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The expected value and variance under quota share reinsurance follow by scaling the exponential distribution:The retained stop loss moments for an exponential aggregate follow from the payout $Y=\min (S,M)$, its survival function equals $e^{-x/{\mu }_S}$ for $0\le x<M$ and zero for $x\ge M$. The tail integral formula for moments gives, with $m=M/{\mu }_S$ and $r=e^{-m}$,Consequently the retained moments areMatching the two expected values forces $\alpha =1-r$, which lies strictly between zero and one. The difference of the variances simplifies toIndeed $h(m)=m-1+e^{-m}$ has $h(0)=0$ and $h^{\prime }(m)=1-e^{-m}\ge 0$ for $m\ge 0$. Because $M>0$, the difference is actually positive. At equal retained expected value, aggregate stop loss reinsurance reduces the variance more than quota share reinsurance.

An exact observed event contributes its event-time density:For exponential-rate estimation from censored exposure, the event-model survival likelihood under independent censoring is ${\prod }_{i}f(x_i{)}^{v_i}F(x_i{)}^{1-v_i}$. For the exponential distribution, put $d={\sum }_i{v}_i$ and $X={\sum }_i{x}_i$. ThenIf $d>0$ and $X>0$, ${\ell }^{\prime }(\theta )=d/\theta -X$ vanishes atThe second derivative $-d/{\theta }^2$ is strictly negative, so this is the unique global maximum-likelihood estimate. The likelihood also tends to zero at both ends of the positive parameter range. If $d=0$ and $X>0$, the likelihood decreases on $\theta >0$; its supremum is at $\theta \downarrow 0$, with no attained positive estimate. Thus every censored time belongs in the exposure denominator.

Condition on the bulb being alive at age $\tau$. The next hour's survival likelihood contribution isIt does not depend on the unknown age $\tau$. This is memorylessness of the exponential distribution: the residual lifetime given survival has the same exponential distribution as a new lifetime.

For the Erlang loss formula, assume a Poisson process of arrivals with rate $\lambda$, independent holding times with exponential distribution of mean $1/\beta$, unit capacity usage per accepted call, and rejection without waiting when all $C$ units are occupied. Here $C$ is a positive integer and the offered load is $a=\lambda /\beta$. The call population on $\{0,\dots ,C\}$ is a birth-death process; detailed balance givesBy Poisson arrivals see time averages, the fraction of arrivals rejected is the time probability of full occupancy. ThereforeThe paper's $\nu$ is the offered load, rather than necessarily an arrival rate. Exponential holding times suffice for this derivation; no unproved insensitivity of loss networks assumption is needed.

A loss network with fixed routing has capacities $C_j$, call types $r$, and a link-route incidence matrix $A$ with $A_{jr}=1$ exactly when a type-$r$ call requires link $j$. An accepted call holds one unit on every link of its route for its holding time; if any required link is full, the call is rejected everywhere. Take independent Poisson processes of calls, independent holding times, and route offered loads ${\nu }_r$. For the finite-variable formulation below, take $C_j\ge 1$; a zero-capacity link and all routes using it may be removed first.

The Erlang fixed point approximation treats blocking at different links as independent and thins the traffic offered to link $j$ by acceptance at the other links. Its reduced offered load isThe equality uses $A_{jr}\in \{0,1\}$: only routes with $A_{jr}=1$ contribute. Treating this reduced stream as Poisson leads to $B_j=E(a_j(B),C_j)$. Independence and Poisson thinning here are the approximation, not exact properties of a general loss network.

To construct the convex potential for the Erlang fixed point, write $e_C(a)=E(a,C)$ and $m_C(a)=a[1-e_C(a)]$, the carried load of an Erlang loss resource, equivalently the mean occupancy in the single-link stationary distribution. Differentiating its finite sums gives, for $a>0$,The variance is positive because the finite occupancy law is nondegenerate for $a>0$. Furthermore $e_C$ maps $[0,\infty )$ increasingly onto $[0,1)$, while $m_C$ increases from zero to $C$. SetIn particular $U(0,C)=0$, $U(\cdot ,C)$ is continuous and strictly increasing, and $U(z,C)\to C$ as $z\to \infty$.

Forthe first sum is convex, and each integrated strictly increasing function is strictly convex. Thus $F$ is a strictly convex function on $y\ge 0$. Since $U$ tends to a positive limit on every coordinate, $F$ is a coercive function there; it has a unique minimizer. Its coordinate derivative isThe Erlang fixed point equation is exactly ${\partial }_{j}F=0$. For an unused link, $a_j=0$ and the optimum is $y_j=0$; for a used link with positive offered load, the derivative at zero is negative and the optimum is positive. Hence every fixed point gives the unique minimizer, and the minimizer gives a fixed point. The Erlang fixed point is unique under these fixed-routing assumptions.

A convergent cyclic substitution for the Erlang fixed point updates one coordinate at a time, repeatedly cycling through all links:Always use the most recently available values of the other coordinates. Start, for example, with $B=0$. In $y$ coordinates this is exact coordinate descent for $F$: with other coordinates fixed, the displayed update is the unique coordinate minimizer.

Here is a convergence proof. All iterates stay in the initial compact sublevel set of the coercive function $F$, and their potential values decrease to a limit. Each single-coordinate update is continuous, since its reduced load is continuous and finite and $E(a,C)<1$. Thus a whole-sweep map $T$ is continuous. If a full-sweep subsequence converges to $z$, continuity gives $F(Tz)=F(z)$, because both consecutive potential values converge to the same limit. Strict coordinate convexity means equality can occur only when no coordinate changes during that sweep. Thus $z$ minimizes every coordinate, satisfies the nonnegative-orthant first-order conditions, and is the unique global minimizer of $F$. Every subsequential limit is therefore the same point, proving convergence of the full-sweep iterates and, by continuity of the updates, all intermediate iterates. This argument applies to sequential substitution; it does not assume simultaneous updates converge.

A feasible per-flow allocation satisfies $x_r>0$ for each active route and ${\sum }_r{A}_{jr}{n}_r{x}_r\le C_j$. It has proportional fairness if every feasible competitor obeysBy the first-order inequality for the strictly concave logarithm, this is equivalent to maximizing weighted logarithmic utility ${\sum }_{r:n_r>0}{n}_r\log x_r$ over the feasible allocations. Inactive-route rates are immaterial; set them to zero as a convention.

Write $a=n_{\{1\}}$, $b=n_{\{2\}}$, $c=n_{\{1,2\}}$, and $N=a+b+c$. The capacity constraints are $a{x}_1+c{x}_0\le 1$ and $b{x}_2+c{x}_0\le 1$. For $c>0$, put $q=c{x}_0$. Each active local route uses its remaining capacity $1-q$. Omitting constants, the utility isIts maximum has $q=c/N$; when $a+b=0$, the maximum is the endpoint $q=1$. Hence for active routesThese formulas also give $x_1=1/a$ and $x_2=1/b$ when $c=0$. When $N=0$ there is no allocated service. Equivalently, the aggregate route rates for $N>0$ areThis is the proportionally fair allocation on a two-resource linear network for a two-resource linear flow network.

For independent Poisson processes of documents and independent document sizes with exponential distribution of rate ${\mu }_r$, the memorylessness of the exponential distribution makes the population a continuous-time Markov chain. Its transition intensities areThe departure intensity sums the $n_r$ independent completion hazards ${\mu }_r{x}_r$. There is no departure on an inactive route. Set ${\rho }_r={\nu }_r/{\mu }_r$, with ${\mu }_r>0$.

To prove the proposed stationary distribution, define the balance function of a flow-level networkFor an active local route, $\Psi (n-e_r)/\Psi (n)=(a+b)/N$, while for the through route this ratio is $c/N$. Thus ${\Lambda }_r(n)=\Psi (n-e_r)/\Psi (n)$. The weightssatisfy detailed balance on every arrival-departure pair:Each total jump rate is bounded by ${\sum }_r({\nu }_r+{\mu }_r)$, so the chain is a nonexplosive Markov chain; normalized detailed-balance weights therefore give its stationary distribution, the stationary law of a two-resource linear flow network.

Use the negative binomial series to sum first over $c$:The remaining two geometric series giveAll summands are nonnegative. Consequently the sum is finite exactly under the stability conditions for a two-resource linear flow networkThese say that the offered load on each unit-capacity resource is strictly below one. Equality is insufficient: at least one geometric series then diverges. If all arrival rates are positive, the chain is irreducible and the normalizable law is its unique equilibrium law.

Finally put $q_i={\rho }_i/(1-{\rho }_0)$, $i=1,2$. Summing over $c$ yieldsThus the two local-route counts are independent, each with geometric distribution on the nonnegative integers. The conditional through-route count has negative binomial distributionFor positive traffic rates this depends on the local counts, so all three route counts are not independent. More quantitatively, $\mathrm{Cov}(a,c)={\rho }_0\mathrm{Var}(a)/(1-{\rho }_0)>0$, and likewise for $b$. If zero arrival rates are allowed, the exceptional independent cases are ${\rho }_0=0$, or ${\rho }_1={\rho }_2=0$, when the relevant counts are deterministic zero. This distinction is local-count independence with through-flow dependence.

Starting in state $1$, the onset waiting time has the exponential distribution of rate $\lambda$. Thus the probability of onset before the next screen is $1-e^{-\lambda t}$. This counts individuals who have already progressed to clinical disease as well as those still pre-clinical.

Conditional on onset at time $u<t$, the Markov property makes the remaining progression time exponential with rate $\nu$. Thus the probability of clinical progression before the screen, conditional on that onset time, is $1-e^{-\nu (t-u)}$.

Integrating over the onset density gives the unconditional clinical probabilityThe probability of being pre-clinical at the next screen is insteadTogether with $P_{11}(t)=e^{-\lambda t}$ and the exponential distribution for a patient initially in state $2$, the transition matrix isThis equals the matrix exponential $e^{Qt}$. The assumed unequal intensities avoid a zero denominator; at $\lambda =\nu$ the continuous limit is $P_{12}(t)=\lambda t{e}^{-\lambda t}$.

For the stochastic SIR model, the state $(S,I,R)$ is a continuous-time Markov chain on nonnegative integer triples summing to $N+1$, initially $(N,1,0)$. Its only transitions areRates vanish when a proposed transition would leave the state space. Equivalently, with independent unit-rate Poisson processes $A,B$, the Poisson time-change representation of a Markov chain isThe left limits describe the state immediately before each jump. In the Gillespie algorithm, wait an exponential distribution time with rate $\beta si+\gamma i$, then choose infection or recovery in proportion to these two rates.

Put $U=H(T)$ and $V=H(C)$, so $Y=\min (U,V)$ and $0\le \mathbb{E}Y\le \mathbb{E}U=1$. Its expectation is not generally one and depends on censoring. Under independent censoring, condition on $V=v$: the exponential distribution givesWriting $\Delta =1_{T\le C}$ for the observed event indicator, it follows that $\mathbb{E}Y=\mathbb{E}\Delta$. The exponential mean imputation under independent censoring therefore gives a modified Cox–Snell residual with known mean:On censoring, the added one is the expected remaining transformed lifetime, by memorylessness of the exponential distribution. This gives a mean-one variable, not necessarily another exponential variable. Equivalently, $M=\Delta -Y=1-Y^{\ast }$ is a martingale residual with mean zero. Independent censoring, or its appropriate conditional version, is essential for these identities; arbitrary informative censoring does not justify the correction.

Fit a model using the relevant covariates, calculate ${\hat{M}}_i=v_i-{\hat{H}}_i(x_i)$, and plot these martingale residuals against a candidate explanatory variable or against an included variable's value. A smooth systematic trend away from zero can indicate an omitted effect or an unsuitable functional form, such as a nonlinear age effect. After fitting an appropriate effect, the residual trend should diminish. The equivalent plot of ${\hat{Y}}_i^{\ast }$ has mean-one reference rather than zero.

Choose an individual uniformly from the population. The law of total probability gives the population density and survivor function as mixtures:A hazard function is a ratio of density to survival, so averaging the densities does not average those ratios. Instead,These are the proportions of the original individual types among those still event-free at $t$, rather than their equal starting proportions. For example, a mixture of different exponential distributions progressively favours the lower-rate individuals. This population hazard of a survival mixture generally differs from $n^{-1}{\sum }_i{h}_i(t)$.

Use the allowed ratio-of-expectations approximation where $G(u)>0$. The unconditional numerator and denominator from part (b) giveThe common censoring survivor function cancels in this approximation. By differentiating the logarithm, its value is the population integrated hazard:This is not generally the arithmetic mean of the $H_i(t)$; it is the cumulative hazard function of the population mixture from part (a). The cancellation requires a common independent censoring law and adequate observation of the time range.

The approximation establishes approximate unbiasedness, not exact finite-sample unbiasedness. For a direct counterexample, take one uncensored individual with exponential distribution of rate $\lambda$. The Nelson–Aalen estimator is then $1_{T\le t}$, whose expectation is $1-e^{-\lambda t}$, whereas $\overline{H}(t)=\lambda t$. Thus it is not an unbiased estimator in general.

More explicitly, for independent individuals with a common event hazard $h$, the exact expected estimator isThis exhibits finite-sample bias of Nelson–Aalen estimation and dependence on $G$ when empty risk sets are possible. In a heterogeneous population, replacing the random ratios by ratios of their means introduces an additional approximation. On adequately observed intervals with large risk sets, the population-mixture expression is the appropriate large-sample target.

First work with the filtration ${\mathcal{G}}_t={\mathcal{F}}_t^W\vee \sigma (1_{\{u\le \tau \}}:0\le u\le t)$, which records the Brownian motion and whether default has occurred. On $\{s>\tau \}$ all subsequent defaultable stock prices vanish. On $\{s\le \tau \}$, the memoryless property of the exponential distribution gives survival from $s$ to $t$ with conditional probability $e^{-\lambda (t-s)}$. The residual survival event is independent of the future independent increments of the Brownian motion. Thus, for deterministic $s<t$,Also $\mathbb{E}{\hat{S}}_t=e^{\lambda t}{e}^{-\lambda t}{S}_0=S_0$, so this is a true martingale, with no localization needed. Its natural filtration is contained in $\mathcal{G}$; applying the tower property of conditional expectation to the displayed identity yields the requested natural-filtration martingale property:The printed convention keeps the stock alive at $t=\tau$. It is a left-continuous convention at that one random time. Replacing it by $1_{\{t<\tau \}}$ gives the usual càdlàg zero-recovery default model; since $\mathbb{P}(\tau =t)=0$ at each fixed $t$, the fixed-time martingale calculations above are unchanged.

In a loss network with fixed routing, calls of type $r$ arrive as independent Poisson processes with rates ${\nu }_r$. Every accepted call occupies $A_{jr}$ units of resource $j$ throughout its holding time. The link-route incidence matrix $A$ has nonnegative integer entries, $C$ is the resource-capacity vector, $n_r$ counts calls currently present, and $e_r$ is the $r$th coordinate unit vector. The feasible occupancy space isThe inequalities are coordinatewise. A call that would leave this space is rejected immediately, without queuing or changing $n$. With independent unit-rate exponential distributions for holding times, the memoryless property makes $n$ a continuous-time Markov chain, whose nonzero off-diagonal transition rates areThere are $n_r$ independent clocks of rate one for the second transition. Rejected arrivals are not jumps and do not appear in the off-diagonal transition rates.

Let $a(n)={\prod }_r{\nu }_r^{n_r}/n_r!$. For every feasible adjacent pair,Thus detailed balance for a continuous-time Markov chain gives the product-form stationary distribution of a loss network:In this convention $G(C)$ is the reciprocal normalizing sum. If every route consumes some resource, finite capacities make $S(C)$ finite. Even zero-resource routes cause no normalization difficulty: the sum is at most $\exp ({\sum }_r{\nu }_r)$. With positive arrival rates the reachable feasible class is an irreducible Markov chain, since calls can depart to the empty state and any feasible occupancy can be built from it. The stationary distribution is therefore unique on that class.

The vector $m=C-An$ records the free units of each resource. Its probability distribution ${\pi }^{\prime }(m)$ sums $\pi (n)$ over all occupancies with the same free resources. The map from $n$ to $m$ generally loses information: different call mixtures may consume exactly the same resources.

For ${\pi }^{\prime }(m)>0$, the conditional expectation can be computed by removing one call. The factorial identity $n_r\pi (n)={\nu }_r\pi (n-e_r)$ givesIndeed, $n\mapsto n-e_r$ is a bijection from the occupancies in this sum with $n_r>0$ to occupancies using $C-m-A{e}_r$ units. Conversely, adding one call to any such occupancy is feasible because its final resource use is $C-m$. Consequentlyand the conditional expectation is zero otherwise. Conditioning the identity ${\sum }_r{A}_{jr}{n}_r=C_j-m_j$ and multiplying by ${\pi }^{\prime }(m)$ now yieldsAlternatively, multiply the preceding unnormalized factorial identity by $A_{jr}$ and sum over $r$ directly; this proves the same formula even when ${\pi }^{\prime }(m)=0$, without dividing by a zero probability. This gives both the conditional expectation argument and the direct summation argument.

Use an ideal carrier-sense multiple access scheme on the interference graph. An active station finishes its transmission after an independent unit-rate exponential distribution waiting time. An inactive station has an independent attempt clock with an exponential distribution of rate $e^{{\theta }_r}$; an attempt succeeds only when all its neighbors are inactive. Attempts that are blocked do not change the state. Thus, on the independent sets represented by $S$, the continuous-time Markov chain has transition ratesThe empty independent set can be reached by successive deactivations, and any other independent set can be reached from it by successive activations. Hence this finite chain is an irreducible Markov chain. The weight $a_{\theta }(n)=e^{\theta \cdot n}$ satisfiesBy detailed balance for a continuous-time Markov chain, its unique stationary law isThis is a finite exponential family. Its natural parameter of an exponential family is $\theta$ and its mean parameter of an exponential family is the vector of active fractions.

The throughput region of an interference graph is exactly $\Lambda =\mathrm{conv}S$. To see the possibly less immediate inclusion, suppose a random feasible schedule has mean $v\ge \lambda$. Retain each of its active vertices $r$ independently with probability ${\lambda }_r/v_r$, interpreting zero coordinates as always deleted. A subset of an independent set remains an independent set; the thinned schedule has mean exactly $\lambda$. Thus $\lambda$ lies in the convex hull of $S$. The opposite inclusion follows from the definition using equality. Moreover $0,e_1,\dots ,e_R\in S$, so this convex hull has full dimension.

Now choose $\epsilon >0$ such that $\mu =\lambda +\epsilon 1$ remains in the interior of $\mathrm{conv}S$. Use interior moment matching in a finite exponential family. Consider the convex functionThere is a ball of radius $\delta >0$ around $\mu$ contained in $\mathrm{conv}S$. Maximizing a linear functional over the convex hull is the same as maximizing over its generating set, so, for every $\theta$,The last inequality uses the point $\mu +\delta \theta /\Vert \theta \Vert$ when $\theta \ne 0$. Hence $F$ is a coercive function and has a finite minimizer ${\theta }^{\ast }$. Differentiating the finite normalizing sum givesThis covariance matrix is a positive-definite matrix: a linear functional constant on all states of positive weight must be constant on $0,e_1,\dots ,e_R$, so its coefficients vanish. Thus $F$ is a strictly convex function; its minimizer is unique. At that minimizer the desired strict service margins areFor a backlogged station that transmits at unit speed whenever active, the long-run throughput is ${\mathbb{E}}_{\theta }[n_r]$, by the ergodic theorem for a finite continuous-time Markov chain. Therefore the ideal carrier-sense multiple access scheme can supply a strict service margin for every arrival vector in the interior of the throughput region of an interference graph. The local attempt rates can realize the whole interior capacity region in this sense. Boundary points may require parameters tending to infinity, and the argument gives no uniform delay or mixing-time bound near the boundary.

Use the rate convention for the exponential distribution. By convolution of independent random variables, the probability density function of $S=X+Y$ is, for $s\ge 0$,It is zero for $s<0$. The expression is nonnegative whichever rate is larger, and defines a two-stage hypoexponential distribution. As a check, its limit as $\mu \to \lambda$ is ${\lambda }^{2}s{e}^{-\lambda s}$, the shape-two gamma distribution.

For $M=\min \{X,Y\}$, independence gives the survival function $\mathbb{P}(M>s)=e^{-\lambda s}{e}^{-\mu s}$ for $s\ge 0$. Differentiating its complement givesThus the minimum has an exponential distribution whose rate is the sum of the rates.

The conditional density givesusing either integration by parts or the expected value of an exponential distribution. HenceThis is a finite value for almost every realization of $X$, but not an integrable random variable: $\mathbb{E}[1/X]={\int }_0^{\infty }{e}^{-x}/xdx=\infty$. Since $Y\ge 0$, its conditional expectation remains well-defined in the nonnegative, extended-expectation sense; the tower property of conditional expectation then gives $\mathbb{E}Y=\infty$.

Integrate the joint probability density over $y$ to obtain the marginal density of $X$:Thus $X$ has an exponential distribution of rate one. Dividing the joint probability density by this marginal density gives the conditional densityEquivalently, $Y\mid X=x$ has an exponential distribution of rate $x$. The value at $x=0$ can be assigned arbitrarily: it is a probability-zero conditioning value and the density ratio there is not meaningful.