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The part of the log-likelihood depending on $\theta$ isCompleting the square separates the sum into a term independent of $\theta$ and $n(\overline{X}-\theta {)}^T{\Sigma }^{-1}(\overline{X}-\theta )$. Positive definiteness makes its unique minimum occur at $\overline{X}$. Since linear combinations of independent multivariate normal vectors are normal,The Hessian matrix $-n{\Sigma }^{-1}$ is negative definite, confirming the global likelihood maximum.

Use the convention that the likelihood ratio is maximized alternative likelihood divided by null likelihood. The alternative excludes zero, but its supremum agrees with the unrestricted maximum, even on the exceptional sample $\overline{X}=0$. Completing the square givesUnder the null, $\sqrt{n}\overline{X}\sim {\mathcal{N}}_d(0,I_d)$, so $2\log \Lambda \sim {\chi }_d^2$ exactly for every $n$. Equivalently $\Lambda$ is distributed as $e^{Y/2}$ for $Y\sim {\chi }_d^2$. With the reciprocal likelihood-ratio convention the statistic is $-2\log \Lambda$ instead.

Wilks theorem predicts a limiting chi-squared distribution with degrees of freedom equal to the difference of parameter dimensions, here $d-0=d$, under regularity and an interior null point of the unrestricted model. This example strengthens the asymptotic conclusion to an exact finite-sample identity.

Assume incompressible inviscid layers, gravity $-g{\mathbf{e}}_y$, no surface tension, and irrotational perturbations of the uniform base flows. The full bulk equations are $\nabla \cdot {\mathbf{u}}_j=0$ and ${\partial }_t{\mathbf{u}}_j+{\mathbf{u}}_j\cdot \nabla {\mathbf{u}}_j=-{\rho }_j^{-1}\nabla p_j-g{\mathbf{e}}_y$. At the rigid walls $u_{1y}(h)=u_{2y}(-h)=0$. At the material interface $y=\eta (x,t)$ each layer obeys $u_{jy}={\eta }_t+u_{jx}{\eta }_x$, and $p_1=p_2$. There is no tangential-velocity matching condition across this vortex sheet.

Write ${\mathbf{u}}_j=(U_j,0)+\nabla {\varphi }_j$, with $U_1=U$, $U_2=-U$. Then ${\nabla }^2{\varphi }_j=0$, and the wall conditions are ${\partial }_y{\varphi }_1(h)={\partial }_y{\varphi }_2(-h)=0$. Linearizing the interface conditions at zero givesThe second follows from linear Bernoulli pressure $p_j^{\prime }=-{\rho }_j({\partial }_t+U_j{\partial }_x){\varphi }_j$ and the hydrostatic pressure shift $-{\rho }_{j}g\eta$.

For $k>0$, use amplitudes ${\varphi }_1=B_1\cosh k(y-h)e^{ikx+\sigma t}$ and ${\varphi }_2=B_2\cosh k(y+h)e^{ikx+\sigma t}$. Kinematics gives $B_1=-(\sigma +ikU)A/[k\sinh kh]$ and $B_2=(\sigma -ikU)A/[k\sinh kh]$. The dynamic condition becomesCompleting the square yieldsA positive real part occurs precisely whenThe function $\tanh (kh)/k$ decreases strictly from $h$ at $k=0$ to zero at infinity. Indeed its derivative has numerator $kh{\mathrm{sech}}^2(kh)-\tanh (kh)<0$, since $\sinh (2kh)>2kh$. Therefore the minimum speed producing instability at every finite nonzero wavenumber isAt equality every $k>0$ still satisfies the strict growth inequality, since $\tanh (kh)<kh$; the limiting infinite-wavelength mode is marginal. A strict uniform long-wave margin requires $U>U_{\mathrm{crit}}$.

By the Markov property, all factors in the stationary likelihood function that involve $X_2$ are the transitions from $X_1$ to $X_2$ and from $X_2$ to $X_3$. Hence conditioning on later observations adds no information once the two neighbors are known. Writing $u=X_2-\mu$, $u_1=x_1-\mu$ and $u_3=x_3-\mu$, its conditional density is proportional toThe quadratic coefficient of $u$ is $1+{\varphi }^2$ and the linear coefficient is $-2\varphi (u_1+u_3)$. Completing the square therefore gives the Gaussian AR1 bridgeHere ${\sigma }^2$ is the innovation variance, not the marginal stationary variance.

A Bogomolny bound expresses a static field energy as nonnegative squares plus a term fixed by the topological charge or boundary data. Setting the squares to zero gives the first-order Bogomolny equations. Their solutions minimize the energy in that sector and satisfy the second-order Euler-Lagrange field equations, although a general stationary solution need not attain the bound.

For one real scalar in one spatial dimension, take the Lagrangian densityA finite-energy field configuration in this static sector must approach scalar-field vacua ${\varphi }_{\pm }$ with $W^{\prime }({\varphi }_{\pm })=0$ at the two ends. Completing the square gives the square completion for a one-dimensional kink:Choose the sign for which the boundary term is nonnegative. ThusTaking the derivative of the equality equation gives ${\varphi }_{xx}=W^{\prime }{W}^{\prime \prime }=d([W^{\prime }{]}^2/2)/d\varphi$, the static Euler-Lagrange field equation. The boundary term is invariant under deformations keeping the asymptotic scalar-field vacua fixed; it is not a contribution from the local shape of the kink.

For a phi-four kink, let $c>0$, $W(\varphi )=c^2\varphi -{\varphi }^3/3$, and select the sector ${\varphi }_{-}=-c$, ${\varphi }_{+}=c$. Then $W^{\prime }=c^2-{\varphi }^2$, and the increasing Bogomolny equation is ${\varphi }_x=c^2-{\varphi }^2$. Separating variables yieldsThe integration constant $X$ is the translational collective coordinate. Directly, ${\varphi }_K^{\prime }=c^2{\mathrm{sech}}^2(c(x-X))$ and its energy density is $c^4{\mathrm{sech}}^4(c(x-X))$, whose integral is $4c^3/3$. The decreasing antikink has the reversed boundary values, profile $-{\varphi }_K$, and the same energy. A Lorentz boost produces the exact uniformly moving kink $c\tanh [c\gamma (x-X_0-vt)]$, with Lorentz factor $\gamma =(1-v^2{)}^{-1/2}$ and energy $\gamma M$.

In two spatial dimensions, choose the Abelian Higgs model at critical coupling, with $D_i={\partial }_i-i{A}_i$, $B={\partial }_1{A}_2-{\partial }_2{A}_1$, and energyThe gauge covariant derivative is used throughout this normalization. Integration by parts, using $[D_1,D_2]=-iB$, gives $\int |D_i\varphi {|}^2=\int |(D_1+i{D}_2)\varphi {|}^2+\int B|\varphi {|}^2$, with a vanishing boundary divergence for the decaying vortex fields. Combining this identity with the magnetic and potential terms gives the Bogomolny square completion for an Abelian Higgs vortex:Finite-energy field configurations have $|\varphi |\to 1$ at infinity, and $D_i\varphi \to 0$ makes the magnetic flux equal to the phase winding $2\pi N$. For $N>0$,These are the Bogomolny vortex equations for an Abelian Higgs vortex. Opposite signs give antivortices and the bound $E\ge \pi |N|$. The coefficient $\pi$ follows from the explicit energy normalization above; other conventions can give $2\pi$. Static solutions of fixed positive vortex number have equal energy independent of their positions, giving the Abelian Higgs vortex moduli space used for slow dynamics.

For $U(y)=c^2-y^2$, completing the square gives the Bogomolny boundConsequentlywith equality exactly when the Bogomolny equationholds everywhere. Separating variables, or differentiating the proposed form directly, gives all solutions with the required limits:These are the translated phi-four kinks.

Differentiating the first-order equation yieldswhich is precisely the Euler-Lagrange equation from part (a). Thus every configuration saturating the first-order bound is automatically a stationary solution of the second-order variational equation.

Write $a=(2,1{)}^T$, so the data map is $a^{T}u$. The Gaussian likelihood and prior distribution giveCompleting the square, or using Gaussian conjugacy for a normal linear model, identifies a multivariate normal distribution with precision matrix $C_{\delta }^{-1}=I_2+{\delta }^{-2}a{a}^T$. Since $a^{T}a=5$, its posterior mean and covariance matrix areThe parameters areFor example, $(I_2+{\delta }^{-2}a{a}^T)C_{\delta }=I_2$, which checks the square completion.

The orthogonal vectorsare eigenvectors of the covariance matrix. Their respective variances areThey are independent under the posterior distribution, by independence of uncorrelated jointly normal variables. Thus uncertainty is different in the two directions even for positive noise: the data constrain $e_{\parallel }$, whereas $e_{\perp }$ spans the null space of the data map.

As $\delta \to 0$, the posterior distribution converges to the law ofThis is supported on $2u_1+u_2=m$, and its posterior mean is the minimum-norm least-squares solution. The prior distribution still determines the distribution along the unobserved null space; it has not disappeared. This example of a Gaussian posterior in the zero-noise limit distinguishes exact recovery of an observed component from recovery of the whole unknown.

Use the shape-rate convention for the Gamma distribution. DefineThe zero null space assumption makes $B$ a positive-definite matrix, and $u_{\ast }$ is the unique ordinary least squares solution. The Gaussian likelihood contributes ${\gamma }^{k/2}$, which must be retained because $\gamma$ is unknown. Multiplying it by the improper prior and the Gamma distribution gives the normal-gamma posterior with a flat prior:Here the constant density of the improper prior cancels on normalization. PutThe normal equation $A^{T}r=0$ makes the residual orthogonal to the range of $A$. The orthogonal decomposition by a closed subspace into that range and its orthogonal complement givesIntegration over $u$ by the Gaussian integral, followed by integration over $\gamma$ using the Gamma distribution, yieldsThus the posterior distribution is proper, despite the improper prior; here $k\ge d$ follows from the zero null space assumption.

Completing the square in $u$ and collecting the powers of $\gamma$ give the two conditional distributions:Explicitly their probability density functions areFor a joint maximum a posteriori estimate relative to $dud\gamma$, first fix $\gamma >0$. The unique maximizing $u$ is $u_{\ast }$. The remaining log density is $(a-1)\log \gamma -b_0\gamma$ plus a constant. Consequently, if $a>1$,The stated assumptions do not always imply $a>1$. If $a=1$, the supremum occurs as $\gamma \downarrow 0$ but is not attained on $\gamma >0$; if $a<1$, the density is unbounded there. In either case there is no joint maximum a posteriori estimate in the stated parameter space. Calling zero a boundary mode does not make it an admissible positive precision.

If “MAP estimators” instead means separate modes of the marginal distributions, integration givesThus the marginal maximum a posteriori estimate of $u$ is always $u_{\ast }$, and that of $\gamma$ is $(s-1)/b_0$ when $s>1$, with the same nonattainment at zero when $s\le 1$. Joint and marginal maximization are different operations; the usual joint interpretation gives the preceding boxed pair.

Multiplying the Gaussian likelihood and prior and completing the square givesTherefore normal-normal conjugacy gives

Finite energy requires $|\varphi |\to 1$ and $D\varphi \to 0$ on the circle at spatial infinity. Writing $\varphi \sim e^{i\chi }$ there gives $A\sim d\chi$. The vortex number is the winding numberBy Stokes theorem, the magnetic flux is quantized:

For the rotationally symmetric Abelian Higgs vortex ansatz, $A=f(r)d\theta$ and $\varphi =h(r)e^{ik\theta }$ giveAfter the angular integration, the Abelian Higgs model energy becomesCompleting the square in the two pairs of terms givesRegularity at the origin and approach to the vacuum at infinity requireMore precisely, $h(r)=O(r^k)$ and $f(r)=O(r^2)$ near the origin. The boundary term is $k$, soThe bound is saturated exactly when both squares vanish, giving the radial Bogomolny vortex equationsThe asymptotic value $f(\infty )=k$ also makes the flux $2\pi k$, so the ansatz has vortex number $N=k$.

The terms involving $\mathbf{p}$ can be written by completing the square:Translation of the integration variable in the Gaussian functional integral over $\mathbf{p}$ contributes only a $\varphi$-independent determinant. The effective free energy is thereforewhereThis is the surfactant renormalization of the square-gradient coefficient.

Use completing the square in each coordinate:Comparing with the standard circle equation gives

Multiplying the polar equation by its denominator gives $Ar-Bx=1$. Squaring and using $r^2=x^2+y^2$ yieldsTake $C=\sqrt{A^2-B^2}>0$. Completing the square givesThe standard ellipse equation is consequentlywhose centre is $\boxed{(B/C^2,0)}$. Squaring has introduced no unwanted branch: on this ellipse the minimum of $1+Bx$ is $A/(A+B)>0$, so $Ar=1+Bx$ follows with the correct positive sign.

Use completing the square after taking out the leading coefficient:Thus $a=3$, $b=-2$ and $c=-1$. Since a real square is nonnegative, the global minimum is $-1$, attained at $x=2$.

Use completing the square by adding and subtracting $3^2$:

The tangent half-angle substitution $t=\tan (x/2)$ gives $\sin x=2t/(1+t^2)$ and $dx=2dt/(1+t^2)$. The endpoints become zero and one, soUse completing the square, $t^2+t+1=(t+1/2{)}^2+3/4$, and integrate:Thus both integrals equal $\boxed{\pi /(3\sqrt{3})}$.

Use completing the square to obtainThe locus is therefore a circle with centre $(1,0)$ and radius $2$. It meets the horizontal axis at $(-1,0)$ and $(3,0)$ and the vertical axis at $(0,\pm \sqrt{3})$; its highest and lowest points are $(1,\pm 2)$.

Factor out $-1/2$ and use the method of completing the square inside the bracket:

For $c\ne 0$, completing the square gives the conic sectionThe possible levels follow fromFor $0<c<1$, the contours are ellipses centred at $(0,1/c)$, shrinking to the maximum as $c\uparrow 1$. The zero contour is the parabola from part (c). For $-1<c<0$, the contours are vertical-opening hyperbolas. At $c=-1$ they degenerate into the intersecting straight lines, and for $-2<c<-1$ they are horizontal-opening hyperbolas. These descriptions specify the remaining contour families and their symmetry about the $y$ axis. The lower bound $-2$ is approached as $|x|\to \infty$ at fixed $y$ but is never attained.

Completing the square under the square root gives $\sqrt{(x-1{)}^2}=|x-1|$. Thus the equation is $|x-1|=2$, and the definition of absolute value gives $x-1=\pm 2$. Hence