Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 36 (Semiparametric Statistics)

A statistical path through $P$ is a family $t\mapsto P_t\in \mathcal{P}$, defined on an open interval containing zero, with $P_0=P$. Write $p_t=d{P}_t/d\mu$ and $p=p_0$. Its differentiability in quadratic mean at zero means that there is a square-integrable function $g\in L^2(P)$ such thatThe function $g$ is the score function of the statistical path. Thus it is the derivative of the square-root probability density function, multiplied by two and divided by $\sqrt{p}$ where $p>0$. The score function is determined only $P$-almost everywhere. The displayed definition also controls any probability mass entering a region where $p=0$; a pointwise derivative of the log probability density function alone would not do that. The required derivative is in the square-root-density $L^2$ norm.

Put $r_t=\sqrt{p_t}-\sqrt{p}-(t/2)g\sqrt{p}$. By differentiability in quadratic mean, $\Vert r_t{\Vert }_{L^2(\mu )}=o(|t|)$ andin L2 space. The second limit follows because the first gives $\Vert \sqrt{p_t}-\sqrt{p}{\Vert }_2=O(|t|)$.

Both probability density functions integrate to one. Consequently the L2 inner product satisfiesThe Cauchy-Schwarz inequality permits passage to the limit in this inner product, givingThus every score function is a mean-zero function. This proof uses square-root differentiability in quadratic mean, without requiring differentiation under the density integral.

Fix $P=P_{\theta ,\eta }$. Let $\mathcal{N}$ be the nuisance tangent space: the closed linear span in $L^2(P)$ of score functions of statistical paths that vary only the nuisance parameter $\eta$. By the preceding argument, $\mathcal{N}$ is contained in the mean-zero L2 space $L_0^2(P)$.

Let ${\Pi }_{\mathcal{N}}$ 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.

The map $h\mapsto Ph$ is a bounded linear functional on L2 space, since $|Ph|\le \Vert h{\Vert }_2$ by the Cauchy-Schwarz inequality. Its kernel, the mean-zero L2 space, is therefore closed. Every nuisance score function and the parametric score function $\dot{\ell }$ are centered, so both ${\Pi }_{\mathcal{N}}\dot{\ell }$ and $\tilde{\ell }$ are centered. HenceMoreover, the efficient score belongs to the orthogonal complement of the nuisance tangent space. Writing $\dot{\ell }=\tilde{\ell }+{\Pi }_{\mathcal{N}}\dot{\ell }$ givesThis is the efficient-score projection identity; it remains valid when the efficient information is zero.

Choose a collection of statistical paths through $P_f$ that are differentiable in quadratic mean. A statistical tangent set at $P_f$ is the set of their score functions. In particular each member belongs to $L_0^2(P_f)$, by the mean-zero score identity under quadratic-mean differentiability.

A statistical tangent set records which first-order directions the chosen statistical paths can realize. Its closed linear span in L2 space is the statistical tangent space. The tangent set consists of attainable scores; the tangent space also includes their linear combinations and $L^2(P_f)$ limits.

Let $g$ be a bounded function with $P_{f}g=0$. A bounded density tilt realizes this direction:These are nonnegative probability density functions, since $\int f_t=1+t{P}_{f}g=1$. The uniform Taylor expansion of the square root givesThe squared L2 norm of this remainder is $O(t^4{P}_f{g}^4)=o(t^2)$, since $g$ is bounded. Thus the statistical path is differentiable in quadratic mean with score function $g$.

Conversely every score function is centered by the mean-zero score identity under quadratic-mean differentiability. Choosing all these bounded density tilts therefore gives the statistical tangent setThis is a valid choice of statistical tangent set; it does not assert that every possible score function in the unrestricted density model is bounded.

Consider any differentiable-in-quadratic-mean path with score function $g$. Put ${\delta }_t=\sqrt{f_t}-\sqrt{f}=(t/2)g\sqrt{f}+r_t$, where $\Vert r_t{\Vert }_2=o(|t|)$. The quadratic-mean to L1 density derivative follows fromBy the Cauchy-Schwarz inequality,Since $a$ is a bounded function, multiplying this L1 norm bound by $\Vert a{\Vert }_{\infty }$ provesThe last equality uses $P_{f}g=0$. The derivative is a bounded linear functional of $g\in L^2(P_f)$, so the required pathwise differentiability of a statistical functional holds, in particular relative to the statistical tangent set from part (b). Its derivative isFor the explicit bounded density tilts in part (b), this derivative is also obtained by direct integration, with no remainder term.

For a pathwise differentiable statistical functional, an influence-function representer is a mean-zero function $\phi \in L^2(P)$ such that every admissible score function $g$ satisfies $D{\psi }_P(g)=P(\phi g)$. The efficient influence function, also called the canonical gradient, is the unique such representer in the statistical tangent space. Equivalently, it is the orthogonal projection of any representer onto that statistical tangent space. The Pythagorean theorem in an inner-product space shows that it has the smallest squared L2 norm among all representers.

Here the statistical tangent space is all of $L_0^2(P_f)$. To verify the closure explicitly, take $h\in L_0^2(P_f)$, truncate it to $h_m=\max (-m,\min (h,m))$, and set $g_m=h_m-P_f{h}_m$. Then $g_m$ is bounded and centered, and $g_m\to h$ in $L^2(P_f)$, by dominated convergence and the Cauchy-Schwarz inequality. Part (c) supplies a representer already in this space. HenceIts variance is $P_f{a}^2-(P_{f}a{)}^2$.

The closure must be taken in the density-weighted space $L^2(P_f)$. An unweighted reading of $L^2[0,1]$ in the printed hint is false. For example, when $f(u)=2u$, the function $h(u)=u^{-3/4}-8/5$ has $P_{f}h=0$ and $P_f{h}^2=36/25$, so bounded centered truncations converge to it in $L^2(P_f)$; nevertheless $h\notin L^2([0,1],du)$. This illustrates density of bounded centered scores and fixes the measure in the closure statement.

For the bounded density tilt $f_t=f(1+tg)$, differentiate the polynomial in $t$:Since every score function is centered, the centered representer isA bounded baseline $f$ makes this a bounded mean-zero function, hence an element of $L_0^2(P_f)$. Thus it is the efficient influence function for the density fourth-power functional relative to the regular bounded density tilts. Its squared L2 norm is $16\{\int f^7-(\int f^4{)}^2\}$.

A bounded baseline alone does not make this functional differentiable along every quadratic-mean differentiable path. The derivative above is the intended regular-path answer. To see the need for the qualification, let $f_0=1$ and put $b(s)=6s(1-s)$ on $[0,1]$, extended by zero outside. For $0<|t|<1$, defineEach $f_t$ is a nonnegative continuous probability density function, because its narrow bump has mass $t^4$. Moreover,It is therefore a differentiable-in-quadratic-mean path with zero score function. YetThe density fourth-power functional is not even continuous along this statistical path, although every $f_t$ is individually bounded. Consequently no efficient influence function represents derivatives over the unrestricted class of all such paths.

One sufficient additional condition is a common bound $f_t,f\le M$ for all small $t$. Taylor expansion then bounds the fourth-power remainder by $6M^2(f_t-f{)}^2$, whileTogether with the quadratic-mean to L1 density derivative, this gives $\psi (P_{f_t})-\psi (P_f)=t{P}_f({\phi }_{f}g)+o(|t|)$. Under this local bound, or when the chosen statistical paths are the bounded density tilts, the boxed canonical gradient is fully justified. The counterexample is a spike obstruction to density-power differentiability.

Fix any estimator $\hat{\eta }$ and put $A(x)=d(\hat{\eta }(x),f)$, $B(x)=d(\hat{\eta }(x),g)$ and $D=d(f,g)$. The triangle inequality for the metric gives $A+B\ge D$. HenceThe larger of the two risk functions dominates their average. Using nonnegativity to replace both probability density functions by their minimum,The right side is independent of the estimator, so taking the infimum provesThis metric squared-loss two-point bound uses overlap of the observation laws to quantify how hard it is to distinguish the two separated parameters.

Apply the preceding argument to the two product observation laws and the real-valued estimator $T={\hat{\psi }}_n$. Alternatively, directly useThe average of the two mean squared errors is at least $(\theta -\tau {)}^2/4$ times the overlap of the two joint probability density functions. Their overlap equalsTaking the infimum over estimators therefore yieldsHere $R_n^{\ast }$ is the minimax risk for this functional. The factor $\frac{1}{2}\Vert P_f^{(n)}-P_g^{(n)}{\Vert }_1$ is the total variation distance, so the same metric squared-loss two-point bound applies to a functional even when distinct density parameters have the same functional value.

For each $n\ge 1$, choose the continuous probability density functionsThe second is at least $1/2$ and integrates to one. Their expected values differ byWriting $g_n/f=1+{\Delta }_n$ givesThe supplied product bound, also obtained from the chi-squared divergence of product measures, impliesFor the final elementary estimate, compare the exponential series with the geometric series: $e^x\le (1-x{)}^{-1}$ for $0\le x<1$. Thus the overlap of the product laws is at least $1/2$. Part (b) now gives the explicit mean-estimation minimax lower bound for continuous densitiesSo one may take $C=1/1152$, uniformly for all $n\ge 1$. The alternatives are allowed to depend on $n$, because the minimax risk takes a supremum over all continuous probability density functions separately at each sample size.

Write $e=y-g_{\theta }(x)$ and $h_{\theta }(x)=\partial g_{\theta }(x)/\partial \theta$. Holding the nuisance parameter $\eta$ fixed, the parametric score function is the derivative with respect to $\theta$ of the one-observation log-likelihood. The joint probability density function isBecause $\partial e/\partial \theta =-h_{\theta }(x)$, differentiating givesFor instance $E_{\eta }{h}_{\theta }(X{)}^2<\infty$ ensures a square-integrable score function. Its expected value is zero by independence and the centered normal distribution of the error. This is the Gaussian regression score.

Vary only the probability density function of $X$, using ${\eta }_t=\eta (1+tb)$ with bounded $b$ and $E_{\eta }b(X)=0$. The nuisance score function is $b(X)$, giving the statistical tangent setIts closed linear span is the nuisance tangent space of all centered $L^2(\eta )$ functions of $X$, by density of bounded centered scores.

The conditional expectation of the parametric score function given $X$ is zero:It is therefore orthogonal to this nuisance tangent space. Its orthogonal projection onto that space vanishes, so the efficient score is unchanged. By independence and $E{\epsilon }^2={\sigma }^2$,These equal the parametric score function and Fisher information when $\eta$ is known. There is no loss of information from the unknown covariate density. This is adaptivity to an unknown covariate distribution; it follows from score orthogonality, without having to estimate the nuisance parameter.

Now $p_{\theta ,\eta }(x,y)=v(x)f(y-g_{\theta }(x))$. With $h_{\theta }={\partial }_{\theta }{g}_{\theta }$ and the location score $\rho (e)=-f^{\prime }(e)/f(e)$, differentiation of the log-likelihood givesTake the usual regularity conditions $h_{\theta }\in L^2(v)$ and $\rho \in L^2(f)$ so that this score function belongs to L2 space.

For the specified nuisance statistical path, differentiating $\log (1+t\gamma (e))$ at zero gives the score function $\gamma (e)$. Normalization requires $E_f\gamma =0$. There is a second constraint: the statistical path must preserve the model's zero error expected value. ThusThis constraint follows from being a path through the model, rather than from normalization alone. For a centered symmetric logistic distribution, for example, $\gamma (e)=\tanh (e)$ is bounded and has zero expected value, but $E[\epsilon \tanh (\epsilon )]>0$; it would not preserve the mean.

For any $\psi \in L^2(v)$, independence impliesAll these products are integrable by the Cauchy-Schwarz inequality, since $\epsilon \psi (X)$ is square-integrable. Thus every such error-density score function is orthogonal to every $e\psi (x)$. Covariate-density nuisance score functions $b(X)$ are also orthogonal to these functions, since $E\epsilon =0$. This is the mean-preserving error tangent space orthogonality used in the final calculation.

First use the additional admitted form of the efficient score. Write $\tilde{\ell }=e\zeta (X)$ and let ${\tau }^2=E_f{\epsilon }^2>0$. The difference $\dot{\ell }-\tilde{\ell }$ is the orthogonal projection onto the nuisance tangent space. Part (c) makes every $e\psi (X)$ orthogonal to that space. Therefore, for all $\psi \in L^2(v)$,The function in braces belongs to L2 space; choosing it as $\psi$ shows it vanishes $v$-almost everywhere. Hence the required conditional deduction isUnder the regular tail condition $sf(s)\to 0$ at both infinities, integration by parts gives $\int s{f}^{\prime }(s)ds=-1$, and the formula becomes $e{h}_{\theta }(x)/{\tau }^2$. 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 $\{\gamma :E_f\gamma =E_f(\epsilon \gamma )=0\}$. The full nuisance tangent space is the orthogonal sum of this space and the centered functions of $X$.

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 $q$ with $E_f(\epsilon q)\ne 0$. Such a $q$ exists by truncating $\epsilon$, since ${\tau }^2>0$. The correction coefficients tend to zero by the Cauchy-Schwarz inequality, so the corrected truncations converge in L2 space.

Let $m=E_v{h}_{\theta }(X)$. With $E_f\rho =0$ and $E_f(\epsilon \rho )=1$, the orthogonal projection of $h_{\theta }(X)\rho (\epsilon )$ onto the error nuisance space is $m\{\rho (\epsilon )-\epsilon /{\tau }^2\}$; 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, $\rho (e)=e/{\tau }^2$, so this reduces to the admitted formula. It also does so when $h_{\theta }$ is constant.

A concrete counterexample to the generality of the admission is $g_{\theta }(x)=\theta x$, with $X$ uniform on $[-1,1]$ and independent standard logistic distribution error. Here $m=0$ and $\rho (e)=\tanh (e/2)$, so the actual efficient score is $x\tanh (e/2)$. It cannot equal $e\zeta (x)$ because $\tanh (e/2)/e$ is not constant. This score function is already orthogonal to every nuisance score function: its factor $X$ is centered against error-only directions, and its factor $\rho (\epsilon )$ 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.