For active-route per-flow rates $x_r>0$, feasibility means ${\sum }_{r:j\in r}{n}_r{x}_r\le C_j$ for every resource. The defining condition for a proportionally fair allocation is that every feasible alternative $\tilde{x}$ satisfiesThis is equivalent to maximizing weighted logarithmic utility ${\sum }_{r:n_r>0}{n}_r\log x_r$ over the feasible set. Its gradient is $(n_r/x_r)$, so the displayed inequality is precisely the first-order condition for this concave function. Strict concavity gives unique active-flow rates. Routes with $n_r=0$ have no utility term or actual per-flow rate to determine; set their aggregate service to zero. This is the inactive routes in proportional fairness convention.

Abbreviate the route populations as $n_1,n_2,n_3,n_4$ in cyclic order and let $z_r=n_r{x}_r$ denote aggregate route service. The resource constraints areEquivalently, $\max (z_1,z_3)+\max (z_2,z_4)\le 1$. Fix these two maxima as $u,v$. Because utility is increasing in each active allocation, every active odd route attains $u$ and every active even route attains $v$. Dropping terms constant in the populations, the objective becomesIf both groups are nonempty, the maximum has $u+v=1$ and derivative $P/u-Q/(1-u)=0$, giving $u=P/M,v=Q/M$. If one group is empty, every route in the other group gets aggregate service one. For $M>0$, all cases are summarized, on active routes, byAt $M=0$ there are no flows and all aggregate services are zero. In particular $n_1{x}_1=(n_1+n_3)/M$ whenever $n_1>0$, as printed in the PDF. The per-flow rate is not generally $1/M$: opposite routes can use the same resource capacities in parallel.

An exponential document size with parameter ${\mu }_r$ has completion hazard ${\mu }_r{x}_r$ when a flow is served at rate $x_r$. Summing over its $n_r$ flows gives the transition ratesThus an active odd route departs at rate ${\mu }_{r}P/M$, and an active even route at rate ${\mu }_{r}Q/M$. An empty route has departure rate zero even if its opposite route is populated.

Put ${\alpha }_r={\nu }_r/{\mu }_r$ andFor an active odd route and an active even route respectively,Hence aggregate service is the neighboring-state ratio of this balance function of a flow-level network. Consider weights $w(n)=\Phi (n){\prod }_r{\alpha }_r^{n_r}$. For every birth edge,These are detailed balance equations. The total departure rate is bounded by ${\sum }_r{\mu }_r$ and total arrival rate by ${\sum }_r{\nu }_r$, so the process is nonexplosive. Provided the normalizing sum is finite, its stationary distribution is thereforeThe large parentheses in the original PDF represent a binomial coefficient, not a ratio. The verified adjacent-state ratios explain why that distinction matters.

The required stability condition can be established directly from the normalizing sum. Set $a={\alpha }_1,c={\alpha }_3,b={\alpha }_2,d={\alpha }_4$, and $h_p(a,c)={\sum }_{i=0}^p{a}^i{c}^{p-i}$. Grouping opposite-route counts givesIf $A_{\ast }=\max (a,c)$ and $B_{\ast }=\max (b,d)$, then $A_{\ast }^p\le h_p\le (p+1)A_{\ast }^p$, with the same bounds for the other group. Restriction to the routes attaining the maxima bounds $B$ below by ${\sum }_m(A_{\ast }+B_{\ast }{)}^m$. If $A_{\ast }+B_{\ast }<1$, the upper bound is at most ${\sum }_m(m+1{)}^2(A_{\ast }+B_{\ast }{)}^m$, which converges. ThusEquivalently every resource's offered load is strictly below its unit capacity. With positive arrival rates the irreducible chain then has the unique stationary probability distribution above. Without this condition the weights remain an invariant measure but are not normalizable; the question's probability-distribution claim implicitly needs this hypothesis.

For completeness, the normalizing constant has the four-cycle partition function expressionTo derive it when $a\ne c$ and $b\ne d$, substitute $h_p=(a^{p+1}-c^{p+1})/(a-c)$ into the double sum and use ${\sum }_{p,q}\left(\genfrac{}{}{0ex}{}{p+q}{p}\right)u^p{v}^q=(1-u-v{)}^{-1}$. The resulting expression iswhich simplifies to the displayed formula. Continuity within the stability region covers equal opposite loads and zero loads.

Take a finite set of source-sink pairs $S$, with fixed demands $b_s\ge 0$, and finitely many permitted routes $R_s$ for each pair. Every positive demand must have at least one available route. Let ${\nu }_r$ be route flow and ${\rho }_j$ link throughput. The source-sink route incidence matrix and link-route incidence matrix arefor simple routes; if a route traverses a link repeatedly, its entry counts those traversals. Thus $H\nu =b$ conserves each demand and $A\nu =\rho$ aggregates link traffic.

At a Wardrop equilibrium, every route carrying positive flow has minimum delay among routes for the same source-sink pair. Ifthere are numbers ${\tau }_s$ such that $c_r(\rho )\ge {\tau }_s$ for $r\in R_s$, with equality whenever ${\nu }_r>0$. This is a nonatomic routing game: one infinitesimal user's route change does not alter the link costs it sees.

For continuous nondecreasing link delays, define the Beckmann potentialEach term is convex because its derivative is nondecreasing. The feasible route-flow set $\mathcal{F}=\{\nu \ge 0:H\nu =b\}$ is nonempty and compact: $0\le {\nu }_r\le b_s$ on each pair's routes. Hence the continuous function $V(A\nu )$ attains a minimum. Its gradient with respect to route flow isIf $r$ carries positive flow and $r^{\prime }$ serves the same pair, moving a small amount of flow from $r$ to $r^{\prime }$ is feasible. At a minimizer its directional derivative $c_{r^{\prime }}-c_r$ is nonnegative. Thus every used route minimizes delay, proving existence of a Wardrop equilibrium.

Conversely, at such an equilibrium, any feasible competitor $\tilde{\nu }$ satisfiesIndeed, the competitor's cost sum for pair $s$ is at least ${\tau }_s{b}_s$, while the current flow's cost sum equals ${\tau }_s{b}_s$. This first-order variational inequality is sufficient for optimality of a differentiable convex function. ThereforeThe argument proves both directions of the claimed optimization equivalence, including unused routes with costs tied at the minimum.

If each $D_j$ is strictly increasing, each integral is strictly convex, hence so is $V$ as a function of link throughputs. The feasible link-throughput set $A\mathcal{F}$ is convex. Two distinct minimizing vectors would have a feasible midpoint with strictly smaller potential, so the equilibrium link-throughput vector is unique. Strictly convex function in $\rho$ need not give strictly convex function in $\nu$, because $A$ can have a nontrivial kernel.

For an explicit route-flow ambiguity, put two parallel links in each of two consecutive stages, with one unit of demand traversing both stages. Order the four routes by the first and second link choices: $(1,1),(1,2),(2,1),(2,2)$. Set all four link delays to $D_j(z)=z$. Theninduces throughput $1/2$ on each link, and every route has delay one. Every member of this family is an equilibrium, though its route-flow vector changes. This is route-flow nonuniqueness at a Wardrop equilibrium with strictly increasing delays.

For Braess paradox, use vertices $s,u,v,t$ and one unit of $s$-to-$t$ demand. The delays are $D_{su}(z)=z$, $D_{ut}(z)=1$, $D_{sv}(z)=1$, and $D_{vt}(z)=z$. Initially the only routes are $sut$ and $svt$. If the upper route carries $p$, their costs are $1+p$ and $2-p$, so equilibrium has $p=1/2$ and common delay $3/2$.

Add the directed link $u\to v$ with delay zero. Let the upper, lower and new middle routes carry $p,q,c$, where $p+q+c=1$. Their costs becomeIf $p>0$, the middle route is strictly cheaper than the upper route, contradicting equilibrium; likewise $q>0$ is impossible. Thus all flow uses the middle route, with $p=q=0,c=1$, and all three route costs are two. Therefore

The potential for the expanded network isThe old flow remains feasible and has potential $5/4$; the new equilibrium has potential $1$. The enlarged feasible set has indeed lowered the minimized potential. But total travel delay is ${\sum }_j{\rho }_j{D}_j({\rho }_j)$, not this integral potential: it rises from $3/2$ to $2$. This explains the paradox within the optimization framework. The optimal total-delay value cannot worsen when the old flow is still allowed; selfish equilibrium minimizes a different function.

The phenomenon also survives strictly increasing delays on every link. Replace each constant-one delay by $1+\epsilon z$ and the new-link delay by $\epsilon z$. Symmetry gives upper and lower flows $p=\epsilon /(1+3\epsilon )$ after expansion, with middle flow $(1+\epsilon )/(1+3\epsilon )$. The common new delay is $2-\epsilon (1-\epsilon )/(1+3\epsilon )$, while the old delay was $3/2+\epsilon /2$. For $\epsilon =1/10$, these are $251/130>31/20$. Thus the example does not depend on interpreting increasing delays as merely nondecreasing.

Use one particle for each telephone line, including unused lines. Let $f$ be the number of free lines, $h$ the number of callers waiting for or being checked by an operator, and $a$ the number using automated service. Then $f+h+a=N$. These are the three colonies of a closed migration process, with directed cycle $f\to h\to a\to f$ and departure ratesRejected attempted calls leave the state unchanged. They must not be represented as extra customers in the waiting stage.

Equivalently the state is $(h,a)$ with $h+a\le N$. Its transitions areThe directed cycle has equal traffic weights; choosing each weight to be $\nu$ gives the unnormalized product-form stationary distribution of a closed migration processWe can verify global balance for a continuous-time Markov chain directly. Weighted incoming transitions from $(h-1,a)$, $(h+1,a-1)$ and $(h,a+1)$ contribute, respectively,Their sum equals the weight times the outgoing rate. This proves stationarity after normalization, without falsely asserting reversibility of the directed migration cycle.

For $n=h+a$ occupied lines, summing over the automated population givesThe normalizing constant is $G_N={\sum }_{n=0}^{N}H(n)$. The attempted calls form an independent Poisson process, so Poisson arrivals see time averages: the rate of attempted arrivals seeing a full state is $\nu$ times its stationary probability, and division by the total attempted-arrival rate gives the loss proportion. ConsequentlyThe numerator is essential; the PDF contains it even though the converted TeX loses it.

With two independent operators, the total check-completion rate is $\lambda \min (h,2)$. In particular it is only $\lambda$ when one caller is present. DefineThe human-stage weight becomes $(\nu /\lambda {)}^h/D_2(h)$, while the free and automated weights are unchanged. The same global-balance calculation applies using these departure rates. ThusThis is the blocking probability in a finite-line sequential-service network with two operators. It remains valid for $N=0$, when every attempted call is lost; for $N=1$ it coincides with the single-operator answer.

Let $(A_t,B_t)$ be a Markov jump process on integer populations, with rateswhen $a,b>0$, where $\kappa >0$. Make both axes absorbing, so the fight ends when one population reaches zero. Start at $(N,b_N)$, with $b_N=\lfloor \alpha N\rfloor$; when $\alpha N$ is integral this is the printed initial condition exactly. The harmless rounding convention makes the model defined for every integer $N$.

Use the embedded jump chain $(A_j,B_j)$ rather than physical time. Its probabilities of the two losses are respectively $b/(a+b)$ and $a/(a+b)$, independently of $\kappa$. It reaches an axis after at most $N+b_N\le 2N$ steps; hold it constant thereafter. PutBefore absorption its two increments are $-2a+1$ and $2b-1$, soAfter absorption set ${\mu }_j=0$. Consequentlyis a discrete-time martingale. All populations are at most $N$, so each martingale difference is bounded in absolute value by $2N+2\le 4N$. The total drift up to step $2N$ is at most $2N$. This is the squared-population near-invariant for stochastic attrition.

The two-sided Azuma-Hoeffding inequality states that a martingale with $m$ differences bounded by $C$ satisfies $\mathbb{P}(|M_m|\ge z)\le 2\exp (-z^2/(2m{C}^2))$. Its exponential estimate follows by applying the conditional Hoeffding lemma to each centered difference, iterating, and optimizing the exponential Markov bound. With $m=2N$ and $C=4N$ it gives

Let $c=\sqrt{1-{\alpha }^2}>0$, and let $X^N=A_{2N}/N$ be the final proportion. If the first population survives, then the second is zero and $D_{2N}/N^2=(X^N{)}^2$. Hence $|X^N-c|>\delta$ impliesIf the first population is extinguished, $X^N=0$ and $D_{2N}\le 0$. This can satisfy the indicated deviation event only when $\delta <c$, and then $|D_{2N}-c^2{N}^2|\ge c^2{N}^2>c\delta N^2$. Thus the same implication holds in both cases.

The initial rounding error obeys $|D_0-c^2{N}^2|=|{\alpha }^2{N}^2-b_N^2|\le 2N$. Combining it with the drift bound shows that the deviation event forcesfor all sufficiently large $N$. Set $\eta =c\delta /2$ in the preceding concentration inequality. We concludefor sufficiently large $N$, and thereforeThe symmetric stochastic attrition process thus has exponential concentration of terminal stochastic attrition survivors. The deterministic equations conserve $a^2-b^2$ and predict the same surviving fraction, but the martingale argument supplies the stronger exponential probability estimate required here.

The original PDF specifies a bounded solution class $C_b^2$, rather than the $C_0^2$ transcribed in the TeX. Write $\varphi (x)=\min (1,\max (0,-x))$ and fix a time $t>0$. For Brownian motion starting at $x$, considerThe Itô formula and ordinary product rule give a drift proportional to $-u_t+\frac{1}{2}{u}_{xx}-\lambda \varphi u$, which vanishes by the equation. Therefore $F$ is a local martingale. It is bounded on this horizon because $u$ is bounded and $\varphi \ge 0$, so the bounded local martingale criterion makes it a true martingale. Its endpoint values are $F_0=u(t,x)$ and $F_t=e^{-\lambda {\int }_0^t\varphi (B_r)dr}$, using $u(0,\cdot )=1$. ConsequentlyThis proves the required Feynman-Kac formula with a bounded potential directly, and also gives uniqueness within the stated bounded solution class.

Let $I={\int }_0^t\varphi (B_s)ds$. By dominated convergence, $e^{-\lambda I}\to 1_{\{I=0\}}$ in expectation. Continuity of $B$ and $\varphi$ shows that $I=0$ exactly when $B_s\ge 0$ throughout $[0,t]$: a negative value at any time produces a nontrivial interval with positive integrand. For $x<0$ such an interval occurs immediately, so the limiting probability is zero. For $x>0$, the Brownian reflection principle and the atom-free distribution of the running minimum giveFor $x=0$ this probability is zero: the reflection formula for levels $-\epsilon$, followed by $\epsilon \downarrow 0$, shows that Brownian motion goes below zero before $t$ almost surely. ThusThis is the soft killing limit for Brownian nonnegative survival. The assertion is pointwise for positive time, as requested; no claim of uniform convergence across the initial-time boundary is needed.

For the first equation use an integrating factor. The Itô product rule gives $d(e^{\lambda t}{X}_t)=\sigma e^{\lambda t}d{B}_t$, soThe Gaussianity of deterministic Brownian stochastic integrals follows by approximating its deterministic integrand by step functions and using the Itô isometry. HenceThe variance is positive also when $\lambda <0$, since numerator and denominator then both have negative sign. This is the explicit Ornstein-Uhlenbeck solution with initial value one.

For the second equation, take independent standard Brownian motions $U,V$ and define $R_t=\sqrt{(1+U_t{)}^2+V_t^2}$. The associated planar Brownian motion does not hit the origin. To check this rather than silently differentiate at zero, stop its radius upon hitting $\epsilon$ or $L$, with $0<\epsilon <1<L$. The function $\log r$ is harmonic on that annulus, so the optional stopping theorem givesLet $\epsilon \downarrow 0$: hitting zero before $T_L$ has probability zero. Taking integer $L\to \infty$ covers every possible finite hitting time, since continuous paths are bounded on compact intervals. Thus planar Brownian motion avoids a fixed point and $R$ stays positive.

The radial Itô formula is now legitimate and givesThe sum of the two martingale terms has bracket $t$ and starts at zero, so it is a standard Brownian motion $\beta$ by Lévy characterization of Brownian motion. Thus $dR=d\beta +(2R{)}^{-1}dt$ with $R_0=1$. The law-determination assumption permitted in the question identifies the law of $Y$ with that of this dimension-two Bessel process. For independent standard normal random variables $G_1,G_2$,This is the two-dimensional Bessel transition law and answers the requested distribution without needing a special-function density.