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For $m\ge 2$, $f^{\prime \prime }(x)=m(m-1)x^{m-2}\ge 0$ on $[0,\infty )$, so $f$ is a convex function; for $m=1$ it is linear and also convex. The equally weighted random variable has expected value $EX=N^{-1}{\sum }_i{x}_i=1/N$. Applying Jensen's inequality givesMultiplying by $N$ provesFor $m>1$ equality requires all $x_i=1/N$, by strict convexity; for $m=1$ equality holds for every admissible vector.

Put $U=X-EX$ and $V=Y-EY$. Their second moments are the finite positive variances of $X,Y$. The Cauchy-Schwarz inequality for square-integrable random variables givesDivide by the positive product of variances and take square roots. The correlation bound is $\boxed{-1\le \rho (X,Y)\le 1}$. As usual, non-constant here means not almost surely constant, so that the denominator is nonzero.

Zero correlation means that the variables are uncorrelated: $E[(X-EX)(Y-EY)]=0$, or equivalently $E(XY)=EXEY$. It does not imply independence. For example, if $X$ is uniform on $[-1,1]$ and $Y=X^2$, symmetry gives $EX=E(X^3)=0$, hence zero covariance, although $Y$ is determined by $X$. Both random variables are non-constant and have finite positive variances.

Use the uniform probability measure on the symmetric group $S_n$ and the unnormalized Hamming distanceLet $f:S_n\to \mathbb{R}$ have Lipschitz constant $K>0$ for this distance. We will prove the one-sided concentration inequalitiesConsequently $\mathbb{P}(|f-\mathbb{E}f|\ge t)\le 2e^{-t^2/(2n{K}^2)}$ for $t>0$. A zero Lipschitz constant makes $f$ constant, so that case is immediate.

Let $\sigma$ be a uniform permutation, reveal $\sigma (1),\dots ,\sigma (k)$ successively, and define the Doob exposure martingaleFix the first $k-1$ images. For an available value $a$, let $h(a)$ be the conditional mean when $\sigma (k)=a$. Given two available values $a,b$, left-composition with the transposition exchanging $a,b$ is a bijection from completions with next image $a$ to completions with next image $b$. It fixes the previously exposed images and changes exactly two coordinates of each complete permutation. Since conditional completions are uniform, averaging the Lipschitz condition over this pairing gives $|h(a)-h(b)|\le 2K$.

Thus, conditionally on the first $k-1$ images, $M_k-M_{k-1}$ has mean zero and takes values in an interval of length at most $2K$. We need the range-length version of the Hoeffding lemma: if a real random variable $Z$ has mean zero and lies in an interval of length $\ell$, thenHere is a proof. For $\psi (\lambda )=\log \mathbb{E}{e}^{\lambda Z}$, differentiation gives ${\psi }^{\prime \prime }(\lambda )={\mathrm{Var}}_{\lambda }(Z)$ under the exponentially tilted probability distribution. Any distribution on an interval of length $\ell$ has variance at most ${\ell }^2/4$, because its squared distance from the interval midpoint is at most ${\ell }^2/4$. Since $\psi (0)={\psi }^{\prime }(0)=0$, integrating this second-derivative bound gives the lemma for both signs of $\lambda$. The same argument holds for every conditional distribution.

Apply it conditionally to each martingale difference and use the tower property of conditional expectation repeatedly:The exponential Markov bound now gives, for $\lambda >0$,Choose $\lambda =t/(n{K}^2)$ to obtain the upper tail in the box, and apply the argument to $-f$ for the lower tail. This proves concentration of Lipschitz functions on the symmetric group. Using only the bound $|M_k-M_{k-1}|\le 2K$ in the absolute-increment Azuma-Hoeffding inequality would lose a factor four in the exponent; the conditional range length is what yields the constant needed in part (ii).

Let $i(G)$ be the number of independent sets of the biregular graph $G$. Choose one such set $I$ uniformly and write $X_v=1_{v\in I}$. All information entropies below use logarithms to base two, soWe will prove $\boxed{i(G)\le (2^r+2^s-1{)}^{m/s}}$.

We state and justify the entropy facts used. For a finite-valued random variable $Y$, $H(Y)=-{\sum }_y{p}_y{\log }_2{p}_y$ with $0{\log }_{2}0=0$. Its conditional entropy is the mean of this quantity over the conditioning variable. Factoring the joint probabilities proves the chain rule for information entropy, $H(Y,Z)=H(Y)+H(Z\mid Y)$. Conditioning reduces entropy: $H(Y\mid Z,T)\le H(Y\mid Z)$. To see this, condition on $Z$ and use concavity of $-x{\log }_{2}x$ to compare the mixture of the conditional distributions given $T$ with their mean entropy.

The precise Shearer inequality needed here is: if subsets $S_j$ of the indices of a finite random vector $Y$ cover every index at least $d$ times, then $dH(Y)\le {\sum }_{j}H(Y_{S_j})$. To prove it, order the indices. The chain rule for information entropy expandsThe inequality is conditioning reduces entropy. Summing over $j$ counts each nonnegative conditional term at least $d$ times and proves the assertion.

For every $w\in W$, its vertex neighbourhood $N(w)\subseteq U$ has $s$ vertices. Every $u\in U$ lies in exactly $r$ of these neighbourhoods. Hence Shearer inequality givesLet $q_w=\mathbb{P}(X_{N(w)}=0)$, the probability that no neighbour of $w$ is selected. Conditional on a particular selected subset of $U$, a vertex $w$ with a selected neighbour is forbidden, and all other vertices of $W$ are available. Because there are no edges within $W$, every subset of the available vertices is a valid completion. Uniform choice of $I$ makes these completions equally likely. The available coordinates are therefore independent fair binary choices, each of entropy one, and the forbidden coordinates are zero. Averaging their number givesBy the chain rule for information entropy,It remains to bound each bracket. Let $Y=X_{N(w)}$, a binary vector of length $s$, with distribution $(p_y)$ on $\{0,1{\}}^s$, and put $Z=2^r+2^s-1$. Assign weight $w_0=2^r$ to the all-zero vector and weight $w_y=1$ to every nonzero vector. ThenThe first inequality is Jensen's inequality for the concave logarithm; the weights sum to $2^r+(2^s-1)$. This explicitly handles zero probabilities and does not assume the neighbourhood indicators are independent.

There are $n$ right vertices, so combining the bounds and using $rm=sn$ givesExponentiating proves the required independent-set bound for biregular graphs. The proof uses both degree constraints: $r$ supplies the neighbourhood-cover multiplicity and the conditional-completion weight, while $s$ determines the number of neighbourhood patterns.

For $q>p$, let $M={\sup }_{n\ge 0}{S}_n$, including $S_0=0$. It is a nonnegative integer-valued random variable, possibly infinite before the following estimate. The tail-sum formula for expectation and part (i) giveThus $\boxed{\mathbb{E}\underset{n\ge 0}{\sup }{S}_n\le p/(q-p)}$, and the supremum is finite almost surely. The sum identity follows by writing $M={\sum }_{k\ge 1}{1}_{\{M\ge k\}}$ and applying monotone convergence theorem. If the printed supremum is read literally over $n\ge 1$, its positive part has exactly the same positive-level tails and its negative part is bounded by one. The same upper bound on its expected value remains valid.

A real Gaussian process $(Z_t{)}_{t\in T}$ is a family of random variables such that every finite vector $(Z_{t_1},\dots ,Z_{t_d})$ has a multivariate normal distribution, allowing degenerate distributions. Equivalently every finite real linear combination is a Gaussian random variable. Its mean and covariance kernel are $m(t)=\mathbb{E}{Z}_t$ and $C(s,t)=\mathbb{E}[(Z_s-m(s))(Z_t-m(t))]$.

For any finite set of indices, these functions give the whole mean vector and covariance matrix. The characteristic function of the vector isThe standard uniqueness theorem for characteristic functions therefore determines every finite-dimensional distribution, including the singular cases. Cylinder events form a generating class for the product sigma-algebra on ${\mathbb{R}}^T$ and are closed under finite intersections, so agreement on all cylinder events gives agreement of the process laws by the uniqueness theorem for probability measures. Thus the mean and covariance determine the law on this coordinate sigma-algebra even when $T$ is uncountable. This statement does not claim uniqueness on an arbitrary finer sigma-algebra of path properties.

Let $T_{\pi }$ permute coordinates $0,\dots ,m$, fixing the later coordinates. For any integrable random variable $V$, its group averageis invariant under this group, hence ${\mathcal{S}}_m$-measurable. If $C\in {\mathcal{S}}_m$, invariance of $C$ and measure preservation of each $T_{\pi }$ give $\mathbb{E}[1_{C}V\circ T_{\pi }]=\mathbb{E}[1_{C}V]$. Thus $A_{m}V$ satisfies the integral definition of $\mathbb{E}[V\mid {\mathcal{S}}_m]$. This proves symmetrization as conditional expectation.

Apply it to $V=1_{A_k}(X_k)$ for $m\ge k$. Under a uniform permutation of $m+1$ coordinates, its selected coordinate is uniform among all of them, soThe reverse martingale convergence theorem says that, for decreasing sigma-algebras ${\mathcal{H}}_m$ and an integrable $V$, these conditional expectations converge almost surely and in $L^1$ to $\mathbb{E}[V\mid {\bigcap }_m{\mathcal{H}}_m]$. Here ${\mathcal{S}}_m\downarrow \mathcal{S}$, because every finitely supported permutation belongs to one of the finite groups. ThereforeIntersect these probability-one events for the finitely many $k=0,\dots ,p$ and multiply the limits. Expanding the product of finite sums givesThe sum here includes repeated indices. No independence of the original coordinates was assumed; exchangeability and reverse conditional-expectation convergence are precisely what replaced it.

A state-price density is a strictly positive random variable $Z$ withOn the stock-state space this means a positive integrable function $z(s)$ satisfying ${\int }_0^{C}z(s)ds=C/R$ and ${\int }_0^{C}sz(s)ds=C$. Equivalently $dQ/dP=RZ$ is an equivalent probability measure with ${\mathbb{E}}_Q{S}_1=R$. Such measures exist because $0<R<C$, for example exponential tilts of the uniform law with mean tuned to $R$.

By the fundamental theorem of asset pricing, any arbitrage-free claim price $p$ is represented by such a density in the enlarged market. For payout $S_1^2$,with strictness because equivalence preserves the stock's nondegenerate distribution. Also $S_1^2<C{S}_1$ almost surely, so $p<C$. Hence $R<p<C$.

When $C=3R$, nonnegative wealth for all stock states requires $-1/2\le \pi \le 1$. At $\pi =1$ wealth is $S_1$. The tangent inequality for the strictly concave square root gives, for $s>0$,Uniform integration gives $\mathbb{E}\sqrt{S_1}=2\sqrt{C}/3$ and $\mathbb{E}{S}_1^{-1/2}=2/\sqrt{C}$, so the expected linear term vanishes exactly when $C=3R$. Strict concavity makes equality possible only for $\pi =1$. Thus the unique optimal holding is ${\pi }^{\ast }=1$, including this boundary optimum without assuming an interior maximizer.

The marginal utility price density isIt is integrable and satisfies both pricing constraints: $\mathbb{E}{Z}_{\ast }=1/R$ and $\mathbb{E}(Z_{\ast }{S}_1)=C/(3R)=1$. Therefore

For any nonnegative random variable $X$ with finite expectation and any $a>0$, Markov's inequality statesIndeed $X\ge a{1}_{\{X\ge a\}}$, and taking expectations proves the bound. Nonnegativity is essential.

For a nonnegative integer-valued random variable, the probability generating function isinitially for $|s|\le 1$, with $0^0=1$. For the Poisson distribution, insert its probability mass function and sum the exponential series:Thus

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 a nonnegative integer-valued random variable, the probability generating function iswith $0\le s\le 1$ always allowed. Its derivative at $1$ from below gives the expectation when finite.

The waiting time $N$ has the positive-integer geometric distribution $P(N=j)=p{q}^{j-1}$. For the capped geometric waiting time, $X=j<k$ exactly when $N=j$, while $X=k$ occurs whenever $N\ge k$. Thus the terminal mass is $q^{k-1}$, givingThe finite sum is a polynomial; any apparent singularity of the rational expression is removable. This includes $k=1$, where $G_X(s)=s$.

Differentiating the rational form at $s=1$ yieldsAs a second interpretation, the expectation is the sum of the tail probabilities $P(X\ge j)=q^{j-1}$ for $1\le j\le k$. Their finite geometric sum gives the same answer.

The moment-generating function is $M_X(t)=\mathbb{E}[e^{tX}]$ at real $t$ for which that expectation is finite; $M_X(0)=1$. For independent random variables, the joint law factorizes, soThis is the transform form of independence. Its ordinary finite-valued interpretation applies where both factors are finite, with the transform of a zero multiple equal to $1$.

The Burton-Keane theorem says that for independent bond percolation on the hypercubic lattice ${\mathbb{Z}}^d$, $d\ge 2$, at every fixed $p\in [0,1]$, there is almost surely at most one infinite percolation cluster. It does not assert that such a cluster exists, nor does it assert absence of percolation at the critical parameter. The cases $p=0$ and $p=1$ are immediate; assume $0<p<1$ for the proof.

Let $N_{\infty }$ be the number of infinite percolation clusters. This random variable is translation invariant. By translation ergodicity of Bernoulli percolation, it is almost surely a constant $k\in \{0,1,2,\dots ,\infty \}$. One way to see ergodicity is to approximate an invariant event by events on finite sets of edges, translate one approximation far enough to make the two approximations independent, and then let the approximation error vanish. Its probability equals its square, hence is zero or one.

The other probabilistic input is the finite-energy property of Bernoulli percolation. Conditionally on all edges outside a finite set, every prescribed inside configuration has positive probability, namely a product of powers of $p$ and $1-p$. Thus an event of positive probability can be followed by a specified finite modification and still gives a positive-probability image event. Outside infinite arms are unaffected.

First exclude a finite $k\ge 2$. Some finite box meets at least two infinite percolation clusters with positive probability. Open all edges in and incident to that box. These clusters merge. Opening finitely many edges cannot create an infinite percolation cluster solely out of finite clusters, since it joins only finitely many such clusters. Hence the number of infinite percolation clusters decreases with positive probability, contradicting its almost-sure constant value. This rules out every finite multiplicity greater than one.

Now suppose $k=\infty$. With positive probability a finite box meets at least three distinct infinite percolation clusters. Retain an exterior infinite ray from each of three of them. Replace the finitely many interior and boundary edges by a minimal tree joining their three attachment graph vertices, and close all unused edges there. The exterior rays belong to distinct original clusters, so have no outside connections to each other. The resulting joined cluster has a branch graph vertex $v$ with exactly three incident open edges, and deleting $v$ leaves three infinite components. If a terminal lies in the middle of the interior connecting path, its exterior ray supplies the third branch there. Such a graph vertex is a trifurcation vertex in percolation. A finite box has only finitely many candidate locations and modifications; therefore some fixed graph vertex has positive trifurcation probability. By translation invariance, that probability is a common number $\tau >0$ at every graph vertex.

The decisive deterministic input is the trifurcation boundary-counting lemma. Let $T$ be all these degree-three trifurcation vertices in percolation in a finite box $B_n$. In each infinite percolation cluster meeting $T$, delete the graph vertices of $T$, contract each remaining connected component of a graph adjacent to $T$ to a node, and retain the incidences with the deleted graph vertices, including direct edges between graph vertices of $T$. This is a finite tree: a cycle would connect two neighbours of an original trifurcation vertex in percolation without passing through it, contradicting its three-way separation. Every deleted graph vertex has degree three. Every leaf component is infinite, since a finite leaf would give a finite branch after removing its adjacent trifurcation vertex in percolation. The elementary tree degree identity therefore supplies at least $|T\cap C|+2$ infinite leaf components for a cluster containing $T$.

Each such infinite component reaches the outer graph vertex boundary ${\partial }^{+}{B}_n$, and different components reach different boundary graph vertices. Summing over the clusters yields, in particular,For a nearest-neighbour lattice box,Taking expectations and using the common trifurcation probability givesThis contradicts $\tau >0$ and excludes infinitely many infinite percolation clusters. Thus only $k=0$ or $k=1$ remains:The proof explains the role of the amenable boundary growth of lattice boxes: finite-energy property of Bernoulli percolation would make branching points have positive volume density, while their disjoint infinite branches can only be supplied through a boundary of lower order. It uses no planar duality and works in every indicated dimension.

Let ${ϵ}_{\mathrm{loc}}(\mathbf{x})=2\nu S_{ij}{S}_{ij}$ and define a coarse-grained energy dissipation at scale $r$, for example by a ball average,This is the random variable denoted ${ϵ}_{\mathrm{AV}}(r)$. The refined similarity hypothesis uses the local supply of energy available to motions of size $r$, rather than replacing it everywhere by a global mean. In its inertial range form,At high local Reynolds number, $V$ has a universal conditional distribution independent of ${ϵ}_r$, forcing details and $r$ within the inertial range. This givesFor signed odd orders the coefficients carry the appropriate sign; absolute moments use their own coefficients. Universality is retained in the normalized conditional velocity statistics, while the distribution of ${ϵ}_r$ can retain integral-scale information. This distinction resolves the particular averaging objection: its nonuniversal viscous dissipation moments are now explicitly present.

The quoted second viscous dissipation moment does not by itself determine every longitudinal structure function exponent. The extra lognormal intermittency model assumes $Y_r=\log ({ϵ}_r/ϵ)$ is Gaussian. Write its variance as ${\sigma }_r^2$. Normalization of the mean requires $⟨Y_r⟩=-{\sigma }_r^2/2$, soThe second-moment estimate gives $\exp ({\sigma }_r^2)=B(\ell /r{)}^{\mu }$. Hence, with $q=p/3$,In particular ${\zeta }_3=1$, ${\zeta }_2=2/3+\mu /9$ and ${\zeta }_6=2-\mu$.

The most problematic step is the extrapolation of a lognormal viscous dissipation law to rare events and all moment orders. It is not entailed by a measured second moment. Its quadratic exponents eventually decrease with $p$ and become negative for $p>3+6/\mu$. If the velocity has finite uniformly bounded moments, the bound $⟨|\Delta v{|}^p⟩\le 2^p⟨|u_x{|}^p⟩$ prohibits such divergence as $r\to 0$. Thus the model cannot be an unrestricted high-order inertial range law. This is a criticism of the lognormal closure, rather than an algebraic contradiction in refined similarity itself.

Relative to a filtration $({\mathcal{F}}_t)$, progressive measurability means that, for every $T<\infty$, the mapis measurable for the product sigma-algebra $\mathcal{B}([0,T])\otimes {\mathcal{F}}_T$ and the Borel sigma-algebra on $\mathbb{R}$.

Fix $T>0$ and divide $[0,T]$ into $2^m$ equal subintervals with mesh $h_m=T/2^m$. DefineSince $X$ is adapted, every random variable $X_{k{h}_m}$ is ${\mathcal{F}}_{k{h}_m}$-measurable, hence ${\mathcal{F}}_T$-measurable. Each approximation is consequently $\mathcal{B}([0,T])\otimes {\mathcal{F}}_T$-measurable. For $t<T$, its sampling time lies strictly to the right of $t$, tends to $t$, and never exceeds $T$. Right continuity implies $X_t^{(m)}\to X_t$; at $T$ equality is exact. Thus $X$ is the pointwise limit of measurable functions on this product space. As $T$ was arbitrary, $X$ is progressively measurable. This is the theorem that right-continuous adapted processes are progressively measurable.

The right-endpoint approximations need not themselves be adapted at their intermediate times. What the proof requires is their joint measurability with respect to the single terminal sigma-algebra ${\mathcal{F}}_T$. The proof uses the pathwise càdlàg convention. If path regularity is assumed only almost surely, under a completed filtration setting the stochastic process to zero on its common exceptional null event gives an indistinguishable progressively measurable version. Arbitrary values on that null event need not make the original stochastic process progressively measurable: even with a complete filtration, a null sample point may be assigned a non-Borel time function. This is why almost sure path regularity does not ensure progressive measurability.

Fix $t$ and put $\sigma =t\wedge \tau$. The stopping time property implies that $\sigma$ is an ${\mathcal{F}}_t$-measurable random variable with values in $[0,t]$: for $s<t$,and for $s\ge t$ the event is the whole space. By part (c), the restriction of $X$ to $[0,t]\times \Omega$ is $\mathcal{B}([0,t])\otimes {\mathcal{F}}_t$-measurable.

The evaluation map $\omega \mapsto (\sigma (\omega ),\omega )$ is measurable from $(\Omega ,{\mathcal{F}}_t)$ to this product space: the inverse image of a measurable rectangle $A\times C$ is $\{\sigma \in A\}\cap C$. Composing it with the jointly measurable stochastic process gives an ${\mathcal{F}}_t$-measurable random variableSince this holds for every fixed $t$, the stopped process is adapted. If path regularity holds only almost surely, first apply the proof to its pathwise regular representative; with a completed filtration, the original stopped variable differs only on a null event and is also ${\mathcal{F}}_t$-measurable. The adaptedness of a stopped right-continuous process requires neither boundedness of $\tau$ nor a martingale assumption; $\tau =\infty$ is harmless because $t\wedge \tau \le t$.

Let $W_n$ be the interpolation in part (a). The operationis continuous, since $|I(f)-I(g)|\le \Vert f-g{\Vert }_{\infty }$. By the Donsker invariance principle and the continuous mapping theorem,The exact trapezoidal integral of the linear interpolation isTherefore the statistic in question differs from $I(W_n)$ by $S_n/(2n^{3/2})$. Using independence, zero means, and unit variances givesThis error tends to zero in $L^2$, hence in probability. The Slutsky theorem now proves the integrated random-walk limit:The limiting law can also be made explicit. The time integral is a Gaussian random variable, as a mean-square limit of linear combinations of a Gaussian process. It is centered, and the covariance identity $\mathbb{E}(B_s{B}_t)=\min (s,t)$ givesThus the terminal value of integrated Brownian motion here has law $N(0,1/3)$.

Use the natural filtration ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$. The first exit time is a stopping time. If $r\le x$, it is zero, so its expectation is already finite. Now assume $0<x<r$.

The positive-increment hypothesis gives a number $h>0$ with $\mathbb{P}(X_1>h)>0$. Choose an integer $m$ so large that $r/m<h$, and put $p=\mathbb{P}(X_1>r/m)>0$. A block of $m$ increments all exceeding $r/m$ has probability $p^m$. From any point still in $(0,r)$, that block forces an upper exit before the block ends.

By independence, conditional on ${\mathcal{F}}_{km}$ and survival to time $km$, the next block has this same probability. Thus the geometric tail bound from a uniform escape probability givesUsing the tail-sum formula for the expected value of a nonnegative integer-valued random variable,This is the random-walk exit bound from a positive-increment block. In particular, the exit occurs with probability one. The mean-zero assumption is not needed for this first bound.