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Here projection means Hermitian orthogonal projection: $P=P^{\ast }=P^2$ with $\mathrm{tr}P=k$, as indicated by the unitary conjugation in the question. These matrices represent the complex Grassmannian as orthogonal projections, a compact manifold of complex dimension $k(n-k)$. The constants $c_i$ are real. Let $A^{\ast }=-A$ and use the given tangent directions $[A,P]$. Cyclicity of the matrix trace givesThe commutator $B=[P,C]$ is skew-Hermitian. If this derivative vanishes for every skew-Hermitian $A$, choose $A=B$ to obtainso $B=0$. Conversely $[P,C]=0$ makes every tangent derivative zero. Because the diagonal entries of $C$ are distinct, a commuting matrix $P$ is diagonal. Its idempotent entries are zero or one. Thus the critical projections are exactlywith $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ critical points and values $f(P_I)={\sum }_{i\in I}{c}_i$.

To compute the Hessian matrix, write $E=\mathrm{span}\{e_i:i\in I\}$ and $E^{\perp }=\mathrm{span}\{e_j:j\notin I\}$. A neighboring plane is the graph of a complex linear map $Z:E\to E^{\perp }$. With $W=\left(\genfrac{}{}{0ex}{}{I}{Z}\right)$, its orthogonal projection is $P(Z)=W(I+Z^{\ast }Z{)}^{-1}{W}^{\ast }$. Writing $C_E,C_{E^{\perp }}$ for the diagonal blocks, we getExpanding $(I+Z^{\ast }Z{)}^{-1}=I-Z^{\ast }Z+O(\Vert Z{\Vert }^4)$ yieldsEach complex coordinate has two real components, each with Hessian coefficient $2(c_j-c_i)$. Every coefficient is nonzero, proving that $f$ is a Morse function. Its negative directions correspond to $j<i$. For $I=\{i_1<\cdots <i_k\}$ there are $i_a-a$ complementary indices below $i_a$, givingThis is the diagonal trace Morse function on a complex Grassmannian. Its critical values need not be distinct; ties between sums do not make its critical points degenerate.

All Morse indices are even. Choose a Morse-Smale gradient flow and form its integral Morse chain complex. The odd-degree chain groups vanish, so every differential vanishes. Equivalently the Morse handle-attachment theorem supplies only even-dimensional cells. The integral homology of complex Grassmannians is thereforewhere $N_r$ counts the subsets with ${\sum }_a(i_a-a)=r$, and $0\le r\le k(n-k)$. In particular there is no torsion. The nondecreasing sequence $i_a-a$ lies between zero and $n-k$, so reversing it identifies these subsets with partitions of an integer $r$ fitting in a $k$ by $(n-k)$ rectangle. The Poincaré polynomial is explicitlyFor the permitted example $\mathrm{Gr}(2,4)$, the subsets $12,13,14,23,24,34$ have indices $0,2,4,4,6,8$. ThusThe general calculation above applies throughout $2\le k\le n-2$; it does not reduce to any of the excluded rank-one cases.

The same pricing relation gives $x^{\top }{A}_0\ge 0$ whenever the terminal payment is nonnegative. Therefore negative initial cost also contradicts a strictly positive state-price density, even if the terminal payment is identically zero.

This second form is needed for directions invisible in the terminal payments. Let $L$ be the linear span of the essential values of $A_1$, equivalentlyIf $A_0\notin L$, let $v$ be its nonzero orthogonal projection on $L^{\perp }$ and set $x=-v$. Then $x^{\top }{A}_1=0$ almost surely while $x^{\top }{A}_0=-\Vert v{\Vert }^2<0$. Hence the second form exists in this case. Part (b) handles all remaining cases.

With the divisor-$n$ sample covariance matrix defined above, the maximum-likelihood estimators areTheir joint sampling distribution is specified byHere $W_p(\nu ,V)$ uses the Wishart distribution convention $\mathbb{E}W=\nu V$. Thus the covariance maximum-likelihood estimator has expected value $(n-1)V/n$; the unbiased estimator of the covariance matrix is $nS/(n-1)$. These distributions follow from orthogonal projections of the jointly multivariate normal observations onto the constant-observation direction and its orthogonal complement, though no proof is required here. The covariance maximum in the positive-definite parameter space requires $S$ nonsingular, which holds almost surely when $n>p$ and $V$ is positive definite. Otherwise the rank of a centered sample covariance matrix shows why the usual unrestricted finite maximum does not exist.

Parameterize the two line segments byPut $w=A-C$ and abbreviate $a=u\cdot u$, $b=u\cdot v$, $c=v\cdot v$, $d=u\cdot w$, and $e=v\cdot w$. Minimizing the Euclidean distance is equivalent to minimizing the convex quadratic functionover a closed square. For nonzero nonparallel segment directions, $\Delta =ac-b^2=\Vert u\times v{\Vert }^2>0$. An interior minimum must satisfywhose unique solution isKeep it as a candidate only if both parameters lie in $[0,1]$.

Every remaining minimum lies on the boundary. Define $[x{]}_{[0,1]}=\max (0,\min (1,x))$. The four edge minima are the point-to-segment orthogonal projectionsEvaluate $F$ at those four pairs and at the admissible interior pair, and return a pair with the smallest value. This solves closest points on two line segments completely: continuity guarantees a minimum, and every interior or boundary location has been tested. Do not independently clamp $s_0$ and $t_0$: the mixed term couples them. For example, $A=(0,0)$, $B=(1,0)$, $C=(2,-2)$ and $D=(3,-1)$ give the line minimizer $(s_0,t_0)=(4,2)$. Its independently clamped pair $(1,1)$ has squared distance $5$, whereas the correct edge minimizer $(1,1/2)$ has squared distance $9/2$.

Without general position, explicitly handle these configurations:

In floating-point arithmetic, almost-parallel directions require a condition-aware or higher-precision solve; they must not be declared exactly parallel solely because the computed determinant is small. The returned distance is $\boxed{\sqrt{\min F}}$, and the returned points are obtained by substitution into $P(s),Q(t)$.

The orthogonal projection condition is $(f-s^{\ast },N_i)=0$ for all $i$. Since $M_i=k{N}_i/(t_{i+k}-t_i)$, it is equivalentlyThe mixed-normalization spline Gram matrix is invertible: $G=D^{-1}H$, where $D_{ii}=(t_{i+k}-t_i)/k>0$ and $H_{ij}=(N_i,N_j)$ is the positive-definite ordinary Gram matrix of the basis. Nonnegativity and the unit-integral normalization of a B-spline give $|b_i|\le \Vert f{\Vert }_{\infty }{\int }_0^1{M}_i=\Vert f{\Vert }_{\infty }$. The subpartition of unity for B-splines gives, on all of $[0,1]$,ConsequentlysoThis is the maximum-norm bound for spline projection. Using a subpartition matters when the basic knot interval is a proper subinterval of $[0,1]$. As in the paper's assertion that the image lies in $C[0,1]$, the spline space here must consist of continuous functions; order-one step splines or full-multiplicity interior knots do not have that property.

Split the vector $x$ into its normal and tangential orthogonal projections:A reflection in the plane negates the first component and preserves the second. Thus $Hx=x-2(n\cdot x)n$, giving the Householder reflectionSince $n\cdot n=1$ and $n\cdot u=n\cdot v=0$,The columns of a matrix representation of a linear map are the coordinates of the images of the ordered basis vectors. Consequently, in the orthonormal basis $(u,v,n)$,Equivalently, with the orthogonal matrix $Q=(uvn)$, the change of basis formula is $N=Q^{T}MQ$.

For the given orthonormal basis, the orthogonal projection isIt belongs to $W$, and $⟨v-\pi v,e_j⟩=0$ for every basis vector. Thus $v-\pi v\in W^{\perp }$. For any $w\in W$, decompose $v-w=(v-\pi v)+(\pi v-w)$. The summands are orthogonal, so the Pythagorean theorem in an inner-product space for the inner product givesEquality holds exactly when $w=\pi v$. Hence $\pi v$ is the unique closest point of $W$ to $v$. This proves the minimizing property as well as the formula, including $W=\{0\}$ and $W=V$.

princomp(): principal component analysis. Supply observations in rows and numeric variables in columns. After centering, let $Z$ denote the resulting matrix and $C=Z^{T}Z/n$ its sample covariance matrix, with an explicitly stated normalization. WriteThe first direction maximizes the Rayleigh quotient $q^{T}Cq$ over unit vectors; successive directions maximize it subject to orthogonality to the earlier directions. The principal component scores are $z_{ij}=q_j^T(x_i-\bar{x})$, with variances ${\lambda }_j$ and zero pairwise sample covariance. Zero covariance alone does not imply independence. Each principal component loading describes a direction in variable space; its sign is arbitrary, and repeated eigenvalues leave the basis inside the corresponding eigenspace arbitrary.

The function returns principal component loadings, component standard deviations sdev and, when requested, principal component scores. By default it uses a covariance matrix; cor=T instead performs principal component analysis on a correlation matrix, standardizing nonconstant variables to variance one. This choice matters when variables have different measurement units. These interface details are documented in the S-Plus help for princomp.

A scree plot shows the ordered eigenvalues; a plot of the first two principal component scores can reveal groups, unusual observations or a dominant gradient. The first $k$ components explain ${\sum }_{j\le k}{\lambda }_j/{\sum }_j{\lambda }_j$ of the total variance. Reconstructing each centered observation by its orthogonal projection onto these directions minimizes average squared reconstruction error among $k$-dimensional linear subspaces. Thus PCA gives an optimal linear compression for squared reconstruction error, with scaling and the retained dimension chosen explicitly. It is an unsupervised learning method, rather than a classification rule fitted to supplied labels.

tree(): classification trees and regression trees. A call such as tree(response ~ predictors, data=dat) fits a recursive partitioning model. At a current node, test numerical predictors against candidate thresholds, or divide the levels of a categorical predictor into two sets. Select a split that most improves the fitting criterion, then repeat within its children. Numerical splits give axis-aligned regions. A categorical response produces a classification tree; a numeric response produces a regression tree, as specified in the S-Plus tree documentation.

For a regression tree, the constant prediction in a terminal region is its sample mean, minimizing the sum of squared residuals there. For a classification tree, estimate class probabilities by the observed proportions in a leaf and predict a most probable class under equal misclassification costs. The classification-tree deviance of a node iswith $0\log 0=0$; splitting maximizes the decrease from parent to children. This fitting criterion differs from the number of wrong classifications. The middle sketches show the same three-region partition as a binary tree: first split on variable one, then split the right-hand branch on variable two.

Growing to small leaves can overfit. Use cost-complexity tree pruning, balancing fitted loss $R(T)$ against leaf count,and select complexity by cross-validation. Plotting the fitted tree and annotating nodes displays the rules; prediction routes each new observation through the fitted splits. A tree gives an interpretable, piecewise-constant prediction rule; training fit alone does not measure prediction accuracy.

The rpart() alternative. This also fits classification and regression trees by recursive partitioning, with squared-error splitting for ordinary numeric responses and, by default for classification, Gini impurity $1-{\sum }_r{p}_r^2$. Classification settings can instead use information-based splitting and specify prior probabilities and a loss matrix. It supports surrogate splits to approximate a node's primary left/right assignment when the primary predictor is missing. Its complexity-parameter path and cross-validated errors support cost-complexity tree pruning; for example, select a complexity parameter and apply prune(fit, cp=...). The maintainer's rpart documentation describes the criteria and the distinction in handling surrogate predictors. The partition and tree sketches apply to either implementation; the fitted tree need not be the same because splitting criteria and controls can differ.

cmdscale(): classical multidimensional scaling. Its input is a dissimilarity matrix or stored distance object. It outputs coordinates for the observations in a chosen dimension, usually two for visualization. Let $D^{(2)}$ contain the squared entries of $D$, and putFor centered Euclidean coordinates $X$, the distance identity $d_{ij}^2=\Vert x_i{\Vert }^2+\Vert x_j{\Vert }^2-2x_i^T{x}_j$ gives $G=X{X}^T$: multiplying by $H$ removes the row and column norm terms. Diagonalize this Gram matrix and retain its $k$ largest positive eigenvalues. The coordinates areThus classical scaling exactly recovers Euclidean distances when all positive eigenvalues are retained. Reducing dimension approximates the centered Gram matrix; it does not in general minimize squared discrepancies in the original distances. Negative eigenvalues identify a failure of the Euclidean distance matrix criterion, so their size matters when interpreting a display. The historical S-Plus cmdscale help describes the dimension argument, eigenvalue output and optional additive adjustment to dissimilarities.

Plot the two coordinate columns on equally scaled axes and label observations. Nearby plotted points suggest small dissimilarities only insofar as the dimension reduction fits the input well. The bottom sketches give an exact Euclidean example: a configuration and its independently recovered two-dimensional coordinates have identical pairwise distances, despite different coordinate axes. Classical scaling converts distances into geometry; translations, rotations and reflections do not change the represented distances. For Euclidean distances between observations, its coordinates agree with the corresponding principal component scores up to these harmless choices of axes.

Conjugating the coordinate reflection by the rotations supplies reflections in the other two coordinates. The quarter-turns, together with those reflections, also supply coordinate transpositions. Consequently the represented group consists of all signed permutation matrices, the full symmetry group of a cube of order $2^3\cdot 3!=48$.

Let $C$ be any real or complex matrix commuting with the group representation. Commutation with the three independent coordinate reflections forces all off-diagonal entries of $C$ to vanish. Commutation with the quarter-turn about the first axis equates its second and third diagonal entries, and the quarter-turn about the second axis equates its first and third entries. ThereforeFor completeness, if the complexified group representation had a proper invariant subspace, its orthogonal complement would be invariant too, since the generating matrices are unitary. The orthogonal projection onto that subspace would be a nonscalar commuting matrix, contradicting the calculation. Thus the complexification is irreducible, proving absolute irreducibility of the real group representation.

For $\mathbf{b}\ne 0$, define the orthogonal projection of $\mathbf{a}$ onto its span byThe first is parallel or antiparallel to $\mathbf{b}$, or zero, and ${\mathbf{a}}_{\perp }\cdot \mathbf{b}=0$. This proves the decomposition. If $\mathbf{a}\ne 0$ as well, comparison with ${\mathbf{a}}_{\parallel }=|\mathbf{a}|\cos \theta \hat{\mathbf{b}}$ yieldsIndeed $|\mathbf{a}{|}^2=|{\mathbf{a}}_{\parallel }{|}^2+|{\mathbf{a}}_{\perp }{|}^2$ shows that the ratio lies in $[-1,1]$. There is a unique $\theta \in [0,\pi ]$ with this cosine. The definition is symmetric in the two vectors, unchanged by orthogonal coordinate changes and positive rescaling, and agrees with ordinary plane geometry in their span: projecting a unit vector onto the other gives its cosine. Parallel, perpendicular and antiparallel directions give $0,\pi /2,\pi$ respectively. The zero vector has no defined angle; if $\mathbf{b}=0$, the nonzero-axis projection formula itself must not be used.

The vertices of the centered hypercube are sign vectors $\mathbf{u}\in \{-1,1{\}}^n$. A body diagonal identifies $\mathbf{u}$ and $-\mathbf{u}$, hence there are $2^{n-1}$ such diagonals. If two sign vectors disagree in $h$ positions, their dot product is $n-h-h=n-2h$. For odd $n$ this integer is odd and cannot be zero. Thus no two body diagonals are perpendicular in odd dimension, by the orthogonality of hypercube body diagonals.

In four dimensions, mutually orthogonal nonzero vectors are linearly independent, so at most four mutually perpendicular diagonals are possible. Four are achieved by the directionsEach has squared norm four and every distinct pair has zero dot product; these are the rows of a Hadamard matrix. Therefore the maximum is four.

For distinct diagonals, the Hamming distance is $h=1,2,3$; $h=0,4$ would represent the same line. Oriented vertex directions therefore give cosines $1/2,0,-1/2$. The angles areFor unoriented diagonals, the usual smaller angle between the lines is $\pi /3$ or $\pi /2$; the supplementary intersection angle $2\pi /3$ also occurs when the two rays are chosen oppositely. Stating this convention avoids counting a line and its reverse as different diagonals.

Take $Q_j$ nonzero of exact degree $j$, so $Q_0,\dots ,Q_n$ are an orthogonal basis of the polynomials of degree at most $n$. Their orthogonal projection isIndeed its residual is orthogonal to every such polynomial. For another polynomial $q=p_n+r$ of degree at most $n$, the squared error issince the cross term vanishes. It is uniquely minimized at $r=0$. The approximation need not have exact degree $n$ when the highest coefficient happens to vanish.

Expanding the ordinary least squares objective givesFor a nonzero vector $v$, $v^T{X}^{T}Xv=\Vert Xv{\Vert }^2>0$, since the design matrix has matrix rank $p$. The Hessian matrix is therefore positive-definite, and $Q$ is strictly convex. Solving its normal equations gives the unique least-squares estimator:Indeed, expansion about this stationary point and $X^T(Y-X\hat{\beta })=0$ give $Q(\beta )=Q(\hat{\beta })+(\beta -\hat{\beta }{)}^T{X}^{T}X(\beta -\hat{\beta })$, which also proves the global minimum directly.

Define the hat matrix $H=X(X^{T}X{)}^{-1}{X}^T$. It is a symmetric matrix and an idempotent linear map, since $H^T=H$ and $H^2=H$, and its image is the column space of $X$. Thus $H$ is the orthogonal projection onto the fitted mean space. The fitted values are $HY$ and the regression residual vector is $(I-H)Y$. Because $I-H$ is also a symmetric matrix and an idempotent linear map,These minimization and projection identities do not need errors with a normal distribution; the normal linear model becomes relevant to the distributional calculations.

Let $\Phi (z)=1-{\varphi }_{1}z-\cdots -{\varphi }_p{z}^p$. The causality root criterion for an autoregressive model,is a sufficient condition for a weakly stationary process solution, and gives the causal solution $X_t={\sum }_{j\ge 0}{\psi }_j{\epsilon }_{t-j}$. Indeed $1/\Phi$ is analytic in a disk of radius greater than one, so its Taylor coefficients ${\psi }_j$ decay geometrically. The series therefore has mean-square convergence and defines a linear filter of a stationary time series; multiplying by $\Phi$ verifies the equation.

For the specified order-two model, put $\phi =(1+\sqrt{5})/2$. Since $\phi -{\phi }^{-1}=1$,When $\alpha \ne 0$, its roots are $1/(\alpha \phi )$ and $-\phi /\alpha$. Both have modulus greater than one exactly when $|\alpha |<1/\phi$. The case $\alpha =0$ is white noise and satisfies the same inequality. ConsequentlyThis is the causal stationarity condition selected above. Without causality, a two-sided autoregressive model can also have a noncausal stationary autoregression when its polynomial has no unit-circle zeros. Thus the inequality is not a necessary condition for all conceivable weak stationarity solutions; the causal interpretation is the one required by the subsequent innovations and forecasting questions.

To find the Wold representation, expand $1/\Phi (z)={\sum }_{j\ge 0}{\psi }_j{z}^j$. Matching coefficients gives ${\psi }_0=1$, ${\psi }_1=\alpha$ and ${\psi }_j=\alpha {\psi }_{j-1}+{\alpha }^2{\psi }_{j-2}$. HenceHere $F_1=F_2=1$ are the Fibonacci numbers. The coefficient formula also follows by partial fractions in the displayed factorization. Absolute summability follows from $|\alpha \phi |<1$ and $|\alpha /\phi |<1$. The past observation span equals the past noise span, because the causal series gives one inclusion and ${\epsilon }_t=X_t-\alpha X_{t-1}-{\alpha }^2{X}_{t-2}$ gives the reverse inclusion. Thus these noises are the linear innovations, and there is no additional deterministic component. This is the Fibonacci-coefficient autoregression.

The best linear prediction from an infinite past is the orthogonal projection onto the closed span of $X_T,X_{T-1},\dots$. Future white noise is orthogonal to that span. Substitution in the recursion therefore givesThe respective errors are ${\epsilon }_{T+1}$ and $\alpha {\epsilon }_{T+1}+{\epsilon }_{T+2}$, so their mean squared errors are ${\sigma }^2$ and $(1+{\alpha }^2){\sigma }^2$. No normal distribution assumption is needed for these linear forecasts.

Assume $I,J\ge 2$. The additive two-way model has dimension $1+(I-1)+(J-1)=I+J-1$; the null model, with no exam effects, has dimension $I$. Write row, column and grand means as ${\overline{X}}_{i\cdot },{\overline{X}}_{\cdot j},{\overline{X}}_{\cdot \cdot }$. Orthogonal projection onto the full model gives fitted values ${\overline{X}}_{i\cdot }+{\overline{X}}_{\cdot j}-{\overline{X}}_{\cdot \cdot }$, while the null fit is ${\overline{X}}_{i\cdot }$. The added exam sum of squares and full residual sum of squares areThe preceding orthogonal-projection result gives independent null chi-squares with $J-1$ and $(I-1)(J-1)$ degrees of freedom. ThereforeReject the no-exam-effect hypothesis for a large value. The printed alternative repeats the identifying constraint ${\sum }_j{\beta }_j=0$; a genuine alternative additionally has at least one nonzero ${\beta }_j$. No interaction term is being fitted; the independent equal-variance Gaussian error assumption is what makes this exact test valid.

Define the hat matrix $H=X(X^{T}X{)}^{-1}{X}^T$. It is symmetric and idempotent, the orthogonal projection onto the column space of $X$. Since $HX=X$,The covariance is ${\sigma }^{2}H{H}^T={\sigma }^{2}H$. This normal distribution can be singular when $p<n$, since all fitted vectors lie in the $p$-dimensional model subspace.

For $y\ne 0$, the squared Euclidean norm of the orthogonal projection residual is nonnegative:This proves the Cauchy-Schwarz inequality $|x\cdot y|\le |x||y|$. Equality holds exactly when $x$ is a scalar multiple of $y$. If $y=0$, equality is automatic. Equivalently, equality means the two vectors are linearly dependent, including the cases with a zero vector.

For a point $C$ equidistant from the coordinate vertices,must be independent of $i$. Thus all $C_i$ are equal, and their sum is one. The centroid is thereforeFor every $x$ in the base probability simplex, the Cauchy-Schwarz inequality gives $1=(1,\dots ,1)\cdot x\le \sqrt{n}|x|$. Equality forces the same coordinates as $C$. Thus $C$ is the unique closest point of a probability simplex to the origin.

The angle $\alpha$ between $OC$ and any edge from the origin to $e_i$ satisfiesConsequently

A positively oriented planar rotation matrix acts byFor a unit vector $n$ in three dimensions, the parallel and perpendicular orthogonal projections areOn the plane perpendicular to $n$, the map $v\mapsto n\times v$ preserves lengths and turns vectors through a positive right angle about $n$. It squares to minus the identity there. Thus $\cos \theta x_{\perp }+\sin \theta (n\times x_{\perp })$ rotates that component, while the parallel component remains fixed. This proves the Rodrigues rotation formulaIts component form uses the Levi-Civita symbol:Since $n_i{n}_i=1$ and ${ϵ}_{iik}=0$, its matrix trace is $3\cos \theta +1-\cos \theta =1+2\cos \theta$.

For the given matrix, the matrix trace is $1+2r$ with $r=1/\sqrt{2}$, so $\cos \theta =1/\sqrt{2}$. In the stated angular range, $\theta =\pm \pi /4$. The antisymmetric entries giveTherefore the two axis-angle pairs areReversing the axis and the angle leaves the Rodrigues rotation formula unchanged, so these describe the same rotation.

Use a strictly positive bank account as numéraire, and write all prices and payoffs in discounted units. There are $d$ risky assets with deterministic initial price vector $S_0$, terminal price vector $S_1\in L^2(P)$, and gains $Y=S_1-S_0$. A self-financing portfolio with initial capital $x$ and constant risky holdings $\theta$ has discounted terminal wealth $x+{\theta }^{T}Y$. The initial amount in the bank account is $x-{\theta }^T{S}_0$, so $x$ denotes total initial capital, not just the cash holding. Assume the usual absence of arbitrage, and hence existence of an equivalent martingale measure; wherever a general probability space is used below, explicitly assume that such a measure exists and makes the asset payoffs integrable.

The attainable claims are precisely the linear subspaceThe finite-dimensional subspace is closed, so every square-integrable contingent claim $H$ has a unique orthogonal projection onto $\mathcal{A}$. This is its optimal one-period least-squares hedging payoff. Holdings are unique after redundant assets have been removed. Putand suppose the remaining covariance matrix $C$ is invertible. Expanding the squared error givesFirst minimize over $x$, then complete the square in $\theta$. ThusThe residual $L=H-x_{\ast }-{\theta }_{\ast }^{T}Y$ satisfiesIndeed, every other squared error equals this minimum plusThis proves optimality, and also shows that zero error is equivalent to attainability of a European contingent claim. If initial capital $x$ is prescribed instead, the normal equations for linear least squares give $\mathbb{E}[Y{Y}^T]\theta =\mathbb{E}[Y(H-x)]$; one must not use the free-capital formula without this adjustment.

A dominated martingale measure is a probability measure $Q\ll P$ such that ${\mathbb{E}}_{Q}Y=0$. Its Radon-Nikodym derivative $Z$ satisfies $Z\ge 0$, ${\mathbb{E}}_{P}Z=1$, and ${\mathbb{E}}_P[ZY]=0$. An equivalent martingale measure additionally requires $Z>0$ almost surely. Every integrable attainable claim $H=x+{\theta }^{T}Y$ then has price $x={\mathbb{E}}_{Q}H$, independent of which such measure is chosen. For an unattainable contingent claim, different dominated martingale measures can give different prices; least-squares approximation chooses a projection criterion rather than exact replication.

Write $Y=m+M$, where $M=Y-m$ is the centered martingale part of the one-period gains. The minimal martingale measure in a one-period market changes the means of the traded gains to zero while preserving all square-integrable mean-zero directions $L$ orthogonal to $M$. Its candidate density isDirect computation gives ${\mathbb{E}}_P{Z}_{\ast }=1$ and ${\mathbb{E}}_P[Z_{\ast }Y]=m-C{C}^{-1}m=0$. If ${\mathbb{E}}_{P}L=0$ and ${\mathbb{E}}_P[ML]=0$, then ${\mathbb{E}}_P[Z_{\ast }L]=0$, as required. Conversely, an $L^2$ density with this preservation property is orthogonal to the orthogonal complement of $\mathrm{span}\{1,M_1,\dots ,M_d\}$, so it lies in that span. Its normalization and martingale constraints force exactly $Z_{\ast }$. Moreover,Every other square-integrable signed pricing density has the form $Z_{\ast }+D$, with $D$ orthogonal to $1$ and $Y$, hence to $Z_{\ast }$. Thereforeso the minimal density is also the minimum-norm one-period pricing weight. This coincidence concerns one period; it is not an unrestricted assertion about dynamic hedging models.

Positivity needs checking. If $Z_{\ast }>0$ it defines an equivalent martingale measure; if merely $Z_{\ast }\ge 0$, it defines a dominated martingale measure. Otherwise it defines only a signed martingale measure. For example, take $S_0=2$ and $Y=-1,1,3$ with physical probabilities $1/100,89/100,10/100$. Then $m=59/50$, $C=1019/2500$, and $Z_{\ast }=-4350/1019$ in the third state. There is nonetheless an equivalent martingale measure assigning probabilities $3/5,3/10,1/10$, whose mean gain is zero. Thus absence of arbitrage does not guarantee positivity of the least-squares density, and a least-squares initial cost need not be an arbitrage-free price for an unattainable payoff.

Finally prove completeness and uniqueness of dominated martingale measures. Suppose first that every bounded contingent claim is attainable. For any event $A$, replicate $1_A=x_A+{\theta }_A^{T}Y$. Under any dominated martingale measure, $Q(A)=x_A$. Thus all such measures agree on every event. Existence follows from our no-arbitrage assumption, giving uniqueness.

Conversely, fix an equivalent martingale measure $Q_0$, and let $k=\dim \mathcal{A}$. If the probability space admits $k+1$ disjoint events $A_1,\dots ,A_{k+1}$ of positive probability, choose a basis $a_1,\dots ,a_k$ of $\mathcal{A}$, including $1$. The $k\times (k+1)$ matrix with entries ${\mathbb{E}}_{Q_0}[a_i{1}_{A_j}]$ has a nonzero null vector $(c_j)$. Consequently the bounded, nonzero random variablesatisfies ${\mathbb{E}}_{Q_0}h=0$ and ${\mathbb{E}}_{Q_0}[hY]=0$. For $0<|\epsilon |<1/\Vert h{\Vert }_{\infty }$, defineThese are distinct equivalent martingale measures, because their densities are positive and their gain moments remain zero. Thus uniqueness rules out such a partition. A probability space with no $k+1$ disjoint positive-probability events consists, modulo null sets, of at most $k$ atoms: repeatedly split any non-atomic positive event to obtain a larger finite partition. On $N$ atoms the payoff space has dimension $N$, while $\mathcal{A}$ has dimension $k\le N$. Hence $N=k$ and $\mathcal{A}$ is the whole payoff space. Every contingent claim is attainable. This provesHere completeness means all bounded claims, or all square-integrable claims in the $L^2$ formulation. The existence/no-arbitrage hypothesis matters: on a one-state space, cash and a stock with a sure positive discounted gain span every payoff, but no dominated martingale measure can make that gain have zero mean.

A Yang-Mills instanton is a localized finite-action Euclidean gauge field with self-dual or anti-self-dual gauge curvature. It is a four-dimensional space-time event, rather than a static three-dimensional finite-energy monopole. Take gauge group $SU(2)$ first, with Skew-Hermitian generators $T_a=-i{\sigma }_a/2$, $[T_a,T_b]={ϵ}_{abc}{T}_c$, and $\mathrm{tr}(T_a{T}_b)=-{\delta }_{ab}/2$. Writeand fix orientation $d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge d{x}^4$. The positive Euclidean action isVarying $F$ gives $\delta F=D\delta A$. Integration by parts makes stationarity equivalent to $D_{\mu }{F}_{\mu \nu }=0$. The Bianchi identity $DF=0$ follows by expanding $F=dA+A\wedge A$, using $d^2=0$, and cancelling the remaining matrix commutator terms. Consequently $F=\pm \ast F$ implies the Yang-Mills equations: $D\ast F=\pm DF=0$. Self-duality of gauge curvature is an extra first-order condition, not a property of every finite-action gauge connection.

For the usual instanton decay conditions, the field is asymptotically pure gauge, $A\sim h^{-1}dh$ on the sphere at infinity. The boundary map $h:S^3\to SU(2)\simeq S^3$ has an integer degree. With the present trace convention define the instanton number byThe identityturns this integral into a boundary winding integral: for $A=h^{-1}dh$, the three-form in parentheses is $-\frac{1}{3}\mathrm{tr}(h^{-1}dh{)}^3$, giving $k=(24{\pi }^2{)}^{-1}{\int }_{S^3}\mathrm{tr}(h^{-1}dh{)}^3$, with the chosen boundary orientation. The normalized invariant three-form computes the degree of $h$, so $k\in \mathbb{Z}$. Equivalently the gauge bundle extends across conformal infinity to a bundle on $S^4$, and this is its Second Chern number convention. A smooth deformation preserving the boundary class cannot change this integer.