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The 36 ordered outcomes are equiprobable. For $2\le s\le 12$, let $c_s=6-|s-7|$ count outcomes with sum $s$. Then $\mathbb{P}(A_s)=c_s/36$ and $\mathbb{P}(B_i)=1/6$ for $1\le i\le 6$. Their joint probability is $1/36$ if $1\le s-i\le 6$, and zero otherwise.

If the joint probability is zero, independence is impossible in these ranges because both marginal probabilities are positive. If it is $1/36$, the independent events condition requires $1/36=c_s/216$, or $c_s=6$. This happens only for $s=7$, and then all six values of $i$ allow the partner $7-i$. Therefore within the possible sums and facesIf arbitrary integer labels outside these ranges are included, an impossible sum or face defines an empty event, which is trivially independent of every event.

First take $s_i>0$, the hypothesis required for the printed ratios and their optimization formula to be well-defined. Since $\sum p_i=1$,The maximizing and minimizing problems therefore have exactly the same solutions.

Let $\tau >0$ and minimize the separable optimization LagrangianEach summand is convex, and its derivative is $-p_i{s}_i/(s_i+x_i{)}^2+\tau$. Its minimizer is $x_i(\tau )=(\sqrt{p_i{s}_i/\tau }-s_i{)}^{+}$. The sum of these coordinates is continuous in $\tau$, decreases from infinity to zero, and is strictly decreasing wherever positive. Thus there is a unique $\tau$ for which the sum is $b$. For every feasible $y$, coordinatewise minimization gives $L(y,\tau )\ge L(x(\tau ),\tau )$; both constraint terms vanish. The Lagrangian sufficiency theorem proves global optimality, not just a necessary stationary condition.

Put $r_i=p_i/s_i$ in decreasing order. A positive bet occurs exactly when $r_i>\tau$. Hence the positive bets form an initial block. To interpret the maximal index in the question, one may include any subsequent indices tied at $r_i=\tau$: their bets are zero but their marginal-return equality still holds. Take $k=\max \{i:r_i\ge \tau \}$. ThenFor this block, $\sqrt{\tau }=({\sum }_{i\le k}\sqrt{p_i{s}_i})/(b+{\sum }_{i\le k}{s}_i)$, with the threshold inequalities determining $k$. Zero-probability horses receive zero bets.

Let $H$ be the set attaining $\rho =\max r_i$, and write $S_H={\sum }_{i\in H}{s}_i$, $P_H={\sum }_{i\in H}{p}_i=\rho S_H$. For sufficiently small $b$, the solution uses only $H$, with $x_i=b{s}_i/S_H$ there and $\tau =\rho [S_H/(S_H+b){]}^2$. If the next ratio is $r_{\ast }<\rho$, this remains valid when $b\le S_H(\sqrt{\rho /r_{\ast }}-1)$; if all outside ratios are zero there is no finite threshold. Put $S={\sum }_i{s}_i$. The fixed pool is $S+b$, so the expected gross payout and net gain areTied best horses receive bets proportional to their existing stakes; betting all on one tied horse is not generally optimal for finite $b$. This is pari-mutuel expected-return allocation.

The literal allowance $s_i=0$ needs qualification. At $s_i=x_i=0$ the printed objective contains $0/0$ and the ordered ratios are undefined. Even defining an unbacked horse's payout to be zero does not always yield an attained optimum. For example, take $b=1$, $p_1=p_2=1/2$, $s_1=0$, $s_2=1$. For $x_1>0$, the normalized expected payout is $1/2+(1/2)(1-x_1)/(2-x_1)$, tending to $3/4$ as $x_1\downarrow 0$ but never attaining it; at $x_1=0$ that zero-payout convention gives $1/4$. Thus the proved allocation formula requires positive existing stakes, or an explicitly modified market model. The zero-stake case cannot be justified by silently dividing by zero.

Let $X_j$ be the number of failures in mix $j$. The total observed failures are $32+2(12)+3(2)+4(1)=66$ among $600$ blocks. Equal marginal failure probabilities give $E(X_j)=6p$, regardless of dependence within a mix. Thus the unbiased estimator based on grouped Bernoulli failure counts isFor a conventional binomial confidence interval, additionally assume independent block outcomes. Then the total $S$ has binomial distribution $\mathrm{Bin}(600,p)$ and the estimated standard error is $\sqrt{\hat{p}(1-\hat{p})/600}=0.01277$. The normal approximation gives the approximate 95 percent intervalOne can instead obtain an exact, generally conservative binomial interval by solving $P_{p_L}(S\ge 66)=0.025$ and $P_{p_U}(S\le 66)=0.025$ for the two endpoints. These equations invert binomial tail tests.

Equal marginal probabilities alone do not imply independent blocks. If the mixes are independent but blocks within a mix can be dependent, use the sample variance of the 100 counts. Here $\sum X_j^2=114$ and $\overline{X}=0.66$, so $s_X^2=(114-100(0.66{)}^2)/99=0.7115$. The estimated variance of $\hat{p}=\overline{X}/6$ is $s_X^2/(100\times 36)$; its standard error is $0.01406$, giving the large-sample interval approximately $(0.0824,0.1376)$. If even between-mix independence is absent, no stated model justifies either numerical coverage claim. The estimate remains unbiased, but the uncertainty calculation needs its sampling assumptions.

Let $G$ be a finite undirected $d$-regular graph on $N$ vertices, with adjacency matrix $A$ and normalized random walk matrix $P=A/d$. Define edge expansion and normalized conductance byAn expander family has orders tending to infinity, uniformly bounded maximum degree, and a uniform positive lower bound on $h$. For a regular graph the constant vector is an eigenvector of $P$ with eigenvalue one. Write its eigenvalues as $1={\mu }_1\ge {\mu }_2\ge \cdots \ge {\mu }_N\ge -1$, and let $\delta =1-{\mu }_2$. This spectral gap measures the energetic cost of departing from a constant function.

For every mean-zero vector $f$, the spectral theorem and the graph's energy identity giveApply this to $f=1_S-|S|1/N$. Its norm squared is $|S|(1-|S|/N)$ and its energy is $e(S,S^c)/d$. ThereforeThus a uniform spectral gap gives uniform edge expansion when $d$ is fixed. This also bounds exterior vertex expansion: each exterior vertex accounts for at most $d$ boundary edges, so $|N(S)\setminus S|\ge e(S,S^c)/d$.

For the converse, take a mean-zero eigenvector $f$ for ${\mu }_2$ and choose its sign so that its positive support has at most $N/2$ vertices. Put $g=f^{+}$, which is nonzero. Truncation decreases energy in the following precise sense: for any two real numbers $a,b$,Summing over edges yields $⟨g,(I-P)g⟩\le ⟨g,(I-P)f⟩=\delta \Vert g{\Vert }_2^2$. For the threshold sets $S_t=\{v:g(v{)}^2>t\}$, every nonempty set lies in that small positive support. Integrating its boundary count gives the layer-cake identityThe Cauchy-Schwarz inequality bounds the left side bybecause the first sum is at most $d\delta \Vert g{\Vert }_2^2$ and the second at most $2d\Vert g{\Vert }_2^2$. Cancelling proves the Cheeger inequality in both directions:Hence bounded-degree regular families have uniform edge expansion exactly when their second-largest normalized eigenvalue stays uniformly below one.

Absolute eigenvalues give additional information. Put $\rho ={\max }_{i\ge 2}|{\mu }_i|$. Centering the indicators of $S,T$ and applying the operator-norm bound on the mean-zero subspace proves the expander mixing lemma:Here $e(S,T)=1_S^{T}A{1}_T$, so internal edges are counted as oriented incidences. A small $\rho$ makes edge counts resemble uniformly distributed edges. It also gives rapid random walk mixing: starting at a vertex, $P^t$ contracts the mean-zero component of its point mass by at most ${\rho }^t$ in Euclidean norm; Cauchy-Schwarz inequality then gives total variation distance at most $\frac{1}{2}\sqrt{N}{\rho }^t$ from the uniform distribution. A bipartite graph has eigenvalue $-1$ and its non-lazy walk alternates sides, even if it has good edge expansion. The lazy walk $(I+P)/2$ has nonconstant eigenvalues between zero and $1-\delta /2$, resolving this periodicity. It is important to distinguish the gap $1-{\mu }_2$ controlling cuts from the absolute bound controlling non-lazy mixing.

Here is a fully specified deterministic construction of an infinite family. Its degree is large but fixed. SetChoose the lexicographically first ordered list $a_1,\dots ,a_d\in W$ satisfyingThis finite deterministic search terminates. To prove existence, choose the list uniformly and independently. For each nonzero $u$, its signs are independent unbiased signs; the exponential-moment estimate $\cosh t\le e^{t^2/2}$ and Markov inequality give an exceptional probability at most $2e^{-d/200}$. A union bound gives total failure probability at mostThus at least one qualifying list exists. Fix it once and for all.

Let $H$ be the Cayley graph on $W$ whose $j$th neighbor of $x$ is $x+a_j$. It is an undirected $d$-regular port-labeled graph, allowing repeated edges and loops in this intermediate construction. Its characters $x\mapsto (-1{)}^{u\cdot x}$ form an orthogonal eigenbasis, with the displayed averages as normalized eigenvalues. Therefore $\rho (H)\le 1/10$. Each loop counts as one transition port; degrees here mean row sums of the adjacency matrix.

We next define the zig-zag product and prove the bound used for iteration. A port labeling of a $D_0$-regular undirected graph $F$ gives an involution ${\mathrm{Rot}}_F(v,a)=(w,b)$, recording the endpoint and reverse port. If a $d$-regular graph $H_0$ has vertex set $[D_0]$, the product has vertices $(v,a)\in V(F)\times [D_0]$. For each ordered pair of small-graph ports, move from $a$ to $a^{\prime }$ in $H_0$, follow the big edge ${\mathrm{Rot}}_F(v,a^{\prime })=(w,b^{\prime })$, and move from $b^{\prime }$ to $b$ in $H_0$. The endpoint is $(w,b)$; reversing the steps reverses the two small-graph port labels. The product is undirected, has $|V(F)|D_0$ vertices, and has degree $d^2$.

Write $B=I\otimes P_{H_0}$ and let $R$ be the permutation matrix of ${\mathrm{Rot}}_F$. The normalized product operator is $BRB$. For a mean-zero unit vector, decompose $f=u+v$ into its cloud-constant part and its cloud-orthogonal part. Then $Bu=u$, $\Vert Bv\Vert \le \rho (H_0)\Vert v\Vert$, and $\Vert R\Vert =1$. The compression of $R$ to cloud-constant vectors is exactly $P_F$, so $|⟨u,Ru⟩|\le \rho (F)\Vert u{\Vert }^2$. Expanding the quadratic form and applying Cauchy-Schwarz inequality givesThe operator is symmetric, so its mean-zero operator norm is bounded by this same quantity. No unproved product spectral estimate is needed.

Start with the loop-inclusive complete graph $G_0$ on $D$ vertices, whose normalized matrix is $J/D$ and whose nonconstant eigenvalues are all zero. Given a $D$-regular $G_t$, its two-step graph $G_t^2$ has degree $D^2$ and normalized operator $P_{G_t}^2$. Label a two-step edge by its ordered pair of ports; reversal reverses the path and its reverse ports. These $D^2$ labels are the vertex set of the fixed graph $H$. DefineIt has degree $d^2=D$, and its number of vertices is $N_{t+1}=N_t{D}^2$, hence $N_t=D^{2t+1}$. By the proved bound,Starting from zero, induction gives $\rho (G_t)\le 1/2$, since $(1/2{)}^2+21/100=46/100<1/2$. The cut-energy argument therefore gives multiplicity-counted edge expansion at least $D/4$ for every $G_t$.

To obtain ordinary simple graphs, delete loops and merge parallel edges. The maximum degree is at most $D$, and any one unordered vertex pair had multiplicity at most $D$. Thus every cut retains at least $1/D$ of its multiplicity-counted boundary, givingThese simple graphs form an expander family. Every choice is deterministic: the finite base search is fixed independently of $t$, and the recursive edge rules determine all subsequent adjacency lists. Once that base is fixed, adjacency lists can be generated successively using the previous lists, with a constant amount of work per output edge. This is squaring and zig-zag iteration yields bounded-degree expanders. The spectral argument supplies the uniform expansion guarantee and explains why squaring improves the eigenvalues while the product restores a constant degree.

Choose a fixed prime $p\ge k$, and put $s=2p-2$. We first derive the polynomial step: a graph $H$ on $N$ vertices and $M>(p-1)N$ edges, with maximum degree at most $2p-1$, contains a nonempty $p$-regular subgraph. Use a Boolean variable $x_e$ for each edge and work over ${\mathbb{F}}_p$. DefineThe first product has degree at most $(p-1)N<M$, while the full squarefree monomial ${\prod }_e{x}_e$ has coefficient $(-1{)}^{M+1}\ne 0$ in the second term with its minus sign. The Combinatorial Nullstellensatz, with every grid $\{0,1\}$, gives a nonzero value of $R$. Since $R(0)=0$, this is a nonempty edge selection. The second product is then zero, so the first is nonzero. Fermat's little theorem makes each selected vertex degree zero modulo $p$. The maximum-degree bound implies that every positive selected degree is precisely $p$. This proves the prime-regular subgraph from Boolean polynomial constraints.

We next turn density into an almost-regular bipartite subgraph, keeping the logarithmic loss explicit. Let the original graph have $n\ge 2$ vertices and $m>0$ edges. Some bipartition retains at least $m/2$ edges: choose the sides independently with probability $1/2$ and average the number crossing. Its average degree is at least $m/n$. Repeatedly delete a vertex of degree at most half the current average degree. Such a deletion cannot lower that average, sinceThe process cannot delete the last positive-density graph, because the maintained average is positive. At termination the minimum degree is an integer $\sigma >m/(2n)$.

Choose the larger side $A$ of this remaining bipartite graph, with opposite side $B$. Keep exactly $\sigma$ edges at each vertex of $A$; choices at different vertices of $A$ do not conflict. Remove any now-isolated vertices of $B$. Thus $|A|\ge |B|$, and every vertex of $A$ has degree $\sigma$.

Such a half-regular graph contains a half-regular restriction with a perfect matching. Choose a minimal nonempty $X\subseteq A$ satisfying $|N(X)|\le |X|$, which exists because the whole side $A$ satisfies this inequality. In fact equality holds: if it were strict, deleting a vertex of $X$ would retain the inequality, contradicting minimality; the singleton case has at least one neighbor. Every proper subset $Y\subset X$ satisfies $|N(Y)|\ge |Y|$. Hall's marriage theorem therefore supplies a perfect matching on $X\cup N(X)$. All $\sigma$ edges incident with each vertex of $X$ remain inside that restriction.

Delete this perfect matching and repeat the restriction argument. The left degrees decrease by one at each step. We obtain $\sigma$ edge-disjoint perfect matchings $M_0,\dots ,M_{\sigma -1}$ on nested nonempty vertex sets, with sizesFor any $i$ with $i+s\le \sigma -1$, the union of $s+1=2p-1$ consecutive matchings in a graph has maximum degree at most $2p-1$ and at least $(s+1)N_{i+s}/2$ edges on $N_i$ vertices. Ifit has more than $(p-1)N_i$ edges, so the polynomial step yields a $p$-regular bipartite subgraph. Every regular bipartite graph decomposes into perfect matchings: for a set in one part, counting incident edges gives Hall's inequality, so a perfect matching exists; remove it and repeat. Choosing $k$ of these $p$ matchings in a graph gives a nonempty $k$-regular subgraph. This is the nested matching criterion for a regular subgraph.

If no $k$-regular subgraph exists, every displayed size ratio must therefore satisfy the reverse inequality. Iterating at indices $0,s,2s,\dots$ gives, with $r=\lfloor (\sigma -1)/s\rfloor$,Consequently $r\le \log n/\log ((s+1)/s)$ andCombining with $\sigma >m/(2n)$ proves the required edge bound. For example the finite constantworks for every $n\ge 2$, and the case $n=1$ has no edges. ThusThe logarithm comes from how often a nested matching in a graph vertex set can shrink by a fixed factor before becoming empty; it is not supplied by the polynomial step alone.

In bond percolation on the square lattice, every nearest-neighbour edge independently has state one, meaning open, with probability $p$, and state zero, meaning closed, with probability $1-p$. Thus the configuration law is the product measure ${\mathbb{P}}_p={⨂}_{e\in E}\mathrm{Bernoulli}(p)$. An open path in bond percolation uses only open edges. An increasing event $A$ satisfies $\omega \in A,{\omega }^{\prime }\ge \omega \Rightarrow {\omega }^{\prime }\in A$: opening further edges cannot destroy the event.

The Harris-FKG inequality for two increasing events isFor disjoint occurrence of increasing events, write $A\square B$ when disjoint sets of open edges separately witness $A$ and $B$, so each event is forced by its own witness regardless of the other edges. The Van den Berg-Kesten inequality statesFor events given by cylinder sets these are finite-coordinate assertions. Connection events have finite open-path witnesses; truncating to finite boxes and taking increasing limits gives the same disjoint-occurrence bound for them. Sharing a vertex between the two witness paths is permitted; sharing an edge is not.

For $1\le m<n$, take a simple open path from the origin to the radius-$n$ boundary and let $x$ be its first vertex on the radius-$m$ boundary. Its initial segment witnesses $0\leftrightarrow x$. If $z$ is its final vertex, then $\Vert z-x{\Vert }_{\infty }\ge n-m$. Consequently the remaining segment reaches $x+\partial {\Lambda }_{n-m}$, and its initial portion until that visit witnesses $x\leftrightarrow x+\partial {\Lambda }_{n-m}$. The two segments use disjoint edges. HenceThe union bound, Van den Berg-Kesten inequality and translation invariance of the product measure now give the weighted BK boundary-splitting estimateFor $m=n$ the same bound follows directly from the union bound, with $g_0=1$. With the conventional $\partial {\Lambda }_0=\{0\}$, $m=0$ gives equality.

Write $C(0)$ for the percolation cluster of the origin. By summing its membership indicators, the percolation susceptibility isIf it is finite, then $a_m\to 0$. For $0<p<1$, choose $m$ with $0<a_m=a<1$. Iterating the boundary-splitting estimate with this fixed $m$ givesFor $k\ge 2m$, $\lfloor k/m\rfloor \ge k/(2m)$, so $g_k\le \exp [-(-\log a)k/(2m)]$. To handle all smaller positive radii without an unspecified prefactor, observe that $g_k\le g_1=1-(1-p{)}^4<1$. ThereforeFor $1\le k<2m$, the second chosen rate gives $e^{-\gamma k}\ge g_1$, and for larger $k$ the first rate applies. At $k=0$ both sides equal one. The endpoint $p=0$ has $g_k=0$ for every $k\ge 1$, so any positive rate works; $p=1$ is excluded by finite susceptibility. This proves exponential one-arm decay from finite susceptibility with the required unit prefactor.

Use the contact process convention that each infected vertex recovers at rate one and sends infection arrows to each neighbour at rate $\lambda$. Starting from a finite set, the process stays finite and is nonexplosive: its total birth rate is at most $\lambda (d+1)$ times its current size, so a linear-rate branching process dominates the number of births. The bound involving $(d-1{)}^{-1}$ presupposes the branching-tree case $d\ge 2$.

For a nonempty finite infected set $A$, put $a=|A|$, let $e(A)$ count its internal edges, and let $b(A)$ count edges with exactly one endpoint in $A$. Each such boundary edge contributes infection rate $\lambda$, even when several lead to the same vacant vertex. Since the induced graph is a forest, $e(A)\le a-1$, and the edge boundary of a finite forest in a regular tree givesA recovery multiplies ${\nu }_{\rho }(A)={\rho }^a$ by ${\rho }^{-1}$, while an infection multiplies it by $\rho$. Therefore its infinitesimal generator isSubstituting the boundary lower bound givesAt the absorbing empty set the generator is zero, rather than this nonempty-set expression. If $\lambda \rho (d-1)\ge 1$, the generator is nonpositive in every state. Both ${\nu }_{\rho }$ and its generator are bounded, since $a{\rho }^a$ is bounded and $b(A)\le (d+1)a$. Applying Dynkin's formula over successive time intervals with the Markov property proves the exponential cardinality supermartingale for the contact process on a tree. In particular,If $\tau$ is the extinction time, $1_{\{\tau \le t\}}\le {\nu }_{\rho }({\xi }_t^A)$, so ${\mathbb{P}}_{\lambda }(\tau <\infty )\le {\rho }^{|A|}<1$. Choosing $\rho \in [1/(\lambda (d-1)),1)$ proves global survival whenever $\lambda >1/(d-1)$. Obtaining a strict improvement of the threshold requires an additional comparison.

Here is a two-vertex branching comparison for contact-process survival. Delete one edge incident to a starting vertex and retain its forward rooted subtree, so every vertex has $d$ children. Partition it into cells as follows: pair each cell root $u$ with one distinguished child $v$; the other $d-1$ children of $u$ and all $d$ children of $v$ become roots of child cells, and repeat recursively. Each cell thus has $2d-1$ child cells.

Suppress arrows back from a child cell into its parent, allow each child cell to be activated only once, and let infection inside each two-vertex cell evolve with the usual recoveries and reinfections until both vertices are healthy. A child cell is activated at its root by the first outgoing arrow to it while the corresponding parent vertex is infected. This process is contained in the original infection process under the graphical representation of the contact process: only infection paths have been removed. The Poisson clocks inside a child cell and on its outgoing edges have not previously been used. The Strong Markov property and independence of the clocks therefore make different cells' offspring counts independent and identically distributed. Dependence among different children of the same cell does not matter. The cell genealogy is a Galton-Watson process.

Let $a$ be the probability that a cell, initially infected only at $u$, sends an arrow along one specified outgoing edge from $u$ before internal extinction. Let $b$ be the corresponding probability for one specified outgoing edge from $v$, and let $c$ be the probability of success for the first target when both cell vertices are initially infected. First-step conditioning on the three nonempty pair states givesFor example, from the state with only $u$ infected, recovery at rate one fails, an internal arrow at rate $\lambda$ leads to the two-infected state, and the target arrow at rate $\lambda$ succeeds. From the state with only $v$ infected, the target cannot be reached until $u$ is infected. Symmetry identifies this latter success probability with $b$. Solving these equations yieldsBy linearity of expectation, the offspring mean is $M(\lambda )=(d-1)a+db$. At the comparison rate ${\lambda }_0=1/(d-1)$,These finite-state probabilities are continuous in $\lambda$. Hence some $0<{\lambda }^{\prime }<{\lambda }_0$ still has $M({\lambda }^{\prime })>1$. The branching-process extinction criterion gives positive probability of infinitely many activated cells. Nonexplosion prevents infinitely many activations in bounded time, and each activation is connected to its ancestor by an infection path. Thus infection survives for all time on that event. The global survival threshold of the contact process consequently satisfiesRetaining reinfection inside pairs is what supplies the strict improvement; a comparison that permits only one infection per vertex would give merely the non-strict bound.

For the local-survival estimate, set $h(x)={\rho }^{g(x)}$ and $w_{\rho }(A)={\sum }_{x\in A}h(x)$. Each vertex has one neighbour of weight ${\rho }^{-1}h(x)$ and $d$ neighbours of weight $\rho h(x)$. Recoveries remove their vertex weights. Ignoring the restriction that an infection target must be vacant only increases the birth contribution, soThese weights have finite expectations on finite time intervals: a birth-only branching process with the same spatial arrows has expected weight at most $w_{\rho }(A)e^{\lambda ({\rho }^{-1}+d\rho )t}$. This justifies localization and the expectation identity for the generator. If $c=1-\lambda ({\rho }^{-1}+d\rho )>0$, Dynkin's formula and the integrating-factor form of the Gronwall inequality giveFor $d\ge 2$, take $\rho =d^{-1/2}$, which minimizes ${\rho }^{-1}+d\rho$ and makes it $2\sqrt{d}$. Starting from $\{x\}$ then givesIn particular the expected total occupation time of $x$ is finite. Its recovery clock has rate one, so the expected number of recovery marks that occur while $x$ is infected equals this occupation-time integral and is finite. Each infection episode ends almost surely at a recovery. Infection at unbounded times would therefore require infinitely many such recoveries, an event of probability zero. This height-weighted local extinction bound for the contact process provesThis is a bound for the local survival threshold of the contact process, not just a bound on single-time infection probabilities. For the degree-two tree $d=1$, the same local conclusion follows by choosing $\rho <1$ sufficiently close to one for each $\lambda <1/2$; the displayed expression $1/(d-1)$ for the global threshold is not a finite numerical assertion in that case.

The joint law is the Edwards-Sokal coupling. For a fixed spin configuration, the sum over each edge state contributes $(1-p)+p{\delta }_e(\sigma )$. Since ${\delta }_e$ is either zero or one, setting $\beta =-\log (1-p)$ givesConsequently the spin marginal distribution is the Potts measureThis follows from a finite sum that factorizes over edges, without any independence assumption on the spin marginal.

For a fixed edge configuration, the joint weight vanishes unless the spins agree across every open edge, equivalently unless they are constant on each open percolation cluster. Each of the $k(\omega )$ clusters, including isolated vertices, may independently be assigned any of $q$ colors. Thus exactly $q^{k(\omega )}$ spin configurations have nonzero weight, and their common weight is the edge-product factor divided by $Z$. The edge marginal distribution is therefore the random-cluster measureIn this convention $Z^{\prime \prime }=Z$. Dividing the joint mass by this positive marginal gives the conditional distributionEquivalently the clusters receive independent uniformly chosen colors. Conversely, conditionally on the spins, edges are independent: an unequal-spin edge must be closed, while an equal-spin edge is open with probability $p$. These two conditional descriptions explain the Edwards-Sokal coupling directly.

If $x$ and $y$ lie in the same open cluster, their conditional spin-equality probability is one. If they lie in different clusters, their two independent uniform colors agree with probability $1/q$. Averaging over the random-cluster measure yieldsand hence the spin-connectivity identity for the Potts modelThe identity also includes $x=y$, when connectivity and spin equality are both certain.

Let the common vertex degree be $r>0$. For each unordered edge place an independent Poisson process of rate $1/r$, and at a mark interchange the two endpoint occupations. If exactly one endpoint is occupied, this moves its particle across that edge at rate $1/r$; if both or neither are occupied, the occupied set stays unchanged. Thus each particle proposes jumps at total rate one, uniformly among its neighbours, with occupied-target proposals suppressed. This is the stated symmetric simple exclusion process. An edgeless graph has the identity dynamics and both conclusions below hold trivially.

In the stirring representation of symmetric exclusion, put a distinct label at every vertex and interchange labels at every edge mark, even when the occupations agree. There are finitely many marks on every finite time interval. Their chronological composition is a random permutation ${\pi }_t$ of the vertex set, and simultaneously for every initial set,Each edge swap is its own inverse. Reflecting the edge-mark times in $[0,t]$ preserves their joint Poisson process law and reverses the order of the swaps. Hence ${\pi }_t^{-1}$ and ${\pi }_t$ have the same distribution. Pathwise,Taking probabilities and then using the inverse-permutation distribution proves the product self-duality of symmetric exclusion in set form:No assumption about disjointness or equal cardinality of $A$ and $B$ is required; impossible inclusions have probability zero on both sides.

For invariance, independently choose the initial occupied set $A$ under the product measure ${\mu }_{\rho }$. The occupations are independent identically distributed Bernoulli distribution variables, so permuting their sites by any deterministic permutation preserves their joint law. Conditional on the independent stirring permutation ${\pi }_t$, the output therefore still has law ${\mu }_{\rho }$. Averaging proves the Bernoulli invariant laws of finite symmetric exclusion:There is also a direct check from the duality. For every $B\subseteq V$,because the exclusion process conserves particle number. These are exactly the joint occupation moments of ${\mu }_{\rho }$; on a finite set they determine the law by the inclusion-exclusion principle. The endpoint densities give the absorbing empty and fully occupied configurations, and the empty-set moment is one also at $\rho =0$.

Normalize the riskless financial asset to $B_0=1$, $B_1=b>0$, and write $S_0,S_1$ for the $s$ risky asset prices. Their discounted gain random vector is $Y=S_1/b-S_0$. A zero-cost portfolio holding $h\in {\mathbb{R}}^s$ in the risky assets and $-h\cdot S_0$ in the riskless asset has terminal value $bh\cdot Y$. Thus an arbitrage meansAn equivalent martingale measure is a probability measure $Q$ with the same null sets as $P$, under which $Y$ is integrable and ${\mathbb{E}}_{Q}Y=0$. The discounted risky prices are then martingales over the single trading period.

If such $Q$ exists, a nonnegative terminal gain has expectation zero under $Q$. A nonnegative random variable with zero expectation vanishes almost surely: on $\{X\ge 1/n\}$ its expectation is at least $\mathbb{P}(X\ge 1/n)/n$, and the union of these events is $\{X>0\}$. Equivalence transfers this conclusion to $P$, excluding arbitrage.

For the converse we prove the needed positive-density separation proof of the one-period asset-pricing theorem. First make the gain integrable without changing its null sets. Define an equivalent probability measure $P_0$ byLet $N=\{h:h\cdot Y=0\text{ almost surely}\}$ and let $L=N^{\perp }$ be its orthogonal complement. Taking a finite basis of $N$ shows $Y\in L$ almost surely. If $L=\{0\}$, the gain is zero and $P_0$ already gives the required measure. Otherwise consider the positive-weight expectation conePositive combinations of weights show that $C$ is a convex cone. It is open relative to $L$. Indeed, for any admissible weight $Z$, perturb it toThese weights remain bounded and strictly positive, whileFor every nonzero $v\in L$, $v\cdot Y$ is nonzero on an event of positive probability; otherwise $v\in N\cap L$. ConsequentlyThe matrix is finite because $\Vert Y{\Vert }^2/(1+\Vert Y\Vert )\le \Vert Y\Vert$. The kernel of $A_Z$ on $L$ is zero, so it defines an invertible matrix on the finite-dimensional vector space $L$. For any sufficiently small $\delta >0$, every $y$ with $\Vert y\Vert <\delta /\Vert A_Z^{-1}\Vert$ is the image of a vector in the radius-$\delta$ ball, since $\Vert A_Z^{-1}y\Vert <\delta$. Hence the perturbations give a neighborhood of zero, proving relative openness of $C$.

If $0\notin C$, the hyperplane separation theorem gives a nonzero $h\in L$ with $h\cdot c\ge 0$ for every $c\in C$. The separating constant can be taken to be zero because $C$ is a cone. Put $A=\{h\cdot Y<0\}$ and use the strictly positive weights $Z=1_A+\epsilon$. Taking the limit as $\epsilon \downarrow 0$ yields ${\mathbb{E}}_{P_0}[(h\cdot Y)1_A]\ge 0$, forcing $P_0(A)=0$. But $h\notin N$, so $P_0(h\cdot Y>0)>0$. This is an arbitrage, a contradiction. Hence $0\in C$: some bounded strictly positive $Z$ satisfies ${\mathbb{E}}_{P_0}[ZY]=0$. Normalize it by $dQ/d{P}_0=Z/{\mathbb{E}}_{P_0}Z$. This gives an equivalent martingale measure with integrable gains, establishing the fundamental theorem of asset pricing here without any further existence theorem.

Now retain only the $m$ positive-probability states of the finite probability space. Let $V\subseteq {\mathbb{R}}^m$ be the linear span of the discounted terminal asset payoffs; it contains the constant vector $1$. A complete market means $V={\mathbb{R}}^m$. An equivalent martingale measure is a vector $q$ of strictly positive state probabilities whose pairing with each asset payoff gives its initial price. By linearity it prices every attainable payoff. If $V={\mathbb{R}}^m$, the indicator functions of individual states are attainable, so these prices determine every $q_j$ uniquely.

If $V$ is proper, choose $0\ne v\in V^{\perp }$. Since $1\in V$, ${\sum }_j{v}_j=0$. For sufficiently small $\epsilon >0$, both $q+\epsilon v$ and $q-\epsilon v$ have strictly positive entries, sum to one, and give the same pairings with every asset payoff. They are distinct equivalent martingale measures. Therefore market completeness is equivalent to uniqueness of the equivalent martingale measure.

Let ${\varphi }_T(y)=(2\pi T{)}^{-1/2}{e}^{-y^2/(2T)}$ be the normal distribution density. The Brownian reflection principle gives the killed endpoint densityIndeed, reflection after the first hit of $a$ maps paths ending below $a$ that have hit the barrier to paths ending at $2a-y$; symmetry gives ${\varphi }_T(2a-y)={\varphi }_T(y-2a)$.

By the Girsanov theorem, weighting the path law by $\exp (\nu W_T-{\nu }^{2}T/2)$ makes the coordinate process a Brownian motion with drift $\nu$. Completing the two squares yieldsIntegration up to $x\le a$ proves the joint endpoint and maximum law for drifted Brownian motion:Here $\Phi$ is the standard normal distribution function. Setting $x=a$ also gives a continuous barrier distribution, so equality at the barrier has zero probability.

For the up-and-in option, use the risk-neutral measure of the Black-Scholes model with interest rate $\rho$ and spot volatility $\sigma >0$. ThenAbove $a$, the barrier has certainly been reached. Below $a$, subtracting the killed density from the unrestricted endpoint density leaves $e^{2a\nu }{\varphi }_T(y-2a-\nu T)$. Thus the discounted expected value of its payoff isIn the second integral put $z=y-2a$, and defineThen $S_0{e}^{\sigma y}=S_0{e}^{\sigma z}/\kappa$, $z\le -a$ means $S_0{e}^{\sigma z}\le \kappa b$, and $e^{2a\nu }={\kappa }^{-\nu /\sigma }$. Both integrals now use the unrestricted terminal stock distribution. This proves that the reflected European payoff for an up-and-in option isTheir initial risk-neutral pricing values agree for payoffs with finite absolute expectations in these integrals. The identity concerns prices; the two terminal payoffs need not agree along individual paths.