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For $x_n=x_{\ast }$, the notation of part (i) gives $b=a$, $d=k_{\ast \ast }$, and $c=k_{\ast \ast }+{\sigma }^2$. Write $v={\mathrm{Var}}_{n-1}(f_{\ast })=k_{\ast \ast }-a^T{A}^{-1}a$. The remaining covariance is then $v$ and the new observation's remaining variance is $v+{\sigma }^2$. ThusEquivalently,The expression is valid also when $v=0$ and ${\sigma }^2>0$. For $v>0$, a noiseless observation, interpreted as the limit ${\sigma }^2\downarrow 0$, removes all remaining variance at the observed target.

The source's assertion that the target input always gives the greatest reduction needs an additional hypothesis. The reduction formula above does not require that assertion. For a counterexample, use the linear kernel $k(x,x^{\prime })=x{x}^{\prime }$ with ${\sigma }^2=1$ and target $x_{\ast }=1$. With no informative previous observations, $v=1$. This can also be arranged with $n-1=1$ by observing first at $x_1=0$, which has zero covariance with all latent values. Observing at the target gives reduction $1/2$, whereas observing at $x_n=2$ givesIndeed this Gaussian process is $f(x)=xZ$ for a standard normal random variable $Z$, so the measurement at $2$ observes the same unknown coefficient with a larger signal relative to its noise.

One sufficient extra condition is that the existing latent posterior variance be maximal at $x_{\ast }$ among all candidate inputs. To prove it, let $w=v_{n-1}(x)$ and $r=\mathrm{Cov}(f(x),f_{\ast }\mid {\mathbf{y}}_{1:n-1},\theta )$. The Cauchy-Schwarz inequality gives $r^2\le vw$. If $w\le v$, thenbecause $w\mapsto w/(w+{\sigma }^2)$ is increasing. Thus target sampling is optimal under this extra condition, but it is not implied by the paper's arbitrary-covariance function assumptions.

Use the normalizationThis is ridge regression; the penalty is the squared Euclidean norm. The centering removes the need for an intercept. Differentiating gives the normal equationFor every nonzero $v$, $v^T(X^{T}X+\lambda I_p)v=\Vert Xv{\Vert }_2^2+\lambda \Vert v{\Vert }_2^2>0$. The coefficient matrix is therefore a positive-definite matrix, even if the design matrix has deficient matrix rank. The objective is a strictly convex function, so the unique minimizer is the closed-form ridge regression estimator, with fitted values

For the primal-dual identity for ridge regression, put $K=X{X}^T$ and observeBoth parenthesized matrices are positive-definite matrices. Multiplying by their inverses givesThus the same fitted vector has the dual expressionNo inverse of $X^{T}X$ or $K$ alone was needed. This is also the kernel-ridge hat matrix for the linear kernel.

Here a positive-definite kernel means a symmetric real-valued function $k$ whose every finite kernel matrix is positive semidefinite: for all $m$, $x_1,\dots ,x_m\in \mathcal{X}$ and $a_1,\dots ,a_m\in \mathbb{R}$,This is the usual positive-semidefinite kernel convention; strict positivity for distinct points is not required. In particular $k(x,x)\ge 0$. Write $a=k(x,x)$, $b=k(x^{\prime },x^{\prime })$ and $r=k(x,x^{\prime })$. Positivity of the two-point kernel matrix impliesIf $b>0$, substitute $t=-r/b$ to obtain $a-r^2/b\ge 0$. If $b=0$, the affine expression $a+2tr$ can be nonnegative for every $t$ only if $r=0$. Both cases give the Cauchy-Schwarz inequality for positive-semidefinite kernels:The zero-diagonal case is essential when the positive-definite kernel is not strictly positive.

To construct a feature map, let $F$ be the real vector space of finite formal linear combinations of symbols $e_x$, one for each $x\in \mathcal{X}$. Define a symmetric bilinear form byThe positive-definite kernel property makes this a positive semidefinite bilinear form. It need not yet be an inner product: distinct formal vectors can have zero squared length. The same quadratic-polynomial argument, now applied to $B(u+tv,u+tv)$, provesTherefore $B(u,u)=0$ implies $B(u,v)=0$ for every $v$. The setis the radical of a bilinear form, hence a linear subspace. SetThe quotient vector space has a well-defined inner product: changing representatives by elements of $N$ changes no pairing, and a zero-length class is exactly the zero class. It is therefore an inner product space.

Define $\varphi (x)=e_x+N$. By construction,Every positive-definite kernel thus has a feature representation, including degenerate kernels and the zero kernel. This is its canonical feature space of a positive-semidefinite kernel. The construction already gives the requested inner product space.

PutThe numerator is the linear kernel, while the reciprocal of the denominator is the kernel from part c applied to $\Vert x{\Vert }^2$ and $\Vert y{\Vert }^2$. The product of positive-semidefinite kernels therefore shows that $a$ is a positive-semidefinite kernel. The Cauchy-Schwarz inequality and the arithmetic-geometric mean inequality give $|a(x,y)|\le 1/2$, soEach power is positive semidefinite by the Schur product theorem, and the convergent sum is positive semidefinite.

Moreover $k_2(x,x)=1$. If $\Phi$ is its canonical feature map, thenThe feature-space norm gives symmetry and the triangle inequality. Finally, $d(x,y)=0$ implies $k_2(x,y)=1$, hence $a(x,y)=1/2$ andwhich is equivalent to $\Vert x-y{\Vert }^2=0$. Thus $d$ is a metric rather than merely a pseudometric.

A symmetric function $k$ is a positive-semidefinite kernel when every finite Gram matrix $(k(x_i,x_j))$ is positive semidefinite.

For any real design matrix $X$ and $\lambda >0$, $(X^{T}X+\lambda I{)}^{-1}{X}^T=X^T(X{X}^T+\lambda I{)}^{-1}$. Multiplication by $X$ identifies the primal ridge regression fitted values with $K(K+\lambda I{)}^{-1}Y$, $K=X{X}^T$, the kernel-ridge hat matrix for the linear kernel. No rank assumption is needed.

A reproducing-kernel Hilbert space is a Hilbert space of functions in which point evaluation is continuous and satisfies $f(x)=⟨f,k(x,\cdot )⟩$.