Artificially Strong Correlations After Regression Imputation

Why Regression Predictions Can Create Stronger Associations

Regression imputation fills a missing outcome with its predicted value from a regression model. A prediction lies on the fitted regression line, so it has no residual deviation from that line. Observed values generally do have residual deviations. When predicted values replace missing observations, the completed data can therefore show a tighter association between the outcome and its predictors than the data support.

This is a property of the completed data, not evidence that the underlying relationship is stronger. It can matter when correlations, regression coefficients, or other summaries are calculated as if every completed value were observed. The earlier material on the law of total variance in regression imputation considers the variance aspect; here the focus is how the association between a variable and its predictor can be altered.

Definition: In deterministic regression imputation, fit a regression model using cases with observed outcome \(Y\), then replace each missing \(Y_i\) with its fitted prediction \(\hat{Y}_i\). For a simple linear regression with predictor \(X\), the imputed value is \(\hat{Y}_i=\hat{\alpha}+\hat{\beta}X_i\). The prediction contains no added residual variation.

The calculation relies on a regression model that is suitable for the relationship being imputed. In the simple linear case, suppose the conditional mean is correctly represented by \(E(Y\mid X)=\alpha+\beta X\), and the observed cases provide a reasonable basis for estimating that relationship. The observed cases must cover the predictor values for which predictions are needed; otherwise, imputation extrapolates beyond the observed data. For inference about the missing outcomes, the missingness process and the information used in the model also matter. Regression imputation does not by itself make missingness ignorable or correct a misspecified model.

How Prediction Changes the Scatterplot

An ordinary regression prediction is the fitted mean at a given predictor value. An observed outcome can be written as its fitted value plus a residual, \(Y_i=\hat{Y}_i+e_i\). In deterministic regression imputation, the completed value for a missing case is just \(\hat{Y}_i\): its residual is set to zero. Thus, in a scatterplot of \(Y\) against \(X\), imputed cases lie exactly on the fitted line, while observed cases may lie above or below it.

A short algebraic decomposition makes this visible. For a completed data set of \(n\) cases, write \(Y_i^*=\hat{\alpha}+\hat{\beta}X_i+e_i^*\), where \(e_i^*\) is the observed residual for an observed case and zero for an imputed case. Let \(S_{xx}=\sum_i(X_i-\bar{X})^2\), and define \(S_{xy}\) and \(S_{yy}\) in the same centered-sum way. Then:

S_{xy}=\hat{\beta}S_{xx}+\sum_i(X_i-\bar{X})e_i^*

The fitted-line component contributes \(\hat{\beta}S_{xx}\). The residual component depends on the observed residuals that remain. When missing cases receive zero residuals, the completed data may have less residual variation around the line than would a data set in which those outcomes had actually been observed. The correlation \(r=S_{xy}/\sqrt{S_{xx}S_{yy}}\) depends on both the cross-product and the two variances, so it can strengthen even when the fitted slope does not change.

Worked Example: One Prediction Increases the Correlation

Consider three hypothetical cases with observed pairs \((X,Y)=(-1,-1),(0,1),(1,0)\). A fourth case has \(X=2\) and missing \(Y\). We first fit the simple linear regression to the three observed cases.

The observed means are \(\bar{X}=0\) and \(\bar{Y}=0\). The centered sums are \(S_{xx}=(-1)^2+0^2+1^2=2\), \(S_{yy}=(-1)^2+1^2+0^2=2\), and \(S_{xy}=(-1)(-1)+(0)(1)+(1)(0)=1\). Therefore the fitted slope is \(\hat{\beta}=S_{xy}/S_{xx}=1/2\), and the intercept is \(\hat{\alpha}=\bar{Y}-\hat{\beta}\bar{X}=0\). The observed correlation is \(r=1/\sqrt{2(2)}=0.5000\).

At \(X=2\), deterministic regression imputation gives \(\hat{Y}=0+(1/2)(2)=1\). The completed pairs are \((-1,-1),(0,1),(1,0),(2,1)\), with \(\bar{X}=0.5\) and \(\bar{Y}=0.25\). Their centered sums are:

S_{xx}=5,\qquad S_{xy}=2.5,\qquad S_{yy}=2.75,\qquad r=\frac{2.5}{\sqrt{5(2.75)}}=0.6742.

The completed-data correlation, about 0.674, is stronger than the observed correlation of 0.500. The added point lies exactly on the fitted line. This example demonstrates that one deterministic prediction can strengthen the correlation; it does not establish that every prediction must do so.

More Imputed Cases and the Visual Pattern

The effect can be more noticeable when a larger share of the outcome values is imputed. The following example uses the same observed regression, so the comparison isolates what happens as additional predicted values are included.

Worked Example: More Predictions Make the Line Look Tighter

Suppose the three observed pairs are \((0,0),(1,2),(2,1)\). Their means are \(\bar{X}=1\) and \(\bar{Y}=1\), with \(S_{xx}=2\), \(S_{yy}=2\), and \(S_{xy}=1\). Hence \(\hat{\beta}=1/2\), \(\hat{\alpha}=1/2\), and the observed correlation is \(1/\sqrt{2(2)}=0.5000\).

Now suppose two more cases have \(X=3\) and \(X=4\), with missing outcomes. Their predictions are \(\hat{Y}=0.5+(0.5)(3)=2\) and \(\hat{Y}=0.5+(0.5)(4)=2.5\). The completed-data means are \(\bar{X}=2\) and \(\bar{Y}=1.5\). Centering each completed pair gives \(S_{xx}=10\), \(S_{xy}=5\), and \(S_{yy}=4\), so:

r=\frac{5}{\sqrt{10(4)}}=0.7906.

The correlation rises from 0.5000 to about 0.791. The fitted slope remains \(S_{xy}/S_{xx}=5/10=0.5000\); what changed is the arrangement and spread of points around the fitted relationship. The following coordinate sketch marks observed cases with O and imputed cases with I:

Scatterplot (vertical axis \(Y\), horizontal axis \(X\))
\(Y=2.5:\quad\) . . . . I
\(Y=2.0:\quad\) . O . I .
\(Y=1.0:\quad\) . . O . .
\(Y=0.0:\quad\) O . . . .
\(\phantom{Y=0.0:}\quad X=0\quad 1\quad 2\quad 3\quad 4\)

The two imputed points fall exactly on the fitted line through the observed data. The observed points do not all fall on that line. In a plot, the imputed points may therefore make the overall pattern appear more linear than the observed values alone suggest.

Extrapolation Can Make the Distortion More Severe

A strong completed-data correlation can also arise when a prediction is made at a predictor value far outside the observed range. Such a prediction is especially difficult to justify: the linear relationship estimated among observed cases may not continue that far. The arithmetic can show how a seemingly orderly point can have substantial influence.

Worked Example: A Prediction Far Beyond the Observed Range

Use the observed data \((0,0),(1,2),(2,1)\) from the previous example. The fitted line is \(\hat{Y}=0.5+0.5X\), and the observed correlation is 0.5000. Suppose one additional case has \(X=10\), far beyond the observed range from 0 to 2, and its missing outcome is imputed as \(\hat{Y}=0.5+(0.5)(10)=5.5\).

For the four completed pairs, \(\bar{X}=3.25\) and \(\bar{Y}=2.125\). The centered sums are \(S_{xx}=62.75\), \(S_{xy}=31.375\), and \(S_{yy}=17.1875\). Therefore:

r=\frac{31.375}{\sqrt{62.75(17.1875)}}\approx 0.9554.

The computed correlation is about 0.955, even though the observed correlation was 0.500. The fitted slope in the completed data is still \(31.375/62.75=0.5000\). The extreme predictor value contributes substantially to the centered sums, and its imputed outcome lies exactly on the fitted line. This calculation is not evidence that the relationship is genuinely that strong: it illustrates why predictions outside the observed predictor range require particular caution.

Assumptions, Diagnostics, and Alternatives

A careful interpretation starts by checking what the regression model is being asked to do. For a continuous outcome and one predictor, a linear conditional mean is a reasonable starting point only when the observed data support an approximately linear relationship over the predictor range used for imputation. A fitted line estimated from observed outcomes must also be relevant to the missing cases. If the missing cases occupy a different predictor range or differ in important measured characteristics, the observed relationship may not transfer to them.

Plot observed outcomes against predictors. Look for curvature, changing spread, clusters, and sparse regions. Predictions in unsupported regions are extrapolations, not ordinary interpolation.
Mark which values were imputed. A plot that distinguishes observed and imputed points can reveal a row of points lying exactly on the fitted line.
Compare observed residuals with imputed cases. Observed residuals show departures from the fitted line; imputed cases have residual zero by construction. That contrast is the source of the artificial alignment.
Track the fraction and location of missing outcomes. A large fraction of predictions, especially at extreme predictor values, can make the completed-data pattern depend heavily on the imputation model.

These checks are diagnostic, not a way to recover the unobserved outcomes. The missing values cannot be checked directly against their true values. A completed-data correlation is also not a diagnostic for whether the assumed model or missingness assumptions are correct.

The main alternatives have different limitations. Complete-case analysis avoids inserting predictions but discards cases with missing outcomes and can be biased or inefficient, depending on the missingness process. Mean imputation does not use the predictor-specific relationship and can distort associations in other ways. Deterministic regression imputation uses predictors but removes residual spread for imputed cases. Stochastic regression imputation, treated next, differs by adding random variation; it is not the same as simply treating fitted values as observed outcomes. None of these choices should be taken as a universal fix for missing data.

Key takeaway: A deterministic prediction is a fitted mean, not an observed outcome with its own residual deviation. When imputed values are included in a completed-data correlation, the resulting association may be artificially strong. Report and interpret it with the imputation method and its assumptions in view.

Common Mistake

Wrong reasoning: “The completed-data correlation is higher, so the relationship between the variables must be stronger.” This treats imputed predictions as if they carried the same information as observed outcomes.

Correction: The correlation is a calculation on the completed values. Because deterministic regression imputation places every imputed point directly on the fitted line, the calculation can reflect the method’s zero-residual assignments rather than stronger evidence in the data. Check the observed-only pattern, mark imputed cases, and examine predictor overlap before interpreting the completed-data correlation.

Check Your Understanding

For each question, answer first, then compare with the solution immediately below it.

A regression prediction replaces a missing outcome. What residual deviation from the fitted line does that imputed value have?
Solution: It has residual deviation zero by construction: the imputed value equals the fitted prediction.
In a hypothetical analysis, the observed correlation is 0.40 and the correlation after deterministic regression imputation is 0.68. Does this alone show that the underlying association is stronger?
Solution: No. The increase may result from imputed points lying exactly on the fitted line. The completed-data correlation alone cannot establish that the underlying association is stronger.
A fitted simple regression has slope \(0.6\), \(S_{xx}=8\), and the completed data have \(S_{xy}=4.8\). What is the completed-data slope \(S_{xy}/S_{xx}\), and does this calculation establish the completed-data correlation?
Solution: The slope is \(4.8/8=0.6000\). No. Correlation also depends on \(S_{yy}\), so \(S_{xy}\) and \(S_{xx}\) alone do not determine it.
Observed predictor values range from 2 to 9. A missing outcome at \(X=25\) is filled using the fitted line. Name the main diagnostic concern.
Solution: The prediction is an extrapolation far beyond the observed predictor range. The observed cases may not support the linear relationship at \(X=25\), so the imputed value and any resulting correlation need particular caution.