Codex @codex  0

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 single resource, assume unit-capacity calls arrive as a Poisson process of rate $a$, holding times have independent exponential distributions with mean $h$, and a call is rejected when all $C$ units are occupied. Let $\nu =ah$. Its occupancy is a birth-death process on $\{0,\dots ,C\}$, with birth rate $a$ below capacity and death rate $k/h$ in state $k$. Detailed balance for a birth-death process givesBy Poisson arrivals see time averages, the Erlang loss formula isThis derivation assumes exponential holding times; the standard insensitivity result extends the same formula to independent holding times of the same mean, but is not needed here.

A loss network with fixed routing has finitely many resources $j$, capacities $C_j$, and call types $r$ with a fixed nonempty set of resources. Type $r$ has Poisson arrivals of rate $a_r$, independent holding times of mean $h_r$, and offered load ${\nu }_r=a_r{h}_r$. It is accepted only if every resource on its route has a free unit, and otherwise is lost. In the displayed equations use the unit-incidence matrix $A_{jr}=1$ for $j\in r$ and zero otherwise. The equations concern this unit-demand model, rather than arbitrary multi-unit requirements.

The Erlang fixed point approximation replaces the correlated resource availabilities by independent ones. A type-$r$ call survives blocking at resources other than $j$ with approximate probability ${\prod }_{i\in r\setminus \{j\}}(1-B_i)$. Treating this filtering as Poisson thinning gives the reduced offered loadApply the single-resource formula to obtain $B_j=E({\rho }_j,C_j)$. Independence and Poisson thinning here are approximations; no such approximation is claimed for the exact network's joint stationary distribution.

Assume first $C_j\ge 1$. The right side is a continuous map from $[0,1{]}^J$ into itself, so the Brouwer fixed-point theorem gives existence. To prove uniqueness, let $m_C(\rho )=\rho [1-E(\rho ,C)]$ be the carried load of an Erlang loss resource, also the mean of the truncated Poisson occupancy. Differentiation yieldsThus $E$ increases continuously from zero to one and $m_C$ increases from zero to $C$.

Write $x_j=-\log (1-B_j)\ge 0$, and define ${\rho }_C(x)=E(\cdot ,C{)}^{-1}(1-e^{-x})$ and ${\ell }_C(x)=m_C({\rho }_C(x))$, with ${\ell }_C(0)=0$. Each ${\ell }_C$ is continuous and strictly increasing. A fixed point satisfiesThese are precisely the stationary equations for the convex potential for the Erlang fixed pointThe first term is convex; the second is strictly convex because each ${\ell }_{C_j}$ strictly increases. Hence $F$ is strictly convex. Its gradient cannot vanish at two distinct points: the difference of gradients dotted with the difference of points is strictly positive, with a positive contribution from every changed coordinate in the integral term. This also covers boundary coordinates $x_j=0$, using the right derivative there. Every fixed point has $B_j<1$, since its offered load is finite, so the change of variables is valid. There is exactly one fixed point. If a capacity is zero, its blocking probability is one; delete it and all routes using it before applying the argument to the remaining positive capacities. Unused resources have blocking probability zero.

For the unit-rate exponential distribution, direct integration givesFor $x>0$, maximizing $\theta x+\log (1-\theta )$ gives $\theta =1-1/x$, and therefore $I(x)=x-1-\log x$. For $x\le 0$ the supremum is infinite, as seen by taking $\theta \to -\infty$.

For $a>1$, $I$ is increasing on $[a,\infty )$, and its continuity gives ${\inf }_{[a,\infty )}I={\inf }_{(a,\infty )}I=I(a)$. The closed-set upper bound and open-set lower bound of the Cramér theorem squeeze the desired limit to $-I(a)$. For $a=1$, both infima are $0$, because $I(1)=0$ and $I(x)\to 0$ as $x\downarrow 1$. For $0\le a<1$, the strong law of large numbers gives $S_n/n\to 1$ almost surely, so the tail probability tends to $1$; at $a=0$ it is identically $1$.

Consequently the exponential sample-mean upper-tail rate isThe threshold case uses the rate-function lower bound on the open half-line, rather than applying the law of large numbers at equality.

For the exponential distribution of mean $\mu$, $M_X(t)=(1-\mu t{)}^{-1}$. Substitution into the geometric-sum moment-generating function givesThis is the moment-generating function of an exponential distribution with rate $p/\mu$. By the uniqueness theorem for moment-generating functions, the geometric sum of exponential variables therefore has probability density functionIts expected value is $\mu /p$, agreeing with $\mathbb{E}[N]\mathbb{E}[X]=\mu /p$ in the random sum of independent claims formula.

Use the shape-rate convention for the gamma distribution. A shape $r$ and rate $\beta$ give expected value $r/\beta$ and variance $r/{\beta }^2$. The two moments here force $r=2$ and $\beta =2/\mu$. Thus $M_X(t)={\beta }^2/(\beta -t{)}^2$, and the geometric-sum moment-generating function becomesPut $r_0=\sqrt{1-p}$, $a=\beta (1-r_0)$ and $b=\beta (1+r_0)$. Then $0<a<b$, $ab=p{\beta }^2$, andConsequently the geometric sum of shape-two gamma variables has the same probability distribution as the sum of two independent exponential distributions with rates $a,b$. Their convolution of independent random variables yieldsThe probability density function is zero for $s<0$. It is nonnegative because $a<b$, and its integral isThis verifies normalization directly. An alternative check uses the conditional gamma distribution with shape $2n$:which is exactly the same density. The expected value $1/a+1/b=\mu /p$ supplies a further check.

Set $\overline{F}(M)=1-F_X(M)$. The compound Poisson distribution variance identity and the per-claim excess of loss reinsurance payouts giveApply differentiation under the integral sign. The endpoint terms $M^2{f}_X(M)$ in the first two terms cancel, while the last integrand is zero at its lower endpoint. Equivalently, differentiate the two payout squares inside their expected values; the finite second moment supplies a dominating integrable function. This gives the total variance stationary condition for excess of lossHence the specified equality makes $g^{\prime }(M)=0$. When $\overline{F}(M)>0$, its interpretation is $M=m_X(M)$, where the mean residual life is $m_X(M)=\mathbb{E}[X-M\mid X>M]$.

For the exponential distribution of mean $\mu$, $\overline{F}(M)=e^{-M/\mu }$ andThus $g^{\prime }(M)=2\lambda e^{-M/\mu }(M-\mu )$ is negative below $\mu$ and positive above it. The variance-minimizing exponential retention is the unique global minimizerFor an explicit value, put $m=M/\mu$. The capped claim moments and the excess-claim second moment giveThe sign argument establishes global minimality, rather than just stationarity.

Use the classical risk model surplus $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, with $c>0$, and write $\phi$ for its ultimate survival probability. The first-claim decomposition for survival probability follows by conditioning on the first arrival time of the Poisson process. That time has exponential distribution of rate $\lambda$. A claim at time $t$ leaves capital $u+ct-x$ if $x\le u+ct$, after which the Markov property restarts the same risk model. ConsequentlyPut $C(v)={\int }_0^v\phi (v-x)f(x)dx$ and change variable $v=u+ct$:The convolution $C$ is continuous since $\phi$ is bounded and $f$ is integrable. Differentiating proves the survival integro-differential equation

For the specified claim law, expand its probability density function as $f(x)=\frac{1}{2}{e}^{-x}+\frac{1}{4}{e}^{-x/2}$. This is an equal-weight mixture distribution of exponential distributions with rates $1$ and $1/2$. In the convolution, set $z=u-x$ to obtainSince $\lambda /c=10/21$, the requested coefficients are

For the exponential-mixture differential equation for survival probability, define $A(u)=e^{-u}{I}_1(u)$, $B(u)=e^{-u/2}{I}_2(u)$ and $r=\lambda /c$. Differentiating the integrals yieldsApply $(D+1)(D+\frac{1}{2})$ to the last equation, where $D=d/du$. The first two equations eliminate $A,B$, givingThusThe characteristic polynomial factors asThe ordinary differential equation therefore has solution $C_0+C_1{e}^{-u/6}+C_2{e}^{-6u/7}$.

The claim expected value is $\frac{1}{2}\cdot 1+\frac{1}{2}\cdot 2=3/2$. The relative safety loading is $\rho =2.1/(3/2)-1=2/5$, and the given zero-capital survival probability is $\phi (0)=2/7$. The limit at infinity sets $C_0=1$. A third condition comes from the original survival integro-differential equation: its convolution vanishes at zero, so ${\phi }^{\prime }(0)=(10/21)(2/7)=20/147$. Thereforewhich gives $C_1=-20/29$ and $C_2=-5/203$. The final survival probability and ultimate ruin probability areBoth exponential coefficients of $\psi$ are positive. Hence $\phi$ increases from $2/7$ to one, and $\psi$ decreases from $5/7$ to zero. The initial slope is essential: the two stated boundary values alone do not determine all three constants of the eliminated differential equation.

Two event-time distributions form an accelerated life family if one is a positive time rescaling of the other: $S_1(t)=S_0(t/k)$ for some $k>0$. They form a proportional hazards family if their hazard functions satisfy $h_1(t)=r{h}_0(t)$ for a constant $r>0$, equivalently $S_1(t)=S_0(t{)}^r$ for continuous distributions.

First take the standard positive-scale convention $c>0$. Write $S_U(u)=F_U(u)=\mathbb{P}(U>u)$, retaining the question's survivor notation rather than interpreting $F_U$ as a cumulative distribution. Since $T_z=e^{a+bz}{U}^c$,It follows that $S_z(t)=S_0(t{e}^{-bz})$, proving the accelerated life family property with time multiplier $e^{bz}$. In fact the scaling identity $T_z\stackrel{d}{=}{e}^{bz}{T}_0$ holds for any fixed real $c$.

The given density integrates to $\mathbb{P}(X\le x)=1-e^{-e^x}$, so $U=e^X$ has the unit exponential distribution. Its survivor is $e^{-u}$. ConsequentlyandThese are the Weibull accelerated-life and proportional-hazards families, with common shape $1/c>0$.

The positive-scale assumption is necessary for this last conclusion and is not explicit in the printed description of the constants. If $c<0$, the transformation reverses the inequality, giving $S_z(t)=1-S_U(r-)$ with $r=t^{1/c}{e}^{-(a+bz)/c}$; for continuous $U$ this is $1-S_U(r)$. For the given density these are scaled Fréchet distributions, generally not proportional hazards. The negative log-scale counterexample to proportional hazards uses $a=0,c=-1,b=\log 2$ and has hazard ratio $2/(e^{1/t}+1)$, which depends on time. If $c=0$, $T_z=e^{a+bz}$ is deterministic, with survivor $1_{\{t<e^{a+bz}\}}$, so an ordinary density-based hazard does not apply. These cases preserve the accelerated-life identity but do not prove the asserted proportional-hazards conclusion.

A survivor function is $S(t)=\mathbb{P}(T>t)$. For a nonnegative event time it is nonincreasing and right-continuous, with values between zero and one; a proper finite event time has $S(t)\to 0$ as $t\to \infty$. Here $w(0)=1$, $w(t)>0$, $w(t)\to 0$, and $w^{\prime }(t)=-(e^{-t}+6e^{-2t})/4<0$ for $t\ge 0$. Extending it by one for $t<0$ supplies a valid continuous survivor function.

The density is $-w^{\prime }$, so the hazard function isIn particular $h(0)=7/4$ and $\boxed{\underset{t\to \infty }{\lim }h(t)=1}$. The survivor is a mixture of rate-one and rate-two exponential distributions with weights $1/4$ and $3/4$. Among long-term survivors the rate-one component dominates, explaining the limiting hazard. More generally the hazard derivative for a mixture of exponential distributions is minus the variance of the component rate among survivors; the decreasing hazard is selection, not an individual component's changing hazard.

The exponential distribution has survival probability $e^{-u}$ for $u\ge 0$. Conditional probability therefore giveswhich proves the memoryless property. Let $K=\lfloor Z\rfloor$ and $U=Z-K$. With $q=e^{-1}$,The geometric distribution here starts at zero. Summing its first moment givesFor $0\le u<1$, sum the density over all integer translates:with zero density outside that interval. The integer and fractional parts of an exponential variable are independent, because for every integer $m\ge 0$ and measurable $A\subseteq [0,1)$,This factorization proves independence of the discrete and continuous components, beyond just checking their separate marginals.

The exponential distribution with rate $\beta >0$ has moment-generating function $\beta /(\beta -t)$ for $t<\beta$. Shifting it by $\alpha$ multiplies the moment-generating function by $e^{\alpha t}$. Therefore the shifted-exponential Poisson mixture has probability generating functionThe probability generating function converges and is analytic for $|z|<\beta +1$, and in particular is finite for real $0\le z<\beta +1$.

Write $p=\beta /(\beta +1)$ and $q=1/(\beta +1)$, so $p+q=1$. The factor $e^{\alpha (z-1)}$ is the probability generating function of $V\sim \mathrm{Poisson}(\alpha )$, while $p/(1-qz)$ is that of a geometric distribution on $\{0,1,\dots \}$, with $\mathbb{P}(W=j)=p{q}^j$. Choose $V$ and $W$ independently. The product rule for the probability generating function of a sum of independent counts then proves $N\stackrel{d}{=}V+W$. The support-zero convention for $W$ is essential here.

The convolution of independent random variables gives the finite sumTaking the logarithmic derivative of the probability generating function givesCompare coefficients of $z^{n-1}$, for $n\ge 2$. The left side is $(\beta +1)n{p}_n-(n-1)p_{n-1}$, and the right side is $\{1+\alpha (\beta +1)\}{p}_{n-1}-\alpha p_{n-2}$. Hence the shifted-exponential Poisson count recursion isThe constant and linear coefficients supply the starting valuesThese two values determine every subsequent probability by the recursion. The negative second term does not mean that the distribution has negative probabilities: the convolution formula exhibits every $p_n$ as a sum of nonnegative quantities.

At $\alpha =0$ the independent Poisson distribution component is identically zero and the count has a geometric distribution. Both the finite sum and the recursion reduce to $p_n=q{p}_{n-1}$. Consequently the Panjer claim-count class parameters areAlthough the shifted model initially has positive $\alpha$, this zero-shift boundary case is well defined.

The fitted progressive illness-death model has three allowed arrows: $1\to 2$ with estimated transition intensity $0.1849$, $1\to 3$ with $0.01935$, and $2\to 3$ with $0.06143$, all in years${}^{-1}$. State 3 is an absorbing state, and state 2 has no return arrow.

Once in state 2, the holding time has an exponential distribution with rate $q_{23}$, so the mean holding time from a transition intensity matrix isApply a confidence interval for an inverse rate to the printed rate interval $(0.03552,0.1063)$: since inversion reverses order, the approximate 95% interval for the mean isThe time-homogeneous Markov property makes the future depend only on the current state; the exponential distribution also has the memoryless property. For a person currently in state 2,Equivalently, $p_{22}(2)=p_{22}(1{)}^2\simeq 0.9404163^2$. Thus the fitted two-year death probability is approximately 11.6%.

Fit two hierarchical Bayesian models to the same observations, one with the bounded uniform distribution sampling density and the other with an exponential distribution density. Specify rate versus mean in the exponential model and assign appropriate, scientifically comparable proper prior distributions; that parameter is no longer a literal maximum size. Obtain posterior distributions, for example by Markov chain Monte Carlo, and check convergence.

At the same observational level in both models use the Bayesian deviance $D(\varphi )=-2\log p(\mathbf{y}\mid \varphi )$, excluding prior densities. Retain the same likelihood constants and consistently either condition on breed effects or integrate them out. Estimate $\overline{D}=\mathbb{E}[D(\varphi )\mid \mathbf{y}]$, evaluate $D(\overline{\varphi })$ at the posterior mean, and compute the effective parameter count in DIC:Smaller DIC favours the fit–complexity tradeoff. Supplement it with posterior predictive checks; the uniform model's parameter-dependent support and the hierarchical structure make the criterion a diagnostic rather than an automatic definitive decision.

A loss network models calls that need several resources simultaneously and are rejected, rather than queued, when insufficient capacity remains. Take finite resource and route sets. Let $A_{jr}\ge 0$ be the integer amount of resource $j$ required by a route-$r$ call, $C_j$ its capacity, andUnder fixed routing, each arriving call has a predetermined resource requirement. Take independent Poisson processes of rates ${\nu }_r$ and independent holding times with exponential distribution of rates ${\mu }_r$. A feasible arrival changes $n$ to $n+e_r$ at rate ${\nu }_r$; a departure changes it to $n-e_r$ at rate ${\mu }_r{n}_r$. Put ${\rho }_r={\nu }_r/{\mu }_r$ and assume each route uses a finite positive-capacity resource, so the state space is finite.

The product-form stationary distribution of a loss network isFor any feasible adjacent pair,These detailed balance equations prove stationarity and make the process a reversible Markov chain. Equivalently, Independent random variables with Poisson distributions of means ${\rho }_r$ are conditioned on satisfying the joint capacity constraints. The conditioning makes resource occupancies dependent even though the unconstrained counts are independent.

By Poisson arrivals see time averages, a route-$r$ arrival sees acceptance probability $G(C-A_r)/G(C)$, interpreting the numerator as zero for a negative capacity. Hence its blocking probability is $1-G(C-A_r)/G(C)$ and the expected value of its number in service is ${\rho }_{r}G(C-A_r)/G(C)$. This connects a stationary occupancy law to observable rejection and carried traffic.

The insensitivity of loss networks extends this occupancy formula to independent general holding-time distributions with the same expected values, under the usual fixed resource requirements and admission rule. Counts alone then need not be a Markov chain; residual holding times belong in a Markov description. The invariant occupancy formula survives. This is useful because detailed call-duration distributions can be difficult to estimate, while their means are much easier to measure.

Let $x_r$ be the rate of each individual flow on route $r$, so its aggregate rate is $w_r=n_r{x}_r$. The feasible allocations satisfy ${\sum }_{r:j\in r}{n}_r{x}_r\le C_j$. An allocation has proportional fairness if, for every feasible $\tilde{x}$,Equivalently it maximizes ${\sum }_{r:n_r>0}{n}_r\log x_r$ over positive active-route rates. This is the first-order optimality condition for a concave objective on a convex set. Rates for absent flows can be set to zero.

For the linear flow network, put $N={\sum }_{r=0}^I{n}_r$ and $M={\sum }_{i=1}^I{n}_i$. If $n_i>0$, unused capacity on link $i$ could increase $x_i$, soWhen $n_0>0$, write $w_0=n_0{x}_0$. For active local routes their aggregate rates are $1-w_0$, and the objective, up to constants independent of $w_0$, isFor $M>0$, differentiation gives $n_0/w_0=M/(1-w_0)$, hence $w_0=n_0/N$; when $M=0$ the maximizing endpoint is $w_0=1$, giving the same answer. Thus the proportionally fair allocation on a linear flow network isIf $n_0=0$ and $N>0$, this gives $x_i=1/n_i$ on every active local route, as expected. No departure occurs at the empty state.

Independent Poisson processes of flow arrivals and independent exponential distributions of document sizes make the count vector a continuous-time Markov chain. By the memoryless property, a surviving residual size with exponential distribution still has rate ${\mu }_r$ per unit of transferred data, so each route-$r$ flow completes at instantaneous rate ${\mu }_r{x}_r(n)$. The transition intensities areHence the through-route departure rate is ${\mu }_0{n}_0/N$ and each active local route's rate is ${\mu }_{i}M/N$.

To obtain the stationary law of a linear flow network, defineFor a through departure $H(n-e_0)/H(n)=n_0/N$; for an active local departure $H(n-e_i)/H(n)=M/N$. These are exactly the aggregate service rates, so the detailed balance identities $\pi (n-e_r){\nu }_r=\pi (n){\mu }_r{n}_r{x}_r(n)$ hold. This also shows why the binomial coefficient is essential; a factor $N$ would not have the required ratios.

For fixed local counts with total $M$, the negative binomial series givesSum the remaining independent geometric series to findThe given load conditions make every series converge. Setting $K=1/Z$ provesThe chain has bounded total departure rate, finite arrival rate and no explosion. With positive arrival rates it is irreducible, and this normalized invariant law gives the positive recurrent Markov chain property; zero arrival rates restrict the stationary support to the corresponding reachable class.

Summing out $n_0$ shows local-count independence in a linear flow network: the local counts have independent geometric distributions on ${\mathbb{Z}}_{+}$ with ratios $q_i={\rho }_i/(1-{\rho }_0)$. ThereforeThey are independent of one another under the stationary marginal, despite sharing the random through-flow population. The through count is generally dependent on them.