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A histogram groups numerical observations into adjacent intervals called bins and draws one rectangle per bin. Its attraction is that a long list of values becomes a visible pattern of concentration, gaps and skewness. Specify the bin edges, the treatment of values on those edges, and what the vertical axis means: count, proportion, or density. Those choices are part of the graphic. ^[ref]

Let $n>0$ observations be covered by disjoint bins $B_j$ with positive widths ${\Delta }_j$ and counts $n_j$, so ${\sum }_j{n}_j=n$. For example, use left-closed, right-open bins, closing the final right edge too. The three common height choices have different meanings:

For the density convention, a histogram is a piecewise constant probability density function estimate:Its normalization follows by adding rectangular areas:If the observations have units, density has reciprocal units; a density height is not a probability and can exceed $1$. With equal bin widths, count heights and density heights differ only by a common scale factor. With unequal widths, count-height areas $n_j{\Delta }_j$ give wider bins extra visual weight unrelated to their counts. Dividing by width removes that distortion. R's histogram documentation explicitly describes area-one normalization for unequal bins. ^[ref]

Reuse the twelve invented values under box plot, with bins $[0,2),[2,4),[4,8),[8,22]$. The counts are $(1,3,6,2)$ and the widths are $(2,2,4,14)$. Hence the density heights and areas are exactlyThe second and third bars have equal density heights even though the third contains twice as many observations: it is also twice as wide. The last bar contains only two observations; its count-height area would be $2\cdot 14=28$, larger than the six-observation bar's $6\cdot 4=24$. Its correct density area is only $1/6$.

Bin choice trades detail against stability. Broad bins merge distinctions: the toy bin $[8,22]$ hides the empty stretch between $8$ and $20$. Narrow bins expose gaps but can make each individual observation dominate a bar. With one-unit bins starting at $0$, the bins $[2,3),[4,5),[6,7)$ each contain two values and appear equally prominent; that apparent pattern depends on grouping the repeated integers together. Moving the bin origin also changes membership without changing the observations. For density estimation, this is a form of the bias-variance tradeoff: averaging over broad bins loses local detail, while narrow bins give more variable density estimates across repeated statistical samples. ^[ref] A useful practice is to compare a few plausible widths and offsets, keeping common edges when comparing groups. Automatic bin selectors are starting points, rather than guarantees of a unique correct picture. ^[ref]

A histogram retains only bin membership. Its density estimate spreads each bin's fraction uniformly across that bin, so our last bar assigns density throughout $[8,22]$ even though the only toy values there are $8$ and $20$. It does not recover within-bin positions or observation order, and a visually convincing shape does not establish a population probability distribution. A box plot provides a complementary compact summary of quartiles; neither graphic replaces the original observations.

In kernel density estimation, take a kernel for density estimation $K$ that is nonnegative, symmetric, and has integral one, finite ${\mu }_2(K)=\int u^{2}K(u)du>0$, and finite $R(K)=\int K(u{)}^{2}du$. Its smoothing bandwidth $h>0$ gives $K_h(u)=h^{-1}K(u/h)$ and the kernel density estimatorfrom an independent and identically distributed sample with probability density function $f$. Nonnegativity and unit mass make the estimate itself a probability density function. We compare kernel for density estimation shapes by their optimized mean integrated squared error, rather than by their performance at an arbitrarily common numerical smoothing bandwidth.

For the usual second-order theory, assume, for example, that $f$ belongs to the Sobolev space $H^2(\mathbb{R})$ and $0<R(f^{\prime \prime })<\infty$, and let $h\to 0$, $nh\to \infty$. The expectation is the convolution $K_h\ast f$. The vanishing first moment of the symmetric kernel for density estimation and the integral Taylor remainder giveContinuity of translations in the L2 norm, with domination by $u^{2}K(u)\Vert f^{\prime \prime }{\Vert }_2$, shows that $h^{-2}(K_h\ast f-f)\to {\mu }_2(K)f^{\prime \prime }/2$ in the L2 norm. Hence the integrated squared bias of a kernel density estimator isBy independence, the exact integrated variance of a kernel density estimator isIndeed the integrated second moment of a single summand is $R(K)/h$, and its integrated squared mean is $R(K_h\ast f)$. The latter quantity is bounded by $R(f)$, by Young's convolution inequality, so its contribution is $O(n^{-1})$. The bias-variance decomposition of mean squared error therefore yields the leading criterionThis asymptotic mean integrated squared error displays the bias-variance tradeoff: a small smoothing bandwidth increases integrated variance, while a large one increases squared bias of an estimator.

Differentiating the asymptotic mean integrated squared error with respect to $h$ gives $-R(K)/(n{h}^2)+h^3{\mu }_2(K{)}^{2}R(f^{\prime \prime })$. It changes sign once, from negative to positive, so its unique minimum isAt this minimum the integrated variance term is four times the squared bias of an estimator term. Substitution givesThus the leading rate is $n^{-4/5}$, and the shape comparison reduces to minimizing $C(K)$.

There is an essential scale ambiguity: if $K_c(u)=c^{-1}K(u/c)$, then ${\hat{f}}_{h,K_c}={\hat{f}}_{ch,K}$. Changing the scale of the kernel for density estimation merely changes the numerical convention for the smoothing bandwidth. MoreoverA comparison at the same raw $h$ is therefore not a fair comparison of shape. One can normalize the second moment to one, but that does not make the optimal smoothing bandwidth independent of shape.

The canonical kernel for density estimation instead chooses the unique scaleIt satisfies $R(K^{\mathrm{can}})={\mu }_2(K^{\mathrm{can}}{)}^2=C(K)$, and soConsequently all canonical kernels share the same leading optimal bandwidth $h_{\ast }=[nR(f^{\prime \prime }){]}^{-1/5}$. Only the multiplicative constant $C(K)$ depends on shape. This is precisely the separation of shape choice and scale choice that makes the canonical kernel normalization useful. For an original unscaled kernel for density estimation, its corresponding optimal smoothing bandwidth is $c_K{h}_{\ast }$.

To prove optimal shape, scale a candidate kernel for density estimation to have second moment one. Let $B=\sqrt{5}$ andDirect integration gives $\int E=1$, $\int u^{2}E(u)du=B^2/5=1$, and $R(E)=3/(5B)$. Thus $E$ is a rescaled Epanechnikov kernel. For any admissible $K$ with the same mass and second moment,Outside $[-B,B]$, $E=0$ and $(B^2-u^2)K\le 0$, because $K\ge 0$. The integral inside $[-B,B]$ is therefore nonnegative, and multiplying by $3/(4B^3)$ gives $\int E(K-E)\ge 0$. Expanding the square now provesEquality forces $K=E$ almost everywhere. At fixed second moment, minimizing $R(K)$ is equivalent to minimizing $C(K)$. Since $C(K)$ is scale invariant, this proves the optimality of the Epanechnikov kernel over all nonnegative symmetric second-order kernels for density estimation, including the equality case up to rescaling.

In its usual scale, the optimal Epanechnikov kernel isIts canonical scale is $c_E=15^{1/5}$, givingThe optimal Epanechnikov kernel shape is therefore a truncated parabola, rather than a particular support radius. The canonical kernel and unit-second-moment version represent the same shape but different scale conventions.

The numerical gains over other reasonable kernels for density estimation are modest. Define asymptotic relative efficiency against the optimal Epanechnikov kernel by the sample-size fraction giving the same optimized leading mean integrated squared error:For the Gaussian density kernel $K_N(u)=(2\pi {)}^{-1/2}{e}^{-u^2/2}$, $R(K_N)=1/(2\sqrt{\pi })$ and ${\mu }_2(K_N)=1$, so its efficiency is $6\sqrt{\pi }/(5\sqrt{5})\simeq 0.9512$. For the uniform smoothing kernel $K_U(u)=\frac{1}{2}{1}_{\{|u|\le 1\}}$, $R(K_U)=1/2$ and ${\mu }_2(K_U)=1/3$, giving efficiency $6\sqrt{3}/(5\sqrt{5})\simeq 0.9295$. The Gaussian density kernel is infinitely smooth, whereas the optimal Epanechnikov kernel has compact support and allows each observation to affect only a finite window. Such computational and smoothness considerations may outweigh a small leading-efficiency difference.

In practice $R(f^{\prime \prime })$ is unknown. A plug-in estimator estimates this roughness functional from a pilot estimate of the second derivative and substitutes it into the optimal smoothing bandwidth formula. Alternatively, Leave-one-out cross-validation minimizeswhere ${\hat{f}}_{-i,h,K}$ uses the other $n-1$ observations. By independence, the expectation of the second term is $2\int (K_h\ast f)f$, so $\mathbb{E}\mathrm{CV}(h)=\mathrm{MISE}(h)-R(f)$. Thus it targets integrated error without knowing $f$. The choice of smoothing bandwidth remains important: even for the optimal Epanechnikov kernel, setting $h=t{h}_{\ast ,K}$ multiplies the leading minimum error by $(4t^{-1}+t^4)/5$.

The preceding optimum concerns nonnegative second-order kernels for density estimation and twice-smooth densities on the real line. Signed kernels of order ell can cancel higher moments for smoother densities and yield different rates, but may produce negative density estimates. A support boundary or a density jump also requires separate bias of a kernel density estimator analysis. These settings do not contradict the proved second-order optimality of the Epanechnikov kernel.

For ordinary homoscedastic fixed-design nonparametric regression, take observations $Y_i=g(x_i)+{ϵ}_i$ at known distinct sites, with independent mean-zero errors of common variance. A degree-$p$ local polynomial estimator at a target $u$ solves the finite-dimensional weighted least squares problemHere $K$ is a nonnegative regression kernel, $h>0$ is the smoothing bandwidth, and sufficient local design rank ensures uniqueness. The estimate of the regression function is $\hat{g}(u)={\hat{\beta }}_0(u)$. Each target has its own locally fitted polynomial; this is not one global polynomial fit. The bandwidth controls the localization and the bias-variance tradeoff.

A natural cubic smoothing spline instead solves the global penalized problemThe Sobolev space here consists of functions with square-integrable weak derivatives through order two. The minimizer is a natural cubic spline with knots at the observation sites and linear tails outside the extreme sites. Equivalently one may perform the minimization on that finite-dimensional spline space: the minimum roughness property of the natural cubic spline interpolant shows that replacing a candidate by the natural spline with the same site values preserves the data loss and decreases the second-derivative penalty. The smoothing parameter $\lambda$ controls curvature; it differs from the local bandwidth $h$.

For the interval-average observation problem, the printed upper limit in the loss is an indexing error: there are only $n-1$ responses and intervals, so a term with $i=n$ would involve the undefined $Y_n$ and $x_{n+1}$. The coherent loss uses the upper limit $n-1$. Put ${\Delta }_i=x_{i+1}-x_i$ and define the linear observation functionalsThe intended criterion on $C^1[a,b]$ is thereforeUnlike the ordinary cubic smoothing spline, this uses interval averages as data and a first-derivative penalty; the resulting spline degree will be two.

Fix any candidate $v\in C^1[a,b]$. By the supplied interpolation property, there is a unique quadratic spline $q\in C^1[a,b]$ with knots $x_1,\dots ,x_n$, constant on both exterior intervals, such that $L_{i}q=L_{i}v$ for every $i$. Put $d=v-q$. ThenOn each interior knot interval $q^{\prime \prime }$ is a constant $c_i$, and on the two exterior intervals $q^{\prime }=0$. Integrate $q^{\prime }{d}^{\prime }$ by parts separately on the intervals. At every knot the boundary contributions cancel because both $q^{\prime }$ and $d$ are continuous; the contributions at $a$ and $b$ vanish because $q^{\prime }$ is zero there. ThusThis orthogonality is the key to the quadratic smoothing spline for interval averages. Expanding the roughness penalty givesThe data-fit terms are equal because the interval averages are equal. Consequently,Equality can occur only if $d^{\prime }=0$ everywhere: the derivative is continuous and its squared integral is zero. Then $d$ is constant, and any of its zero interval averages forces that constant to be zero. Therefore every minimizer must equal its matching quadratic spline $q$.

We conclude that the minimizer is a continuously differentiable quadratic spline with knots at the specified sites and constant tails on both exterior intervals. Existence and uniqueness also follow from the supplied interpolation property. The interval-average map identifies the finite-dimensional space of these quadratic splines with ${\mathbb{R}}^{n-1}$, and the criterion becomes a strictly convex quadratic function of that average vector: its data-fit term is $\Vert Y-y{\Vert }^2$, while its penalty is nonnegative and quadratic. It has a unique minimizer, and the strict equality argument above proves uniqueness in the full function class.

To isolate the bias-variance tradeoff of the smoothing step, fix the three neighbor locations and write $Y_{(k)}=m_k+{ϵ}_{(k)}$, with independent zero-mean errors of common variance ${\sigma }^2$. At a target $x$, the local estimate isThe bias and variance of a three-neighbour weighted smoother areConsequently the variance is minimized by equal weights, $w=1/3$, when it equals ${\sigma }^2/3$. It decreases as $w$ increases from zero to $1/3$, and increases thereafter toward ${\sigma }^2$ as $w\to 1$. In particular, variance is not increasing over the entire interval $0<w<1$.

Giving more weight to the closest location reduces the influence of more distant responses and often reduces smoothing bias. If the closest location is the target itself, $m_1=m(x)$, put $\Delta =(m_2+m_3)/2-m(x)$. Then the squared bias is $(1-w{)}^2{\Delta }^2$, andThus a locally flat signal favors equal averaging, while stronger local smoothing bias favors greater nearest-point weight. At a general target the bias is affine in $w$, and its absolute value need not decrease: farther neighbors can cancel one another's approximation errors. Even without that cancellation, a smoothness-based worst-case bias bound decreases with $w$ because the first neighbor is closest. The exact tradeoff is therefore lower variance near balanced weights versus potentially lower bias near the closest-point fit. These formulas describe the elementary smoother; fitted partial residuals, centering and repeated backfitting can introduce dependence, so the variance of the final additive fit requires its full fitting operator rather than this single-step calculation.

An orthonormal multiresolution analysis is a sequence of closed subspaces $V_j\subset L^2(\mathbb{R})$ satisfying $V_j\subset V_{j+1}$, $\overline{{\bigcup }_j{V}_j}=L^2(\mathbb{R})$, ${\bigcap }_j{V}_j=\{0\}$, and $f\in V_j$ iff $f(2\cdot )\in V_{j+1}$. A scaling function $\varphi$ has integer translates forming an orthonormal basis of $V_0$, and its scaled translates ${\varphi }_{j,k}(x)=2^{j/2}\varphi (2^{j}x-k)$ span $V_j$. Thus $V_j$ contains information at resolution about $2^{-j}$.

The detail space $W_j=V_{j+1}\cap V_j^{\perp }$ is generated by the corresponding wavelet functions ${\psi }_{j,k}(x)=2^{j/2}\psi (2^{j}x-k)$. The orthogonal splitting $V_{j+1}=V_j\oplus W_j$ iterates toAccordingly a square-integrable signal has a coarse-and-detail wavelet series,and its wavelet projection onto $V_J$ keeps the coarse coefficients and details below level $J$. Increasing $J$ adds finer-scale oscillations and localized features. On the real line there are infinitely many translates at each level; for a bounded observation interval, with an appropriate interval or periodic construction, the relevant approximation dimension satisfies $d_J\asymp 2^J$. For the Haar wavelet on a unit interval this is exactly $2^J$, corresponding to a separate mean on each dyadic subinterval.

This hierarchy gives a concrete meaning to wavelet approximation complexity. Coarse resolution uses few coefficients and can miss sharp features; fine resolution uses more coefficients and can fit noise. By orthogonality, projection bias is the energy of the omitted details,For an orthonormal sequence model with independent coefficient noise of variance ${\sigma }^2/n$, estimating the $d_J$ retained coefficients without shrinkage givesThe first term decreases with resolution while the second increases, making the bias-variance tradeoff explicit. Equivalently, a fixed orthogonal regression projection of rank $d_J$ has effective degrees of freedom $d_J$ and prediction optimism $2{\sigma }^2{d}_J/n$ under independent equal-variance errors.

Wavelets also permit local rather than only global complexity control. Thresholding small estimated detail coefficients, for example by soft thresholding, retains strong local structure while suppressing fine-scale noise elsewhere. A sparse retained coefficient set can represent localized features much more economically than keeping every coefficient through a high level. Its locations and scales matter as well as its count. If that set is selected using the same noisy responses, however, selection adds adaptivity: its raw coefficient count is not automatically its covariance-based degrees of freedom. Resolution or threshold tuning should therefore account for prediction error, using an appropriate risk estimate or cross-validation, rather than only decreasing training error.

Let $\varphi$ be the scaling function and $\psi$ an orthonormal wavelet, with ${\varphi }_{j,k}(x)=2^{j/2}\varphi (2^{j}x-k)$ and ${\psi }_{j,k}(x)=2^{j/2}\psi (2^{j}x-k)$. For $f\in L^2(\mathbb{R})$, the wavelet series at a coarse level $j_0$ isThe series converges in the L2 norm. Its coarse part captures broad variation and successive wavelet levels add finer detail. Truncating at levels $j<J$ produces the orthogonal projection onto $V_J$; Parseval identity makes its squared approximation error the sum of the omitted squared coefficients. Merely square-integrable functions need not have pointwise convergence.

For the Haar wavelet, use the cell convention $I_{j,k}=(k{2}^{-j},(k+1)2^{-j}]$ and let $I_j(x)$ be the unique such interval containing $x$. The Haar scaling functions are $2^{j/2}{1}_{I_{j,k}}$, so the Haar projection isFor a merely locally integrable function this cell-average formula remains defined, even if a global $L^2$ orthogonal projection is unavailable. If $\Vert f^{\prime }{\Vert }_{\infty }=L$, the mean value theorem gives Lipschitz continuity and, since $|t-x|\le 2^{-j}$ on the cell,Thus the requested bound holds with $\boxed{c=\Vert f^{\prime }{\Vert }_{\infty }}$, by the Haar projection error for a Lipschitz function.

To estimate a density, replace each scaling or wavelet coefficient $\int f{\varphi }_{j,k}$ or $\int f{\psi }_{j,k}$ by the sample average of that basis function. The finite-resolution Haar density estimator is thereforeIt is the histogram on the dyadic cells, is nonnegative, and integrates to one. Its expectation is $K_{j}f(x)$. A density with bounded derivative is bounded: the Lipschitz density height bound gives $f(x{)}^2\le L$ by integrating the triangular lower envelope $(f(x)-L|t-x|{)}_{+}$. The case $L=0$ cannot be a probability density on the whole real line. Hence $B=\Vert f{\Vert }_{\infty }<\infty$.

The cell probability is $p_j(x)={\int }_{I_j(x)}f\le B{2}^{-j}$. The sample count has a binomial distribution, soThe Cauchy-Schwarz inequality bounds the mean centered absolute deviation by the square root of this variance. Adding the projection bias givesChoose $j_n=\lfloor ({\log }_{2}n)/3\rfloor$, so $2^{j_n}\asymp n^{1/3}$. The two terms have the same order, andThis is a bias-variance tradeoff: refining the cells decreases the averaging bias but increases the sampling variance.

The Haar projection averages this normal distribution density on each cell; the finer cells follow its variation more closely. A Haar density estimator replaces these exact cell probabilities by empirical frequencies and therefore also introduces sampling noise.

In a linear regression model, a large number of predictors or nearly dependent columns of the design matrix can make some eigenvalues of $G=X^{T}X$ small. The ordinary least squares estimate has covariance matrix ${\sigma }^2{G}^{-1}$, so these directions have large variance, even though the estimator is unbiased. Ridge regression reduces this variance by shrinking the estimate towards zero, at the cost of introducing bias. This bias-variance tradeoff can improve the mean squared error when the signal is not too large in the shrunken directions.

For the usual preprocessing with a regression intercept, separate the constant column from the predictors and leave the intercept unpenalized. Subtract the sample mean from the response and from each nonconstant predictor, then divide each centered predictor column by its positive scale, for example its Euclidean norm. Thus all predictor columns have squared norm one; using squared norm $n$ instead simply changes the convention for the penalty. Centering is performed on the slope columns, not on the constant column. A full-rank design containing an intercept retains full rank on its centered slope columns. A model without an intercept should retain its specified mean structure unless an intercept is deliberately introduced.

Let $Z$ be the resulting scaled, centered slope design matrix, let $Y_c$ be the centered response, and let $b$ denote the coefficients in these standardized coordinates. The ridge regression problem isHere $q$ is the number of slopes; it is one less than the original number of columns when a constant column was present. The gradient of this objective is $2(Z^{T}Z+\lambda I)b-2Z^T{Y}_c$. Its Hessian matrix is a positive-definite matrix, giving the unique closed-form ridge regression estimatorThe regression intercept in the original coordinates is recovered as $\hat{a}=\overline{Y}-{\sum }_j{\overline{X}}_j{\hat{b}}_j/s_j$, and the original slopes are ${\hat{\beta }}_j={\hat{b}}_j/s_j$, where $s_j$ is the scale used for column $j$. This is the unpenalized intercept in ridge regression convention. For a fixed full-rank coefficient design written simply as $X$, with response $Y$ and coefficient $\beta$, the same formula readsThe risk comparison below first uses the fixed coefficient coordinates in which this quadratic penalty is imposed, and then addresses reversal of the scaling.

Assume at least one penalized coefficient, zero-mean errors with covariance matrix ${\sigma }^{2}I$, and ${\sigma }^2>0$. No normal distribution assumption is needed for this mean squared error calculation. Put $G=X^{T}X$ and $A_{\lambda }=(G+\lambda I{)}^{-1}$. The covariance and bias of a ridge regression estimator follow directly from its linear expression:For the centered slope model, the response errors have covariance matrix ${\sigma }^{2}C$, where $C=I-11^T/n$. Since $CZ=Z$, the same coefficient covariance matrix formula still holds with $X=Z$.

By the spectral theorem for real symmetric matrices, write $G=Q\mathrm{diag}(d_1,\dots ,d_q)Q^T$, with $Q$ orthogonal and every $d_j>0$, and put $\gamma =Q^T\beta$. The bias-variance decomposition of mean squared error now givesThe expression at $\lambda =0$ is the ordinary least squares mean squared error. Its right derivative isTherefore, for every fixed coefficient vector, all sufficiently small positive penalties strictly improve mean squared error.

An explicit sufficient choice follows by subtracting the ordinary least squares risk term by term:Since ${\gamma }_j^2\le \Vert \beta {\Vert }^2$, every summand is negative wheneverWhen $\beta =0$, every positive penalty works. This is a pointwise result: the permitted penalty may depend on the true coefficient vector. A fixed positive penalty cannot improve risk for all arbitrarily large signals, because its squared bias grows quadratically with the signal. The same distinction appears in uniform directional risk improvement by ridge regression and unbounded directional risk of fixed ridge shrinkage. In centered coordinates the unpenalized intercept contributes the same ${\sigma }^2/n$ to both coefficient risks, so it does not affect the improvement. Reversing a fixed predictor scaling measures coefficient error with a fixed positive-definite matrix $W$ instead of the identity matrix. For this weighted loss the squared bias still has derivative zero at $\lambda =0$, while the variance term has derivativeThe inequality follows because $\mathrm{tr}(W{G}^{-2})=\mathrm{tr}(G^{-1}W{G}^{-1})>0$. Thus sufficiently small penalties also improve the original-coordinate mean squared error, though the explicit sufficient bound above was for the unweighted coordinates. Recovering an uncentered intercept adds a fixed nonnegative quadratic form to the slope loss, and its sample-mean error is uncorrelated with the centered slope errors, so the same weighted argument applies. A model with only an unpenalized intercept gives identical estimators. If the noise variance were zero, the strict improvement assertion would fail; positive noise variance is essential.

With the same squared-error normalization as in Question 2, the ridge estimator solvesThe unpenalized intercept in ridge regression is $\hat{a}=\overline{Y}$ because the columns of $X$ are centered. Differentiating the remaining quadratic objective gives $(X^{T}X+n\lambda I_p){\hat{\beta }}_{\lambda }^R=X^T{Y}_c$. Since $\lambda >0$, the left matrix is positive definite even when $X$ is rank-deficient. Thus the closed-form ridge regression estimator isThe second equality again uses centering. If the penalty convention is $\lambda \Vert b{\Vert }_2^2/2$ with an unnormalized squared-error term $\Vert Y_c-Xb{\Vert }_2^2/2$, replace $n\lambda$ by $\lambda$ throughout; the limiting assertion is unchanged.

For the full-column-rank case, take a singular value decomposition $X=UD{V}^T$ with positive singular values $d_1,\dots ,d_p$. The ordinary least squares estimator, which is the slope maximum-likelihood estimator in this normal linear model, and the ridge regression estimator are respectivelyConsequently principal-component shrinkage by ridge regression multiplies the least-squares coordinate in direction $v_j$ by $d_j^2/(d_j^2+n\lambda )$. Ridge shrinks most strongly in directions with the smallest singular values. These are combinations of predictors that the data distinguish least well.

Near multicollinearity creates small singular values, and the factor $1/d_j$ in ordinary least squares greatly amplifies noise in those directions. Its coefficient variance there is ${\sigma }^2/d_j^2$, while the ridge variance iswhich is smaller. The price is bias: the expectation of the ridge coordinate is $d_j^2/(d_j^2+n\lambda )$ times the corresponding true coefficient. This bias-variance tradeoff can substantially reduce estimation and prediction error when the poorly identified directions do not contain an excessively large signal. It is not a guarantee of improvement for every coefficient vector. Adding $n\lambda$ to every eigenvalue also improves the condition number of the normal matrix and makes numerical inversion more stable.

To handle every rank and every relation between $n$ and $p$, put $G=X^{T}X$ and use its spectral theorem for real symmetric matrices. Choose an orthonormal eigenbasis $v_1,\dots ,v_p$ with eigenvalues ${\gamma }_j\ge 0$, and put $z=X^{T}Y$. ThenIf ${\gamma }_j=0$, then $\Vert X{v}_j{\Vert }_2^2=v_j^{T}G{v}_j=0$, so $X{v}_j=0$ and $v_j^{T}z=(X{v}_j{)}^{T}Y=0$. Thus these terms are exactly zero for every positive $\lambda$; there is no divergent component in the null space. Each remaining term converges to $(v_j^{T}z/{\gamma }_j)v_j$. By the definition of the Moore-Penrose pseudoinverse,and thereforeThis proves the vanishing-penalty ridge limit without an invertibility assumption. The limit is the minimum-norm least-squares solution: every other least-squares solution differs by a vector in $\ker X$, orthogonal to the displayed solution, and therefore has at least as large a Euclidean norm.

For the unit-width box kernel, the convolution and kernel density estimator areandIn particular $\mathbb{E}{\hat{f}}_{n,h}^K=g_h$. The scaling gives $\Vert K_h{\Vert }_2^2=h^{-1}\Vert K{\Vert }_2^2=h^{-1}$. Even if $f$ is not square-integrable, Young's convolution inequality gives $g_h\in L^2$ because $f\in L^1$ and $K_h\in L^2$.

Using independence to eliminate the cross terms in the centred estimator, and Tonelli theorem to integrate the nonnegative variance, gives the exact integrated variance of a kernel density estimator:The Cauchy-Schwarz inequality now yieldsThis constant comes from the unit $L^2$ norm of the unscaled unit-width box kernel.

For the piecewise constant probability density function, integrability forces the constants on both unbounded outer intervals to be zero. Thus $f$ is bounded, has compact support, and has finitely many jumps. Let ${\Delta }_r=f(x_r+)-f(x_r-)$ denote the jump at $x_r$. For $h$ smaller than the least gap between consecutive breakpoints, the smoothing regions $[x_r-h/2,x_r+h/2]$ do not overlap. Outside these regions the convolution equals $f$.

Within a region, put $v=x-x_r$. For $-h/2<v<0$, the bias is ${\Delta }_r(1/2+v/h)$; for $0<v<h/2$, it is $-{\Delta }_r(1/2-v/h)$. Endpoint values do not affect an $L^2$ norm. Integrating these two triangular errors givesThis is the box-kernel bias of a piecewise constant density. By the triangle inequality, with $C_f=({\sum }_r{\Delta }_r^2/12{)}^{1/2}$,Balancing this bias-variance tradeoff with $h_n=n^{-1/2}$ gives, for all sufficiently large $n$,