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Leverage and Influential Observations · Tutorial 552 of 991

DFFITS and Casewise Prediction Changes

Use DFFITS to interpret how much an observation’s fitted value changes when it is left out of an ordinary least-squares regression.

Intermediate 10 min read

What You'll Learn

  • Define DFFITS using the full-fit and case-deleted fitted values
  • Identify the residual scale and leverage used to standardize the change
  • Interpret the sign of DFFITS in the response’s units and direction
  • Calculate DFFITS from a hypothetical regression diagnostic summary
  • Use a screening cutoff as a rule of thumb, not a deletion rule

Why Look at a Casewise Prediction Change?

A regression fit can be sensitive to particular observations. One way to describe that sensitivity is to ask what happens to the fitted value at an observation’s predictor settings when that observation is removed and the model is refitted. DFFITS expresses this casewise prediction change on a scale that accounts for residual uncertainty and leverage.

DFFITS is an influence diagnostic, not a test of whether an observation is erroneous or should be removed. Its sign gives a direction of change; its magnitude indicates how large the change is relative to its scale. The calculation is for the fitted value at the case’s predictors, not for the observed response itself.

Definition and Model Conditions

Consider an ordinary least-squares regression with \(n\) observations and \(p\) fitted coefficients, including an intercept if the model has one. Write \(y_i\) for the observed response, \(\hat{y}_i\) for the fitted value from the full data, \(e_i=y_i-\hat{y}_i\) for the residual, and \(h_{ii}\) for the observation’s diagonal hat value. Let \(\hat{y}_{(i)}\) be the fitted value at observation \(i\)’s predictor settings after observation \(i\) is omitted and the model is refitted. Let \(s_{(i)}\) be the residual standard error from that deleted fit.

The conventional DFFITS calculation uses an unweighted, full-rank OLS model. The usual model conditions are a correctly specified linear mean, independent errors with mean zero and common variance, and a design for which deleting the case leaves the model estimable. Normal errors are additionally used for exact finite-sample distributional results for studentized quantities; the DFFITS calculation itself is used descriptively.

Definition: For an observation with \(0<h_{ii}<1\) and a defined deleted-fit residual standard error, its DFFITS value is $$ \operatorname{DFFITS}_i=\frac{\hat{y}_i-\hat{y}_{(i)}}{s_{(i)}\sqrt{h_{ii}}}. $$ The numerator is the full-fit prediction minus the prediction from the fit with case \(i\) deleted. The denominator puts this change on a dimensionless scale.

The factor \(s_{(i)}\) measures residual uncertainty in the deleted fit. The factor \(\sqrt{h_{ii}}\) makes the scale specific to the observation’s leverage. DFFITS therefore is not simply a raw change in response units: it describes that change relative to a leverage-scaled residual standard error.

Reading the Sign and Magnitude

If DFFITS is positive, then \(\hat{y}_i>\hat{y}_{(i)}\): including the observation produces a higher fitted value at its predictor settings than omitting it does. If DFFITS is negative, the full-data fitted value is lower. Because \(s_{(i)}\sqrt{h_{ii}}\) is positive, the sign of DFFITS is the sign of \(\hat{y}_i-\hat{y}_{(i)}\).

A value near zero indicates a small casewise fitted-value change relative to its scale. A larger absolute value indicates a more substantial standardized change. This is a diagnostic description, not a p-value or a universal judgment that a case is problematic.

Worked Example: A Positive DFFITS Value

Suppose a hypothetical OLS diagnostic summary for one observation reports \(h_{ii}=0.20\), \(e_i=4.00\) response units, \(\hat{y}_i=50.0\) response units, and \(s_{(i)}=5.00\) response units. Use the OLS deletion relation \(\hat{y}_i-\hat{y}_{(i)}=h_{ii}e_i/(1-h_{ii})\) to find the deleted-fit prediction and DFFITS.

First calculate the fitted-value change: $$ \hat{y}_i-\hat{y}_{(i)} =\frac{0.20(4.00)}{1-0.20} =\frac{0.80}{0.80} =1.00\text{ response unit}. $$

Thus \(\hat{y}_{(i)}=50.0-1.00=49.0\) response units. The DFFITS value is $$ \operatorname{DFFITS}_i =\frac{50.0-49.0}{5.00\sqrt{0.20}} =\frac{1.00}{2.236} =0.4472. $$

The positive sign means the full-data fitted value is higher. The change is \(0.4472\) on the DFFITS scale; it is not a change of \(0.4472\) response units. The raw change is \(1.00\) response unit.

A Useful Formula and Its Justification

The OLS case-deletion relation gives the difference between the two fitted values in terms of the full-fit residual and leverage: $$ \hat{y}_i-\hat{y}_{(i)}=\frac{h_{ii}e_i}{1-h_{ii}}. $$ Substituting this relation into the definition gives a convenient equivalent expression. The externally studentized deleted residual is \(t_i=e_i/[s_{(i)}\sqrt{1-h_{ii}}]\), so the same DFFITS value can be written in terms of that residual.

$$ \begin{aligned} \operatorname{DFFITS}_i &=\frac{h_{ii}e_i/(1-h_{ii})}{s_{(i)}\sqrt{h_{ii}}}\\ &=\frac{e_i}{s_{(i)}\sqrt{1-h_{ii}}} \sqrt{\frac{h_{ii}}{1-h_{ii}}}\\ &=t_i\sqrt{\frac{h_{ii}}{1-h_{ii}}}. \end{aligned} $$

This shows why residual size alone does not determine DFFITS: the studentized deleted residual is multiplied by a factor that grows with leverage. It also shows that, when the quantities are defined, DFFITS has the same sign as the residual \(e_i\). The displayed relation is useful for calculation and interpretation; DFFITS itself remains the standardized change in the fitted value.

Worked Example: A Negative DFFITS Value

For a second hypothetical case, suppose \(h_{ii}=0.40\), \(e_i=-6.00\) response units, \(\hat{y}_i=72.0\) response units, and \(s_{(i)}=4.00\) response units. Find the deleted prediction and DFFITS.

The fitted-value change is $$ \hat{y}_i-\hat{y}_{(i)} =\frac{0.40(-6.00)}{1-0.40} =\frac{-2.40}{0.60} =-4.00\text{ response units}. $$ Therefore \(\hat{y}_{(i)}=72.0-(-4.00)=76.0\) response units. Then $$ \operatorname{DFFITS}_i =\frac{72.0-76.0}{4.00\sqrt{0.40}} =\frac{-4.00}{2.530} =-1.581. $$

The negative sign says the full-data fitted value is lower than the deleted-fit prediction. The absolute value describes a substantial standardized change in this hypothetical calculation. It does not, by itself, establish why the case has that influence or whether the case should be excluded.

Magnitude and Screening

A commonly used screening guideline flags cases for closer examination when $$ |\operatorname{DFFITS}_i|>2\sqrt{\frac{p}{n}}, $$ where \(p\) is the number of fitted coefficients and \(n\) is the full sample size. This is a rule of thumb, not a significance test: it does not give a probability that a case is influential, and it is not a rule for deleting observations. The guideline’s value depends on the model size and sample size, so the cutoff should be reported with those quantities if used.

Worked Example: Applying a Screening Guideline

Suppose a hypothetical regression has \(n=30\) observations and \(p=3\) fitted coefficients, and a case has \(\operatorname{DFFITS}_i=-1.581\), as in the previous calculation. Compute the guideline and interpret the comparison.

The cutoff is $$ 2\sqrt{\frac{p}{n}} =2\sqrt{\frac{3}{30}} =2\sqrt{0.10} =0.6325. $$

Since \(|-1.581|=1.581>0.6325\), the case exceeds this screening guideline and merits investigation. This comparison does not prove that the observation is an error, nor does it prescribe removing it. Its data source, measurement, and role in the analysis should be checked before deciding how to report or handle it.

Leverage Can Amplify a Prediction Change

Two cases with similar residual sizes can have different DFFITS values because their leverages differ. The following hypothetical diagnostic summaries illustrate the calculation. In each row, the reported full-fit residual, leverage, and deleted-fit residual standard error are used with the same deletion relation.

Case\(h_{ii}\)\(e_i\) (response units)\(s_{(i)}\) (response units)Fitted-value change (response units)DFFITS
A0.640.503.000.8890.370
B0.043.003.000.1250.208

For case A, the change is \(0.64(0.50)/(1-0.64)=0.889\) response units, and DFFITS is \(0.889/[3.00\sqrt{0.64}]=0.370\). For case B, the change is \(0.04(3.00)/(1-0.04)=0.125\) response units, and DFFITS is \(0.125/[3.00\sqrt{0.04}]=0.208\). Case B has the larger residual, but case A has the larger DFFITS value because its greater leverage amplifies the casewise prediction change.

Conditions, Limits, and a Common Pitfall

The definition requires a meaningful full and deleted fit and a positive deleted-fit residual standard error. If \(h_{ii}=1\), the factor \(1-h_{ii}\) in the deletion relation is zero; deleting that case may make the fit or prediction unavailable. Such a case does not have an ordinary finite DFFITS value from this formula. More generally, leverage close to one can make the deletion change very large, so calculations should be checked against the design and the actual refitted model.

A common mistake is to interpret a large absolute DFFITS value as proof that the observation is a data error and then remove it. That reasoning confuses a diagnostic flag with a decision. DFFITS identifies a case whose inclusion is associated with a comparatively large change in its own fitted value. The case may be valid, and its influence may be relevant to the scientific question. Check its provenance and measurement, examine the model context, and report the sensitivity responsibly rather than deleting it solely because of a cutoff.

Key takeaway: DFFITS is the full-fit fitted value minus the case-deleted fitted value, divided by \(s_{(i)}\sqrt{h_{ii}}\). Its sign gives the direction of that change, and its absolute value gives its size relative to a leverage-scaled residual uncertainty. A screening guideline prompts investigation; it is not a deletion rule.

Check Your Understanding

Use the definitions and calculation principles above to answer the following questions.

  1. In words, what two fitted values are subtracted to form the numerator of DFFITS?
  2. A case has a negative DFFITS value. Which fitted value is lower: the full-data fitted value or the deleted-fit prediction?
  3. For a hypothetical case with \(h_{ii}=0.25\), \(e_i=2.00\) response units, and \(s_{(i)}=4.00\) response units, calculate the fitted-value change and DFFITS.
  4. Why can two cases with equal residual sizes have different DFFITS values?
  5. What does exceeding \(2\sqrt{p/n}\) indicate, and what does it not justify by itself?