Efficient score 2026-10-07
The efficient score is the residual after orthogonal projection of the parametric score function onto the nuisance tangent space. It is centered and orthogonal to every nuisance direction. Its squared L2 norm is the efficient information.
With unknown independent covariate and centered error distributions, take the regular mean-preserving error tangent space, let , and . Under , and finite second moments, orthogonal projection removes from . Thus the efficient score and efficient information are and . The reduction to is valid for a normal distribution of errors or for constant , but need not hold otherwise. Centered covariates and logistic distribution errors give the counterexample .
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 36 1 c Solution Created 2026-10-03 Updated 2026-10-07
Fix . Let be the nuisance tangent space: the closed linear span in of score functions of statistical paths that vary only the nuisance parameter . By the preceding argument, is contained in the mean-zero L2 space .
Let denote orthogonal projection onto this closed subspace of a Hilbert space. The efficient score and scalar efficient information areThe efficient score is the component of the parametric score function that cannot be reproduced by changing the nuisance parameter. The efficient information is its squared L2 norm; it can be zero, so positivity must not be assumed in the definition.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 36 1 d Solution Created 2026-10-03 Updated 2026-10-07
The map is a bounded linear functional on L2 space, since by the Cauchy-Schwarz inequality. Its kernel, the mean-zero L2 space, is therefore closed. Every nuisance score function and the parametric score function are centered, so both and are centered. HenceMoreover, the efficient score belongs to the orthogonal complement of the nuisance tangent space. Writing givesThis is the efficient-score projection identity; it remains valid when the efficient information is zero.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 36 4 d Solution Created 2026-10-03 Updated 2026-10-07
First use the additional admitted form of the efficient score. Write and let . The difference is the orthogonal projection onto the nuisance tangent space. Part (c) makes every orthogonal to that space. Therefore, for all ,The function in braces belongs to L2 space; choosing it as shows it vanishes -almost everywhere. Hence the required conditional deduction isUnder the regular tail condition at both infinities, integration by parts gives , and the formula becomes . This also confirms the sign.
The admitted product form is an extra restriction; it does not follow for every independent-error regression model. To locate the restriction precisely, suppose the error-density nuisance statistical paths preserve both normalization and mean to first order. Their closed mean-preserving error tangent space is . The full nuisance tangent space is the orthogonal sum of this space and the centered functions of .
For completeness, bounded functions satisfying the two constraints are dense in the error space. Truncate an arbitrary element, subtract its expected value, and then subtract a multiple of a fixed bounded centered function with . Such a exists by truncating , since . The correction coefficients tend to zero by the Cauchy-Schwarz inequality, so the corrected truncations converge in L2 space.
Let . With and , the orthogonal projection of onto the error nuisance space is ; its projection onto the covariate nuisance space is zero. Thus the general efficient score in independent-error regression isThe orthogonality of the two terms gives the displayed efficient information. For a normal distribution of the error, , so this reduces to the admitted formula. It also does so when is constant.
A concrete counterexample to the generality of the admission is , with uniform on and independent standard logistic distribution error. Here and , so the actual efficient score is . It cannot equal because is not constant. This score function is already orthogonal to every nuisance score function: its factor is centered against error-only directions, and its factor is centered against covariate-only directions. The requested formula is valid under its stated additional admission, with the general independent-error formula above explaining its limits.