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For an aggregate claims model, let $X$ denote one claim size, with expected value $\mu$ and variance ${\sigma }^2$. The total $S$ is zero when the claim count is zero. Given $N=n$, independence givesThe law of total expectation and law of total variance, together with $\mathbb{E}N=\mathrm{Var}(N)=\lambda$ for a Poisson distribution, implyThe raw second moment appears because the random count itself contributes $\lambda {\mu }^2$ to the variance.

Conditional on $N$, the moment-generating function of the sum is $M_X(t{)}^N$. Averaging with the Poisson distribution therefore gives the compound Poisson distribution transformThis is valid wherever $M_X(t)$ is finite. Finite first and second moments alone do not ensure positive exponential moments; for positive claims the corresponding Laplace transform always exists.

For the independent portfolios put $\Lambda ={\lambda }_1+{\lambda }_2>0$ and $w_i={\lambda }_i/\Lambda$. Their Poisson distribution counts add to a Poisson distribution with parameter $\Lambda$. By Poisson-multinomial conditioning, conditional on the total count $n$ the first risk count has a binomial distribution with parameters $n,w_1$, and the second is the remaining count. Thus one can generate the same total loss by drawing $n$ independent risk labels with these weights, then drawing each claim from its label's law. The merged severity has mixture distributionIt follows that $T$ has a compound Poisson distribution with count parameter $\Lambda$ and this severity law. This is the fixed-year version of Poisson superposition of insurance portfolios. Its expected value and variance areAlternatively, multiplying the two independent aggregate Laplace transforms yields $\exp \{\Lambda (w_1{M}_{X_1}(t)+w_2{M}_{X_2}(t)-1)\}$ wherever finite, confirming the same compound Poisson distribution.

For the retained compound Poisson aggregate under per-claim reinsurance, replace each claim $X$ by its retained payment $g(X)$. Assume the retention is measurable, with $0\le g(x)\le x$ as usual. The count parameter remains $\Lambda$, and the severity law is the pushforward measure of the mixture severity under $g$. Thus the retained compound Poisson aggregate $T_I$ has a compound Poisson distribution with that transformed severity andIts moment-generating function is $\exp \{{\sum }_i{\lambda }_i(M_{g(X_i)}(t)-1)\}$ on its finite domain. For a general retention $g$, zero retained payments can occur and are allowed as compound-Poisson marks; they may equivalently be removed by Poisson thinning. The two specified contracts retain strictly positive payments for strictly positive claims.

Conditional on $\lambda$, a Poisson distribution has both conditional expectation and conditional variance equal to $\lambda$. For the uniform distribution on $(1,3)$,Thus $K=v/a=6$ and the one-year credibility factor is $Z_1=1/7$. With the observed count equal to one, the Bühlmann credibility estimate of the next year's conditional claim expected value isThe estimate gives relatively little weight to one year because the expected process variance is six times the variance of hypothetical means.

Under quadratic loss, the Bayes estimator under squared error loss of the latent mean $\lambda$ is its posterior mean. For the Poisson-uniform posterior mean, the prior distribution has density $1/2$ on $(1,3)$ and the one-count Poisson distribution likelihood is $\lambda e^{-\lambda }$. Consequently the Bayesian posterior density is proportional to $\lambda e^{-\lambda }$ on that interval, with normalizing integral ${\int }_1^3\lambda e^{-\lambda }d\lambda$.

By conditional independence, $\mathbb{E}(X_2\mid \lambda ,X_1)=\lambda$, so the law of total expectation makes the posterior predictive expected value equal to this same posterior mean. It isThe required antiderivatives are $-(\lambda +1)e^{-\lambda }$ and $-({\lambda }^2+2\lambda +2)e^{-\lambda }$. Evaluating at the two endpoints givesThis estimate differs from $13/7$: the Bühlmann credibility estimate is an optimal affine rule, while the Bayes estimator under squared error loss optimizes over all rules and uses the truncated uniform prior through its exact Bayesian posterior.

An exponential dispersion family has density or mass function, relative to a fixed measure,Here $\theta$ is the natural parameter, $b$ is the cumulant function, and $\varphi$ is the dispersion parameter. Assume the natural parameter lies in the interior of its domain and derivatives can pass through the normalizing integral. Differentiating normalization once and twice givesThe variance function is $V(\mu )=b^{\prime \prime }((b^{\prime }{)}^{-1}(\mu ))$, where the mean-to-natural-parameter inverse exists. The dispersion parameter may be fixed, as in a Poisson distribution, rather than estimated.

For $y=0,1,2,\dots$ and $\lambda >0$, the Poisson distribution has massMatch this to the exponential dispersion family withThen $\mu =b^{\prime }(\theta )=e^{\theta }=\lambda$ and $V(\mu )=b^{\prime \prime }(\theta )=\mu$. The canonical link function expresses the natural parameter in terms of the mean, so the Poisson canonical link is $g(\mu )=\log \mu$. Its conditional variance equals its conditional mean.

The coefficient labelled yr2 identifies year as a factor. With $z_i=0$ in the first year and $z_i=1$ in the second, the Poisson regression assumes independent daily counts conditional on year,The unknown mean statistical parameters are the first-year log daily rate ${\beta }_0$ and the second-year log rate ratio ${\beta }_1$; the dispersion parameter is fixed at one. Approximate 95% Wald confidence intervals areExponentiating yields the first-year fitted daily mean $5.978$, with confidence interval $(5.732,6.234)$, andThe second-year fitted daily mean is $e^{1.78810-0.09816}=5.419$. Thus this Poisson regression estimates a 9.35% fall, with an approximate interval for the percentage fall from 3.66% to 14.70%. These confidence intervals rely on the Poisson distribution and independence assumptions.

The Quasi-Poisson regression retains the same conditional mean but permitsThis is a mean–variance function specification through quasi-likelihood; it does not assign a full probability distribution to each count. For independent observations, the quasi-score equation is proportional to ${\sum }_i{x}_i(Y_i-{\mu }_i)=0$, so the mean statistical parameter estimates equal those from Poisson regression. The output estimates $\hat{\varphi }=3.515351$ through the Pearson dispersion estimator, and inflates the standard errors by approximately $\sqrt{\hat{\varphi }}=1.875$.

The large residual deviance relative to 728 residual statistical degrees of freedom also signals substantial overdispersion. Daily weather, traffic and other omitted conditions may produce greater count variation than a homogeneous Poisson distribution allows. The Quasi-Poisson regression accounts for that extra marginal variance. It still requires a correct conditional mean and an appropriate independence assumption; a common dispersion parameter alone does not repair serial correlation.

Assume conditional independence of the Poisson distributions, and write $u_i={\mu }_{i1}$, $S_i={\sum }_k{e}^{{\eta }_{ik}}$. The correct Poisson mass function has $e^{-\mu }$; the positive sign in the PDF's reminder is erroneous. The group likelihood isMultiply by the shape–rate Gamma distribution prior $b^a{u}_i^{a-1}{e}^{-b{u}_i}/\Gamma (a)$ and integrate. The gamma integral givesIndependent baseline priors give the product over groups. For fixed $a,b$, the leading factors do not depend on $\beta$, proving the requested Gamma-integrated baseline Poisson likelihood.

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.

In the slotted ALOHA model, a station retains one packet until it successfully transmits. Conditional on $(S_t,N_t)=(S,N)$, take fresh independent trials with a Bernoulli distribution of parameter $1/S$ for the $N$ backlogged stations. Let arrivals be independent across slots and independent of these trials, with Poisson distribution of mean $\nu$. New arrivals join the next slot's backlog. An idle slot serves nobody, exactly one attempt serves one packet, and a collision serves nobody. Consequently the backlog changes by arrivals minus the success indicator.

Under these independence assumptions, $(S_t,N_t)$ is a time-homogeneous Markov chain: the feedback probabilities and the independent arrival law depend only on its current state. Merely specifying Poisson arrival marginals would not suffice. For example, take $S_t=1$, $a=b=c=0$, $N_0=0$, and arrivals $Y_0=H$, $Y_1=K$, $Y_2=H$ for independent random variables $H,K$ with Poisson distributions. Histories with $(N_1,N_2)=(0,2)$ and $(2,2)$ both have current backlog two, but the next backlog is respectively two and four. The usual model therefore includes fresh arrivals as an assumption.

Write $p_0=(1-1/S{)}^N$ and $p_1=(N/S)(1-1/S{)}^{N-1}$. Set $p_1=0$ for $N=0$, and use their exact binomial distribution probabilities at $S=1$. Away from the clipping boundary, the conditional drifts areNear $S=1$, the latter must use $\max (1,S+d)-S$ for each update $d$, rather than $d$ itself.

When both coordinates are large with $N/S\to \kappa$, the Poisson limit theorem approximates the attempt count, which has a binomial distribution, by a Poisson distribution of mean $\kappa$. Thus $p_0\to e^{-\kappa }$ and $p_1\to \kappa e^{-\kappa }$. Rescaling state by $K$ and slot time by $K$ motivates the fluid approximation of slotted ALOHAThis is an interior fluid approximation of slotted ALOHA, not the exact conditional drift at a clipped boundary.

Here is one explicit set of sufficient conditions, independent of the arrival rate within the subcritical range:Put $c_0=2/(e-2)$. ThenHence $g$ is negative below one and positive above one: the controller decreases the attempt denominator when offered contention is too small, and increases it when contention is too large.

Use the ratio time change for a homogeneous fluid model, $du/dt=1/s$. The equations becomeFor $\kappa \ge 1$,since $g(\kappa )\ge 0$ and $A(1+c_0)>1$. Also $h(1)=\nu -e^{-1}<0$. Therefore $h$ is negative on $[1,\infty )$, and by continuity on $[\bar{\kappa },\infty )$ for some $\bar{\kappa }<1$. Its value at zero is $\nu \ge 0$, so nonnegative backlog is preserved.

The ratio remains bounded by $K=\max (\kappa (0),1)$ and eventually enters $[0,\bar{\kappa }]$: on the compact interval $[\bar{\kappa },K]$, its derivative is bounded above by a strictly negative number. Once inside it cannot cross upward. The bounded ratio and smooth coefficients make the transformed equations exist for all $u\ge 0$. In this region $g(\kappa )\le g(\bar{\kappa })<0$, so $s(u)$ decreases at least exponentially in $u$. The original time satisfies $t(u)={\int }_0^{u}s(v)dv$ and has a finite limit $T$; boundedness of $\kappa$ gives $n(u)=\kappa (u)s(u)\to 0$ as well. Thus every nonnegative interior fluid trajectory drains to the origin in finite fluid time. At the origin the ratio equation is undefined; the usual stopped fluid trajectory is held there afterward. A state with $s=0<n$ enters the interior under the continuous limiting boundary drift $\dot{s}=c>0$; the same argument then applies.

This proof supplies sufficient conditions, not a characterization of every stabilizing triplet. It also does not assert the positive recurrent Markov chain property of the original stochastic chain solely from the heuristic ODE.

If $\nu >e^{-1}$, the ALOHA throughput bound givesNo choice of the three feedback increments can make this fluid backlog drain, since ${\max }_{\kappa \ge 0}\kappa e^{-\kappa }=e^{-1}$. For the displayed controller, the supercritical fluid trajectory is global and grows instead of draining. At $\nu =e^{-1}$ the ray $n=s>0$ consists of stationary fluid states, explaining why the strict load inequality matters.

For $\lambda \ge 0$, a Poisson distribution has mass $\mathbb{P}(X=k)=e^{-\lambda }{\lambda }^k/k!$ for $k=0,1,\dots$; at $\lambda =0$ it is concentrated at zero. Its moment-generating function isFor independent $X,Y$, the moment-generating function of their sum is the product, $\exp [(\lambda +\mu )(e^s-1)]$. Uniqueness of the moment-generating function near zero therefore gives $X+Y\sim \mathrm{Poisson}(\lambda +\mu )$. Repeating the argument gives $S_n={\sum }_{i=1}^n{X}_i\sim \mathrm{Poisson}(n)$ when every parameter is one.

The displayed sequence is $a_n=\mathbb{P}(S_n\le n)$. The classical central limit theorem states that independent identically distributed variables with finite mean $m$ and nonzero finite variance ${\sigma }^2$ have standardized sum $(S_n-nm)/(\sigma \sqrt{n})$ converging in distribution to the standard normal distribution. Here $m={\sigma }^2=1$, soConvergence of the distribution functions at the continuity point zero justifies the last step. This is the Poisson cumulative probability at its growing mean.

Put $S=X_1+X_2+X_3$. Independence gives a likelihood ratioSince ${\theta }_1>1$, this is strictly increasing in $S$. A threshold chosen between its values at $S=5$ and $S=6$ gives precisely the stated upper-tail critical region, so it is a likelihood-ratio test.

The sum has a Poisson distribution with mean $3\theta$. Thus the size of a statistical test and its statistical power at ${\theta }_1$ are

For the grouped-area problem, let $S_1={\sum }_{i=1}^m{Y}_i$, $S_2={\sum }_{i=m+1}^{2m}{Y}_i$, $T=S_1+S_2$. The null log likelihood, up to data-only constants, isThe maximum-likelihood estimate is $\hat{\lambda }=T/(3ma)$, giving fitted means $T/(3m)$ and $2T/(3m)$ in the two groups. Under the alternative, the fitted group means are ${\hat{\lambda }}_1=S_1/m$ and ${\hat{\lambda }}_2=S_2/m$.

Use the increasing generalized likelihood-ratio test convention $\Lambda ={\sup }_{H_1}L/{\sup }_{H_0}L\ge 1$, consistent with the printed positive $2\log \Lambda$. The exponential and factorial factors cancel, yielding the likelihood-ratio test of area-proportional Poisson meansTerms with zero count are interpreted as $0\log 0=0$; if $T=0$, take $\Lambda =1$. If the inverse ratio is used, the reported test statistic is instead $-2\log \Lambda$.

Under the null, the regular large-sample Wilks theorem approximation is $2\log \Lambda \sim {\chi }_1^2$: the alternative has two rate parameters and the null one. The value $15.67$ far exceeds the usual $1\%$ critical value, about $6.63$, and has approximate tail probability $7.5\times 10^{-5}$. Thus reject proportionality of the expected counts to area: the two groups show evidence of different territory densities. The statistic alone does not say which group has greater density; that requires $S_1/m$ versus $S_2/(2m)$. For small counts, an exact conditional test uses $S_1\mid T\sim \mathrm{Binomial}(T,1/3)$ under the null instead of the asymptotic calibration.

A real Lévy process starts at zero with probability one, has stationary increments and independent increments, is stochastically continuous, and is taken in its càdlàg version. Stationarity means $X_{t+s}-X_s$ has the law of $X_t$; independence means increments over disjoint ordered time intervals are independent. Stochastic continuity means $X_s\to X_t$ in probability as $s\to t$. One can equivalently impose starting at zero, stationary independent increments and stochastic continuity first, and then take a càdlàg modification.

A Poisson random measure with sigma-finite intensity $\nu$ on a measurable space $(E,\mathcal{E})$ is a countably additive integer-valued random measure $N$ such that $N(A)$ has Poisson distribution with parameter $\nu (A)$ whenever $\nu (A)<\infty$, and counts on disjoint measurable sets are independent. A set of infinite intensity has infinite count with probability one. For jump processes the space is often time times a mark space, with intensity $dt\nu (dz)$. These conditions specify both the marginal count laws and their joint independence.

Use the gamma distribution shape-rate convention: $\Theta \sim \mathrm{Gamma}(\alpha ,\beta )$ with $\alpha ,\beta >0$. The conditional Poisson distribution has mean and variance both equal to $\Theta$, so the Bühlmann–Straub model parameters areFor $W={\sum }_j{m}_j$ and $y={\sum }_j{y}_j$, the Bühlmann–Straub credibility factor is $Z=W/(W+\beta )$. Thusand the required expected-count estimate isThe total exposure determines how much information the observed count carries; the number of years alone is not the appropriate denominator.

Introduce a latent indicator $Z_i=1$ for a structural zero. In this parametrization $P(Z_i=1)={\pi }_i$, and conditional on $Z_i=0$ the response has a Poisson distribution with mean ${\mu }_i$. The zero-inflated Poisson regression usesIts observed probability mass function is ${\pi }_i+(1-{\pi }_i)e^{-{\mu }_i}$ at zero and $(1-{\pi }_i)e^{-{\mu }_i}{\mu }_i^{y_i}/y_i!$ at a positive count. The zero-mixing probability here is explicitly ${\pi }_i$; it is distinct from the weight of the count component.

For the expectation-maximization algorithm, start with positive ${\mu }_i$ and $0<{\pi }_i<1$. At iteration $k$, the E-step computesThis distinguishes structural zeros from Poisson-generated zeros. Up to terms independent of the new parameters, the expected complete-data log-likelihood isIt separates into a fractional-response logistic fit and a weighted Poisson fit. Because treatment is binary, each component is saturated over its two treatment groups, so the M-step has closed forms.

For $j=0,1$, let $I_j=\{i:x_i=j\}$, $n_j=|I_j|$, $S_j={\sum }_{i\in I_j}{\tau }_i$, and $C_j={\sum }_{i\in I_j}{y}_i$. The M-step score equations giveHere $\sum (1-{\tau }_i)y_i=C_j$, since every positive count has ${\tau }_i=0$. The explicit coefficient updates areIterate the E- and M-steps until the observed log-likelihood and parameter estimates stabilize. This is the EM algorithm for zero-inflated Poisson regression with explicit binary-group updates; merely naming two regression routines would not supply these expressions. Both treatment groups must be represented for both contrasts to be identifiable. Zero fitted group means or endpoint mixing probabilities are boundary solutions, interpreted through limits of the log or logit coefficients. As with other mixture models, multiple starts help distinguish competing stationary solutions; EM increases the likelihood but does not guarantee a global maximum from an arbitrary start.

The likelihood function for the ordered arrival times of a Poisson process, including the absence of further arrivals before $T$, isThe arrival-time constraint does not depend on $\lambda$. Equivalently, the Poisson distribution of the total count gives $L(\lambda )=e^{-\lambda T}(\lambda T{)}^n/n!$, which differs by a parameter-independent factor. Conditional on the count, the Poisson process conditional arrival times have density $n!/T^n$, independent of $\lambda$. Thus the total count contains all the information about the rate when the exposure is known; it is a sufficient statistic.