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Write ${\mathcal{W}}_n$ for the deterministic set of $n$-step self-avoiding walks from $0$. Each uses $n$ distinct edges, so independence givesFor $n\ge 1$, the function $t\mapsto t^{1/n}$ on $[0,\infty )$ is a concave function. The Jensen inequality gives the root-moment bound for open self-avoiding walksEquivalently, the Lyapunov moment inequality compares exponents $1/n$ and $1$, both strictly positive. Taking the limit superior and the connective constant limit provesThis includes $p=0$ and $n=1$. No independence between the different self-avoiding walk indicators is asserted or needed.

Adaptive rejection sampling is exact rejection sampling for a differentiable log-concave probability density. Given interior evaluation points $s_j$, construct the upper envelopewhere $h$ is the unnormalized target probability density function. The concave function $\ell$ lies below all its tangent lines, so $e^u\ge h$. Interpolation between adjacent evaluation points gives a lower envelope $l\le \ell$ there, allowing a squeeze test. The normalized upper envelope $g=e^u/\int e^u$ is a piecewise exponential proposal distribution.

Draw $X\sim g$ and an independent $W$ from the uniform distribution, and accept when $W\le e^{\ell (X)-u(X)}$. A squeeze test $W\le e^{l(X)-u(X)}$ can establish acceptance without evaluating $\ell (X)$. Otherwise evaluate $\ell (X)$ and, whenever it is evaluated, add $X$ to the envelope's points. Repeat until acceptance. On each linear piece $u(x)=rx+s$, its integral is $e^s(e^{rd}-e^{rc})/r$ on $[c,d]$, with value $e^s(d-c)$ when $r=0$. These integrals select a piece; inverse transform sampling samples within it.

Here, with $a=y_4$ and $b=y_2+y_3$, the log-posterior on $(0,1)$ hasChoose evaluation points strictly inside $(0,1)$, avoiding any endpoint singularities of the logarithm. Every finite set of tangent lines gives a finite upper-envelope integral on this bounded interval; slope conditions needed for unbounded supports are unnecessary. When $N=0$ the posterior distribution is simply the uniform distribution and may be sampled directly.

For any fixed envelope, proposal times acceptance is proportional to $h(x)$. Conditional on all past evaluations and rejections, the next accepted value therefore has the same normalized posterior density. To see that adaptation does not spoil this, at every attempted draw its accepted mass is a scalar times that fixed posterior density; summing over possible rejection histories preserves it. The acceptance probability stays bounded below by its positive initial-envelope value because insertion only lowers the upper envelope. Thus acceptance occurs almost surely. With fresh independent randomness, successive accepted values are exact independent samples from the posterior distribution, even when the envelope is retained and improved between them.

Differentiation of the Hubble parameter gives $\dot{H}+H^2=\ddot{a}/a$. The strong energy condition and the acceleration equation therefore implyAssume the present universe is expanding, $H_0>0$, and normalize the scale factor by $a(t_0)=1$. Since $\ddot{a}\le 0$, the past expanding branch has $\dot{a}(t)\ge \dot{a}(t_0)>0$. On this branch,Integrating from a past time $t$ to $t_0$ gives $H(t{)}^{-1}\le H_0^{-1}+t-t_0$. Thus, where the denominator is positive,The printed inequality needs this past-time qualification: for future times with $H>0$, the inequality reverses. Independently, a concave function lies below its tangent, soPositivity of $a$ prevents the past branch extending before the zero of this tangent. If $t_B$ is the beginning of the expanding universe, its age satisfies $\boxed{t_0-t_B\le H_0^{-1}}$. Without the normalization $a(t_0)=1$, the tangent bound instead reads $a(t)\le a(t_0)[1+H_0(t-t_0)]$.

The Lagrange sufficiency theorem says that if a feasible point $x^{\ast }$ and nonnegative multipliers satisfy complementary slackness and $x^{\ast }$ globally maximizes the Lagrangian function in constrained optimization, then $x^{\ast }$ globally solves the constrained maximization problem. Indeed, for every feasible $x$, the signs of the constraints give $f(x)\le L(x)\le L(x^{\ast })=f(x^{\ast })$. Concavity and the Karush-Kuhn-Tucker conditions are a common way to verify the global Lagrangian maximum.

For fixed $y$, A maximizessubject to $x_i\ge 0$ and ${\sum }_i{x}_i=X$. This is a concave function of $x$. On every active coordinate, the multiplier equation isThus A's optimal allocation is given by the water-filling algorithm:where $\lambda >0$ is chosen so that ${\sum }_i{x}_i=X$. Coordinates with zero expression receive no resource. This is a water-filling algorithm. The formula gives the optimum when every $y_i>0$. If $y_i=0$, any $x_i>0$ already gives A the whole market $u_i$ in that region, so A assigns such coordinates arbitrarily small positive amounts and applies water filling to the remaining budget; without a convention for $0/0$, the resulting value can be a supremum rather than an attained maximum.

For a fixed occupancy $n$, let $x_r>0$ be the per-flow rate on every active route, so feasibility means ${\sum }_{r:j\in r}{n}_r{x}_r\le C_j$. A proportionally fair allocation is a feasible allocation satisfying, for every other feasible $\tilde{x}$,The sum counts the proportional changes over individual flows. By the first-order condition for a concave function, it is exactly the allocation solvingThis is weighted logarithmic utility optimization, and its negative objective is a strictly convex function, which gives unique rates on active routes. Routes with $n_r=0$ have no per-flow rate to determine.

Index the four routes by $1=\{1,2\}$, $2=\{2,3\}$, $3=\{3,4\}$, $4=\{4,1\}$, and let ${\Lambda }_r=n_r{x}_r$ be each route's total service. WriteThe four capacity constraints say ${\Lambda }_1+{\Lambda }_2\le 1$, ${\Lambda }_2+{\Lambda }_3\le 1$, ${\Lambda }_3+{\Lambda }_4\le 1$, and ${\Lambda }_4+{\Lambda }_1\le 1$. Equivalently,For fixed maxima $u,v$, every active route in the first group optimally takes total service $u$, and every active route in the second takes $v$, because its logarithmic objective is increasing. Up to the constant $-{\sum }_r{n}_r\log n_r$, the objective therefore becomes $P\log u+Q\log v$ with $u+v\le 1$. Its maximizer is $u=P/N$, $v=Q/N$ when $N>0$, with the evident endpoint choices when one group is empty. Hence the proportionally fair allocation on a four-cycle isSet ${\Lambda }_r=0$ on inactive routes. If $N=0$, all service totals are zero. In particular, for $n_1>0$, $n_1{x}_1=P/N$, as required. Two disjoint active routes in the same group can each receive the same total service; service totals need not sum to one over the entire network.

With independent document arrivals given by Poisson processes of rates ${\rho }_r$ and independent unit-mean exponential distributions for document sizes, each of the $n_r$ active documents has completion rate $x_r$. The flow-level network model is consequently a continuous-time Markov chain withAt $n=0$ all downward rates are zero. The total departure rate is at most four, so the bounded total jump rate ensures no explosion.

The original PDF's displayed factor is a binomial coefficient, not the plain quotient produced by the TeX extraction. The candidate stationary distribution for this reversible four-cycle flow model isUse $\left(\genfrac{}{}{0ex}{}{0}{0}\right)=1$. For $r\in \{1,3\}$ and $n_r>0$,so $\pi (n)q(n,n-e_r)=\pi (n-e_r){\rho }_r$. For $r\in \{2,4\}$ the corresponding ratio is ${\rho }_{r}N/Q$. Thus detailed balance for a continuous-time Markov chain holds for every adjacent pair, proving the displayed stationary distribution whenever $B<\infty$.

For completeness, its existence condition can be made explicit. Put $\alpha =\max ({\rho }_1,{\rho }_3)$ and $\beta =\max ({\rho }_2,{\rho }_4)$. At fixed $P,Q$,Therefore the sum over states with $N=m$ is at most $(m+1{)}^2(\alpha +\beta {)}^m$, using the binomial theorem. This is summable if $\alpha +\beta <1$. Conversely, restricting to the single maximizing route in each group leaves the series ${\sum }_{m\ge 0}(\alpha +\beta {)}^m$, which diverges if $\alpha +\beta \ge 1$. HenceThese are exactly the four strict resource-load inequalities. With positive arrival rates the Markov chain is irreducible, so the normalized stationary distribution is unique.

Using the representation $X=\mu +\sigma Z$ and differentiation under the integral sign,because $U$ is strictly increasing. Similarly, part (a) givesbecause a twice differentiable concave function has $U^{\prime \prime }\le 0$. ThereforeIn mean-variance optimization, an expected-utility investor consequently prefers a larger mean at fixed standard deviation and a smaller variance at fixed mean. This is the usual risk-averse ordering of Gaussian payoffs.

The Lagrangian sufficiency theorem for a maximization problem says the following. Suppose the constraints are $g_i(q)\ge 0$ and $h_j(q)=0$, and defineIf a feasible $q^{\ast }$ satisfies complementary slackness ${\lambda }_i{g}_i(q^{\ast })=0$ and globally maximizes $L(\cdot ,\lambda ,\mu )$, then $q^{\ast }$ globally maximizes $f$. Indeed, for every feasible $q$,This proves the theorem. In a concave maximization problem, stationarity and the boundary optimality conditions ensure the required global maximum of the Lagrangian.

For the problem at hand, retain $x,z\ge 0$ as the domain and attach a multiplier $\lambda$ to the equality:For $\lambda \ge 1$, this is concave in $(x,y,z)$ on the domain. Its separate maximizers satisfywhile the $x$ term is $(1-\lambda )x$. Thus $x$ may be positive only when $\lambda =1$; when $\lambda >1$, its maximizing value is $x=0$.

Iftake $\lambda =1$. Then $y=1$, $z=a^2-1$, and feasibility fixesThe Lagrangian sufficiency theorem gives the maximum

Ifthen $\lambda >1$ and $x=0$. The equality constraint becomessoConsequentlyThe assumptions $a\ge 1$ and $b\ge 1/2$ ensure $z\ge 0$. At the stationary point $\sqrt{1+z}=ay$, so the maximum value isThe two formulas agree at $b=a^2-1/2$.

Let $w=N(U)/N(R)$, so $1-w=N(V)/N(R)$. Since class counts add,The function $g(p)=p(1-p)$ is concave function on $[0,1]$. Jensen inequality therefore giveswhich is precisely $Q\le 0$. Thus an axis-aligned split cannot increase the weighted empirical Gini impurity.

In independent bond percolation on a graph, let $c_n$ be the number of $n$-step self-avoiding walks from a fixed graph vertex and ${\kappa }_n$ the number that are open. Then $\mathbb{E}{\kappa }_n=c_n{p}^n$. The Jensen inequality for the concave function $t^{1/n}$ gives $\mathbb{E}({\kappa }_n^{1/n})\le p{c}_n^{1/n}$. No independence between different self-avoiding walk indicators is needed. When $c_n^{1/n}\to \mu$, the limit superior is at most $p\mu$.

The tangent line to a differentiable function at $a$ is its affine first-order approximation $f(a)+f^{\prime }(a)(x-a)$. For a differentiable concave function, it is an upper bound throughout a convex interval: the secant-slope inequality and passage to the derivative give $f(x)\le f(a)+f^{\prime }(a)(x-a)$. This upper bound constructs valid envelopes in adaptive rejection sampling.