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The first departure from the no-symptoms state has an exponential distribution with rate $\lambda$. Spending the entire interval there means no departure at all, soThis is the holding time survival probability. The transition probability $p_{11}(t)$ only says that the patient is symptom-free at the endpoint, allowing an onset and recovery in between, and is therefore not the answer. The formula also covers $\lambda =0$, when state 1 is absorbing.

For a treated patient the recovery transition intensity is ${\mu }_T=0.25\times 2=0.5$ per year. A symptoms spell is a holding time with an exponential distribution of rate ${\mu }_T$, soThe treatment multiplier for onset affects how often symptoms begin, not the duration of a spell once the patient is in state 2, under this homogeneous continuous-time Markov chain.

Put $a=y_4$ and $b=y_2+y_3$. Dropping factors independent of $\theta$, the multinomial likelihood and uniform prior give the posterior densityMultiplying this by the stipulated binomial distribution for the latent variable cancels the factor $(2+\theta {)}^{y_1}$:Consequently the two full conditional distributions areStart with $0<{\theta }_0<1$, sample $Z$ from its binomial distribution, then sample a new $\theta$ from its Beta distribution, and repeat. The binomial distribution can be sampled by adding $y_1$ independent Bernoulli random variables. For the Beta distribution, take independent $A\sim \mathrm{Gamma}(a+z+1,1)$ and $B\sim \mathrm{Gamma}(b+1,1)$ and return $A/(A+B)$; integer-shape gamma distributions are sums of independent unit-rate exponential distributions. Both shapes are positive even when some observed counts vanish.

Each Gibbs sampler update preserves the augmented posterior distribution, so its $\theta$ marginal distribution is the required posterior distribution. The positive full conditional distributions on the interior allow exploration of the entire support. These iterates are generally dependent; they are not the independent exact draws constructed in the later parts.

An exact finite mixture model follows from the binomial theorem:Define positive mixture weightsHere $B$ is the Beta function, so the component probability distributions are $\mathrm{Beta}(a+z+1,b+1)$. All coefficients are computable from the observed integer counts.

Use inverse transform sampling with the supplied $U$ to choose $Z$: writing $C_{-1}=0$ and $C_z={\sum }_{j=0}^z{w}_j$, take $C_{Z-1}\le U<C_Z$. Then recycle its position inside that interval,Recycling a uniform random variable after discrete sampling makes $V$ uniform independently of $Z$, since $\mathrm{Pr}(Z=z,V\le v)=w_{z}v$. It is also independent of all supplied random variables with gamma distributions. Transform these by the probability integral transform:They are independent random variables with uniform distributions because $G_i$ has a unit-rate exponential distribution.

Conditional on $Z=z$, put $k=a+z+b+1$ and $r=a+z+1$. Form the $k$ values $V,W_1,\dots ,W_{k-1}$ and return their $r$th order statistic. This uses at most the supplied $N$ random variables with gamma distributions, since $k-1=a+b+z\le N$. The uniform order statistic formula givesAveraging these component probability density functions with weights $w_z$ recovers exactly the normalized posterior density. Unused inputs can be discarded. When $N=0$, $Z=0$, $k=r=1$, and the output is $U$, as required. Interval endpoints and ties have probability zero; any consistent convention there is harmless. Exactness here concerns the mathematical algorithm; finite-precision arithmetic can be stabilized with logarithmic mixture weights.

Under the frozen-age exponential distribution convention, the mean waiting time is the reciprocal of the outgoing transition intensity. The beliefs therefore specifyBoth rates double over ten years. ConsequentlyThese calibrate the age-specific constant-rate means; they are not exact mean ages of future events when age-dependent rates continue increasing throughout the waiting time.

For a literal continuously ageing interpretation, a rate $q{e}^{\beta u}$ gives a Gompertz distribution for the waiting time. For $\beta >0$, its mean isHere $E_1$ is the exponential integral. Solving $m(a,{\beta }_{12})=10$ and $m(a{e}^{10{\beta }_{12}},{\beta }_{12})=5$, and the analogous pair with means 2 and 1, instead giveswith rates and slopes in inverse years. These are an alternative continuous-age calibration, not the frozen-age answer boxed above.

The minimum exceeds $t$ exactly when both times exceed $t$. By independence,Differentiating its survivor function givesThe tail integral formula for moments then givesEquivalently, integration by parts in ${\int }_0^{R}t{f}_X(t)dt$ gives $-R{F}_A(R)F_B(R)+{\int }_0^R{F}_A(t)F_B(t)dt$. The assumed $t{F}_B(t)\to 0$ makes the boundary term vanish, since $0\le F_A\le 1$.

The stated tail condition does not itself guarantee a finite mean. For example, the proper survivor functionssatisfy $t{F}_B(t)\to 0$, but their product behaves as $1/[t\log t\log \log t]$, whose integral diverges. Thus the displayed expectation is valid as an extended nonnegative value; finiteness needs integrability of the product. The exponential distribution in the next part guarantees that finiteness.

For the exponential distribution with mean $\mu$, $f_A(t)=F_A(t)/\mu$ and $F_A(t)=e^{-t/\mu }$. ThereforeSince ties have probability zero, $\mathbb{P}(X=T_B)=1-\mathbb{E}X/\mu$. Taking expected values in the definition of $U$ givesThere is also an imputation interpretation. If A is observed first, $T_A=X$; if B censors A, memorylessness of the exponential distribution and independence imply that the remaining mean lifetime is $\mu$. Thus $U$ is the conditional expectation of $T_A$ given the observed minimum and censoring indicator. The tower property of conditional expectation gives the same equality. This does not say that $U$ and $T_A$ have the same distribution.

Consider a triangle of three links, each with capacity $C$. Calls for each pair of its three nodes arrive as independent Poisson processes at rate $\lambda$ and have independent holding times with an exponential distribution of mean one. Try the direct link first; if it is full, try the two-link path through the third node; if either alternative link is full, reject the call. This is alternative routing in a loss network.

In the symmetric Erlang fixed point approximation, let $B$ be each link's blocking probability. A given link receives direct offered traffic $\lambda$. Each of the other two call types contributes overflow traffic $\lambda B$, screened by the availability $1-B$ of the other link on its alternative path. ThusThe extra term expresses a feedback: blocking sends calls onto longer paths, which consume more total capacity and can create still more blocking.

Here is an explicit finite-capacity example, not just a limiting argument. Take $C=1000$ and $\lambda =950$, and let $F(B)=E(950[1+2B(1-B)],1000)-B$. Evaluating the Erlang loss formula givesThe displayed decimals are rounded, but the four signs can be checked exactly using the rational recursion $E(a,0)=1$, $E(a,k)=aE(a,k-1)/(k+aE(a,k-1))$. By the intermediate value theorem, there is a fixed point in each of $(0,0.01)$, $(0.01,0.1)$, and $(0.1,0.4)$. Numerically these three areThus alternative routing can produce multiple Erlang fixed points. This does not imply multiple stationary distributions for the exact finite continuous-time Markov chain, which is irreducible and has a unique stationary distribution; the multiplicity belongs to the approximation. The symmetric alternative-routing model is also discussed in Kelly's review of fixed point models of loss networks.

A loss network models calls that require several resources simultaneously and are rejected if any required capacity is unavailable; rejected calls do not queue. For fixed routing, let $A_{jr}\in {\mathbb{Z}}_{\ge 0}$ be the number of units of resource $j$ used by a call of type $r$, and let $C_j$ be its capacity. Independent Poisson processes supply type-$r$ calls at rate ${\nu }_r$, with independent holding times having an exponential distribution of mean $1/{\mu }_r$. Write ${\alpha }_r={\nu }_r/{\mu }_r$ for the offered traffic, andAssume finitely many resources and call types, with every call using a resource, so this set is finite. The occupancy continuous-time Markov chain has ratesIts stationary distribution isIndeed, for each feasible upward transition,which is detailed balance for a continuous-time Markov chain. Thus this is a reversible Markov chain. With positive arrival rates it is irreducible on $\mathcal{N}(C)$: departures reach the empty state, and any feasible state can be assembled by arrivals. Its stationary distribution is consequently unique.

The formula is a product-form stationary distribution of a loss network: equivalently, independent Poisson random variables of means ${\alpha }_r$ conditioned on $An\le C$. The conditioning couples the occupancies, so the resources are generally not independent. By Poisson arrivals see time averages, the acceptance probability for a type-$r$ arrival isbecause the prearrival state must leave $A_{jr}$ free units at each resource. Set the numerator to zero if its capacity vector has a negative entry. This exact formula is often expensive to evaluate, which motivates the Erlang fixed point approximation.

The same occupancy stationary distribution extends to independent general holding-time distributions with these means by insensitivity of loss networks; the exponential assumption above makes the occupancy process itself a continuous-time Markov chain and permits the direct detailed balance proof.

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.

First include the mean-gain calculation preceding the numbered clauses. At each electron-multiplying CCD stage one electron contributes an expected $1+p$ electrons to the next stage. If $N_j$ is the charge after stage $j$, conditional expectation gives $\mathbb{E}[N_{j+1}\mid N_j]=(1+p)N_j$. Iteration from $n$ input electrons yieldsFor small $p$, $g\simeq e^{rp}$ if corrections of order $r{p}^2$ are negligible.

In the high-gain approximation the output from one input electron has an exponential distribution of scale $g$. Independent input electrons produce independent cascades, and their charges add. By convolution of independent random variables, assuming the formula for $n$ inputs,Together with the single-electron base case this provesIt is a gamma distribution with shape $n$ and scale $g$, so it is normalized and has expected value $ng$ and variance $n{g}^2$. The formula is a continuous approximation to discrete output charge, not an exact integer-valued branching distribution; for $n=0$ there is a point mass at zero instead.

Independence of random variables and the two unit-rate exponential distributions giveThe linear transformationhas determinant $2$ and inverseThe inequalities $x,y\ge 0$ become $8v/3\le u\le 3v$. By the change-of-variables formula for probability densities,The variables $U$ and $V$ are not independent: the support of their joint density is a wedge rather than a Cartesian product of one-dimensional supports. For example, conditioning on a positive value of $V$ constrains $U$ to a finite interval depending on that value.

Proton and Prune are censored after the estimated survivor curve has already crossed one half. Learning their later death times may add jumps in the far tail but cannot change the first crossing, so the nonparametric median remains seven years. An exponential distribution instead estimates one constant hazard from the total event count divided by total time at risk. Replacing censoring by observed later deaths changes both quantities and therefore changes the fitted rate and its median $\log 2/\hat{\lambda }$.

Conditional on $Y_m=i$, the holding time $V_m=U_m/g(i)$ has an exponential distribution of rate $g(i)$, and after that holding time the next state is $j$ with transition probability $p_{ij}$. The memoryless property of exponential random variables and the Markov property of $(Y_m)$ therefore make $(X_t)$ a continuous-time Markov chain.

Moreover $g(i)<1/\epsilon$, so $V_m\ge \epsilon U_m$. The strong law of large numbers gives ${\sum }_m{U}_m=\infty$ almost surely, hence $J_m\to \infty$ and the process is nonexplosive. Its Q-matrix isEquivalently, if $D_g$ is the diagonal matrix with entries $g(i)$, then

Let $S_n={\sum }_{i=1}^n{X}_i$. The log-likelihood iswith score function $n/\theta -S_n$ and second derivative $-n/{\theta }^2<0$. The unique maximum-likelihood estimator is thereforeFor an exponential distribution of rate $\theta$,The central limit theorem givesApplying the delta method to $h(x)=1/x$, for which $h^{\prime }(1/\theta )=-{\theta }^2$, yields the exponential-rate maximum-likelihood estimator limit

Because $0\le U\le 1$, for fixed $t$ and sufficiently large $n$,ThereforeEquivalently, $n(1-\hat{\theta }/\theta )$ converges in distribution to an exponential distribution of rate one.

Put $c=\log (1/\alpha )$. The limiting probabilityand the certain inequality $\hat{\theta }\le \theta$ suggest the approximate confidence intervalSince $\mathbb{E}[\hat{\theta }]=n\theta /(n+1)$, its expected length isas required.

Write $q_i=-g_{ii}$ for the total rate of leaving $i$. The jump chain of a continuous-time Markov chain $X$ is the discrete-time chain $Y=(Y_n)$ recording the states occupied after successive jumps. Its transition probabilities arewhenever $q_i>0$.

The chain $X$ is irreducible if every state can be reached from every other state with positive probability. A state $i$ is recurrent if, after leaving $i$, the process returns to $i$ with probability one; an irreducible chain is recurrent when one, and hence every, state is recurrent.

The continuous-time Markov chain and its jump chain have exactly the same successive states. If $Y$ is transient, it visits each state only finitely often, so $X$ is transient. If $Y$ is recurrent, it visits a fixed state $i$ infinitely often. At those visits, $X$ has independent exponential holding times of rate $q_i<\infty$. Their sum diverges almost surely, so explosion cannot occur before all those returns. Hence $X$ returns to $i$ infinitely often and is recurrent. Thus, for an irreducible chain,

A renewal-reward process consists of i.i.d. regenerative cycles $({\xi }_n,R_n)$, where ${\xi }_n>0$ is the $n$th cycle length and $R_n$ is the reward accumulated during that cycle. If $0<\mathbb{E}{\xi }_1<\infty$ and $\mathbb{E}|R_1|<\infty$, the renewal-reward theorem states that the cumulative reward $Z(t)$ satisfiesalmost surely, with the usual negligible-remainder condition for an unfinished cycle.

For the machine, let $X\sim \mathrm{Exp}(\lambda )$ be its working lifetime after a repair. The next repair occurs at the first inspection following failure, so one repair-to-repair cycle has lengthwhile its working-time reward is $R=X$. Since an exponential distribution of rate $\lambda$ has mean $1/\lambda$,By the tail-sum formula for the positive integer-valued variable $\lceil X/m\rceil$,Thereforeand the long-run proportion of time for which the machine is working is

An M-M-1 queue has Poisson process arrivals of rate $\lambda$, independent exponential service times of rate $\mu$, and one server. The two letters $M$ indicate the memoryless property of the interarrival and service-time distributions. Its queue length is a birth-death process with birth rate $\lambda$ and death rate $\mu$ away from zero. When $\rho =\lambda /\mu <1$, its normalized stationary distribution isthe powers ${\rho }^n$ quoted in the question form the corresponding unnormalized invariant measure.

For independent exponential distributions, the numerator from part a(ii) iswhile the denominator is the expression in braces. Their ratio is