\) in both periods. Suppose the outcomes are \(y=2,4,6\), respectively, in both periods.
The unit means equal the observed values. Hence, for every observation, \(x_{it}-\bar x_i=0\) and \(y_{it}-\bar y_i=0\). The one-regressor denominator is
The fixed-effects slope formula would divide by zero, so it does not yield a unique estimate. This is not a computational inconvenience that can be fixed by using more decimal places. In the original model, any proposed slope \(\beta\) can be paired with an intercept \(\alpha_i=2x_i-\beta x_i\), producing the same fitted outcome \(2x_i\) for each unit. The data cannot distinguish among those slopes.
The raw data display a perfect between-unit relationship: units with larger \(x\) also have larger \(y\). But that relationship cannot identify the fixed-effects slope because each unit’s \(x\) never changes. The conclusion is nonidentification, not evidence that the slope equals zero.
Multiple Regressors Need Independent Changes
With several regressors, it is not enough for each one to change within units. Their changes must also be sufficiently independent. If two regressors always move in exact proportion within units, the transformed data cannot tell which regressor accounts for the outcome change. The matrix rank condition captures both requirements: nonzero variation and no exact linear dependence among the demeaned regressors.
Worked Example: Separating Two Within-Unit Slopes
Suppose two hypothetical units are each observed in three periods. The following table gives the variables after subtracting each unit’s time average. Both units have the same demeaned patterns.
| Unit | Period | \(x_{1,it}-\bar x_{1,i}\) | \(x_{2,it}-\bar x_{2,i}\) | \(y_{it}-\bar y_i\) |
|---|---|---|---|---|
| A | 1 | -1 | -1 | -1 |
| A | 2 | 0 | 2 | -4 |
| A | 3 | 1 | -1 | 5 |
| B | 1 | -1 | -1 | -1 |
| B | 2 | 0 | 2 | -4 |
| B | 3 | 1 | -1 | 5 |
For each unit, the two regressor-deviation vectors have squared lengths \(1+0+1=2\) and \(1+4+1=6\). Their cross-product is \(1+0-1=0\). Since the pattern occurs twice, the within cross-product matrix is
Its rank is 2, so the two slopes can be separately identified. The outcome deviations satisfy \(y_{it}-\bar y_i=3(x_{1,it}-\bar x_{1,i})-2(x_{2,it}-\bar x_{2,i})\). The two cross-products with the outcome are
Thus the coefficient vector is obtained by solving
This example shows why the rank check concerns the full set of within-regressor patterns. If instead \(x_{2,it}-\bar x_{2,i}=2(x_{1,it}-\bar x_{1,i})\) for every observation, the two columns would be perfectly dependent. The transformed outcome could identify only the combination \(\beta_1+2\beta_2\), not \(\beta_1\) and \(\beta_2\) separately.
Identification, Precision, and Interpretation
A nonzero within sum of squares is enough for algebraic identification with one regressor, but the amount of within variation also matters for precision. If a regressor changes only slightly, or changes in very few units, the slope can be identified while remaining imprecisely estimated. Identification asks whether a unique coefficient is determined in principle by the model and regressor patterns; precision asks how uncertain its estimate is given the disturbances and available data.
Likewise, a full-rank within matrix does not establish that the slope is causal. Fixed effects remove time-constant unit characteristics, including those associated with the regressors, but they do not automatically remove time-varying omitted variables, reverse causation, or measurement error. The exogeneity assumptions must fit the application. When they do not, the coefficient may describe a within-unit association without representing a causal effect.
Common Mistakes
- Using between-unit association as evidence of a fixed-effects slope. Units with high regressor averages may have high outcome averages, but those averages vanish under demeaning. Check within-unit deviations instead.
- Concluding that no within change means a zero slope. If the regressor is constant within every unit, the slope is not identified; the data do not establish that its value is zero.
- Checking regressors one at a time in a multiple-regression model. Each regressor may vary and still be inseparable from another if their within-unit changes are perfectly dependent. Check the rank of the full within cross-product matrix.
- Equating identification with causality. Full rank ensures the transformed regression has a unique coefficient estimate. It does not establish the exogeneity assumptions needed for a causal interpretation.
Check Your Understanding
Use the within-transformation and rank condition to answer each question.
- A regressor differs greatly across units but is constant over time within every unit. Can it identify a fixed-effects slope? Explain.
- For one regressor, what sample condition makes the fixed-effects slope denominator positive?
- Why does a unit’s average regressor value not contribute to the fixed-effects slope formula?
- Two regressors both change within units, but the demeaned second regressor is always three times the demeaned first. Which coefficient information is identified?
- Does full rank of the within-regressor matrix by itself justify a causal interpretation? State one additional type of assumption that is needed.