Why Polynomial Terms Must Follow Their Predictor
A regression model may allow a continuous predictor to have a curved relationship with an outcome by including its square or cube. If the predictor has missing values, the polynomial terms are missing too: they are not separate measurements that can be filled in independently. Each must equal the corresponding function of the predictor value in that completed data set.
This tutorial focuses on that calculation within chained equations. The earlier material on passive imputation establishes the general idea: a derived variable is updated from its inputs rather than receiving its own stochastic imputation. Here the derived variables are \(X^2\) and \(X^3\), where \(X\) is an incomplete continuous predictor. The goal is to maintain the exact polynomial relationship after every new draw of \(X\).
The superscripts \((2)\) and \((3)\) label the derived terms; they do not denote different imputation draws. If \(X_i\) is observed, its original value is retained and the terms are calculated from it. If \(X_i\) is missing, the imputation procedure draws a value for \(X_i\), and both terms are calculated from that draw. Neither \(X^2\) nor \(X^3\) gets an independent random draw.
The Update Rule and Its Justification
Let \(t\) denote a point in the chained-equations procedure at which the predictor has a current completed value \(X_i^{(t)}\). The polynomial terms must be synchronized with that value. After the next conditional draw updates \(X_i^{(t)}\) to \(X_i^{(t+1)}\), the update is:
This rule follows directly from the definition of a deterministic transformation. For any fixed value \(x\), there is exactly one value \(x^2\) and exactly one value \(x^3\). Once a new \(x\) is drawn, retaining the terms computed from an older value would violate those defining identities. For example, if an earlier draw was 2 and a later draw is 3, the corresponding terms change from 4 and 8 to 9 and 27.
A polynomial analysis model might use a linear predictor such as \(\beta_0+\beta_1X+\beta_2X^2\), or \(\beta_0+\beta_1X+\beta_2X^2+\beta_3X^3\). Passive recomputation guarantees that each completed data set contains columns that satisfy the requested algebraic definitions. It does not by itself guarantee that the conditional imputation models are compatible with this analysis model, correctly specified, or adequate for valid inference. Those are separate modeling questions.
Worked Example: Refreshing a Square After an Imputation Draw
Suppose a hypothetical study records a continuous predictor \(X\), measured in tens of units, for four people. The recorded values are 1.2, missing, 2.0, and 2.5. For the missing value, suppose the imputation procedure produces the hypothetical draw \(X=1.6\). Calculate the completed \(X^2\) column.
The units of \(X^2\) are squared tens-of-units. The observed values stay as recorded. For the second person, use the imputed predictor value and square it:
Thus, the completed square column is 1.44, 2.56, 4.00, and 6.25. The imputation draw \(1.6\) is not replaced by a separate draw for the square; its derived value is exactly 2.56. If another imputation draw later changes this person’s \(X\) to 1.9, the square must be refreshed to \(1.9^2=3.61\).
Raw and Centered Polynomial Terms
Polynomial terms can be formed from the raw predictor or from a centered predictor. With a fixed reference value \(c\), define \(U=X-c\), then use \(U^2\) and, if required, \(U^3\). The same passive rule applies: after every update of \(X\), calculate \(U\), \(U^2\), and \(U^3\) from that current value. The reference \(c\) must be specified consistently across completed data sets; changing it between imputations would make the derived columns represent different transformations.
For a model containing all powers up to degree three, centering changes the coefficient representation but not the set of curves the model can express. This follows by expanding the centered terms:
Each centered power is a combination of the raw powers and the intercept. Conversely, \(X=U+c\), so raw powers can be expressed as combinations of \(1,U,U^2,U^3\). Therefore the two bases span the same polynomial model when the corresponding lower-order terms and intercept are included. This algebraic fact does not make the coefficients numerically identical, nor does it justify changing the imputation procedure.
Worked Example: Recomputing a Centered Square and Cube
For a hypothetical predictor measured in years, choose the fixed center \(c=10\). A missing value receives the illustrative imputation draw \(X=12.4\) years. Calculate the centered value and its square and cube.
First, \(U=X-c=12.4-10=2.4\) years. Then:
If a later draw changes \(X\) to 8.5 years, the centered value becomes \(U=8.5-10=-1.5\) years. Its refreshed terms are \(U^2=(-1.5)^2=2.25\ \text{years}^2\) and \(U^3=(-1.5)^3=-3.375\ \text{years}^3\). The negative cubic value is correct: an odd power retains the sign of its input. Keeping 5.76 and 13.824 after the change would leave terms tied to the old draw, not the current predictor.
Where the Refresh Belongs in Chained Equations
In a chained-equations iteration, an incomplete \(X\) is updated by drawing values from its specified conditional imputation model. The polynomial columns should then be refreshed before they are used as current derived values elsewhere in the iteration or in the completed data set. If several conditional updates occur, the terms must track the latest value of \(X\), not merely the value at the start of the iteration.
For each missing entry of \(X\), obtain a draw from the chosen conditional imputation model. Retain observed \(X\) values.
For every row, including rows where \(X\) was observed, compute the required powers from the current completed value. For centered terms, first apply the fixed centering constant.
Verify that the stored square and cube match the current \(X\) values, subject only to numerical storage precision.
Do not treat stale derived columns as current predictors. Retain the refreshed terms in each completed data set used for analysis.
The imputation model for \(X\) still needs to be chosen and assessed for the scientific setting. In particular, algebraic consistency of the columns is not evidence that the conditional distribution used to draw missing \(X\) adequately reflects the outcome model or other relevant associations. Analysis-model compatibility, missingness assumptions, and the quality of the imputation model require their own scrutiny.
Worked Example: Detecting a Stale Cubic Term
In a hypothetical completed data set, one person’s predictor was first imputed as \(X=1.5\), and the stored cubic term was calculated as \(1.5^3=3.375\). A later update replaces the predictor draw with \(X=1.2\). What should the current cubic term be, and what error occurs if the stored value is left unchanged?
The refreshed term is:
The stored value 3.375 is stale because it equals the cube of the old draw, not the current one. The discrepancy is \(3.375-1.728=1.647\). A regression using the stale column would not use the intended cubic transformation of the current predictor. Recomputing the term restores the exact identity; it does not, by itself, diagnose or correct any other imputation-model problem.
Diagnostics, Alternatives, and Limitations
The primary diagnostic is a direct row-by-row identity check in every completed data set. For each row, compare the stored square with the square of the current predictor, and the stored cube with its cube. A useful numerical check compares the absolute discrepancy with a small tolerance appropriate to the software’s storage precision. With centered terms, check against \(X-c\), not against \(X\). Also inspect whether values are missing where the predictor is missing after completion, whether a transformation was inadvertently left unchanged after an update, and whether the analysis uses the intended raw or centered terms.
| Approach | What it does | Main concern |
|---|---|---|
| Passive recomputation | Calculates each power from the current imputed predictor. | Preserves the deterministic definitions, but does not ensure compatible conditional imputation models. |
| Independent imputation of powers | Assigns stochastic values to \(X^2\) or \(X^3\) separately from \(X\). | Can create rows where the powers are not powers of the recorded \(X\). |
| Omitting polynomial terms | Uses a model without the selected nonlinear terms. | May fail to represent the intended nonlinear association if curvature is scientifically relevant. |
A common mistake is to impute \(X\) and \(X^2\) as though they were two unrelated incomplete variables, arguing that both columns need completed values. The correction is that \(X^2\) is determined once \(X\) is known: independently imputing it can contradict the definition. The converse mistake is to conclude that passive recomputation solves every nonlinear-imputation problem. It only enforces the algebraic relationship; it does not establish that the imputation model preserves the relevant analysis-model structure.
Other limitations remain. Extreme imputed values can produce very large squares or cubes, especially when the predictor is on a large numerical scale. Scaling or centering can make numerical calculations more stable, but does not repair an inappropriate conditional model. A cubic term is also highly related to lower powers in many data sets, which can make coefficient estimates unstable; that is a modeling and data-information issue, not a reason to impute the terms independently. Finally, any fixed centering constant and units must be applied consistently in every imputation and in the analysis.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- A missing predictor receives the draw \(X=-2.5\). What are \(X^2\) and \(X^3\), and which term is negative?
- A centered cubic uses the fixed reference \(c=4\). For a current value \(X=6\), calculate \(U\), \(U^2\), and \(U^3\), where \(U=X-c\).
- After an update, \(X\) changes from 3 to 2. What should happen to the square and cube, and why?
- Does exact equality between every stored polynomial term and its transformation of \(X\) prove that the imputation model is compatible with the analysis model? Explain.
Solutions. 1. \(X^2=(-2.5)^2=6.25\), and \(X^3=(-2.5)^3=-15.625\); only the cube is negative because the input is negative and the power is odd. 2. \(U=6-4=2\), \(U^2=2^2=4\), and \(U^3=2^3=8\). 3. Recalculate the terms from the new current value: the square changes from \(3^2=9\) to \(2^2=4\), and the cube changes from \(3^3=27\) to \(2^3=8\). The derived terms must correspond to the current predictor. 4. No. The identity checks establish algebraic consistency only; they do not establish that the conditional imputation models are compatible with the substantive analysis model or otherwise adequate.