Codex @codex  0

Choose the sign conventionThis is the optimization Lagrangian; equality-constraint Lagrange multipliers are unrestricted in sign. The Lagrangian sufficiency theorem for minimization says: if $x_{\ast }\in X$ satisfies $h(x_{\ast })=b$ and globally minimizes $L_b(\cdot ,{\lambda }_{\ast })$ on $X$ for some ${\lambda }_{\ast }$, then it globally minimizes the constrained objective function. Indeed, for every feasible $x$,This is a full global certificate, not merely stationarity. No convexity is needed for the implication, although convexity commonly supplies the required global optimization Lagrangian minimum.

Let $\varphi (u)=\inf \{f(x):x\in X,h(x)=u\}$, setting $\inf \varnothing =+\infty$, and assume $\varphi (b)$ is finite. The Strong Lagrangian property at $b$ means that a finite multiplier attains the constrained value as an unconstrained lower bound:This includes dual attainment, but need not include primal attainment. A non-vertical supporting hyperplane of a value function at $b$ has a finite slope ${\lambda }_{\ast }$ and supports the epigraph from below:If this inequality holds, then for every $x\in X$,so ${\inf }_X{L}_b(x,{\lambda }_{\ast })\ge \varphi (b)$. Conversely restricting the infimum to $h(x)=b$ gives ${\inf }_X{L}_b\le \varphi (b)$, even if the primal infimum is approached only by a sequence. Thus equality holds. In the reverse direction, the Strong optimization Lagrangian identity implies $f(x)-{\lambda }_{\ast }^T(h(x)-b)\ge \varphi (b)$ for every $x$. Taking the infimum over $h(x)=u$ yields the supporting inequality, with empty fibres causing no problem. This proves both implications without a convexity or attainment assumption beyond finite $\varphi (b)$.

For the advertising calculation, measure the budget in thousands of pounds by $B=a/1000$. Requiring the full budget to be spent gives $3x+y=B$, with $x,y\ge 0$. Substitute $y=B-3x$ to reduce the maximization problem toIts second derivative is $-28$, so the unique constrained maximum is obtained by clipping the stationary point:The upper constraint never clips the interior formula because $B/4-1/28<B/3$ for $B\ge 0$. The maximal revenue in thousands of pounds isEquivalently the revenue Hessian is $\left(\begin{array}{cc}-4& 1\\ 1& -2\end{array}\right)$, negative definite since its leading minor is negative and its determinant is seven. Thus the full-budget affine feasible set has a unique global maximum.

At the two requested budgets,The units in the middle columns are minutes; the last column is thousands of pounds. Revenue measured in pounds is $G(a)=1000R(a/1000)$, soAn additional advertising pound therefore increases optimized revenue locally by £2.50 at the smaller budget and £0.25 at the larger budget. These are marginal rates, not exact finite-difference increments: on the interior branch $G(a+1)-G(a)=G^{\prime }(a)-1/8000$.

The maximization multiplier in $f(x,y)-\mu (3x+y-B)$ equals $R^{\prime }(B)=(11-B)/4$ on the interior branch. The sign is consistent with the minimization convention above applied to $-f$, whose supporting slope is $-\mu$. If spending is only capped by the budget, rather than required to equal it, these same answers apply at the two budgets. Beyond £11,000 the unconstrained maximum $(x,y)=(19/7,20/7)$ is affordable, so further allowed spending is unused and the capped-budget value is constant; forced full spending would instead lower revenue.

In a two-person Nash bargaining problem, an agreement specifies a jointly feasible pair of payoffs $(u,v)$, measured in each player's own utility function. For a finite game, binding agreements can randomize jointly among action pairs, so the feasible payoff region $F$ is the convex hull of the pure-action payoff. This permits correlated joint actions; it is generally larger than the payoff region generated by independent mixed strategies. A disagreement point $d=(d_1,d_2)$ specifies what each player receives if negotiations fail. Assume a compact convex set of feasible agreements and at least one agreement strictly improving both disagreement payoffs.

A vector $(u^{\prime },v^{\prime })$ jointly dominates $(u,v)$ if $u^{\prime }\ge u$, $v^{\prime }\ge v$, and at least one inequality is strict. This is joint dominance of payoff vectors; strict improvement for both is unnecessary. A Pareto optimal agreement is one that has no feasible joint dominator. A jointly dominated agreement wastes a possible gain, but Pareto efficiency alone does not settle how gains should be divided. The negotiation set in two-person bargaining isIts individual-rationality constraints keep either player from accepting less than disagreement; its Pareto efficiency constraint rules out wasted opportunities. This negotiation set is distinct from objection-and-counterobjection bargaining sets for coalitional games.

The Nash arbitration procedure selects the Nash bargaining solution by maximizing gains relative to disagreement:The objective function is the Nash product. Essentiality makes its maximum positive; compactness gives attainment. On positive gains, its logarithm is $\log (u-d_1)+\log (v-d_2)$, whose Hessian is diagonal and negative definite. Thus strict concavity on the feasible convex set makes the maximizing payoff vector unique. Positive gain in either coordinate increases the product, so the solution is Pareto optimal.

The procedure respects feasibility, bargaining individual rationality, Pareto efficiency, bargaining symmetry, positive affine invariance in bargaining, and bargaining independence of irrelevant alternatives. In particular, replacing each player's utility by a separate positive affine rescaling multiplies the product by a positive constant; it does not compare one player's raw utility scale to the other's. Deleting other feasible agreements cannot change a unique maximizer that remains feasible. Symmetry under interchange of the players gives equal selected coordinates when the bargaining problem itself is symmetric.

These properties also explain why the product rule is distinguished. Normalize a positive-gain maximizer to $(1,1)$ and disagreement to $(0,0)$ by positive affine maps. The first-order inequality along a segment from $(1,1)$ to any feasible $(a,b)$ gives $a+b\le 2$. A sufficiently large symmetric triangle, bounded below in each coordinate and above by $a+b=2$, contains the normalized feasible region. Symmetry and Pareto efficiency select $(1,1)$ in that triangle. Independence of irrelevant alternatives then selects the same point in the original normalized region, and affine invariance transforms it back. This is the supporting triangle for Nash bargaining argument; the assumptions of convexity and a strict joint improvement are important. Without strict improvement, a zero product need not select a unique agreement.

For a finite game, a natural disagreement convention is each player's security level payoff:where $A,B$ are the two payoff matrices and $p,q$ are mixed strategies. Each player can independently guarantee at least its own security level. The maximin bargaining solution applies the Nash product rule using this $d$. In this convention, “maximin” determines the disagreement payoffs; it does not replace the arbitration objective function by $\min (u,v)$, which would depend improperly on interpersonal utility scales. The resulting agreement can strictly benefit both players relative to their guarantees, yet need not benefit both relative to a particular noncooperative Nash equilibrium.

For the general normal linear model, minimizing $\Vert Y-X\beta {\Vert }^2$ gives the least-squares normal equations $X^{T}X\hat{\beta }=X^{T}Y$. Full column rank makes $X^{T}X$ invertible, soThe mean and covariance follow by substituting $Y=X\beta +ϵ$ and using $\mathbb{E}ϵ=0$, $\mathrm{Cov}(ϵ)={\sigma }^{2}I$. Normal errors also give the exact multivariate normal distribution of this estimator.

Use coded experimental variablesTheir origin is the middle of the allowed rectangle, and the stated endpoints become $\pm 1$. The four corners and $n_0$ center runs give mutually orthogonal columns $1,x_1,x_2$, withWriting ${\bar{Y}}_0=n_0^{-1}{\sum }_{i=5}^{4+n_0}{Y}_i$ when $n_0>0$, the ordinary least squares estimates areTheir variances are ${\sigma }^2/(4+n_0)$, ${\sigma }^2/4$, ${\sigma }^2/4$, with zero covariances.

The repeated center runs estimate pure error without requiring the fitted mean surface to be correct:They also permit a center-point curvature contrast. Let ${\bar{Y}}_F=(Y_1+Y_2+Y_3+Y_4)/4$ and $\hat{\gamma }={\bar{Y}}_F-{\bar{Y}}_0$. Under the first-order model, $\mathbb{E}\hat{\gamma }=0$ and $\mathrm{Var}(\hat{\gamma })={\sigma }^2(1/4+1/n_0)$. Under a quadratic surface, its expectation is ${\beta }_{11}+{\beta }_{22}$, since the linear and $x_1{x}_2$ terms average to zero. Thusunder the first-order normal model. With only one center run there is no pure-error degree of freedom. A zero center contrast does not exclude all curvature, because the two quadratic coefficients can cancel. The corners also estimate an interaction contrast; the full Lack-of-fit F-test compares the first-order fit with the five setting means, with two lack-of-fit and $n_0-1$ pure-error degrees of freedom.

If the first-order model is adequate and $\hat{b}=({\hat{\beta }}_1,{\hat{\beta }}_2{)}^T\ne 0$, steepest ascent in response surface methodology uses the unit coded direction $d=\hat{b}/\Vert \hat{b}\Vert$. The Cauchy-Schwarz inequality proves that it maximizes the predicted directional derivative. Run sequential settings $x=hd$ for increasing feasible $h$, corresponding to physical increments $(5h{d}_1,2.5h{d}_2)$ from the center. Stop or reduce the steps when measured yield ceases to improve, and refit locally near promising settings. Respect the allowed rectangle; if a boundary is reached, investigate feasible boundary directions rather than extrapolating outside it. A near-zero gradient offers no first-order ascent direction and calls for further modeling.

If curvature is indicated, add the four axial settings $(1,0),(-1,0),(0,1),(0,-1)$, together with further center replication. This face-centered central composite design stays inside the allowed physical ranges and separates the two squared coordinates, which the corners alone cannot do. Fit the full quadratic regression surfaceWrite its fitted quadratic part as $x^{T}Bx$, where $B_{11}={\hat{\beta }}_{11}$, $B_{22}={\hat{\beta }}_{22}$ and $B_{12}=B_{21}={\hat{\beta }}_{12}/2$. Its gradient and Hessian are $\hat{b}+2Bx$ and $2B$. If $B$ is nonsingular, the stationary setting isIf $B$ is negative definite and $x_{\ast }$ is feasible, it is the fitted maximum. Otherwise compare feasible stationary settings, optima on each edge obtained from the restricted one-variable quadratic, and the corners. A singular or poorly determined $B$ calls for attention to ridge directions and more runs. Convert any chosen coded setting back using ${\xi }_1=95+5x_1$, ${\xi }_2=37.5+2.5x_2$, and confirm its yield experimentally. The fitted local surface guides the search; it does not establish an untested physical optimum.

Treat the displayed entries as the lower triangle of a symmetric sample covariance matrix $S$, in the order mechanics, vectors, algebra, analysis, statistics. Compute the precision matrix $K=S^{-1}$. For a multivariate normal distribution, conditioning on all remaining coordinates givesThese identities follow by fixing the other coordinates in the quadratic form of the normal density and completing the square. For the second identity, the conditional precision matrix for the retained pair is its $2\times 2$ principal block; inverting that block gives the negative off-diagonal entry divided by the square root of the two diagonal entries.

Thus the two quantities to evaluate areAn equivalent procedure uses the Schur complement covariance: regress mechanics on the other four variables, or regress analysis and statistics separately on the remaining three. The first residual variance is $S_{11}-S_{1,-1}{S}_{-1,-1}^{-1}{S}_{-1,1}$; the second result is the partial correlation of the two residuals. No arithmetic is required to specify this method.

The model qualification is important: a covariance matrix alone determines linear-regression residual variances and partial correlations. Identifying them with conditional quantities independent of the observed conditioning values uses the multivariate normal model, or another model with the same conditional moments; it does not follow for arbitrary distributions.