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Integrating out the second coordinate gives the probability density function of the marginal distributionThus the ratio has the normalized target probability density function. The factor $s$ from the Jacobian determinant is essential; merely transforming the rectangular coordinates as if they remained uniform would give a different probability density function. This is the basis of the ratio-of-uniforms method.

Independently generate $U$ uniform on $(0,1)$ and $V$ uniform on $(b_1,b_2)$, using the bounds above. Accept the proposal ifotherwise generate a fresh pair. On acceptance return $X=V/U$. The bounding rectangle contains the entire ratio region: $U\le \sqrt{h(X)}\le a$, and $V=XU$ lies between the extrema of $x\sqrt{h(x)}$. Conditioning a uniform rectangle point on acceptance makes it uniform on $C_h$, so the preceding marginal distribution calculation proves that the returned value has the desired Laplace distribution.

For this target $H=4$, so the region has area two. The rejection sampling acceptance probability isEndpoints have zero probability and can be excluded to avoid divisions by zero in implementation.

Take a random variable $Z$ with the Rademacher distribution, and setBoth marginal distributions equal the law of $Z$ for every $n$. Hence $X_n$ and $Y_n$ each have weak convergence of random variables to $Z$; one may take $X=Y=Z$ on this probability space. But their sums alternate:Indeed, at $t=\pi /2$ their characteristic functions are $-1$ for even $n$ and $1$ for odd $n$, so the sums have no weak limit at all. Separate marginal weak convergence does not imply convergence of the sum. The marginal weak convergence does not control sums counterexample changes the dependence while leaving the marginals fixed. Unlike convergence in probability, separate marginal weak convergence of random variables contains no information about the joint coupling.

Label the full siblings at the bottom $X,Y$ and their cousin parents $F,M$. Let $F$ have parents $A,B$ and $M$ parents $C,D$, where $A,C$ are siblings with parents $U,V$. Include the other great-grandparents as $P,Q$, parents of $B$, and $R,S$, parents of $D$. This gives fourteen people, with six independent founders in a pedigree $U,V,P,Q,R,S$.

The pedigree graphical model has the following Directed acyclic graph, with both members of each parent pair pointing to every listed child:Let $f(g)$ be the founder genotype law under Hardy-Weinberg equilibrium, and let $t(g\mid h,k)$ be the Mendelian segregation kernel. The joint probability of all genotypes isFor a biallelic locus with allele-1 frequency $p$, $f(11)=p^2$, $f(12)=2p(1-p)$ and $f(22)=(1-p{)}^2$. The kernel $t$ assigns probability $1/4$ to each labelled pair of parental transmissions and adds these probabilities when they produce the same unordered genotype. Summing this Bayesian network factorization over unobserved genotypes gives their marginal distribution or a pedigree likelihood.

After the top card is fixed, precisely three of the remaining fifty-one cards have the same rank. Thus a fresh uniformly shuffled deck has matching probabilityIn each independent round, the algorithm repeats with probability $r=(1-a)p=16p/17$ and stops with probability $1-r$. A match always stops. Summing the probabilities over the possible number of repeated rounds givesFor $0\le p\le 1$ the stopping probability per round is at least $1/17$, so termination is almost sure.

Permutation of the thirteen rank labels preserves both the uniform shuffle and the stopping rule. The two marginal distributions are consequently uniform:To establish independence, it is necessary to check the joint probability distribution, not just these marginals. For a specified rank $x$, the fresh-shuffle diagonal probability is $(1/13)(3/51)=1/221$. For different specified ranks $x,y$, it is $(1/13)(4/51)=4/663$. Reweight by the acceptance probability and sum over repeated rounds to obtainThe diagonal must equal $1/169$ for independence, forcing $17-16p=13$ and thereforeAt this value the off-diagonal expression is also $1/169$, proving sufficiency for every ordered pair. This is an example of match-biased reshuffling preserves uniform rank marginals: state-dependent rejection sampling changes dependence even though both individual rank distributions remain uniform.

Take the coordinate-selection matrixThen $A^{T}X=X_1$, $A^T\mu ={\mu }_1$ and $A^T\Sigma A={\Sigma }_{11}$. The permitted result on a linear image of a multivariate normal vector therefore givesThis is the marginal distribution obtained by discarding the other coordinates. The conclusion holds also for a positive-semidefinite marginal covariance: in that case the Gaussian law may be degenerate and need not have a full-dimensional density. If the original covariance is positive definite, its principal block ${\Sigma }_{11}$ is positive definite as well.

We prove Kantorovich duality theorem here by a positive-functional extension argument; it produces the minimizing transport plan at the same time as equality of the values. Let $Y=X\times X$ and $c(x,y)=d(x,y)$. This is a bounded continuous function because $X$ is a compact metric space. Write $C(Y)$ for the real space of continuous functions on a compact space, and letOn this vector subspace define $\ell (s)=\int fd\mathbf{P}+\int gd\mathbf{Q}$. It is well-defined: two representations differ by a constant in the first variable and the opposite constant in the second, and both measures have mass one. It is positive, since $s\ge 0$ implies $\min f+\min g\ge 0$ and hence $\ell (s)\ge 0$. Also $\ell (1)=1$.

Define the majorant envelope, which is a sublinear functional:Constants majorize every $h$, and positivity gives $\min h\le p(h)\le \max h$, so this quantity is finite. Taking approximate minimizing majorants proves subadditivity; rescaling majorants proves positive homogeneity. Moreover $p(s)=\ell (s)$ for $s\in S$, and subtracting $s$ from a majorant provesThe dual value in the question is exactlySubadditivity at $c+(-c)=0$ gives $m_d\le p(c)$.

On $S+\mathbb{R}c$, assign the value $\ell (s)+t{m}_d$ to $s+tc$. If $c\notin S$ the representation is unique. If $c\in S$, then $m_d=\ell (c)$, so the assignment is still well-defined. It is dominated by $p$: for $t\ge 0$, use $t{m}_d\le tp(c)=p(tc)$; for $t<0$, positive homogeneity gives $p(tc)=(-t)p(-c)=t{m}_d$. Together with the translation identity, these verify domination in both cases.

The real Hahn-Banach theorem extends this functional to $T:C(Y)\to \mathbb{R}$ with $T\le p$. If $h\ge 0$, then $p(-h)\le \ell (0)=0$, so $T(h)\ge 0$. Thus $T$ is a positive linear functional, with $T(1)=1$ and $T(c)=m_d$. Positivity also gives $|T(h)|\le \Vert h{\Vert }_{\infty }$, so $T$ is continuous. The Riesz-Markov-Kakutani representation theorem supplies a Borel probability measure ${\pi }_0$ on the compact space $Y$, satisfying $T(h)={\int }_{Y}hd{\pi }_0$.

For every $f\in C(X)$, its pullback $f(x)$ belongs to $S$, so $\int f(x)d{\pi }_0=\int fd\mathbf{P}$. Likewise $\int g(y)d{\pi }_0=\int gd\mathbf{Q}$. Uniqueness in the Riesz-Markov-Kakutani representation theorem shows that these are exactly the two marginal distributions. Finally, every feasible pair $f+g\le c$ gives a lower bound for the cost of every transport plan, by integration. Our constructed plan achieves that bound:This Kantorovich duality by positive extension establishes the requested existence and equality without assuming an optimal plan in advance.

The metric structure additionally gives the Kantorovich–Rubinstein theorem formulation. For a feasible pair define $h(x)={\inf }_y\{d(x,y)-g(y)\}$. The triangle inequality for $d$ makes $h$ one-Lipschitz continuous, $h\ge f$, and $h(y)\le -g(y)$. Hence the dual objective is at most $\int hd\mathbf{P}-\int hd\mathbf{Q}$. Conversely $(h,-h)$ is feasible for every one-Lipschitz $h$. ThereforeNormalizing $h(x_0)=0$ for a fixed $x_0$ gives a uniformly bounded equicontinuous family; it is closed and compact in the uniform topology by the Arzelà-Ascoli theorem. The objective is uniformly continuous on this family, so the supremum is attained as well. In particular, $m_d$ is the first Wasserstein distance between the two measures.

The law of iterated expectation yieldsOne may obtain $\mathrm{Var}(p_T)=v$ either from its unchanged marginal distribution proved below or directly from the law of total variance: the variance of its conditional mean is $nv/(s+n)$ and its expected conditional variance is $sv/(s+n)$. Hence the correlation coefficient is exactlyAlso $\mathbb{E}[(p_T-p_M{)}^2]=2vs/(s+n)\to 0$. The two probabilities become close while keeping their original beta marginals.

Using the conditional beta distribution expected value and the law of iterated expectation,Equality of the expected values alone does not prove equality of the marginal distributions; the later exchangeability argument does.

Two exchangeable random variables satisfy $(X,Y)\stackrel{d}{=}(Y,X)$: their joint probability distribution is unchanged by swapping the coordinates. Thus for every measurable set $A$,Exchangeability implies identical marginal distributions, but does not imply independence.

For each auxiliary count $x$, the correctly normalized joint expression above is symmetric in $p_M,p_T$. Summing it over $x=0,\dots ,n$ preserves that symmetry, so these are exchangeable random variables. Their marginal distributions are therefore identical. Since $p_M$ was generated from a beta distribution,The Beta-binomial exchangeable coupling changes the dependence, while leaving both prior distributions unchanged.

Write $F=F_N$ and integrate the Kac master equation over $v_2,\dots ,v_N$. For a pair $i,j\ge 2$, the rotation acts only on integrated variables. Its unit Jacobian determinant makes the integrated gain identical to the integrated loss, so all those pairs cancel.

The only remaining pairs are $(1,j)$, $j=2,\dots ,N$. For such a pair, first integrate over every variable except $v_1$ and $v_j$. This yields the corresponding two-coordinate marginal distribution evaluated at the rotated pair. Permutation symmetry of $F_N$ makes all $N-1$ resulting integrals identical to the one for $(1,2)$. The coefficient isConsequently the Kac marginal evolution equation isThe time argument has been suppressed on the right. The loss is consistent with normalization, since $\int {\Pi }_2(v_1,v_2)d{v}_2={\Pi }_1(v_1)$. Under the printed definition $k<N$, this use of ${\Pi }_2$ requires $N\ge 3$. For $N=2$ the same formula holds with the natural extension ${\Pi }_N(F_N)=F_N$.

This identity is exact and generally unclosed. Replacing the two-coordinate marginal distribution by the product of one-coordinate marginals would produce the quadratic collision equation associated with Kac chaos. Permutation symmetry alone does not imply that product approximation.

Let $\bar{\pi }$ be the normalized target and ${\bar{\pi }}_{-i}$ its marginal distribution. A coordinate-$i$ kernel $P_i$ leaves $x_{-i}=x_{-i}^{\prime }$ and has the joint old-new measureThis is symmetric in the old and new state: the common remaining coordinates are fixed and the two conditional factors exchange places. Thus each coordinate kernel satisfies detailed balance. Averaging them with the fixed weights $1/p$ provesIntegrating out the old state proves invariance of $\bar{\pi }$. The common normalization cancels, so detailed balance can also be written with the unnormalized target. This establishes detailed balance of a random-scan Gibbs sampler and a reversible Markov chain. Fresh independent random choices make the update a Markov chain, as explained above; invariance is not an assertion that an arbitrarily initialized chain starts in stationarity.

Interpret the contests as standard all-pay auctions: the highest effort among entrants wins, ties are shared uniformly, and an unentered contest awards no prize. The printed question does not specify a prize allocation rule; the highest-effort convention is the one used for standard all-pay contests in the course author's 2014 lecture slides. Under this convention, we can characterize the unique symmetric participation probabilities and bid marginals. Uniqueness of the entire joint mixed strategy requires a further restriction on dependence, as explained below.

Let $q_j$ be the probability that a player omits contest $j$. Since every player enters exactly two contests, $q_1+q_2+q_3=1$. Let $G_j(b)$ be the probability that a rival is absent from contest $j$ or enters it with effort at most $b$. Rivals' strategy draws are independent between players. For a positive bid $b$ outside a measure atom, the expected payoff from this contest isAn all-pay auction cannot have a positive-effort measure atom in a symmetric equilibrium: slightly overbidding that measure atom gives a positive discrete increase in the winning probability at an arbitrarily small extra cost. Nor can active bids have a measure atom at zero when entry has positive probability, because a small positive bid beats tied zero bids. Gaps inside the active effort support are impossible: moving a bid from the top of a gap to just above its bottom preserves its winning probability and lowers its cost. The effort support starts at zero, because lowering its positive lower endpoint would preserve the chance that every rival is absent.

The all-pay indifference equation with random entry consequently gives the maximal per-contest payoffEvery $q_j$ lies strictly between zero and one. First, if $q_j=1$, the other two contests have certain entry and zero per-contest payoff, while deviating into the unused contest wins a positive prize. This contradicts equilibrium. Next, if $q_j=0$, the remaining omission probabilities sum to one and neither can equal one, so both are positive. Contest $j$ has zero per-contest payoff, whereas both other contests have strictly positive payoffs. Every pair containing $j$ is then worse than omitting it and entering the other two, contradicting certain entry in $j$.

Every omission therefore occurs with positive probability. The three entered pairs must give the same maximal payoff, which forces $u_1=u_2=u_3=u>0$. Normalizing the omission probabilities gives the two-of-three all-pay participation equilibrium:Conditional on entering contest $j$, the effort has distribution functionextended by zero below this interval and one above it. The inequalities $q_1\le q_2\le q_3$ show that the larger prizes are entered more often.

To construct an equilibrium, omit $j$ with probability $q_j$, then draw the two active efforts independently with their respective conditional distributions $H_k$. Every bid in a contest's effort support earns $u$; a bid above $w_j-u$ earns at most $w_j-b<u$. Thus no effort deviation or choice of a different pair improves on total payoff $2u$. This proves existence and verifies the Nash equilibrium without relying only on the indifference equations. The arguments above also prove uniqueness of the omission probabilities and the per-contest marginal distributions.

For the full joint strategy law, however, the printed uniqueness claim is too strong. Given an entered pair $k,\ell$, either use two independent uniform random variables $U,V$ and bids $(H_k^{-1}(U),H_{\ell }^{-1}(V))$, or use a single uniform $U$ and bids $(H_k^{-1}(U),H_{\ell }^{-1}(U))$. These are different copulas with the same conditional marginals. A fixed deviation's expected additive payoff only uses the rivals' per-contest marginal distributions, so both constructions remain Nash equilibria. This is the marginal-equivalent equilibria in additive contests phenomenon. The participation probabilities and bid marginals are unique; the full joint mixed strategy is not unique unless a dependence convention is imposed. Independent conditional sampling gives one canonical representative.

Put $\beta =k\xi \mu$. The random phase screen producesInterpret the normality assumption as a jointly Gaussian process, not merely Gaussian one-point marginal distributions. Write its normalized autocorrelation as $\rho (\zeta )=⟨W(z+\zeta )W(z)⟩$. The difference $W(z+\zeta )-W(z)$ is a zero-mean Gaussian random variable with variance $2(1-\rho (\zeta ))$. Its characteristic function therefore gives the Gaussian phase-screen correlationWithout joint Gaussianity the autocorrelation alone does not determine this expectation.

For an acoustic pressure amplitude $\psi$, with background mass density ${\rho }_0$, the time-averaged acoustic energy flux isAn acoustic potential convention changes the dimensional prefactor. The physical field includes the carrier $e^{ikx}$: a transverse Fourier transform mode has axial wavenumber ${\kappa }_{\nu }=k-{\nu }^2/(2k)$ in the parabolic wave equation. Thus its spectral acoustic flux is proportional to ${\kappa }_{\nu }|\hat{E}(x,\nu ){|}^2$, or to $k|\hat{E}(x,\nu ){|}^2$ at leading paraxial order. Differentiating the reduced field alone would omit the carrier contribution.

The infinite stationary illumination has a spectral measure of a stationary random field, rather than a finite ordinary value of $⟨|\hat{E}{|}^2⟩$ at each $\nu$. To specify the normalization, truncate the screen to an aperture of length $L$, propagate this truncated initial field, and setThe propagation multiplier has modulus one. Using the Gaussian phase-screen correlation and the overlap of the two aperture intervals givesMultiplication by ${\kappa }_{\nu }/(2{\rho }_0\omega )$ gives the ensemble-averaged pressure-normalized modal acoustic energy flux for this finite-aperture convention.

The infinite-aperture spectral normalization consistent with the specified Fourier transform isThe integral and limit are understood as spectral measures of a stationary random field when necessary. By Parseval's identity, this is power per unit transverse length: ${\int }_{\mathbb{R}}{S}_E(d\nu )=C_E(0)=1$. The ensemble-averaged axial spectral acoustic flux is independent of $x$:Here the notation includes Dirac delta distributions in the spectral measure of a stationary random field; the leading-order total flux is $I_{\mathrm{inc}}$. The ${\kappa }_{\nu }$ correction is meaningful within $|\nu |\ll k$, the range of the paraxial approximation.

If $\rho (\zeta )\to 0$ and $e^{{\beta }^2\rho (\zeta )}-1$ is integrable, the stationary phase-screen power spectrum separates into coherent and diffuse wave fields:The coherent fraction is $e^{-{\beta }^2}$ and the diffuse fraction is $1-e^{-{\beta }^2}$. If additionally $|\beta |\ll 1$, the stationary phase-screen power spectrum is $(1-{\beta }^2)\delta +{\beta }^2\hat{\rho }+O({\beta }^4)$ with the same Fourier transform convention. Small $\mu$ alone does not imply small accumulated phase $k\xi \mu$.

At each point the Gaussian process has a standard normal marginal distribution. Its characteristic function gives the coherent screen fieldIt is constant in $z$, so its Fourier transform is $m\delta (\nu )$. Applying the Fresnel propagator to this zero-transverse-wavenumber component gives the ensemble-averaged reduced spectrumThus the reduced mean is independent of $x$. The physical mean $⟨\psi (x,z)⟩=e^{ikx}{e}^{-{\beta }^2/2}$ still has its carrier phase. The coherent attenuation by a Gaussian phase screen takes place at the screen, with no further coherent attenuation in homogeneous space.

The full spectral acoustic flux depends on the second moment, not the square of this mean. For a stationary process, the generalized cross-spectral correlation isIndeed its propagation factors cancel on $\nu ={\nu }^{\prime }$. This explains why the ensemble-averaged spectral acoustic flux in (b) is also independent of $x$, even for nonzero transverse wavenumbers whose individual complex amplitudes change phase. The coherent atom has weight $|m{|}^2=e^{-{\beta }^2}$; the full stationary phase-screen power spectrum also contains fluctuations. Under the decay assumptions in (b), their integrated weight is $1-e^{-{\beta }^2}$. One must use these weights in a spectral measure of a stationary random field, rather than square a Dirac delta distribution. A phase-only screen preserves the leading-order total acoustic energy flux, although it reduces its coherent part.

A coupling of probability distributions is a joint probability distribution of random variables $(X,Y)$ with the specified marginal distributions: $\mathbb{P}(X\in A)=\mu (A)$ and $\mathbb{P}(Y\in A)=\nu (A)$. No independence is required.

For every set $A$, the difference of the two indicator functions vanishes when $X=Y$. ConsequentlyMaximizing over $A$ gives the coupling inequality for total variation:

Put $a=y_4$ and $b=y_2+y_3$. Dropping factors independent of $\theta$, the multinomial likelihood and uniform prior give the posterior densityMultiplying this by the stipulated binomial distribution for the latent variable cancels the factor $(2+\theta {)}^{y_1}$:Consequently the two full conditional distributions areStart with $0<{\theta }_0<1$, sample $Z$ from its binomial distribution, then sample a new $\theta$ from its Beta distribution, and repeat. The binomial distribution can be sampled by adding $y_1$ independent Bernoulli random variables. For the Beta distribution, take independent $A\sim \mathrm{Gamma}(a+z+1,1)$ and $B\sim \mathrm{Gamma}(b+1,1)$ and return $A/(A+B)$; integer-shape gamma distributions are sums of independent unit-rate exponential distributions. Both shapes are positive even when some observed counts vanish.

Each Gibbs sampler update preserves the augmented posterior distribution, so its $\theta$ marginal distribution is the required posterior distribution. The positive full conditional distributions on the interior allow exploration of the entire support. These iterates are generally dependent; they are not the independent exact draws constructed in the later parts.