DFFITS for Influence on an Observation’s Fitted Value

What DFFITS Measures

A regression fit can change when one observation is omitted. DFFITS focuses on one particular consequence of that change: how much the fitted value at the omitted observation’s predictor values changes. It compares the fitted value from the full data with the fitted value obtained after deleting that case, then standardizes the difference so it can be judged relative to the model’s residual scale.

DFFITS is a case-influence diagnostic, not a test that an observation is erroneous. A large value signals that the case has a substantial effect on its own fitted value under the fitted model. It does not, by itself, establish why the case has that effect or whether it should be removed.

Definition: In an ordinary least squares regression with $n$ observations and $p$ fitted coefficients, including the intercept, let $\hat{y}_i$ be the fitted value for observation $i$ using all observations. Let $\hat{y}_{i(i)}$ be the fitted value at observation $i$'s predictor values from a model fitted after deleting observation $i$. If $h_{ii}$ is its leverage and $s_{(i)}$ is the residual standard error from the deleted-case fit, then $$ \operatorname{DFFITS}_i = \frac{\hat{y}_i-\hat{y}_{i(i)}}{s_{(i)}\sqrt{h_{ii}}}. $$ The sign records the direction of the change from the deleted-case fit to the full fit; the magnitude records its standardized size.

Here $s_{(i)}$ estimates the residual standard deviation after deletion. If the full-model residual sum of squares is $\operatorname{SSE}$, the ordinary least squares residual for case $i$ is $e_i=y_i-\hat{y}_i$, and the model remains estimable after deletion, then $$ s_{(i)}^2 = \frac{\operatorname{SSE}-e_i^2/(1-h_{ii})}{n-p-1}. $$ The denominator has $n-1$ observations and $p$ fitted coefficients, so its residual degrees of freedom are $n-p-1$. This calculation requires $h_{ii}<1$, positive deleted-model residual degrees of freedom, and a full-rank design matrix after case $i$ is removed.

Why the Residual-and-Leverage Formula Works

DFFITS can be calculated without fitting every deleted-case model separately. Let $x_i^\mathsf{T}$ be row $i$ of the design matrix and let $\hat{\boldsymbol{\beta}}_{(i)}$ be the coefficient estimate after deleting case $i$. The deleted-case residual at that predictor setting is $e_{i(i)}=y_i-x_i^\mathsf{T}\hat{\boldsymbol{\beta}}_{(i)}$. The standard case-deletion identity for ordinary least squares is $e_{i(i)}=e_i/(1-h_{ii})$. Since $\hat{y}_i=y_i-e_i$ and $\hat{y}_{i(i)}=y_i-e_{i(i)}$, subtracting gives the fitted-value change.

\begin{aligned} \hat{y}_i-\hat{y}_{i(i)} &= (y_i-e_i)-(y_i-e_{i(i)})\\ &= \frac{e_i}{1-h_{ii}}-e_i\\ &= \frac{h_{ii}e_i}{1-h_{ii}}. \end{aligned}

The externally studentized deleted residual is $t_i=e_i/[s_{(i)}\sqrt{1-h_{ii}}]$. Dividing the fitted-value change by the DFFITS standardizing factor therefore gives an equivalent expression:

\operatorname{DFFITS}_i = \frac{h_{ii}e_i}{(1-h_{ii})s_{(i)}\sqrt{h_{ii}}} = t_i\sqrt{\frac{h_{ii}}{1-h_{ii}}}.

This form shows why DFFITS reflects both response discrepancy and leverage. A large externally studentized residual can make DFFITS large, but leverage also magnifies the effect. Conversely, a noticeable residual at a low-leverage point need not produce a large DFFITS. The standardization uses $s_{(i)}$, estimated without the case being assessed; using the full-fit residual standard error instead is not the stated DFFITS definition.

Calculating and Interpreting DFFITS

Use either the deleted-fit definition or the residual-and-leverage form, depending on the quantities supplied. When $\operatorname{SSE}$, $e_i$, $h_{ii}$, $n$, and $p$ are given, first calculate the deleted residual standard error, then calculate the fitted-value change and divide by $s_{(i)}\sqrt{h_{ii}}$. Keeping the signed change matters: a negative DFFITS means the full-data fitted value is below the deleted-data fitted value at that case’s predictors.

Worked Example: A Moderate Change at a Claim-Size Prediction

A hypothetical insurer fits an ordinary least squares model to predict claim size from two rating variables and an intercept. The model has $n=20$ observations and $p=3$ coefficients. Assume the errors are independent and normally distributed with mean zero and unknown variance $\sigma^2$, and the design matrix remains full rank after deletion. For case $i$, the full-model $\operatorname{SSE}$ is $200$, the residual is $e_i=4$, and the leverage is $h_{ii}=0.20$. The full fitted claim size is $\$100$. Calculate DFFITS and check the deleted fitted value.

Method. Calculate the deleted-case residual standard error, then standardize the full-versus-deleted fitted-value change. This avoids treating the residual itself as the fitted-value change.

1. Deleted-case residual sum of squares. The amount removed from the full-model $\operatorname{SSE}$ is $e_i^2/(1-h_{ii})=16/0.80=20$. Thus the deleted-case $\operatorname{SSE}$ is $200-20=180$.

2. Deleted-case residual standard error. There are $20-3-1=16$ residual degrees of freedom after deletion. Hence $s_{(i)}^2=180/16=11.25$, so $s_{(i)}=\sqrt{11.25}=3.3541$.

3. Fitted-value change and DFFITS. The change is $h_{ii}e_i/(1-h_{ii})=(0.20)(4)/0.80=1$. Therefore $\hat{y}_{i(i)}=100-1=99$, and $\operatorname{DFFITS}_i=1/[3.3541\sqrt{0.20}]=1/1.5000=0.6667$.

Check. The deleted-fit identity gives $e_{i(i)}=4/0.80=5$, so the deleted fitted value is $y_i-5$, while the full fitted value is $y_i-4$; their difference is $1$, as calculated. The result is positive because the full-data fit is higher. The common mistake is to divide the residual $4$ directly by a standard error; DFFITS standardizes the fitted-value change, which is $1$, not $4$.

Worked Example: A High-Impact Case and a Screening Guideline

A hypothetical model for monthly equipment-repair costs uses three predictors and an intercept, so $p=4$, with $n=24$. Assume independent normal errors with mean zero and unknown variance $\sigma^2$, and a full-rank design after deletion. For one observation, $\operatorname{SSE}=400$, $e_i=-12$, and $h_{ii}=0.25$. Its full fitted cost is $\$80$. Calculate DFFITS and compare its magnitude with the guideline $2\sqrt{p/n}$.

Method. Re-estimate the residual scale after deletion and use the signed fitted-value change. The guideline is a screening aid, not a formal deletion rule.

1. Deleted residual scale. The deleted $\operatorname{SSE}$ is $400-(-12)^2/(1-0.25)=400-144/0.75=208$. The deleted fit has $24-4-1=19$ residual degrees of freedom. Thus $s_{(i)}=\sqrt{208/19}=3.3087$.

2. Change in fitted value. The change is $(0.25)(-12)/0.75=-4$. The deleted-case fitted cost is consequently $80-(-4)=84$.

3. Standardize and screen. The denominator is $3.3087\sqrt{0.25}=1.6544$, giving $\operatorname{DFFITS}_i=-4/1.6544=-2.418$. The guideline is $2\sqrt{4/24}=0.8165$. Since $2.418>0.8165$, the case is flagged for closer investigation. The negative sign means the full-data fit is lower than the deleted-data fit.

Check. The deleted residual is $-12/0.75=-16$, consistent with the deleted fitted value being $84$ when the observed cost is $68$. The common mistake is to reverse the subtraction and report $+2.418$. That would preserve the magnitude but lose the direction specified by $\hat{y}_i-\hat{y}_{i(i)}$.

Worked Example: The Same Residual Signal at Different Leverages

Two cases from a hypothetical ordinary least squares model have the same externally studentized deleted residual, $t_i=1.5$, but different leverages: $0.10$ and $0.50$. Assume the regression errors are independent normal variables with mean zero and unknown variance $\sigma^2$, and that both deleted fits are estimable. The model has $n=20$ observations and $p=3$ coefficients. Calculate both DFFITS values and compare them with $2\sqrt{p/n}$.

Method. Use $\operatorname{DFFITS}_i=t_i\sqrt{h_{ii}/(1-h_{ii})}$. This isolates the leverage contribution while holding the deleted-residual signal fixed.

Case A. With $h_{ii}=0.10$, DFFITS is $1.5\sqrt{0.10/0.90}=1.5(1/3)=0.5000$.

Case B. With $h_{ii}=0.50$, DFFITS is $1.5\sqrt{0.50/0.50}=1.5000$.

Compare. The screening guideline is $2\sqrt{3/20}=0.7746$. Case A is below it; Case B exceeds it. The same studentized residual signal can therefore correspond to different influence on the case’s own fitted value. This comparison is not evidence that Case B is erroneous; it identifies a case whose effect merits examination.

Screening, Assumptions, and Common Traps

A frequently used screening guideline is $|\operatorname{DFFITS}_i|>2\sqrt{p/n}$, where $p$ counts all fitted coefficients, including the intercept. It is a rule of thumb for identifying cases to review, not a universal significance cutoff. Its usefulness depends on the setting, sample size, model, and purpose of the analysis. A flagged case should prompt checks of data quality, model specification, and the case’s substantive context—not automatic deletion.

The algebraic DFFITS value is defined for the ordinary least squares fit when the deleted model is estimable. Interpreting the standard error in the usual way relies on the regression model’s variance assumptions; for familiar normal-theory interpretations, the errors are also assumed normal. If deletion makes the design matrix rank-deficient, or leaves no residual degrees of freedom, the deleted fit or its residual standard error is not available as defined.

Diagnostic feature	What it tells you	What it does not establish
Large absolute DFFITS	The case substantially changes its own fitted value relative to the deleted-fit residual scale.	That the case is a recording error or should be removed.
Positive DFFITS	The full-data fitted value is higher than the deleted-data fitted value at that case’s predictors.	That the observed response is necessarily unusually high.
High leverage with modest residual signal	Leverage can still increase DFFITS through the factor $\sqrt{h_{ii}/(1-h_{ii})}$.	That high leverage alone guarantees a large DFFITS.

Several calculation errors recur. First, using $e_i$ instead of $h_{ii}e_i/(1-h_{ii})$ confuses a response residual with a fitted-value change. Second, using $\sqrt{1-h_{ii}}$ as the DFFITS denominator confuses this measure with residual standardization: the DFFITS denominator is $s_{(i)}\sqrt{h_{ii}}$. Third, replacing $s_{(i)}$ with the full-model residual standard error changes the statistic. Finally, a cutoff is a prompt for investigation, not a decision rule for excluding data.

Key takeaway: DFFITS is the signed change in a case’s fitted value after including that case, divided by $s_{(i)}\sqrt{h_{ii}}$. Equivalently, it is the externally studentized deleted residual multiplied by $\sqrt{h_{ii}/(1-h_{ii})}$. Interpret its direction and magnitude, and treat screening guidelines as flags for review rather than automatic deletion instructions.

Check Your Understanding

For each problem, use $p$ as the number of fitted coefficients including the intercept. When requested, compare with the guideline $2\sqrt{p/n}$. Give DFFITS to four decimal places.

A hypothetical cost model has $n=18$, $p=3$, $\operatorname{SSE}=150$, $e_i=5$, and $h_{ii}=0.20$. Find $s_{(i)}$, the fitted-value change, and DFFITS. (Answer: $s_{(i)}=\sqrt{118.75/14}=2.9124$, change $=1.25$, DFFITS $=0.9597$.)
In a model with $n=40$ and $p=5$, a case has $t_i=1.2$ and $h_{ii}=0.30$. Calculate DFFITS and determine whether it exceeds the guideline. (Answer: DFFITS $=1.2\sqrt{0.30/0.70}=0.7856$; guideline $=0.7071$, so it exceeds the guideline.)
A hypothetical model has $n=22$, $p=4$, $\operatorname{SSE}=280$, $e_i=-6$, and $h_{ii}=0.10$. Calculate the deleted-case $\operatorname{SSE}$, $s_{(i)}$, the fitted-value change, and DFFITS. (Answer: deleted $\operatorname{SSE}=240$, $s_{(i)}=\sqrt{240/17}=3.7573$, change $=-2/3$, DFFITS $=-0.5610$.)
Two cases have the same $t_i=2.1$, but leverages $0.08$ and $0.32$. For $n=25$ and $p=4$, calculate both DFFITS values and identify which exceeds the guideline. (Answer: $0.6193$ and $1.4406$, respectively; the guideline is $0.8000$, so only the second exceeds it.)
A case in a model with $n=15$, $p=3$ has $e_i=3$, $h_{ii}=0.60$, and $s_{(i)}=2.5$. Calculate DFFITS using the deleted-residual form and compare it with the guideline. (Answer: $t_i=3/[2.5\sqrt{0.40}]=1.8974$, DFFITS $=1.8974\sqrt{0.60/0.40}=2.3238$; guideline $=0.8944$, so it exceeds the guideline.)

What You'll Learn