Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 42 (Statistical Theory)

The profile likelihood for the parameter of interest replaces the nuisance parameter by its constrained best fit: $L_p(\psi )={\sup }_{\lambda }L(\psi ,\lambda )$, with ${\ell }_p(\psi )=\ell (\psi ,{\hat{\lambda }}_{\psi })$. It is not in general a normalized sampling density. A conditional likelihood instead uses the sampling law conditional on a statistic carrying the nuisance dependence. In the canonical exponential family here, conditioning on $S$ removes $\lambda$ because the factor involving $\lambda$ is constant over the conditioning fibre.

Let ${\bar{T}}_1=n^{-1}{\sum }_i{\tau }_1(y_i)$ and let the observed $S$ equal $s$. For the cumulant-generating function of ${\tau }_2(Y)$ at fixed $\psi$,The saddlepoint solves $K^{\prime }(\hat{t})=s$, equivalently $d_{\lambda }(\psi ,{\hat{\lambda }}_{\psi })=s$ with ${\hat{\lambda }}_{\psi }=\lambda +\hat{t}$. This is also the constrained likelihood equation, since ${\ell }_{\lambda }=n(s-d_{\lambda })$. Assume an interior saddle, $d_{\lambda \lambda }>0$, and the usual smooth nonlattice conditions under which $S$ has a density. Exponential tilting to ${\hat{\lambda }}_{\psi }$ makes $s$ the tilted mean. A local normal approximation gives tilted density $\{n/(2\pi d_{\lambda \lambda }(\psi ,{\hat{\lambda }}_{\psi })){\}}^{1/2}$ at that mean; undoing the tilt gives the saddlepoint density approximationThe factor $n^{1/2}$ is for the sample mean, rather than the sum. A transform-based derivation gives the same expression by expanding $K$ to second order at its stationary point.

The joint log-likelihood is $n\psi {\bar{T}}_1+n\lambda s-nd(\psi ,\lambda )-{\sum }_{i}Q(y_i)$. Subtracting $\log {\hat{f}}_S(s;\psi ,\lambda )$ cancels every appearance of the original $\lambda$, leavingAny change from the ambient density to a conditional density on $S=s$ contributes only a data-dependent Jacobian determinant. Thus, up to a constant independent of $\psi$, the saddlepoint conditional likelihood adjustment isEquivalently $B=\frac{1}{2}\log j_{\lambda \lambda }(\psi ,{\hat{\lambda }}_{\psi })$ up to the constant $-\frac{1}{2}\log n$, where $j_{\lambda \lambda }=n{d}_{\lambda \lambda }$ is the nuisance observed information. The adjustment has a positive sign. The printed exponential-family form alone does not guarantee a density for $S$: for example a continuous $Y$ can have ${\tau }_2(Y)=1_{\{Y>0\}}$. In a lattice case one must approximate the probability mass of the sum instead; the displayed density calculation uses the regular continuous interpretation.

A transformation model consists of a group $G$ acting measurably and invertibly on the sample space, an induced group action on the parameter space, and a family $\{P_{\theta }\}$ satisfying $Y\sim P_{\theta }\Rightarrow gY\sim P_{g\theta }$. In a generated, transitive model, every $P_{\theta }$ is the image of a reference distribution under some transformation. In the density formulation, the transformed density includes the appropriate sample-space Jacobian determinant; the model is a statement about whole probability laws, not merely about changing parameter names.

A maximal invariant $A$ has two properties: $A(gy)=A(y)$ for every $g$, and $A(y)=A(z)$ implies $z=gy$ for some $g$. It is therefore a coordinate for the group orbits, retaining exactly the information unaffected by the transformations. An equivariant estimator obeys $T(gy)=gT(y)$, where the right action is the induced parameter action.

For the normalization construction, suppose the estimator determines a group element $t(y)\in G$ with $t(gy)=gt(y)$. Define $A(y)=t(y{)}^{-1}y$. ThenIf $A(y)=A(z)$, rearranging gives $z=t(z)t(y{)}^{-1}y$. This is an actual transformation in $G$, so the normalized sample is a maximal invariant. This group-valued assumption matters: an estimator of a parameter with a nontrivial stabilizer does not necessarily determine a unique normalization. For a transitive parameter space $G/H$, choose a transformation carrying the estimated parameter to the reference value and take the resulting normalized sample modulo the remaining $H$-action. Different choices then have the same residual orbit. Under a free action $H$ is trivial and the simpler formula applies.

For a location-scale model, $g_{a,b}(y_1,\dots ,y_n)=(a+b{y}_1,\dots ,a+b{y}_n)$, $b>0$, with composition $(a,b)(c,d)=(a+bc,bd)$. Its parameter action is $(\mu ,\sigma )\mapsto (a+b\mu ,b\sigma )$. Choose location and scale estimates satisfyingFor instance, on nonconstant samples take the sample mean and the positive square root of $n^{-1}{\sum }_i(y_i-\bar{y}{)}^2$. The standardized vectoris invariant. Conversely, equality of these vectors gives $z_i=c+d{y}_i$ for all $i$, with $d=\hat{s}(z)/\hat{s}(y)>0$ and $c=\hat{m}(z)-d\hat{m}(y)$. Hence it is maximal, not just invariant. Degenerate constant samples have zero fitted scale and form a separate orbit; define a separate invariant value there. The construction is on labelled sample vectors: one must not discard their order unless permutations are also included in the transformation group.

Put $S_1={\sum }_i{Y}_i$, $S_2={\sum }_i{Y}_i^2$ and $R=\sqrt{S_2}$. Apart from a data-only constant, the log-likelihood isThe natural parameters in the ambient two-parameter exponential family of normal distributions are ${\eta }_1=1/\mu$ and ${\eta }_2=-1/(2{\mu }^2)$, restricted to the nonaffine curve ${\eta }_2=-{\eta }_1^2/2$, ${\eta }_1>0$. Its parametrization has rank one. Thus this is a one-dimensional curved exponential family inside a two-dimensional full family.

The Fisher-Neyman factorization theorem makes $(S_1,S_2)$ sufficient. For two samples $y,z$, the parameter-dependent part of their log density ratio is $\Delta S_1/\mu -\Delta S_2/(2{\mu }^2)$. A polynomial in $1/\mu$ is constant on $(0,\infty )$ only if both nonconstant coefficients vanish. The likelihood-ratio criterion for minimal sufficiency therefore proves $(S_1,S_2)$ is minimal sufficient.

Write $Y_i=\mu Z_i$ with independent $Z_i\sim N(1,1)$. Since $\mu >0$, $a=\sqrt{n}\sqrt{{\sum }_i{Z}_i^2}/{\sum }_i{Z}_i$ has a probability distribution independent of $\mu$: it is an ancillary statistic. The denominator is nonzero almost surely. The Cauchy-Schwarz inequality gives $|a|\ge 1$; on the requested positive branch, $S_1>0$ and $a\ge 1$.

The likelihood equation is $n{\mu }^2+S_1\mu -S_2=0$. Its positive root isThere is exactly one positive root. The sign of the score function changes from positive to negative there, and $\ell \to -\infty$ at both ends of $(0,\infty )$ when $S_2>0$, so it is the unique maximum-likelihood estimator. The useful identities areSubstitution yields, dropping the constant $-n/2$,Differentiating twice and using $q/a=q^2-1$ at the fitted value gives the observed informationThe scalar P-star approximation uses $\sqrt{j(v)}\exp \{\ell (\mu ;v,a)-\ell (v;v,a)\}$. Consequently, with its normalizing factor explicitly retained,Here $C_n(a)$ is chosen to make the density integrate to one, and is independent of $\mu$. To see this without assuming an approximate normalizer, defineThe integral is finite and positive, and normalized $p^{\ast }$ is exactlyIndeed $C_n(a)=\{\sqrt{n(1+q^2)/(2\pi )}\exp [n{q}^2/2-nq/a]J_n(a){\}}^{-1}$.

There is a direct check of this conditional scale density in a curved Gaussian family. In polar coordinates $y=Ru$, the sample density times volume element is proportional to ${\mu }^{-n}{R}^{n-1}\exp [-R^2/(2{\mu }^2)+R{\sum }_i{u}_i/\mu ]dRd\omega (u)$. Conditioning on $a$ fixes ${\sum }_i{u}_i=\sqrt{n}/a$. The remaining angular factor is independent of $R,\mu$, while $R=q\sqrt{n}v$. Thus the exact conditional radial density has the same normalized form. For $n=1$ the positive ancillary branch fixes the positive sign and the same one-dimensional radial calculation applies. In this scale model the normalized $p^{\ast }$ expression equals the exact conditional density, not merely its leading approximation.

For $t>0$, compute its conditional distribution by one-dimensional quadrature:It is zero for $t\le 0$; with the conditional-density calculation just given, this normalized formula is exact for almost every admissible $a$. Numerically, subtract the maximum log integrand before exponentiating to avoid overflow. The sampling probability still depends on the stipulated true $\mu$; replacing it by a fitted value would be a separate plug-in inference step.

A scalar M-estimator is a selected root of $n^{-1}{\sum }_i\psi (Y_i,\hat{\theta })=0$, or equivalently a minimizer of a differentiable empirical objective whose derivative is this estimating equation. Its population statistical functional $T(F)={\theta }_F$ solves $\int \psi (x,{\theta }_F)dF(x)=0$. Assume local uniqueness, consistency of the sample root, a valid differentiable local expansion, finite ${\mathbb{E}}_F{\psi }^2$, andThese conditions include domination sufficient to differentiate under the integral and the required local uniform control of the sample derivative.

For $F_{\epsilon }=(1-\epsilon )F+\epsilon {\delta }_z$, differentiate the population estimating equation at $\epsilon =0$. Since its uncontaminated integral is zero,Thus the influence function is $\mathrm{IF}(z;T,F)=-\psi (z,{\theta }_F)/A$, with a proportionality constant independent of $z$.

Expanding the sample equation at its population root givesThe summands have mean zero. The central limit theorem and Slutsky's theorem therefore yield the scalar sandwich variance of an M-estimatorThis $V$ is the limiting variance of the $\sqrt{n}$-scaled error; the estimator's variance is approximately $V/n$. In a correctly centred location family with $\psi (x,\theta )=\psi (x-\theta )$, let $m={\mathbb{E}}_F{\psi }^{\prime }(X-\theta )$. Then $A=-m$, so the influence function is $\psi (z-\theta )/m$ and $V=\mathbb{E}{\psi }^2/m^2$. The minus sign from differentiating $x-\theta$ is essential.

The clipped Huber score in the subsequent subparts requires $0<b<\infty$. The PDF only displays an upper bound: at $b=0$ the estimating function vanishes identically, and at $b<0$ it is a constant with no root. A positive clipping threshold is the regular interpretation.

At the standard normal distribution, symmetry and the odd Huber score give population root zero. For $0<b<\infty$, the derivative of the clipped score is $1_{\{|x|<b\}}$ except at its two corners, which have zero normal probability. Put $m_b=2\Phi (b)-1>0$ and write $\varphi$ for the standard normal density. The sandwich variance of an M-estimator has numeratorUsing ${\varphi }^{\prime }(z)=-z\varphi (z)$ and integration by parts,ThereforeAs checks, it tends to $1$ as $b\to \infty$, the sample mean limit, and to $\pi /2$ as $b\downarrow 0$, the sample median limiting variance. The estimator variance for a sample of size $n$ is approximately this quantity divided by $n$.

The root calculation for the Huber location estimator givesSince $|{\psi }_b(z)|\le b$ and equality is attained for $|z|\ge b$, its gross-error sensitivity isThus it is a B-robust estimator at the normal reference distribution. At a general centred distribution the same conclusion requires $m={\mathbb{E}}_F{\psi }_b^{\prime }(X)>0$, in which case the bound is $b/m$. Boundedness of the score alone does not rescue a zero sensitivity denominator.

Let $X_i=Y_i-{\theta }_0$. Under the symmetric continuous null hypothesis, positive and negative signs have probability $1/2$ and zeros have probability zero. For the sign test, take $N_{+}={\sum }_i{1}_{\{X_i>0\}}$ and reject for small values: $N_{+}\sim \mathrm{Bin}(n,1/2)$ under the null. Choose $k$ with ${\mathbb{P}}_0(N_{+}\le k)\le \alpha$; randomization at the next value attains exact size if desired. The lower tail is appropriate for $\theta <{\theta }_0$.

For the Wilcoxon signed-rank test, rank $|X_i|$ increasingly and put $W_{+}={\sum }_i{R}_i{1}_{\{X_i>0\}}$. Continuous observations have no absolute-value ties almost surely. Conditional on the absolute values, the signs are independent fair signs because symmetry makes the sign independent of magnitude. Hence $W_{+}$ has the distribution of ${\sum }_{r=1}^{n}r{B}_r$ for independent Bernoulli random variables $B_r$ with parameter $1/2$. Again reject in the lower tail, using this exact subset-sum distribution or a normal approximation. Independence gives the requested null moments:For a symmetric kernel $h$ of fixed order $m$, a one-sample U-statistic is $U_n={\left(\genfrac{}{}{0ex}{}{n}{m}\right)}^{-1}{\sum }_{i_1<\cdots <i_m}h(X_{i_1},\dots ,X_{i_m})$. State the U-statistic central limit theorem: if $\mathbb{E}{h}^2<\infty$, $\eta =\mathbb{E}h$, and ${\zeta }_1=\mathrm{Var}(\mathbb{E}[h\mid X_1])>0$, then $\sqrt{n}(U_n-\eta )\Rightarrow N(0,m^2{\zeta }_1)$. The nondegeneracy condition is important.

For $h(x,y)=1_{\{x+y>0\}}$, its order-two U-statistic satisfies the exact Wilcoxon signed-rank statistic identityTo prove it, in an unordered pair the sign of the sum is the sign of whichever observation has larger absolute value. Counting the pairs for which a positive observation is larger contributes its rank minus one; its own positive-sign indicator supplies the remaining one.

Write $F_0$ for the distribution function of $X$ under the null. Symmetry gives $\eta =1/2$ andThe probability integral transform makes $F_0(X_1)$ uniform on $(0,1)$, so ${\zeta }_1=1/12$. Therefore $\sqrt{n}(U_n-1/2)\Rightarrow N(0,1/3)$. The centred $N_{+}$ is $O_p(\sqrt{n})$ and is negligible on the scale $n^{3/2}$ of $W_{+}$. Combining the exact decomposition and its mean givesThe lower-tail approximate level-$\alpha$ rejection rule compares this standardized statistic with ${\Phi }^{-1}(\alpha )$; a half-unit continuity correction can refine discrete-tail calculations.

An Edgeworth expansion refines the central limit theorem by incorporating higher cumulants. For iid centred observations with variance ${\sigma }^2$, standardized sum $Z_n={\sum }_i{X}_i/(\sigma \sqrt{n})$, standardized third cumulant ${\gamma }_1={\kappa }_3/{\sigma }^3$ and fourth cumulant ${\gamma }_2={\kappa }_4/{\sigma }^4$, its central-region density expansion isunder sufficient moment and smoothness/nonlattice conditions. Here $H_3=z^3-3z$, $H_4=z^4-6z^2+3$, $H_6=z^6-15z^4+45z^2-15$ are probabilists' Hermite polynomials. Expanding the characteristic exponent gives the cubic and quartic terms, while squaring the cubic term in the exponential gives the $H_6$ correction. Integrating the first correction, using $(\varphi H_{k-1}{)}^{\prime }=-\varphi H_k$, givesThus skewness determines the first departure from the normal approximation; inversion gives a Cornish-Fisher expansion for quantiles. These are asymptotic expansions, not guaranteed nonnegative densities in extreme tails. Lattice data require continuity/lattice corrections, and moment assumptions alone do not justify a uniform smooth-density expansion.

Parameters $\psi$ and $\lambda$ are orthogonal parameters if the cross block of the expected Fisher information matrix is zero throughout the relevant parameter region:This is a statement about expected information, not a requirement that every observed cross derivative vanish. For regular maximum-likelihood estimators, $\sqrt{n}(\hat{\theta }-\theta )$ has limiting covariance equal to the inverse per-observation Fisher information matrix. Block diagonality therefore gives asymptotically independent fitted interest and nuisance parameters. The efficient information $I_{\psi \psi }-I_{\psi \lambda }{I}_{\lambda \lambda }^{-1}{I}_{\lambda \psi }$ reduces to $I_{\psi \psi }$ at an orthogonal parametrization, although nuisance estimation can still matter at higher orders.

For scalar interest and nuisance, retain $\psi$ and write the old nuisance as $\lambda =\lambda (\psi ,\eta )$. The transformed cross information is ${\lambda }_{\eta }(I_{\psi \lambda }+{\lambda }_{\psi }{I}_{\lambda \lambda })$. Hence an orthogonal nuisance coordinate can locally be found fromsubject to local solvability and a nonsingular coordinate change. Global coordinates need not follow from this local construction. For the normal distribution with separately varying mean and variance parameters $(\mu ,v)$, the scores are $(Y-\mu )/v$ and $-1/(2v)+(Y-\mu {)}^2/(2v^2)$; their product has zero expectation by symmetry. Thus the per-observation information is diagonal with entries $1/v$ and $1/(2v^2)$.

For $I_n={\int }_{{\mathbb{R}}^p}{e}^{nh(\theta )}a(\theta )d\theta$, suppose $h$ has a unique interior maximum $\hat{\theta }$, $H=-h^{\prime \prime }(\hat{\theta })$ is positive definite, the amplitude is smooth there, and contributions away from that maximum are negligible. Set $z=\sqrt{n}(\theta -\hat{\theta })$. The quadratic expansion $nh(\theta )=nh(\hat{\theta })-z^{T}Hz/2+o(1)$ yields the Laplace approximationThe determinant comes from diagonalizing the Gaussian integral; it is an inverse square root, not an inverse determinant. Under standard higher smoothness and domination, the relative error is $O(n^{-1})$.

For a regular model and a fixed smooth positive proper prior $\pi$, apply this to the marginal likelihood $m(y)=\int e^{\ell (\theta ;y)}\pi (\theta )d\theta$. With $j(\hat{\theta })=-{\ell }^{\prime \prime }(\hat{\theta })$,As $j$ is of order $n$, this produces the $p\log n$ complexity term in the Bayesian information criterion after multiplying log marginal likelihood by $-2$. A posterior mode can also be used with the corresponding combined log-density curvature. Boundary modes, singular information and multiple dominant modes invalidate the single interior Gaussian formula; distinct nondegenerate dominant modes must have their contributions added.

Let $W=2[\ell (\hat{\theta })-\ell (\tilde{\theta })]$ be a regular likelihood-ratio test statistic for $r$ smooth restrictions, so its leading null law is ${\chi }_r^2$. Suppose a higher-order calculation gives ${\mathbb{E}}_{0}W=r+a/n+O(n^{-2})$. The Bartlett correction rescales the statistic:Its mean is $r+O(n^{-2})$, eliminating the first-order mean error. In regular likelihood models satisfying the usual Bartlett expansion, the same scaling removes the leading $n^{-1}$ distributional error against the chi-square reference law, leaving an $O(n^{-2})$ error for fixed tail probabilities. Matching a mean alone would not establish this distributional conclusion for an arbitrary statistic.

The coefficient depends on the model, null restrictions and nuisance parameters; it can be obtained from higher derivatives and cumulants of the log-likelihood, or estimated by null simulation when an analytic expression is inconvenient. Nuisance parameters require their appropriate null values or estimates. A correction improves finite-sample calibration within regular asymptotic theory; it does not turn a boundary-parameter or unidentifiable problem into a regular chi-square problem.

In statistical decision theory, the statistical invariance principle says that a procedure should respect transformations that leave the inferential problem unchanged. If a group action preserves the model and carries each of the null and alternative parameter sets into itself, an invariant test has rejection probability function $\phi (gy)=\phi (y)$. A maximal invariant identifies its orbits, so every invariant test factors through that statistic: define its value at an orbit using any representative; invariance makes the definition independent of the representative. If the group is transitive on the null parameter set, the null law of any invariant statistic is free of that nuisance parameter.

For example, testing a normal mean $\mu ={\mu }_0$ with unknown scale, positive transformations $y_i-{\mu }_0\mapsto b(y_i-{\mu }_0)$ preserve a one-sided mean alternative. The statistic $\sqrt{n}(\bar{Y}-{\mu }_0)/s$, with the usual unbiased sample-variance scale $s$, is invariant and has the Student t-distribution with $n-1$ degrees of freedom under the null. For estimation the corresponding requirement is an equivariant estimator, together with a loss function invariant under simultaneous transformation of decision and parameter. Invariance alone does not prove unrestricted optimality or supply a unique best procedure.

There is a distinct probabilistic use of the term: the Donsker invariance principle states that centred iid partial sums, normalized by $\sigma \sqrt{n}$, converge as processes on $[0,1]$ to Brownian motion. Its empirical-process version gives a Brownian bridge, used by goodness-of-fit tests in the next note. These functional limit theorems should be distinguished from the group-based inferential principle.

Population influence functions describe infinitesimal contamination; finite-sample analogues examine actual changes to an empirical sample. For its empirical distribution $F_n$, add an observation $z$ and define the sensitivity curveUnder suitable differentiability it approximates the influence function. A replacement version is ${\mathrm{RC}}_{n,i}(z)=n[T_n(y_1,\dots ,z,\dots ,y_n)-T_n(y)]$, approximately $\mathrm{IF}(z;T,F_n)-\mathrm{IF}(y_i;T,F_n)$. Their supremum magnitudes quantify finite one-observation sensitivity.

The empirical influence function instead substitutes $F_n$ directly in $\mathrm{IF}(z;T,F)$. For a differentiable M-estimator this gives $-\psi (z,T(F_n))/[n^{-1}{\sum }_i{\partial }_{\theta }\psi (y_i,T(F_n))]$. For quantile functionals, a discrete unsmoothed empirical law may not admit the required density derivative; a sensitivity curve or a smoothed empirical law is then preferable.

Global contamination is measured by finite-sample maximum bias $b_n(m;y)={\sup }_{z:d_H(z,y)\le m}|T_n(z)-T_n(y)|$, where $d_H$ counts replacements. Under the convention of the first fraction that causes unbounded bias,This is the finite-sample replacement breakdown point. The largest replacement fraction still guaranteed bounded is one grid step smaller, so the conventions must be specified. One replacement sends a sample mean arbitrarily far, giving $1/n$, while the ordinary scalar sample median has first-breaking fraction $\lceil n/2\rceil /n$ with the usual midpoint convention for even samples. Its limiting value is $1/2$. Local influence and global breakdown answer different questions; a bounded influence function is not by itself a proof of a high breakdown point.

For a fully specified continuous null distribution $F_0$, compare the empirical distribution function $F_n$ with $F_0$. The probability integral transform makes $F_0(Y_i)$ iid uniform under the null, giving distribution-free calibration. Three common discrepancies areThese define the Kolmogorov-Smirnov test, Cramér–von Mises criterion and Anderson–Darling test, respectively. Reject for large values. The supremum detects the largest discrepancy, the unweighted integral combines discrepancies throughout the distribution, and the reciprocal weight emphasizes the tails.

The Donsker theorem for empirical distribution functions gives $\sqrt{n}(F_n-F_0)\Rightarrow B^0\circ F_0$ for a Brownian bridge $B^0$, with covariance $\min (u,v)-uv$. Consequently the standard continuous-null limits are ${\sup }_{0\le u\le 1}|B^0(u)|$, ${\int }_0^1{B}^0(u{)}^{2}du$, and ${\int }_0^1{B}^0(u{)}^2/[u(1-u)]du$ for $\sqrt{n}{D}_n,C_n,A_n$, respectively; the weighted last limit also requires control near the endpoints. Exact finite-sample or asymptotic null quantiles provide critical values.

If parameters in $F_{\theta }$ are estimated from the same observations, these fully specified-null critical values generally fail. A first-order expansion isso the estimated parameter changes the limiting process. One must use the fitted-model null law, adjusted asymptotic theory, or a parametric bootstrap that re-estimates the parameters in each replicate.

For a regular iid model with a fixed-dimensional interior parameter ${\theta }_0$, identifiability, sufficient differentiability and moment domination, and nonsingular per-observation Fisher information $I({\theta }_0)$, a consistent maximum-likelihood estimator solves the score equation. Expanding it giveswhere $s_{\theta }={\partial }_{\theta }\log f_{\theta }$. The score has mean zero and covariance $I$ under the regularity assumptions. This proves the usual large-sample information bound and asymptotic efficiency within regular parametric inference. The observed curvature $j(\hat{\theta })/n$ and the empirical score covariance consistently estimate $I$ when the model is correctly specified.

For $r$ smooth constraints $h(\theta )=0$ with full-row-rank derivative $H$, the Wald test usesThe likelihood-ratio test and score test, evaluated using the constrained fit and nuisance-adjusted efficient information, have the same first-order chi-square law. A quadratic local likelihood expansion explains their asymptotic equivalence. Profile likelihood supplies inference about a parameter of interest with fitted nuisance parameters, and the corresponding chi-square cutoffs give approximate confidence regions.

These conclusions depend on the stated regularity: parameters at a boundary, nonidentifiability, singular information, changing dimension, or parameter-dependent support can give different rates and limits. Under misspecification, the fitted parameter generally approaches a pseudo-true value and its variance is a sandwich covariance matrix, not simply inverse Fisher information.