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Use the standard infinite-population slotted ALOHA model with independent fresh arrivals with Poisson distribution $Y_t\sim \mathrm{Poisson}(\lambda )$, $\lambda >0$. Newly arrived packets attempt in the next slot. Given $N_t=n$ old backlogged packets, each independently attempts with fixed probability $f$, so the retransmission count has binomial distribution $R_t\sim \mathrm{Binomial}(n,f)$, independent of $Y_t$ and earlier random choices. Let $Z_t$ record whether $R_t+Y_t$ is zero, one, or at least two. A single transmission succeeds and departs; a collision makes every transmitted packet remain. ThereforeOnly the present backlog determines the law of the next independent choices, so $N_t$ is a time-homogeneous Markov chain.

For $0<f<1$, put $q=1-f$ and $b_n=nf{q}^{n-1}$, with $b_0=0$. Its nonzero transition probabilities areFor example, a single fresh arrival leaves the backlog unchanged only if no old packet transmits; if an old packet also transmits, that fresh arrival becomes a new backlogged packet. These cases explain the extra diagonal term and the suppressed one-step increase at $n=0$. At $f=1$ the same formulas hold with $q^0=1$, $b_1=1$ and $b_n=0$ for $n\ne 1$.

Here is an almost-sure proof of permanent collisions in constant-probability slotted ALOHA for $0<f<1$, rather than just a positive-drift heuristic. For $\theta >0$, the conditional exponential increment isAs $n\to \infty$ this tends to a number strictly below one. Choose $M\ge 1$ so the expression is at most one for $n\ge M$. Stopping $e^{-\theta N_t}$ when the chain first enters $\{0,\dots ,M-1\}$ gives a nonnegative supermartingale. Downward jumps have size at most one, so it hits at $M-1$. Optional sampling theorem for a supermartingale, first at bounded times and then by a limit, yields the exponential-supermartingale escape boundThis is an irreducible Markov chain: every positive state can successively decrease to zero, and from zero Poisson batches reach every state at least two, with state one then reached from two. An irreducible recurrent Markov chain would hit that finite set with probability one from every state, contradicting the bound. Hence every state is a transient state, every finite set is visited only finitely often, and $N_t\to \infty$ almost surely.

The conditional success probability is $p(N_t)=e^{-\lambda }(b_{N_t}+\lambda q^{N_t})\to 0$. The centered success indicators form a martingale difference sequence with bounded increments. The strong law for martingales with bounded increments givesThe conditional means have Cesàro average zero, while the strong law of large numbers gives $t^{-1}{\sum }_{u<t}{Y}_u\to \lambda$. Therefore $N_t/t\to \lambda$ almost surely. Finallydecays exponentially along this linear backlog growth. Its sum over $t$ is almost surely finite. The Conditional Borel-Cantelli lemma then implies that only finitely many slots have zero or one transmission. Thus $\boxed{\mathbb{P}(\exists J:Z_t=\ast \text{ for every }t\ge J)=1}$. At $f=1$, a fresh batch of size at least two eventually occurs, after which at least two old packets retransmit in every slot, giving the same conclusion directly.

The qualification $f>0$ is essential: if $f=0$, then $Z_t$ depends only on the independent Poisson arrivals, and zero- or one-arrival slots occur infinitely often. Positive mean arrival rate alone, without the Poisson model or another assumption allowing collisions to develop, would also be insufficient: Bernoulli fresh arrivals and initial backlog zero never collide.

One alternative is binary exponential backoff: after $k$ collisions a packet selects a waiting time uniformly from $\{0,\dots ,2^k-1\}$ and tries when its counter expires. Its attempt behavior depends on its collision history, rather than a common constant $f$. Another possibility, if a backlog estimate ${\hat{N}}_t$ is available, is to set the attempt probability near $1/{\hat{N}}_t$, keeping the offered retransmission count near one. These descriptions illustrate adaptive access rules; they do not assert that every backoff variant is stable at arbitrary load.

On a measurable space $(E,\mathcal{E})$ with a sigma-finite measure $\mu$, a Poisson random measure $N$ is a random counting measure such that $N(A)$ has the Poisson distribution with parameter $\mu (A)$ for every measurable $A$ of finite intensity, and the counts on finitely many pairwise disjoint measurable sets are independent. The sample measures are countably additive; repeated atoms are allowed if the intensity has atoms.

When $\mu (A)=\infty$, exhaustion by finite-intensity subsets gives $N(A)=\infty$ almost surely. Indeed their counts increase, and the Poisson distribution probability of a count at most any fixed integer tends to zero as the parameter tends to infinity. Independent Poisson counts on disjoint sets, with means prescribed by $\mu$, characterize the Poisson random measure.

For a measurable $A$ with $(\mu +\nu )(A)<\infty$, independence of the two Poisson random measures givesThus its count has the Poisson distribution with the summed parameter. If $A_1,\dots ,A_m$ are disjoint, the variables within each family of counts are independent, and independence of $\Phi$ and $\Psi$ makes the two families independent of each other. Hence the pairs $(\Phi (A_j),\Psi (A_j))$, and therefore their sums, are independent over $j$.

Adding sample counting measures preserves countable additivity. The definition in (a) therefore provesThis proves the required Superposition theorem for Poisson point processes directly from count distributions.

Use the independent-mark interpretation: the Brownian motions attached to the atoms are independent of one another and of the initial Poisson random measure. Fix $t$ and defineConditionally on the initial atoms, each atom is retained as a dangerous atom if its own path meets the target by time $t$. These tests are independent, with position-dependent retention probability $p_t(x)$. The permitted Poisson thinning property makes the dangerous atoms a Poisson random measure of intensity $p_t(x)dx$.

Its total intensity is computed without invoking a marking or displacement theorem. The event in the definition of $p_t(x)$ says that $x$ belongs toBy the Tonelli theorem,Symmetry of Brownian motion gives $-\xi \stackrel{d}{=}\xi$ as processes. It therefore identifies the last expectation with $\mathbb{E}\mathrm{vol}(W(t))$, where the sign of the deterministic path $f$ is unchanged. This swept set is a Wiener sausage with an added deterministic path.

The total intensity is finite on each bounded time interval. To see this explicitly, write $K_t=1+{\sup }_{s\le t}|f(s)|$ and $R_t={\sup }_{s\le t}|\xi (s)|$. The sausage is contained in a ball of radius $K_t+R_t$. On a finite time grid, splitting according to the first crossing of a positive level $u$ by a one-dimensional Brownian motion shows that the chance of ending above $u$, conditional on the crossing, is at least $1/2$: the remaining independent increment is symmetric. Thus the grid maximum exceeds $u$ with probability at most twice the final Gaussian upper tail. Dense grids and continuity give the same bound for the path maximum. Applying this to each coordinate and its negative givesThe last inequality follows from $\mathbb{E}{e}^{\theta N(0,t)}=e^{{\theta }^{2}t/2}$ and minimizing the exponential Markov bound. Integration of this tail gives finite moments of $R_t$ of every positive order, so $\mathbb{E}\mathrm{vol}(W(t))<\infty$.

There are consequently only finitely many dangerous atoms on each compact time interval, almost surely. For closed balls, continuity of the paths makes each finite hitting-time infimum attained. The event $T>t$ is therefore the absence of dangerous atoms up to time $t$.

The open-ball convention has the same survival probability at each fixed $t$. Indeed for any compact path range $K$, the difference between its closed and open radius-one neighborhoods is the shell $\{x:\mathrm{dist}(x,K)=1\}$, which has zero Lebesgue measure. To see this, choose a nearest $y\in K$ for a shell point $x$. For every sufficiently small $\epsilon >0$, the ball of radius $\epsilon /2$ centered at $x-\epsilon (x-y)$ lies in the open neighborhood and inside the ball of radius $3\epsilon /2$ about $x$. Thus the shell has a fixed proportional hole at every scale; the Lebesgue density theorem forces its measure to be zero. By the Tonelli theorem, closed-touch and open-entry retention probabilities have equal integrals. Independent thinning therefore gives no atom that touches only the boundary before this fixed horizon, almost surely. This also excludes a difference at an open-entry infimum equal to $t$.

The zero-count probability of a Poisson distribution is $e^{-\lambda }$, and henceThis survival among independently moving Poisson traps formula has been derived using only independent thinning and the defining Poisson zero-count probability, with the expectation/volume step supplied by Tonelli's theorem.

A Poisson model assumes independent counts between rats conditional on time, a common mean at a given time, and conditional variance equal to that mean. Comparable observation units and consistent counting are needed; unexplained rat-to-rat heterogeneity is excluded by this simple specification. The quadratic Poisson regression isThe printed nested analysis of deviance tests $H_0:{\beta }_2=0$ against $H_1:{\beta }_2\ne 0$. Its likelihood-ratio statistic is $G^2=3.8548$, approximately ${\chi }_1^2$ under the null hypothesis and regular Poisson asymptotics. The reported $p=0.0496$ gives borderline rejection of the log-linear time effect at 5%, conditional on the Poisson assumptions. This is curvature of the log mean, rather than an ordinary quadratic model for the count itself.

There are only three distinct time values, so the intercept, linear term, and quadratic term span every possible set of three log means. This is Poisson mean saturation at three distinct times: the quadratic fit is saturated for the time-group means, which are $14/7=2$, $17/8=2.125$, and $47/7\simeq 6.7143$. A grouped deviance comparison against unrestricted time-group means has zero statistical degrees of freedom: the quadratic functional form cannot be tested at these three times. The individual-observation deviance 24.515 still describes within-group departure from the Poisson distribution, but small means and zeros make the naive ${\chi }_{19}^2$ goodness-of-fit calibration doubtful. Poisson variation itself can be checked using the replicates, preferably by a parametric bootstrap; it is not rendered untestable by the saturated mean specification.

Overdispersion means conditional variance exceeding the variance assumed by the Poisson mean model. For example, rats may have different latent susceptibility, so if $Y\mid \Lambda \sim \mathrm{Pois}(\Lambda )$, the law of total variance gives $\mathrm{Var}(Y)=\mathbb{E}\Lambda +\mathrm{Var}(\Lambda )>\mathbb{E}Y$. A quasi-Poisson model instead uses $\mathrm{Var}(Y_i)=\varphi {\mu }_i$. The Pearson-residual dispersion estimate is $\hat{\varphi }=1.254068$, using $n-p=22-3=19$.

For the conventional quasi-likelihood comparison with estimated dispersion, useIts upper-tail probability is about 0.096, so the evidence for curvature no longer reaches 5%. The F-distribution is an approximate calibration, not an exact small-sample result for arbitrary overdispersed counts; quasi-likelihood does not supply a full Poisson likelihood. With known dispersion one would instead compare $G^2/\varphi$ approximately with ${\chi }_1^2$. The displayed dispersion code uses the undefined name acf.glm2; its fitted values must be those of aom.glm2 for the stated calculation.

Assume $0<\lambda =1-\epsilon <1$, as required for this nontrivial subcritical binomial random graph. Put $\delta =\lambda -1-\log \lambda >0$. A breadth-first exploration of a binomial random graph is dominated by a Galton-Watson process whose offspring counts are independent random variables with law $\mathrm{Bin}(n,\lambda /n)$. If its total progeny $T$ is at least $j$, the first $j-1$ offspring counts sum to at least $j-1$. Their moment-generating function is bounded by that of a Poisson distribution of mean $\lambda (j-1)$. The exponential Markov inequality, optimized at $t=-\log \lambda$, yieldsTaking $j=\lceil (1+\eta )\log n/\delta \rceil$ for any fixed $\eta >0$, a union bound over all starting vertices gives $\mathbb{P}(L_1\ge j)=o(1)$.

For the matching lower bound, let $X_j$ count tree components of order $j$. The tree-component expectation in the Erdős-Rényi model is exactlyFor $j=O(\log n)$, the Stirling formula givesChoose $j=\lfloor (1-\eta )\log n/\delta \rfloor$, with $0<\eta <1$. Then $\mathbb{E}{X}_j\to \infty$. Distinct overlapping vertex sets cannot both be graph components. For disjoint sets, the joint probability differs from the product only because the $j^2$ between-set edges were counted twice as absent. ThusIt follows that $\mathrm{var}{X}_j/(\mathbb{E}{X}_j{)}^2=o(1)$. The second moment method gives $X_j>0$ with high probability. Combining both bounds, and then taking arbitrarily small fixed $\eta$, proves the sharp subcritical largest-component scaleFinally the Taylor expansion of $-\log (1-\epsilon )$ gives $\delta ={\epsilon }^2/2+{\epsilon }^3/3+\cdots$. The endpoint $\lambda =0$ has no edges and is excluded from this logarithmic asymptotic.

The Poisson distribution with unit parameter has mass $e^{-1}/j!$. Its moment-generating function isFor an integer $k\ge 1$, multiply the exponential probability bound by $e$ to obtain ${\sum }_{j=k}^{\infty }1/j!\le \exp (e^t-kt)$. The exponent is minimized at $e^t=k$, hence $t=\log k\ge 0$, including $t=0$ when $k=1$. Consequently

Take two independent rate-one Poisson processes $N^{+}$ and $N^{-}$, and letThe difference of independent Poisson processes starts at zero and has stationary increments and independent increments. Its paths are càdlàg. For an interval of length $h$, the probability of any jump is $1-e^{-2h}\to 0$, proving stochastic continuity. Thus $X$ is a Lévy process. The characteristic function of a Poisson distribution with mean $t$ is $\exp (t(e^{iu}-1))$, so independence givesThe sample paths are integer-valued step functions with jumps $+1$ or $-1$. On every bounded interval there are only finitely many jumps, and independent Poisson arrival times coincide with probability zero. The combined arrival rate is two: holding times are independent exponentials of rate two, and each jump direction has probability $1/2$, independently of the holding times. This is equivalently a Compound Poisson process of rate two with Rademacher distribution jump sizes. Its paths have finite variation on compact time intervals, although there are infinitely many jumps over the whole half-line almost surely.

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.

For a Poisson distribution of parameter $\lambda$, its probability generating function is $\exp \{\lambda (z-1)\}$. Applying the law of total expectation to the Poisson mixture givesThusFor $|z|\le 1$ the absolute value of the integrand is bounded by one, so this identity always exists as a Laplace transform of the positive mixing variable. An extension to $z>1$ requires the corresponding moment-generating function to be finite; the conditional computation itself does not guarantee positive exponential moments.

For excess of loss reinsurance the insurer pays each claim up to its retention level:The cap applies separately to every claim. In particular the retained annual loss is ${\sum }_j\min (X_j,M)$, rather than a single cap on the annual total.

Let $F_i$ be the cumulative distribution function for the claim size on risk $i$, and put ${\overline{F}}_i=1-F_i$. The retained severity on that risk has the original probability density function on $0<y<M$ and an atom of a measure at $M$ of mass ${\overline{F}}_i(M)$. Thus $T_I$ has a compound Poisson distribution with rate ${\lambda }_1+{\lambda }_2$ and the mixture of these capped severity laws. The mixture's mass at $M$ is ${\sum }_i{\lambda }_i{\overline{F}}_i(M)/({\lambda }_1+{\lambda }_2)$.

For the capped claim moments, use the tail integral formula for moments. Since $\mathbb{P}(\min (X_i,M)>x)={\overline{F}}_i(x)$ for $0\le x<M$ and is zero for $x\ge M$,Substitution into the compound Poisson distribution moment formulas givesEquivalently, the integrals are ${\int }_0^{M}x{f}_i(x)dx+M{\overline{F}}_i(M)$ and ${\int }_0^M{x}^2{f}_i(x)dx+M^2{\overline{F}}_i(M)$. The annual variance uses the retained raw second moments; subtracting their squared means would omit the variation in the Poisson distribution count.

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.