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Write $a=\mu /\lambda <1$. The embedded birth-death chain moves upward with probability $\lambda /(\lambda +\mu )$ and downward with probability $\mu /(\lambda +\mu )$ away from zero. The gambler's ruin formula, obtained by solving its two-point difference equation on $\{0,\dots ,N\}$ and then letting $N\to \infty$, gives ${\mathbb{P}}_k(\text{hit }0)=a^k$. Translation gives return probability $a$ from $r+1$ to $r$. From $r-1$ the chain hits $r$ almost surely: before that hit it remains in a finite reflecting interval, from every state of which a uniformly positive finite-step chance of hitting $r$ is available. This also proves that starting at zero it hits each $r\ge 1$ almost surely.

After departure from $r\ge 1$ the return probability is thereforeThe holding rate is $q_r=\lambda +\mu$, so the exponential occupation-time parameter is $q_r(1-r_r)=\lambda -\mu$. Applying the Strong Markov property at the almost-sure first hit givesAt zero the holding rate is $\lambda$ and return probability after departure is $a$, again giving parameter $\lambda -\mu$. From initial state $k$, the expected occupation time at zero is $a^k/(\lambda -\mu )$, since the time is zero unless zero is reached. Averaging over the initial law with probability generating function $G$ yieldsFor $\mu =0$, the interpretation is $a^0=1$, $a^k=0$ for $k>0$, and $G(0)=\mathbb{P}(X(0)=0)$; the same formulas hold.

For each person, infection occurs with probability $\lambda$ and, conditional on infection, a GP visit occurs with probability ${\rho }_G$. Independent Bernoulli thinning therefore gives visit probability $\lambda {\rho }_G$. More explicitly, the probability generating function isThusand the likelihood is

For $X_1\sim \mathrm{Poisson}(\theta )$, its probability generating function isThus $\mathbb{E}[X_1]=G^{\prime }(1)=\theta$ andThe log-likelihood is, up to an additive constant,so the maximum-likelihood estimator is ${\hat{\theta }}_{\mathrm{MLE}}=\overline{X}$. The central limit theorem gives

Multiply the master equation by $s^n$ and sum over $n\ge 0$. For the immigration terms,For the death terms, shifting the first sum givesTherefore the probability generating function obeys

Let $Z_n$ be the size of generation $n$. This is a Galton-Watson process with offspring meanThe expected generation size in a Galton-Watson process is thereforeIts probability generating function isBy the Galton-Watson extinction fixed point, the extinction probability $q$ is the smallest solution in $[0,1]$ of $G(q)=q$. Factoring givesso $q=(\sqrt{29}-5)/2$. The probability of production continuing forever is

Let $N_j(A)$ be the number of particles of gas $j$ in a bounded Lebesgue measurable set $A$. Because Lebesgue measure is diffuse, two independent spatial Poisson point processes have no common particle almost surely. Hence the mixture count isWriting ${\lambda }_j=z_j|A|$, independence gives the probability generating functionThus

If $A_1,\dots ,A_m$ are pairwise disjoint, the familyis independent: counts in disjoint sets are independent within each gas, and the two gases are independent. Therefore the sums $N(A_r)$ are independent. These two facts prove directly that the mixture is a Poisson point process. By the Superposition theorem for Poisson point processes, its activity is

The probability generating function isDifferentiating $n$ times,At $s=0$, only the term $m=n$ remains. Hence

Let $X_t$ be the number of telephone lines in use. The memorylessness of the exponential distribution makes $(X_t{)}_{t\ge 0}$ a continuous-time Markov chain, specifically the M-M-infinity queue with generator rates

Suppose $X_0=n$ and put $p_t=e^{-\mu t}$. Each initial call is still active at time $t$ with the exponential survival probability $p_t$, independently of the others. Their surviving number is thereforeA call arriving at time $u\in [0,t]$ survives until $t$ with probability $e^{-\mu (t-u)}$. By Poisson thinning, the number of surviving new calls is independent of $Y_t$ and has lawHence $X_t=Y_t+Z_t$ in distribution. Multiplying the probability generating functions of the two independent random variables gives, for $-1\le s\le 1$,Equivalently, the transient distribution of an M-M-infinity queue iswith independent summands. As $t\to \infty$, the binomial term converges to zero and the Poisson mean tends to $\lambda /\mu$. Thereforewhich proves the stated large-time approximation.

A Galton-Watson process with a Poisson distribution of offspring of mean $\mu$ has this probability generating function. For $\mu =1+\epsilon >1$ its branching survival probability $r$ satisfies $-\log (1-r)=(1+\epsilon )r$, and $2\epsilon /(1+2\epsilon )\le r\le 2\epsilon$. Indeed $\epsilon ={\sum }_{j\ge 1}{r}^j/(j+1)$ lies between $r/2$ and $r/[2(1-r)]$. These exact inequalities belong to the Poisson branching process, not to every finite binomial branching process.

A renewal equation decomposes a process at its first renewal. For returns of a Markov chain to a fixed state, $f_k$ is the first-return probability and $u_n$ the probability of being at that state at time $n$. Their probability generating functions satisfy $U=1+FU$.

For the shifted-exponential Poisson mixture, the displayed relation holds for $n\ge 2$, initialized by $p_0=e^{-\alpha }\beta /(\beta +1)$ and $p_1=(\alpha +1/(\beta +1))p_0$. It follows by differentiating the probability generating function and comparing coefficients. Nonnegativity follows from the independent Poisson-geometric convolution of independent random variables representation.

For a nonnegative integer claim count $N$ with $p=\mathbb{P}(N=0)<1$, the conditional law given $N>0$ has masses $\mathbb{P}(N=n)/(1-p)$ for $n\ge 1$. Its probability generating function is $(G_N(z)-p)/(1-p)$. Truncation at zero is meaningful only when the positive-count event has positive probability.