Tutorials › Regression Analysis (Undergraduate) › Cook’s Distance from Deleted Fits

Leverage and Influential Observations · Tutorial 550 of 991

Cook's Distance from Deleted Fits

Learn how Cook’s distance summarizes the scaled change in fitted values caused by deleting one observation, and how to calculate it from ordinary least-squares output.

Intermediate 9 min read

What You'll Learn

  • Define Cook’s distance as a scaled sum of squared changes in fitted values after deleting one observation.
  • Derive the fitted-value changes from the least-squares deletion identity.
  • Calculate Cook’s distance directly from full-fit and deleted-fit predictions.
  • Use residual, leverage, model size, and residual mean square to calculate the same quantity.
  • Check how response units affect the calculation and avoid reversing prediction differences.

Why Summarize a Deleted Fit?

A regression fit can change when one observation is removed. The changes may occur in several coefficients and may alter predictions at many predictor settings. Cook’s distance combines those changes into one number: it compares the full-model fitted values with the fitted values from a model refitted without the observation, then scales the total squared change by the model size and the full-model residual mean square.

The key is to compare predictions at the same set of predictor rows in both fits. The sum includes a prediction at the deleted observation’s predictor values as well as predictions at the other observed predictor values. Cook’s distance is a descriptive measure of how much the fitted values change under this deletion; its calculation does not require a p-value or a decision cutoff.

Definition and Least-Squares Setup

Suppose there are \(n\) observations and \(p\) regression coefficients, including the intercept if the model has one. Write the least-squares model in matrix form as \(y=X\beta+\varepsilon\), where \(X\) has full column rank. The full-data coefficient estimate is \(\hat{\beta}\), and the fitted-value vector is \(\hat{y}=X\hat{\beta}\). Let \(e_i=y_i-\hat{y}_i\) be the residual for observation \(i\), and let \(h_{ii}\) be its diagonal hat-matrix value. The full-model residual mean square is \(s^2=\mathrm{SSE}/(n-p)\), where \(\mathrm{SSE}=\sum_{j=1}^n e_j^2\).

For each observation \(i\), \(\hat{y}_{j(i)}\) denotes the fitted value at predictor row \(j\) from the model refitted after deleting observation \(i\). The subscript \(j(i)\) marks that the fit omits \(i\); it does not mean that row \(j\) is deleted.

Definition: Cook’s distance for observation \(i\) is the sum of the squared differences between the full-fit and deleted-fit predictions at all \(n\) original predictor rows, divided by \(p\) times the full-model residual mean square:
$$ D_i=\frac{\displaystyle\sum_{j=1}^{n}\left(\hat{y}_j-\hat{y}_{j(i)}\right)^2}{p\,s^2}. $$
The numerator is measured in squared response units, as is \(s^2\), so \(D_i\) has no units.

This definition uses the ordinary least-squares setup: the same specified predictors are used in both fits, the full design matrix has rank \(p\), \(n>p\), and deleting observation \(i\) leaves a design matrix of rank \(p\). The last condition is equivalent to \(h_{ii}<1\). The formula is an algebraic diagnostic. For the usual statistical interpretation of \(s^2\) as an estimate of error variance, the classical linear model additionally assumes a correctly specified linear conditional mean and errors with mean zero, constant variance, and independence. Normality is not needed for the algebraic identity below.

From Deleted Fits to a Least-Squares Formula

Refitting the model separately for every possible deletion is not necessary. Least-squares algebra gives the change in the coefficients, and therefore in every fitted value, directly from the full-model residual and leverage.

Theorem: If the full design matrix has rank \(p\) and \(h_{ii}<1\), the difference between the full-fit and deleted-fit coefficient estimates is
$$ \hat{\beta}-\hat{\beta}_{(i)} =\frac{(X^\mathsf{T}X)^{-1}x_i e_i}{1-h_{ii}}, $$
where \(x_i\) is the column vector of predictors for observation \(i\), including a 1 for the intercept when present. Consequently,
$$ D_i=\frac{e_i^2}{p\,s^2}\frac{h_{ii}}{(1-h_{ii})^2}. $$

Here is a justification of both parts. The full least-squares normal equations are \(X^\mathsf{T}X\hat{\beta}=X^\mathsf{T}y\). After deleting row \(i\), the matrix and right-hand side in the normal equations become \(X^\mathsf{T}X-x_ix_i^\mathsf{T}\) and \(X^\mathsf{T}y-x_i y_i\). Let \(A_i=X^\mathsf{T}X-x_ix_i^\mathsf{T}\). Applying the deleted normal equations to the full estimate gives

$$ A_i\hat{\beta} =X^\mathsf{T}y-x_i\hat{y}_i, \qquad A_i\hat{\beta}_{(i)} =X^\mathsf{T}y-x_i y_i. $$

Subtracting the first equation from the second gives \(A_i(\hat{\beta}_{(i)}-\hat{\beta})=-x_i e_i\). The hat-matrix diagonal is \(h_{ii}=x_i^\mathsf{T}(X^\mathsf{T}X)^{-1}x_i\). Direct multiplication shows that

$$ A_i\frac{(X^\mathsf{T}X)^{-1}x_i}{1-h_{ii}} = \frac{x_i-x_i h_{ii}}{1-h_{ii}} =x_i. $$

Thus \(A_i^{-1}x_i=(X^\mathsf{T}X)^{-1}x_i/(1-h_{ii})\), which yields the coefficient identity above. The full-minus-deleted fitted-value difference vector is \(X(\hat{\beta}-\hat{\beta}_{(i)})=e_i H_{\cdot i}/(1-h_{ii})\), where \(H=X(X^\mathsf{T}X)^{-1}X^\mathsf{T}\) and \(H_{\cdot i}\) is its \(i\)th column. Since \(H\) is symmetric and idempotent, the squared length of that column is

$$ H_{\cdot i}^\mathsf{T}H_{\cdot i} =(H^\mathsf{T}H)_{ii} =H_{ii} =h_{ii}. $$

Therefore the sum of squared fitted-value changes is \(e_i^2h_{ii}/(1-h_{ii})^2\). Dividing it by \(p\,s^2\) gives the stated Cook’s distance formula.

Worked Calculations

The definition and the shortcut formula describe the same quantity. The first example calculates it from actual full and deleted fits; the next two use least-squares quantities directly.

Worked Example: Calculate Cook’s Distance from Deleted-Fit Predictions

Consider this hypothetical simple linear regression data set, with an intercept and predictor \(x\). The model has \(p=2\) coefficients.

\(x\)\(y\)
\(-1\)8
011
110
215

The full-data least-squares line is \(\hat{y}=10+2x\), giving fitted values \(8,10,12,14\). The residuals are \(0,1,-2,1\), so \(\mathrm{SSE}=0^2+1^2+(-2)^2+1^2=6\). With \(n=4\) and \(p=2\), the full-model residual mean square is \(s^2=6/(4-2)=3\). The design has full rank, and, as calculated below, the deleted observation has leverage less than 1, so its deleted fit is estimable.

Delete observation \(i\) at \(x=2\), for which \(e_i=15-14=1\). The predictor mean is \(\bar{x}=0.5\), and \(\sum_j(x_j-\bar{x})^2=5\). Its leverage is

$$ h_{ii}=\frac{1}{4}+\frac{(2-0.5)^2}{5} =0.25+0.45=0.70. $$

Refitting the line without \((2,15)\) gives \(\hat{y}_{(i)}=9.6667+x\). Thus, in the original row order, the deleted-fit predictions are \(8.6667,9.6667,10.6667,11.6667\). Define the changes as full-fit minus deleted-fit. They are \(-0.6667,0.3333,1.3333,2.3333\), or exactly \(-2/3,1/3,4/3,7/3\). The sum of their squares is

$$ \sum_{j=1}^4(\hat{y}_j-\hat{y}_{j(i)})^2 =\frac{4+1+16+49}{9} =\frac{70}{9} \approx 7.778. $$

The direct Cook’s distance calculation is therefore

$$ D_i=\frac{70/9}{2(3)} =\frac{70}{54} \approx 1.296. $$

As a check, the shortcut formula gives \(D_i=[1^2/(2\cdot3)] [0.70/(1-0.70)^2]=1.296\), agreeing with the direct calculation. Notice the sign of each difference: for example, the deleted-fit prediction at \(x=-1\) is \(8.6667\), which is greater than the full-fit prediction of \(8\), so full minus deleted is negative.

Worked Example: Use Residual, Leverage, and Residual Mean Square

Suppose a hypothetical full-rank regression output has \(p=3\) coefficients, \(s^2=4\) response-units squared, and for observation \(i\), \(e_i=-2\) response units and \(h_{ii}=0.25\). Assume \(n>p\) and that deleting this row leaves the design full rank; \(h_{ii}=0.25<1\) is consistent with that condition.

Substitute into the least-squares formula:

$$ D_i =\frac{(-2)^2}{3(4)} \frac{0.25}{(1-0.25)^2} =\frac{4}{12}\frac{0.25}{0.5625} =\frac{1}{3}\left(\frac{4}{9}\right) \approx 0.1481. $$

The implied sum of squared fitted-value changes can also be recovered from the definition by multiplying by \(p\,s^2\): \(0.1481(3)(4)\approx1.778\) squared response units. Directly, the unscaled numerator is \(e_i^2h_{ii}/(1-h_{ii})^2=4(0.25)/0.5625\approx1.778\), matching the result.

Worked Example: Check the Response Units

Suppose a hypothetical regression has \(p=2\), full-model residual standard deviation \(s=4\) minutes, and for one observation \(e_i=6\) minutes and \(h_{ii}=0.20\). Assume the design remains full rank after deletion. Then \(s^2=16\) minutes squared, and

$$ D_i =\frac{6^2}{2(16)} \frac{0.20}{(1-0.20)^2} =\frac{36}{32}\frac{0.20}{0.64} =1.125(0.3125) =0.3516. $$

The numerator before scaling is \(36(0.20)/0.64=11.25\) minutes squared. Dividing by \(p\,s^2=32\) minutes squared gives the dimensionless value \(0.3516\). If the response were converted from minutes to seconds, both the residual and \(s\) would be multiplied by 60; their squared ratio \(e_i^2/s^2\), and hence Cook’s distance, would be unchanged.

How the Components Work Together

The shortcut formula separates the calculation into identifiable parts:

$$ D_i =\frac{1}{p} \left(\frac{e_i^2}{s^2}\right) \left(\frac{h_{ii}}{(1-h_{ii})^2}\right). $$

The squared residual compares the observation’s vertical discrepancy with the fitted line to the overall residual scale. The leverage factor accounts for the predictor location and for how much the deleted fit can move when that observation is removed. The divisor \(p\) scales the total change relative to the number of fitted coefficients. These pieces explain why Cook’s distance is not merely a residual measure: the residual and leverage jointly determine the fitted-value change.

For a compact equivalent expression, the internally studentized residual is \(r_i=e_i/[s\sqrt{1-h_{ii}}]\). Substitution gives \(D_i=(r_i^2/p)\{h_{ii}/(1-h_{ii})\}\). This is an algebraic rewrite, not a different definition; the deleted-fit version remains the basis for understanding what Cook’s distance measures.

Common Mistake: Reversing the Prediction Differences

A common error is to state the signs of the differences as deleted-fit predictions minus full-fit predictions while describing them as full-fit minus deleted-fit. In the first worked example, the deleted prediction at \(x=-1\) is \(8.6667\), and the full prediction is \(8\). Therefore, deleted minus full is \(+0.6667\), whereas full minus deleted is \(-0.6667\). The signs must match the stated subtraction order.

Reversing every difference does not change Cook’s distance, because \((a-b)^2=(b-a)^2\). But inconsistent signs make the explanation wrong and can conceal an arithmetic error. State the order explicitly, check at least one pair of predictions, and then square the differences. Also use the full-model \(s^2\) and the full-model number of coefficients \(p\) in the standard definition, rather than silently substituting quantities from the deleted fit.

Key takeaway: Cook’s distance is the sum of squared changes in all \(n\) fitted values after deleting observation \(i\), scaled by \(p\,s^2\). Under full-rank least squares with \(h_{ii}<1\), calculate the same quantity efficiently as \(D_i=e_i^2h_{ii}/[p\,s^2(1-h_{ii})^2]\).

Check Your Understanding

Answer each question using the definition and formulas above.

  1. In the definition of Cook’s distance, why are full-fit and deleted-fit predictions compared at the same original predictor rows?
  2. A model has \(p=4\), \(s^2=9\), \(e_i=3\), and \(h_{ii}=0.20\). Calculate \(D_i\) from the least-squares formula.
  3. For the values in question 2, calculate the sum of squared fitted-value changes before scaling.
  4. What condition on \(h_{ii}\) ensures that deleting observation \(i\) leaves the least-squares coefficient estimate available through the deletion identity?
  5. Explain why changing the response unit from minutes to seconds does not change Cook’s distance.