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Fitted values are the predictions at the predictor values used to fit a regression model. In linear regression with an explicit regression intercept,Here $X$ is the slope design matrix, and $r$ is the regression residual vector. The fitted values estimate the regression function at the observed predictors; they need not equal the observed responses, which may also contain random errors. Prediction at a new predictor vector $x_{\ast }$ uses $\hat{\alpha }+x_{\ast }^T\hat{\beta }$ with the same centering and scaling used in fitting. A prediction interval addresses uncertainty in a new response rather than just displaying this point prediction.

For the five constructed observations in linear regression, ordinary least squares gives fitted values $(4/5,9/5,14/5,19/5,24/5)$. The ridge regression fit at $\lambda =10$ instead gives $(9/5,23/10,14/5,33/10,19/5)$. Both fitted vectors have sample mean $14/5$ because the regression intercept is unpenalized; their regression residuals differ and their residual sums of squares are $4/5$ and $33/10$, respectively.

For ordinary least squares, the fitted values are the orthogonal projection of $Y$ onto the column space of the complete design matrix, including its regression intercept column. They remain unique even in rank-deficient ordinary least squares, although the regression coefficients can be nonunique. The regression residuals are then orthogonal to every column.^[ref] With centered slopes, ridge regression instead satisfies $X_c^{T}r=\lambda {\hat{\beta }}_{\lambda }$, so the regression residuals need not be orthogonal to the slope columns. Their sum is still zero because the regression intercept remains free.

For fixed predictors and scales and a fixed positive penalty, ridge regression gives a linear smoother,For $\lambda >0$ with at least one nonzero slope direction, this matrix is not an orthogonal projection matrix: applying it twice shrinks that direction again. Its effective degrees of freedom are $1+{\sum }_{j=1}^r{d}_j^2/(d_j^2+\lambda )$, where $d_j$ are the positive singular values of $X_c$; the extra $1$ counts the free regression intercept. If the penalty is selected using $Y$, the resulting fitting rule need not be a linear smoother, since its chosen $H_{\lambda }$ also depends on $Y$.