Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 33 / 5

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.

The first model's residual deviance is $671.27$ on $197$ residual degrees of freedom, a ratio of about $3.41$. This is far above the scale expected from a well-fitting dispersion-one grouped binomial regression. It suggests an inadequate mean function, overdispersion, or both. The linear age and exposure restrictions may miss nonlinear associations; the generalized additive model allows those shapes to be estimated through penalized cubic regression splines rather than assumed.

The very small exposure Wald p-value in the first fit suggests an association, but does not establish that a linear logit effect is adequate. Nor does the nonsignificant linear age term exclude every possible nonlinear age effect. Allowing estimated scale also addresses the excessive residual variation, which can arise from clustered data because several births belong to the same woman. The second model's estimated scale $3.3506$ confirms that permitting smooth means has not removed all extra variation. The reason to proceed is poor initial fit and possible nonlinear effects, with dispersion-aware inference, not merely the wish to obtain more significant tests. These observational fits alone do not establish causation by the disaster.

The trace of the smoother's influence matrix is its total effective degrees of freedom. Include the unpenalized intercept as well as both centered smooth terms:The Ref.df entries are for approximate significance calibration and must not be summed to obtain this trace. A consistency check using the generalized cross-validation form $nD/(n-\mathrm{tr}A{)}^2$ and scale estimate $D/(n-\mathrm{tr}A)$ gives $\mathrm{tr}A\approx 200(1-3.3506/3.4217)\approx 4.16$; the small difference is explained by rounding in the displayed quantities. The required trace is about $4.15$.

The original PDF's age smooth is nearly flat, its uncertainty band includes zero across the age range, and its approximate p-value is $0.859$. The exposure smooth is increasing and convex: it is fairly flat at low exposure and rises more steeply at high exposure. Its effective degrees of freedom $1.895$ suggest a modest departure from a straight line, rather than a highly oscillatory relationship.

A suitable simpler model is therefore the grouped quasibinomial regressionwith age omitted and dispersion estimated. A positive quadratic coefficient captures the convex exposure relationship; retaining the linear term respects the usual polynomial hierarchy. This model should be refitted, rather than assigning coefficients from the smooth plot. Compare its residual pattern and dispersion-adjusted fit with the additive model to check that the quadratic approximation is adequate. An exposure-only quadratic logit model is a justified parsimonious candidate, while the graph does not justify a sharp threshold or a zero-dispersion model.