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A cubic spline belongs to $C^2[a,b]$ and is a polynomial of degree at most three on each interval separated by the spline knots. A natural cubic spline is additionally an affine function on the exterior intervals $[a,x_1]$ and $[x_n,b]$. On $[x_1,x_n]$ this is equivalent to the endpoint conditions $s^{\prime \prime }(x_1)=s^{\prime \prime }(x_n)=0$; these conditions remain part of the definition when an outer knot coincides with $a$ or $b$.

Let $s$ be the natural cubic spline interpolant of the prescribed values, and let $v$ be any other interpolating $C^2$ function. Set $u=v-s$, so $u(x_i)=0$. On the interval $[x_i,x_{i+1}]$, integration by parts twice givesbecause the fourth derivative of each cubic piece is zero. The second boundary term is zero because $u$ vanishes at the spline knots. After summing the first terms, the interior contributions cancel by continuity of $s^{\prime \prime }$ and $u^{\prime }$, and the two remaining terms vanish by the natural endpoint conditions. On the exterior intervals $s^{\prime \prime }=0$. Thus ${\int }_a^b{s}^{\prime \prime }{u}^{\prime \prime }=0$, and the second derivative roughness penalty satisfiesIf equality holds, the continuous nonnegative function $(u^{\prime \prime }{)}^2$ has zero integral, so $u^{\prime \prime }=0$ everywhere. Hence $u$ is an affine function, and its zeros at two distinct spline knots force $u=0$. This proves the minimum roughness property of the natural cubic spline interpolant and its equality case: the unique minimum-roughness interpolant is $s$.

For the fixed-design nonparametric regression fit, use the penalized least squares criterionLet $s_z$ be the natural cubic spline interpolant with value vector $z=(z_1,\dots ,z_n{)}^T$. Replacing $v$ by $s_z$ for $z_i=v(x_i)$ leaves the residual sum unchanged and reduces the second derivative roughness penalty, strictly unless $v=s_z$. It therefore remains to minimize the finite-dimensional quadratic formThe supplied spline roughness penalty matrix $K$ is positive semidefinite. For every nonzero vector $w$,Consequently $I+\lambda K$ is a positive-definite matrix, and $Q_{\lambda }$ is strictly convex. Its normal equations for linear least squares have the unique solutionThe preceding strict reduction also proves uniqueness over the whole class $C^2[a,b]$, not just over the natural cubic splines. This is the cubic smoothing spline. If the residual criterion is divided by $n$ instead, the corresponding formula is $(I+n\lambda K{)}^{-1}Y$, reflecting a change in the numerical convention for the smoothing parameter.

The independence and specified error mean and variance are not needed for existence or uniqueness of this minimizer. They do imply, with smoothing matrix $A_{\lambda }=(I+\lambda K{)}^{-1}$ and $g_X=(g(x_i){)}_i$, that $\mathbb{E}\hat{z}=A_{\lambda }{g}_X$ and $\mathrm{Cov}(\hat{z})={\sigma }^2{A}_{\lambda }^2$. The null space of $K$ consists of sampled affine functions: these have zero roughness, and conversely zero roughness forces their natural cubic spline interpolant to be an affine function. Thus these functions are reproduced without bias of an estimator, while other components are smoothed.

Consider a competing interpolant $f$ and set $h=f-g$. At each spline knot, $h(x_i)=0$. If its second derivative roughness penalty is infinite there is nothing to prove, so assume $f^{\prime \prime }\in L^2$. Its absolutely continuous first derivative makes $h^{\prime }$ absolutely continuous as well, with $h^{\prime \prime }$ defined almost everywhere.

On each knot interval the natural cubic spline has constant third derivative. Integrating by parts once givesThe last term vanishes because $h$ is zero at both endpoints. Summing over intervals cancels the interior boundary terms, since $g^{\prime \prime }$ and $h^{\prime }$ are continuous at the knots. The exterior terms vanish because $g^{\prime \prime }(x_1)=g^{\prime \prime }(x_n)=0$. Hence $\int g^{\prime \prime }{h}^{\prime \prime }=0$. Expanding the square now givesSince $\alpha >0$, equality requires $h^{\prime \prime }=0$ almost everywhere. Absolute continuity then makes $h$ affine. It has at least two distinct zeros because $n\ge 2$, so $h=0$. The natural cubic spline is the unique minimizer. This proves the minimum roughness property with absolutely continuous first derivatives, covering the weaker regularity in the question rather than assuming competitors are twice continuously differentiable. The case of two knots is included: the minimizing spline is their straight-line interpolant.

For woman $i$, write $m_i$ for the number of births, $Y_i$ for the number with defects, $a_i$ for baseline age, and $e_i$ for exposure. Let $p_i$ be the common marginal defect probability for her births. The first fit is a grouped grouped-binomial logistic regression:It assumes independent women and, conditionally on their covariates, independent births with the same probability within each woman. The supplied birth totals are the binomial denominators; fitting the proportions with weights $m_i$ is equivalent to this grouped-binomial specification. The fitted linear predictor is $-1.74272+0.02030a_i+2.33028e_i$.

The second fit is a generalized additive model with the same logit link function and the mean specificationThe functions $f_a,f_e$ are penalized cubic regression splines with natural boundary conditions. Centering constraints such as ${\sum }_i{f}_a(a_i)={\sum }_i{f}_e(e_i)=0$ separate them from the intercept; the fitted centered intercept is $-0.2031$. Their second derivative roughness penalties control complexity.

The call estimates a common scale rather than fixing it at one. Its working variance specification for the counts isEquivalently the variance of $Y_i/m_i$ is $\varphi p_i(1-p_i)/m_i$. This is the working moment interpretation of an overdispersed binomial generalized additive model: the code uses binomial deviance for fitting and estimated scale for inference. With $\varphi \ne 1$, it is not an exact independent-binomial sampling model. Women are still treated as independent groups, but within-woman dependence or unobserved heterogeneity can motivate the extra dispersion. The quasibinomial regression interpretation states what the scaled analysis assumes without inventing a full probability distribution for its counts.

Use the natural cubic spline convention appropriate to the interior design knots: the spline is linear on $[a,x_1]$ and $[x_n,b]$. We also need $n\ge 2$. These points will be important for uniqueness.

A cubic spline is a function in $C^2[a,b]$ whose restriction to each of $[a,x_1]$, $[x_1,x_2]$, through $[x_n,b]$ is a polynomial of degree at most three. A spline knot is a junction between consecutive pieces. For a natural cubic spline, the two exterior pieces are affine functions; equivalently, one uses the natural endpoint conditions $g^{\prime \prime }(x_1)=g^{\prime \prime }(x_n)=0$ on the spline over $[x_1,x_n]$ and then extends linearly with matching derivatives. Merely prescribing $g^{\prime \prime }(a)=g^{\prime \prime }(b)=0$ on otherwise arbitrary exterior cubic pieces is a different condition and does not provide the unique interpolant invoked in this question.

The natural cubic spline interpolant $N\mathbf{v}$ to values $\mathbf{v}=(v_1,\dots ,v_n{)}^T$ is the natural cubic spline satisfying $(N\mathbf{v})(x_i)=v_i$. Its existence and uniqueness for $n\ge 2$ are the interpolant fact allowed in the PDF. The interpolation map $N$ is a linear map: a linear combination of interpolants satisfies the same spline, boundary and value conditions, so uniqueness identifies it with the interpolant to the corresponding linear combination of value vectors.

Let $g=N\mathbf{v}$, let $\tilde{g}\in C^2[a,b]$ have the same values, and put $r=\tilde{g}-g$. Then $r(x_i)=0$. We prove the orthogonalityApply integration by parts twice on each polynomial interval. On an interior interval, $g^{\prime \prime \prime \prime }=0$, andThe $g^{\prime \prime \prime }r$ terms vanish because $r$ vanishes at the knots. The $g^{\prime \prime }{r}^{\prime }$ terms cancel at internal knots because $g^{\prime \prime }$ and $r^{\prime }$ are continuous, and vanish at $x_1,x_n$ because $g^{\prime \prime }$ is zero there. On the exterior intervals, $g^{\prime \prime }=0$ identically. Summing therefore gives the claimed orthogonality without assuming that $g^{\prime \prime \prime }$ is continuous across knots.

Expanding the square now gives the minimum roughness property of the natural cubic spline interpolantEquality forces $r^{\prime \prime }=0$ everywhere, since $r^{\prime \prime }$ is continuous. Thus $r$ is an affine function. It vanishes at two distinct design points, so $r=0$, proving equality if and only if $\tilde{g}=g$.

For the penalized fit, take the allowed spline roughness penalty matrix $\Gamma$ to be a symmetric matrix with ${\mathbf{z}}^T\Gamma \mathbf{z}\ge 0$; thus it is a positive semidefinite matrix. If its supplied representative were not symmetric, its symmetric part defines the same quadratic form and is all that is needed. For any $\tilde{g}$, let $\mathbf{v}=(\tilde{g}(x_i){)}_{i=1}^n$. Replacing it by $N\mathbf{v}$ preserves every fitted value and can only decrease the second derivative roughness penalty. Therefore it is enough to minimizeover ${\mathbb{R}}^n$. The matrix $A_{\lambda }=I_n+\lambda \Gamma$ is a positive-definite matrix, sinceCompleting the quadratic or differentiating gives the unique solutionThis is the cubic smoothing spline. More explicitly, the normal equations are $(I_n+\lambda \Gamma ){\hat{\mathbf{v}}}_{\lambda }=\mathbf{Y}$, andFor an arbitrary function with fitted vector $\mathbf{v}$, the earlier orthogonality also givesBoth terms are nonnegative. Equality forces $\mathbf{v}={\hat{\mathbf{v}}}_{\lambda }$ and, by the proved interpolation equality case, $\tilde{g}=N\mathbf{v}={\hat{g}}_{\lambda }$. This proves both existence in $C^2[a,b]$ and uniqueness over the entire stated function class, not just over splines.

The argument is deterministic for each observed $\mathbf{Y}$. The fixed-design nonparametric regression and homoscedasticity assumptions motivate the squared-error loss, but Gaussian errors are not required.

At least two distinct design points are required for uniqueness. Question 4 does not explicitly restate $n\ge 2$, and this hypothesis is necessary. With one design point, every function $\tilde{g}(x)=Y_1+c(x-x_1)$ has zero residual and zero second derivative roughness penalty, for any real $c$. Thus the printed unrestricted uniqueness conclusion would be false for $n=1$. The corrected theorem assumes at least two distinct knots and the stated linear-tail convention. At $\lambda =0$, uniqueness over all of $C^2[a,b]$ also fails; the stipulated $\lambda >0$ eliminates that degeneracy.

A penalized function estimate balancing a residual loss against an integrated derivative penalty. For squared-error loss and the second derivative roughness penalty, the estimate is a cubic smoothing spline. Higher derivative orders give other spline degrees.

Let $b_i=N{e}_i$ be the interpolating cardinal basis and set ${\Gamma }_{ij}={\int }_a^b{b}_i^{\prime \prime }{b}_j^{\prime \prime }dx$. Linearity gives $J(N\mathbf{v})={\mathbf{v}}^T\Gamma \mathbf{v}$, and this Gram matrix is a positive semidefinite matrix. Its null space consists precisely of vectors $(\alpha +\beta x_i{)}_i$, by the null-space statement for the second derivative roughness penalty. Thus its rank is $n-2$, including the zero matrix when $n=2$.