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An M-estimator based on an estimating equation chooses a consistent root ofIts population target $T(F)={\theta }_F$ solves ${\mathbb{E}}_F\psi (X,{\theta }_F)=0$. For a scalar parameter, suppose the root is locally unique andFor the contaminated distribution $F_{\epsilon }=(1-\epsilon )F+\epsilon {\delta }_x$, differentiate its population equation at zero contamination. The influence function satisfiesConsequentlyThe constant of proportionality is independent of $x$. A local expansion of the sample estimating equation similarly givesThe central limit theorem yields asymptotic normality with the sandwich variance of an M-estimatorThus the first-order variance of the unscaled estimator is $V/n$. For a vector parameter the same differentiation gives $\mathrm{IF}=A^{-1}\psi$ and $V=A^{-1}B{A}^{-\mathsf{T}}$, with $A=-\mathbb{E}{\partial }_{\theta }\psi$ and $B=\mathbb{E}\psi {\psi }^{\mathsf{T}}$.

For the location problem, take the target location to be zero and let $Z$ have the standard normal distribution, with density $\varphi$ and distribution function $\Phi$. An odd score has $\mathbb{E}\psi (Z)=0$. For a differentiable score, integration by parts givesThe last expression also gives the appropriate derivative of the population equation for bounded monotone scores with jumps, by differentiating the shifted normal density. Assume $A>0$ and put $h=\psi /A$. ThenMultiplying a score by a positive constant changes neither its estimator nor this normalized influence function.

First establish feasibility. The bound impliesso necessarily $C\ge \sqrt{\pi /2}$. For $C>\sqrt{\pi /2}$, let $K>0$ solveThis solution is unique: the ratio tends to $\sqrt{\pi /2}$ as $K\downarrow 0$, tends to infinity as $K\to \infty$, and has positive derivative becauseTake the Huber score ${\psi }_K(x)=\max (-K,\min (x,K))$. Its derivative is one on $(-K,K)$ and zero outside, so $A=p_K$. HenceIt is odd, nondecreasing, meets the bound, and satisfies $\mathbb{E}[Z{h}_K(Z)]=1$.

To prove optimality, put $a=1/p_K$. For each fixed $x$, the minimum of $u^2-2axu$ over $|u|\le C$ is attained by projecting $ax$ onto $[-C,C]$, which is exactly $h_K(x)$. Therefore every competing normalized influence function $h$ satisfiesTaking expectations, the right-hand side vanishes because both functions satisfy $\mathbb{E}[Zh(Z)]=1$. Thus $\mathbb{E}{h}^2\ge \mathbb{E}{h}_K^2$. This proves the optimal bounded influence function for normal location, even over the larger class of all bounded normalized functions, and hence over the required odd nondecreasing class. Its minimum asymptotic variance isThe numerator is $\mathbb{E}[Z^2{1}_{\{|Z|\le K\}}]+K^2\mathrm{Pr}(|Z|>K)$.

At the included boundary $C=\sqrt{\pi /2}$, equality in the feasibility bound forces $h(x)=C\mathrm{sgn}(x)$ almost everywhere. This is the influence function of the sample median, with $V=\pi /2$. It is obtained from the rescaled scores ${\psi }_K/K$ as $K\downarrow 0$. The boundary solution is the sign score, or equivalently the median estimator. Literal substitution of $K=0$ in the clipped score gives the identically zero function and no identifiable estimating equation; the printed clipped formula therefore requires this limiting interpretation at equality.

Partition the statistical parameter as $\theta =(\psi ,\lambda )$, with $\psi$ of interest. For each fixed $\psi$, maximize over the nuisance parameter to obtain ${\hat{\lambda }}_{\psi }$. The profile likelihood and profile log-likelihood areTheir maximizer is the interest component of the joint maximum-likelihood estimator. At a regular nuisance maximum, ${\ell }_p^{\prime }(\psi )={\ell }_{\psi }(\psi ,{\hat{\lambda }}_{\psi })$ and its negative Hessian matrix is the Schur complementThe profile likelihood ratio is unchanged by a one-to-one nuisance reparametrization, and transforms naturally under a one-to-one interest reparametrization. Under the regular conditions of Wilks theorem, $2\{{\ell }_p(\hat{\psi })-{\ell }_p({\psi }_0)\}$ has a limiting chi-squared distribution with dimension of $\psi$ degrees of freedom. Profiling is maximization, rather than integration or conditioning; $L_p$ need not be a normalized data density, and nuisance fitting can cause appreciable higher-order distortion.

The modified profile likelihood corrects for nuisance curvature and sample-coordinate distortion. In suitable coordinates $(\hat{\psi },\hat{\lambda },a)$, with $a$ an ancillary statistic, setholding $\hat{\psi },a$ fixed when differentiating the data coordinates. Up to an interest-independent factor, the Barndorff-Nielsen modified profile likelihood isEquivalently, implicit differentiation of the nuisance score equation gives $J_{\psi }=|j_{\lambda \lambda }{|}^{-1}|{\ell }_{\lambda ;\hat{\lambda }}|$, andThe semicolon denotes sample-coordinate differentiation, not a second parameter derivative. In orthogonal coordinates, the Cox-Reid adjusted profile likelihood uses the information adjustment alone; it agrees with this modification when $J_{\psi }=1$, as happens below.

For the Inverse Gaussian distribution, write $s={\sum }_i{y}_i$, $r={\sum }_i{y}_i^{-1}$ and $\bar{y}=s/n$. Terms independent of both parameters can be absorbed into $c(y)$, givingThe nuisance score function isIt is positive below $\bar{y}$ and negative above it. Thus its global maximum isindependent of $\psi$. Put $D=r-n/\bar{y}$. ThenThe nuisance observed information at this fit isSince the constrained nuisance estimator is literally the full nuisance estimator as a function of the data, $J_{\psi }=1$ in the specified sample coordinates. Hence the modified profile likelihood for inverse Gaussian shape isFor $n\ge 2$ and a nonconstant positive sample, the Cauchy-Schwarz inequality gives $D>0$. The corresponding fits are ${\hat{\psi }}_p=n/D$ and ${\hat{\psi }}_m=(n-1)/D$. A constant sample has $D=0$ and no finite profile maximum; under continuous independent sampling this event has probability zero when $n\ge 2$. In particular, the modified coefficient accounts for the fitted mean nuisance parameter.

Work in a regular statistical model, with fixed support, differentiability under the integral, and nonsingular Fisher information. Let $U=(U_{\psi },U_{\lambda })$ be the score function. The blocks are orthogonal statistical parameters at $\theta$ whenThis is global parameter orthogonality if it holds throughout the parameter domain; otherwise it is an assertion at a specified point.

For an independent size-$n$ sample, the regular maximum-likelihood estimator satisfieswhere $I$ is the information in one observation. Thus the fitted blocks are independent to first order in their limiting normal distribution. The efficient information for $\psi$ is generally the Schur complement $I_{\psi \psi }-I_{\psi \lambda }{I}_{\lambda \lambda }^{-1}{I}_{\lambda \psi }$; under orthogonality it is just $I_{\psi \psi }$. Estimating the nuisance block therefore has no first-order variance cost for the interest block, compared with knowing the nuisance parameter.

This also reduces local movement of the constrained nuisance fit. If $j=-{\ell }^{\prime \prime }$ is the observed information, differentiating its nuisance score equation givesUnder regular independent sampling and orthogonality, $j_{\lambda \lambda }=O_p(n)$ and the centered mixed block is $O_p(\sqrt{n})$. Thus this derivative is $O_p(n^{-1/2})$ locally, and an $O_p(n^{-1/2})$ change in $\psi$ changes the nuisance fit by only $O_p(n^{-1})$. Neither exact finite-sample independence nor vanishing observed cross-information follows from expected-information orthogonality.

For the displayed density family, $\lambda$ is scalar andThe mean-zero score identity gives ${\mathbb{E}}_{\theta }{\nabla }_{\psi }t=0$ for $\lambda \ne 0$. ButsoThis is score factorization implies parameter orthogonality. If zero belongs to the parameter domain, the conclusion extends there by continuity of the regular expected derivatives. At zero the density itself has no dependence on $\psi$, so identifiability and nonsingular information for that block must be treated separately.

Use the two-by-two payoff entries visible in the original PDF. If $q$ is the probability that II uses its first action, I's expected payoffs from its two pure strategies are $4-q$ and $2q$. Their difference $4-3q$ is strictly positive for every $q\in [0,1]$. Thus I's first action strict dominance its second, and every Nash equilibrium has I use the first action with probability one. Against that action, II receives 8 from its first action and 4 from its second, so its first action is uniquely optimal. ThereforeThis excludes nontrivial mixed equilibria as well as the other pure pairs.

For the maximin bargaining solution, first compute the two security level payoffs. If I uses its first action with probability $p$, its payoffs against II's two pure strategies are $2+p$ and $4p$. Both increase with $p$, so its best worst-case payoff is attained at $p=1$ and is $d_1=3$. If II uses its first action with probability $q$, its payoffs against I's two actions are $4+4q$ and $6-6q$. The lower of these is maximized at their intersection,These are separate security calculations, not the expected payoffs from a pair of security strategies.

Cooperation permits jointly chosen lotteries, giving the convex hull of the four payoff. The vector $(3,8)$ dominates $(2,0)$ and $(0,6)$, so the Pareto frontier is the segment from $(3,8)$ to $(4,4)$, with equation $v=20-4u$. Individual rationality relative to $d=(3,24/5)$ restricts this toOn this negotiation set, the Nash product isIts unique maximizer is interior, givingOne implementing binding agreement always uses I's first action and uses II's first action with probability $3/5$, its second with probability $2/5$. This agreement raises both payoffs above their security levels, but II receives $32/5<8$, less than at the noncooperative equilibrium. Thus player II prefers the noncooperative game under the specified maximin bargaining rule. The cooperative feasible set contains the equilibrium payoff; it is the particular arbitration rule, not an inability to cooperate at that payoff, that produces II's lower allocation.

The all-to-one shortest path problem asks for the minimum length $v_i$ of a directed path from every vertex $i$ to the root $n$, with $v_n=0$ and $v_i=\infty$ if no such directed path exists. Positive directed edge lengths allow directed cycles to be deleted, so a shortest directed path, when one exists, uses at most $n-1$ directed edges. Its optimality equations areA Bellman-Ford algorithm initializes $d_n^{(0)}=0$ and other labels to infinity, then repeatskeeping the root label zero. Induction on $k$ proves that $d_i^{(k)}$ is the least length among directed paths using at most $k$ directed edges: either the old best directed path remains, or the first directed edge is followed by a directed path with at most $k-1$ directed edges. Thus $n-1$ rounds give the exact distances. An in-place variant repeatedly relaxes $d_i\leftarrow \min (d_i,c_{ij}+d_j)$; each decreased label triggers further incoming-arc relaxations. It terminates when no finite label can improve. The Bellman inequalities then telescope along each directed path to show the labels are lower bounds on all directed path lengths, while every finite label represents a discovered directed path and is an upper bound on the optimum. This proves correctness. The method is a label-correcting shortest-path algorithm: a processed label can be revised many times. Full scans cost $O(n|A|)$.

Let $d={\min }_{i\ne n}{c}_{in}$ and choose $j$ attaining it. Every directed path from a nonroot vertex to $n$ has a final directed edge costing at least $d$, and all preceding directed edges have positive lengths. Therefore every $v_k\ge d$. The direct directed edge from $j$ to $n$ has length $d$, soIf no incoming root directed edge exists, all these distances are infinite and there is no finite next vertex to settle.

The all-to-one Dijkstra algorithm sets the root's label to zero and all others to infinity. Repeatedly choose an unsettled vertex $j$ with least finite label, make that label permanent, and relax every incoming directed edge $(i,j)$ of an unsettled vertex:Store the corresponding successor $j$ whenever the label improves. Stop when all vertices are settled or every remaining label is infinite. This applies the usual source version to the reversed graph; relaxing outgoing original directed edges would solve the wrong direction of the problem.

For its correctness, inductively assume every settled distance is exact. Any directed path from an unsettled vertex to the root must first enter the settled set at some vertex $w$, through an directed edge $(z,w)$. That directed edge has already been relaxed, so $d_z\le c_{zw}+v_w$. The newly selected label $d_j$ is no greater than $d_z$. The directed path's preceding nonnegative lengths therefore make its total length at least $d_j$. On the other hand, $d_j$ itself is the length of a discovered directed path. Hence $d_j=v_j$, and it can never need correction. This is a label-setting shortest-path algorithm. Straight array selection costs $O(n^2+|A|)$.

For four vertices and root 4, the full symbolic execution is as follows. After settling 4, the tentative labels are $(c_{14},c_{24},c_{34},0)$. Selectset $v_j=c_{j4}$, and update the two other labels to $d_i=\min \{c_{i4},c_{ij}+c_{j4}\}$. Select the smaller of those labels at vertex $k$, set $v_k=d_k$, then for the last vertex $\ell$ setThe stored successors give the corresponding shortest routes; infinity conventions handle missing directed edges and disconnected vertices.

The supplied PDF contains no visible network diagram or directed edge lengths in the space for this example, and the converted TeX omits the diagram as well. Numerical distances and route choices cannot be determined from the supplied data. The symbolic run above gives the exact four-node calculation once those lengths are available; no network has been guessed.