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Mass conservation in the Zeldovich approximation gives $\rho d^{3}r=\bar{\rho }{a}^3{d}^{3}q$. Expanding the trajectory Jacobian determinant to first order therefore yields $\boxed{\delta =-{\nabla }_q\cdot \Psi }$. At the same order, ${\nabla }_r=a^{-1}{\nabla }_q$ andThe last equality uses ${\nabla }_q\times \Psi =0$ and the vector calculus $\nabla \mathrm{div}={\nabla }^2+\nabla \times \nabla \times$.

Split the gravitational potential into its homogeneous-background and perturbed parts. The background has ${\nabla }_r{\Phi }_0=(4\pi G\bar{\rho }/3)r$. For the perturbation, Poisson equation and irrotationality givewith a harmonic or uniform-acceleration contribution set to zero by the perturbation boundary conditions. Its divergence is $-4\pi G\bar{\rho }{\nabla }_q\cdot \Psi =4\pi G\bar{\rho }\delta$, as required. Insert this and the pressure gradient into $\ddot{r}=\ddot{a}(q+\Psi )+2\dot{a}\dot{\Psi }+a\ddot{\Psi }$. The background acceleration cancels using $\ddot{a}/a=-4\pi G\bar{\rho }/3$, leavingTake minus the comoving divergence, and expand $\delta (q,t)={\sum }_k{\delta }_k(t)e^{ik\cdot q}$. Since ${\nabla }_q^2$ acts on a Fourier mode by $-k^2$, each mode satisfiesHere the expansion coordinate is comoving; confusing it with the physical coordinate would lose the factor $a^{-2}$.

The first-order chain rule, with $x,y$ regarded as functions of $u,v$, isApply the product rule when differentiating the first identity in $v$, and apply the chain rule to $f_x,f_y$. Equality of mixed partial derivatives for a $C^2$ function givesThus $\boxed{H=x_{uv},K=y_{uv}}$. These terms record the second derivatives of the coordinate transformation, not of $f$.

For the given transformation,The mixed-derivative coefficient is zero, while $x_u{x}_v=x$, $y_u{y}_v=-y^2/x$, and $1/v^2=y/x$. The required partial differential equation is therefore exactly $g_{uv}=0$. Integrating first in $u$ and then in $v$ on a coordinate rectangle giveswhere $A,B$ are arbitrary twice continuously differentiable functions. The Jacobian determinant of the transformation is $-2u/v$, so this calculation uses patches with $u,v\ne 0$. On a positive-quadrant patch we can take $u=\sqrt{xy}$ and $v=\sqrt{x/y}$, obtainingChanging branches merely redefines the arbitrary functions.

For completeness, the original equation also makes sense in quadrants with $xy<0$, where that particular real square-root transformation is unavailable. Multiplying the equation by $x$ givesOn any quadrant with $xy\ne 0$, put $a=\log |x|$, $b=\log |y|$; then set $\xi =a+b$, $\eta =a-b$. The equation becomes $4f_{\xi \eta }=0$. Thus an equivalent general local answer on every such quadrant isHere $\Phi ,\Psi$ are arbitrary $C^2$ functions on the relevant intervals. If a solution is required across $y=0$, its quadrant expressions must extend with matching derivatives; on that line the equation itself imposes $x{f}_{xx}(x,0)+f_x(x,0)=0$, so its trace is $a_0\log |x|+b_0$ on each interval with $x\ne 0$. The displayed coordinate transformations do not assert invertibility at $y=0$.

For a one-to-one continuously differentiable change of variables with nonzero Jacobian determinant in the interior, the change of variables formula isThe absolute value is necessary even if the transformation reverses orientation.

The original region is the first-quadrant area between the hyperbola and circle, above the horizontal-axis segment. More explicitly,Its vertices are $(1,0)$, $(\sqrt{2},0)$ and $(\sqrt{3/2},1/\sqrt{2})$. Under the squared-radius difference substitution, the inverse isand the image is the triangle $1\le v\le u\le 2$, with vertices $(1,1),(2,1),(2,2)$.

The forward Jacobian determinant isIt is nonzero in the interior; the zero on $y=0$ lies on a measure-zero boundary, so the integral formula applies by taking interior limits. Since $x^{5}y-x{y}^5=xyuv$ and $4x^2{y}^2=u^2-v^2$, the inverse-Jacobian factor cancels $xy$, givingIntegrating in $v$ first and then $u$,Therefore

An oriented atlas is a smooth atlas whose smooth transition maps have positive Jacobian determinants at every point of their domains. Such an atlas supplies an orientation of a vector space on every tangent space, and these choices agree on overlaps.

Let $\omega$ be a nowhere-vanishing smooth top-degree differential form. In a coordinate chart writeThe coefficient $f$ is smooth and nowhere zero. Restrict the coordinate chart to connected neighborhoods; the intermediate value theorem makes the sign of $f$ constant on each. If $f$ is negative, replacing $x^1$ by $-x^1$ changes the sign of the coordinate volume form. Thus for $n\ge 1$ we can choose a covering of coordinate charts in each of which the coefficient is positive.

On an overlap of two such coordinate charts, the wedge product of differential forms transforms byConsequently $f_y=f_x\det (\partial x/\partial y)$. Both coefficients are positive, so the Jacobian determinant is positive, as is that of the inverse transition. A nowhere-vanishing top form therefore determines an oriented atlas. The zero-dimensional case has only zero-dimensional transitions, whose determinant is one, and is automatically orientable. Connectedness of $M$ is not needed for the positive-dimensional construction; only the local sign choice matters.

Here isotropic tensors are invariant under proper rotation matrices, $R^{T}R=I$ and $\det R=1$. The isotropic second-rank tensor and isotropic third-rank tensor forms in three dimensions arewith scalar coefficients $A,B$. One can see completeness directly. A half-turn about each coordinate axis kills every off-diagonal component of a rank-two tensor, and quarter-turns make its three diagonal components equal. For rank three, the half-turns require that each coordinate label occur an odd number of times in a nonzero component. Thus only permutations of $(1,2,3)$ survive. Quarter-turns interchange two labels while reversing one sign, forcing the surviving components to alternate with the parity of a permutation. This gives a multiple of the Levi-Civita symbol.

The tensor transformation law verifies invariance of these forms:The first equality uses orthogonality and the second uses the determinant formula. If reflections were also required while using the ordinary tensor transformation law, the rank-three coefficient would have to vanish. The proper-rotation convention is the one under which $B{ϵ}_{ijk}$ is an isotropic tensor.

The ball is unchanged by any rotation matrix, and rotating coordinates has Jacobian determinant one. Thus changing variables $\mathbf{y}=R\mathbf{x}$ givesThese moments are therefore isotropic tensor integrals. For the Cartesian moments of a three-dimensional ball, the second moment is $\alpha {\delta }_{ij}$. Taking its tensor contraction givesThe third moment vanishes: $\mathbf{x}\mapsto -\mathbf{x}$ preserves the ball and changes the sign of its integrand. Equivalently, its symmetry under index interchange excludes a nonzero multiple of the antisymmetric Levi-Civita symbol.

The fourth moment is symmetric under every permutation of its four indices. The general rank-four isotropic tensor form consequently has equal coefficients, yieldingFor example, comparing $T_{1122},T_{1212},T_{1221}$ shows that the three coefficients agree. Contracting $i=j$ and $k=l$ givesFinally the vector triple product identity gives $\mathbf{x}\times (\mathbf{\Omega }\times \mathbf{x})=r^2\mathbf{\Omega }-(\mathbf{\Omega }\cdot \mathbf{x})\mathbf{x}$. The first term integrates to $3\alpha \mathbf{\Omega }$ and the second to $\alpha \mathbf{\Omega }$, using the second moment. Hence

The characteristic vector field $b=({\nabla }_{v}H,-{\nabla }_{x}H)$ is a divergence-free vector field:The Jacobian determinant $J_t=\det D{\Phi }_t$ obeys ${\dot{J}}_t=(\nabla \cdot b)({\Phi }_t)J_t=0$, with $J_0=1$. Thus the flow preserves phase-space volume, the Liouville theorem in Hamiltonian mechanics. Changing variables $(x,v)={\Phi }_t(\xi ,\eta )$ and using (i) givesThis applies on intervals where the flow is a diffeomorphism of the phase space and the integrals are defined; for finite conservation statements require integrability. For instance compactly supported initial data and $F(0)=0$ give the usual setting. An arbitrary smooth $F$ need not make these integrals finite on infinite-volume space, and merely smooth $H$ does not exclude escape of characteristics to infinity. Equivalently $F(f{)}_t+{\nabla }_z\cdot [bF(f)]=0$, with no flux through infinity, proves the same Casimir invariant of a transport equation conservation.

A fixed comoving volume $d^{3}q$ labels a fixed collection of cold dark matter particles. Their background mass is $a^3\bar{\rho }{d}^{3}q$, which is constant by mass conservation. At the perturbed positions it is $\rho d^{3}r$. The Jacobian determinant of the Zeldovich approximation map isTaking its reciprocal and using mass conservation givesFor the perturbed field the Poisson equation must be ${\nabla }_r^2\Phi =4\pi G\rho =4\pi G\bar{\rho }(1+\delta )$. The displayed background-only right-hand side in the PDF would give no clumping force; it is the background equation, not the full perturbed equation.

Write $\Phi ={\Phi }_b+\phi$, where ${\nabla }_r{\Phi }_b=(4\pi G/3)\bar{\rho }\mathbf{r}$. To linear order,For the longitudinal, curl-free displacement used in the growing Zel'dovich mode, this integrates to ${\nabla }_q\phi =-4\pi G\bar{\rho }{a}^2\mathbf{\psi }$, after removing external homogeneous harmonic fields. More generally the force involves only the longitudinal part ${\mathbf{\psi }}_{\parallel }$, which has the same divergence as $\mathbf{\psi }$.

Differentiate $\mathbf{r}=a(\mathbf{q}+\mathbf{\psi })$ twice at fixed $\mathbf{q}$ and substitute $\ddot{\mathbf{r}}=-{\nabla }_r\Phi$. The terms proportional to $\mathbf{q}+\mathbf{\psi }$ cancel using $\ddot{a}=-(4\pi G/3)\bar{\rho }a$, leavingFor a longitudinal displacement the right side is $4\pi G\bar{\rho }\mathbf{\psi }$. Taking minus the comoving divergence gives the general scalar density contrast equation

For independent random variables, $\{\min (X,Y)>u\}=\{X>u,Y>u\}$ and $\{\max (X,Y)\le v\}=\{X\le v,Y\le v\}$. Factor their probabilities to obtain the distribution functions of independent minima and maxima:These formulas also allow atoms; no probability density function assumption is needed.

For independent unit-rate exponential distributions, $P(U>u)=e^{-2u}$ for $u\ge 0$, so $U\sim \mathrm{Exp}(2)$. To establish the independent minimum and gap of two exponential variables, set $D=V-U$. On either ordering branch, the change of variables is $(X,Y)=(u,u+d)$ or $(u+d,u)$ with unit absolute Jacobian determinant. Both contribute $e^{-2u-d}$; ties have probability zero. Thus the joint probability density isIt factors into normalized exponential probability density functions, proving $U\sim \mathrm{Exp}(2)$, $D\sim \mathrm{Exp}(1)$ and their independence.

The maximum is $V=U+D$. A half-scaled unit-rate exponential has rate two, so the independent pair $(D,U)$ has the same joint law as $(X,Y/2)$. ConsequentlyUsing the expected values, variances and independence of the summands,The spacing factorization explains why an order statistic can be represented as a sum of independent waiting times.

For fixed $u\in (0,1)$, $x=1-u\in (0,1)$ andAs $v$ runs from zero to one, $y$ decreases continuously from one to zero. Thus the map is onto the open unit square and is one-to-one, with smooth inverseIts denominator is positive in the open square. This is a rational square diffeomorphism. Its Jacobian determinant isWriting $A=1-(1-x)y$, the inverse relation gives $1-uv=x/A$. ConsequentlyThe second transformation also maps the open unit square bijectively onto the open unit square: for fixed $w$, $u=(1-t)/(1-wt)$ decreases strictly from one to zero as $t$ goes from zero to one, while $v=1-w$ independently spans $(0,1)$. Its positive Jacobian determinant isTo express this in $x,y$, the composite transformation gives $y=w/[x+(1-x)w]$ and $x=t(1-w)/(1-wt)$. Put $B=1-(1-x^2)y$. Solving these relations givesThe chain rule for Jacobian determinants therefore givesThe requested integrand is exactly the reciprocal of this positive determinant. Apply the change of variables formula using the composite diffeomorphism:Possible singular behavior at the square's boundary does not invalidate the calculation: first integrate over the images of compact interior squares, where all transformations are smooth with nonzero Jacobian determinant, and then increase these domains to the full square. Positivity and monotone convergence theorem justify this limit.

Use $(y,w)$ to label the initial point in phase space. The characteristic equations for a transport equation are $\dot{X}=a(V)$ and $\dot{V}=0$, with $X(s)=y$ and $V(s)=w$. Since the velocity remains fixed, the ordinary differential equations integrate directly:This characteristic flow map exists for all real times, even if $a$ grows rapidly at infinity: each trajectory samples just one fixed velocity. Its inverse is $S_{t,s}$, and its Jacobian matrix is the block triangular matrixConsequently its Jacobian determinant is one. The nonzero Jacobian determinant of $a$ is unnecessary for this construction.

The displayed estimate has two independent defects. Even for $a(v)=v$, its right side cannot use $f(t)$. At $t=1$, choose nonnegative smooth functions $\psi ,\chi$ with compact support and prescribe $f(1,x,v)=\psi (x)\chi (v)$. This is a legitimate transported solution with $f_{\mathrm{in}}(x,v)=\psi (x+v)\chi (v)$. The proposed inequality would requireVelocity dilation makes $\Vert \chi {\Vert }_1/\Vert \chi {\Vert }_{\infty }$ arbitrarily large, while the other factors stay fixed. Thus no universal $C_a$ works with the same-time mixed Lebesgue norm.

There is a second issue after replacing $f(t)$ by $f_{\mathrm{in}}$: the Jacobian determinant bounds give a local diffeomorphism, not necessarily a one-to-one map. Suppose additionally that $a$ is injective. With $y=x-ta(v)$, the change of variables formula and $|\det Da|\ge {\alpha }_1$ giveSurjectivity is not needed because the domain of integration can be enlarged. With injectivity, the corrected estimate has $C_a=1/{\alpha }_1$. The upper bound ${\alpha }_2$ is unnecessary.

More generally, dispersion with a nonlinear velocity map uses the area formula to sum over inverse branches. Define the weighted inverse multiplicityIf ${\mathrm{esssup}}_w{m}_a(w)<\infty$, the same argument gives the corrected estimate with $C_a=\Vert m_a{\Vert }_{\infty }$. At most $N$ inverse branches give $C_a\le N/{\alpha }_1$. Thus the missing global assumption concerns multiplicity, not merely local volume distortion.

For a noninjective map with constant Jacobian determinant giving a counterexample in two dimensions, write $v=(s,r)$ and setThe polar coordinates calculation gives $\det Da=e^{2s}{e}^{-2s}=1$, yet $a(0,2\pi k)=(1,0)$ for every integer $k$. To turn this into a failure of the estimate, take a small open disk $W$ about $(1,0)$ that avoids the origin. For each of $N$ inverse branches over $W$, choose a smooth cutoff ${\chi }_k(v)=\chi (a(v))$ supported on that branch, where $\chi$ has support strictly inside $W$ and is extended by zero. These velocity supports are disjoint. SetThen $\Vert f_{\mathrm{in}}{\Vert }_{L_x^1{L}_v^{\infty }}=\Vert \psi {\Vert }_1\Vert \chi {\Vert }_{\infty }$, independent of $N$, whereas at a fixed $t>0$ and suitable $x$,The last integral is positive and independent of $N$. No finite $C_a$ exists for this fixed $a$, although ${\alpha }_1={\alpha }_2=1$. In one dimension, by contrast, a nonvanishing derivative has a constant sign, so the lower derivative bound makes $a$ a global diffeomorphism.

Let $S_{s,t}(y)$ be the position at time $t$ of the characteristic curve satisfying $X(s)=y$ and $\dot{X}=u(t,X)$. Smooth uniqueness implies $S_{t,s}=S_{s,t}^{-1}$. Differentiating in the initial point gives the variational equationThe Liouville formula for a fundamental matrix, or the derivative formula for a determinant, therefore givesBecause $u=(-{\varphi }_{x_2},{\varphi }_{x_1})$, the mixed second derivatives cancel and $\mathrm{div}u=0$. The initial Jacobian determinant is one, soThus this Hamiltonian flow preserves planar Lebesgue measure. This calculation assumes smoothness as in the part; a rough-flow conclusion later must be justified by approximation rather than by differentiating a merely continuous velocity.

For a smooth one-to-one coordinate transformation with nonsingular derivative, the Jacobian determinant isLocally a small coordinate box maps, to first order, to a parallelepiped whose volume is the absolute determinant of the derivative times the original volume. Summing these local volume approximations gives the change of variables formula, with $|J|$ accounting for either orientation. The usual regularity and nonsingularity hypotheses are part of this substitution theorem.

The region is the upper half of the spherical shell between radii two and three, including its annular flat boundary in the equatorial plane. Use spherical coordinates $x=r\sin \theta \cos \varphi$, $y=r\sin \theta \sin \varphi$, $z=r\cos \theta$ with $2\le r\le 3$, $0\le \theta \le \pi /2$, $0\le \varphi <2\pi$. Their Jacobian determinant is $r^2\sin \theta$; the polar axis and azimuth seam have measure zero, so do not obstruct this integration.

The integrand is $x^2+y^2=r^2{\sin }^2\theta$. Consequently

The divisor determines the bundle through its local equations. Choose an open cover on which $V$ has a reduced holomorphic defining equation $f_i$, using $f_i=1$ on sets disjoint from $V$. Because $V$ is a complex submanifold of codimension of a submanifold one, its ideal is locally generated by a coordinate; consequently $f_i/f_j$ is a nowhere-zero holomorphic function on each overlap. Inside the sheaf of meromorphic functions on a complex manifold, consider the locally free sheaf of rank oneThese local descriptions agree on overlaps. With $e_i=1/f_i$, its transition functions of a vector bundle are specified byThe ratios satisfy the Čech cocycle condition, and the associated holomorphic line bundle associated to a divisor is $[V]$. The section $1=f_i{e}_i$ is holomorphic and vanishes to order one along $V$. Replacing $f_i$ by $u_i{f}_i$, with $u_i$ a holomorphic unit, replaces the frame by $u_i^{-1}{e}_i$ without changing the subsheaf of meromorphic functions. Refining the cover changes no local sections either. This proves choice independence up to isomorphism rather than merely producing a bundle for one cover.

For a complex manifold of dimension $n$, the canonical bundle isIt is the holomorphic line bundle of holomorphic top-degree differential forms. To compute it on Complex projective space, let $S=\mathcal{O}(-1)$ be the complex tautological line bundle, so that $S^{\ast }=[H]=\mathcal{O}(1)$ is the hyperplane line bundle. At a line $\ell \subset {\mathbb{C}}^{n+1}$, the tangent space is $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1}/\ell )$. Tensoring the tautological quotient sequence by $S^{\ast }$ gives the Euler sequence on complex projective spaceTaking the top exterior power gives $\det T^{1,0}{\mathbb{P}}^n=\mathcal{O}(n+1)$; taking its dual bundle yieldsHere a negative tensor power means the corresponding positive tensor power of the dual bundle.

The proposed sections over $U$ form a sheaf: compatible maps into $L$ glue uniquely, and their projections remain $\varphi$. Addition and multiplication by holomorphic functions are defined in the fibres of $L$, making this a sheaf of ${\mathcal{O}}_X$-modules. If $e_i$ is a holomorphic local frame of $L$ on $W_i\subset Y$, every section over $U\subset {\varphi }^{-1}(W_i)$ has the unique expressionThus the sheaf is a locally free sheaf of rank one. Its transition functions of a vector bundle are those of $L$ composed with $\varphi$, so the corresponding pullback vector bundle is precisely ${\varphi }^{\ast }L$. This proves invertibility locally and compatibility globally.

For the local blowup of a complex manifold at a point, work on the projective chart $Y_i\ne 0$ and writeAll the incidence equations reduce to $x_j=u{t}_j$. Therefore $(u,(t_j{)}_{j\ne i})$ gives a holomorphic coordinate chart isomorphic to ${\mathbb{C}}^n$. On an overlap with $Y_{\ell }\ne 0$,which is a biholomorphism where $t_{\ell }\ne 0$. The incidence subset is closed in the product of two Hausdorff spaces and is covered by these charts; hence it is a complex manifold of dimension $n$. In each chart the exceptional fibre is $u=0$, so it is a smooth complex submanifold, with its induced projective coordinates givingAway from the origin the inverse of the projection is $x\mapsto (x,[x])$, which is holomorphic; the projection is therefore a biholomorphism outside $E$.

To find the canonical bundle, pull back the nowhere-zero top form on ${\mathbb{C}}^n$. In the chart above, its Jacobian determinant givesThe sign depends only on the ordering of the coordinates. This is a section of $K_X$ with divisor on a complex manifold exactly $(n-1)E$: it has that vanishing order in every exceptional chart and no zeros elsewhere. A nonzero meromorphic section of a holomorphic line bundle with divisor $D$ identifies its holomorphic line bundle with $[D]$, by sending the local generator $1/f_i$ of $\mathcal{O}(D)$ to the nowhere-zero frame obtained by dividing the section by $f_i$. ConsequentlyThis also includes $n=1$, when the local map is an isomorphism and the exponent is zero.

For a general complex manifold, choose a coordinate neighbourhood $U$ of $P$, with $P$ sent to $0$, and replace $U$ by the inverse image of this coordinate neighbourhood in the local blowup of a complex manifold at a point. Glue this space to $M\setminus \{P\}$ along $U\setminus \{P\}$ using the preceding biholomorphism. The resulting charts give a complex manifold $\tilde{M}$ and a holomorphic map $\varphi$; the exceptional fibre is ${\mathbb{P}}^{n-1}$ and the map is a biholomorphism off that fibre. For completeness, the local projection is proper because its inverse image over a compact subset is closed in that subset times compact Complex projective space. This ensures the gluing is Hausdorff: separate points with distinct images downstairs, and points over $P$ inside the local blowup. The atlas is second countable as well.

The top exterior power of the differential defines a global holomorphic bundle map ${\varphi }^{\ast }{K}_M\to K_{\tilde{M}}$. Equivalently, it is a section of $K_{\tilde{M}}\otimes ({\varphi }^{\ast }{K}_M{)}^{-1}$. The same local Jacobian determinant has vanishing order $n-1$ on $E$, and the differential is invertible elsewhere. Applying the holomorphic line bundle associated to a divisor construction gives the canonical bundle formula for a point blowup

For the Box-Muller transform, draw independent $U,V$ from the uniform distribution on $(0,1)$ and setThe radial density is $r{e}^{-r^2/2}$ for $r>0$, with an independent uniform angle. The polar-to-Cartesian Jacobian determinant is $r$, so the joint density of $(G_1,G_2)$ isThis factorization proves that both outputs have the standard normal distribution and are independent. Repeating the Box-Muller transform with fresh independent uniform pairs gives independent normal outputs; discard one extra output if the desired sample size is odd.

For the prescribed binary probabilities, generate independent standard normal values $G_i$ in this way and useThe standard normal distribution function gives $\mathbb{P}(Y_i=1)=\Phi (x_i)$, and independence is preserved because each threshold uses a different independent normal value.

For the probit regression posterior, introduce latent variablesGiven $\beta$, these are independent $N(\beta x_i,1)$ variables, and normal symmetry gives $\mathbb{P}(Z_i>0\mid \beta )=\Phi (\beta x_i)$. Thus this data augmentation has exactly the observed binary likelihood. With the specified normal prior distribution, the augmented joint posterior is proportional toThe latent-normal Gibbs sampler for probit regression alternates two blocks. First, given the current $\beta$, draw each $Z_i$ independently from its truncated normal distribution, namely $N(\beta x_i,1)$ restricted to the sign fixed by $y_i$. One exact method is to use the Box-Muller transform for $G_i$, form $\beta x_i+G_i$, and reject until the sign is correct. The probability of success is positive at every finite parameter value, so the method is valid, although it can be slow for a rare sign.

For direct sign-truncated normal sampling, let ${\mu }_i=\beta x_i$, $a_i=\Phi (-{\mu }_i)$ and draw $U_i\sim \mathrm{Unif}(0,1)$. An inverse transform sampling implementation isUse suitable tail or survival-function evaluations when floating-point probabilities approach zero or one; the rejection construction remains a valid alternative.

Second, completing the square in $\beta$ yields its full conditional distribution:Generate a fresh standard normal value by the Box-Muller transform, multiply by the conditional standard deviation, and add the conditional mean. Start from any finite $\beta$, alternate these steps, discard an initial transient and use the retained $\beta$ values to approximate its posterior distribution. These are dependent Markov chain Monte Carlo samples, rather than independent posterior draws. Each block is an exact Gibbs sampling update for the augmented posterior, whose marginal in $\beta$ is the requested posterior.

The reversible-jump Markov chain Monte Carlo state comprises the model index and that model's parameter. Its unnormalized posterior density in model $j$ isFor models of equal dimension, choose model $k\ne j$ with probability $s_{jk}(\theta )$ and propose ${\theta }^{\prime }$ using density $q_{jk}({\theta }^{\prime }\mid \theta )$ on the destination parameter space. The equal-dimension reversible-jump acceptance probability isOn acceptance change both model and parameter; on rejection retain both. Choose model proposals connecting all models of positive posterior mass and within-model kernels exploring their supports, with aperiodicity, to obtain an ergodic chain. Include within-model updates targeting its conditional posterior, and use model occupation proportions after the initial transient to estimate posterior model probabilities. The formula includes the model prior probabilities, parameter prior densities, likelihoods, reverse model-selection probability, and reverse parameter-proposal density.

This direct-density formula is Metropolis–Hastings algorithm on a disjoint union of equal-dimensional spaces. If instead a move uses a deterministic bijection $T:(\theta ,u)\mapsto ({\theta }^{\prime },u^{\prime })$ with auxiliary proposal densities $g_{jk}(u\mid \theta )$ and $g_{kj}(u^{\prime }\mid {\theta }^{\prime })$, useThe Jacobian determinant is one for identity matching, but equal model dimensions alone do not make a nonlinear map volume-preserving. For a direct proposal density the change of variables is already included in that density, so no additional Jacobian factor is inserted.