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Write $\mu =1+\epsilon$, $p=\mu /n$, $q=1-p$, and let $T$ be the total progeny of a binomial branching process. The preceding binomial branching survival correction gives $\rho =(2+o(1))\epsilon$. For a branching process conditioned on extinction, the offspring probability generating function is $f((1-\rho )s)/(1-\rho )$, with meanfor sufficiently large $n$. Summing expected generation sizes therefore gives $\mathbb{E}[T;T<\infty ]=(1-\rho )/(1-f^{\prime }(1-\rho ))=O(1/\epsilon )$.

Put $K=\lceil {\epsilon }^{-3}\rceil$ and let $N_{\ge K}$ count vertices in graph components of order at least $K$. A breadth-first exploration of a binomial random graph is dominated by $T$. The Markov inequality applied to finite total progeny givesAlways $L_1\le K+N_{\ge K}$. Since $n{\epsilon }^4\ge n^{1/3}\to \infty$, we have $K=o(\epsilon n)$ and $n/(\epsilon K)=O(n{\epsilon }^2)=o(\epsilon n)$. Hence $\mathbb{E}{L}_1\le (2+o(1))\epsilon n$. This bound controls the expected value directly, including rare large graph components.

For the lower bound, use the breadth-first exploration of a binomial random graph with $U_t$ unseen vertices, $A_t$ active vertices and $t$ explored vertices, starting from $U_0=n$, $A_0=0$. Let $b_t=1_{\{A_{t-1}=0\}}$ indicate a new root. Conditional on the past, the newly discovered count $Z_t$ has a binomial distribution $\mathrm{Bin}(U_{t-1}-b_t,p)$. Put $D_t=Z_t-p(U_{t-1}-b_t)$. The unseen-count recursion givesHere $M_t$ is a martingale. Up to $T_0=\lceil 3\epsilon n\rceil$, martingale-difference orthogonality gives $\mathbb{E}{M}_{T_0}^2\le C\epsilon n$, since each conditional variance is at most $np<2$ and $q^{-2T_0}$ is bounded.

Take $h=\sqrt{\epsilon }+(n{\epsilon }^3{)}^{-1/8}$. Then $h\to 0$, $\epsilon =o(h)$ and $h^{2}n{\epsilon }^3\to \infty$. The Doob L2 maximal inequality impliesThe Taylor expansion, uniformly for $t\le T_0$, gives $F(t)=\epsilon t-t^2/(2n)+O({\epsilon }^{3}n+\epsilon )$. Thus $F(t)\ge h{\epsilon }^{2}n/3$ throughout the integer interval $[\lceil h\epsilon n\rceil ,\lfloor (2-h)\epsilon n\rfloor ]$ for large $n$. The new-root sum is nonnegative, so on the preceding event $A_t>0$ throughout this interval. No graph component finishes there: all these explored vertices belong to one graph component, of order at least $(2-2h)\epsilon n-O(1)$, with high probability. Its expected value is therefore at least $(2-o(1))\epsilon n$. Combining both inequalities provesThe barely-supercritical largest-component expectation uses both a positive exploration window and a finite-progeny bound; a bare convergence-in-probability assertion would not by itself justify this expectation.

For the binomial branching process, the offspring probability generating function is $f(s)=(1-p+ps{)}^n$. The Galton-Watson extinction fixed point gives $1-\rho =f(1-\rho )=(1-p\rho {)}^n$.

The printed finite-binomial upper bound is false without an additional asymptotic qualification. For $n=2$ and $p=(1+\epsilon )/2$, solving the fixed-point equation exactly givesFor instance $\epsilon =1/20$ gives $\rho =80/441>1/10=2\epsilon$. More generally, at every fixed $n>1$ the Taylor expansion gives $\rho =2n\epsilon /(n-1)+O_n({\epsilon }^2)$, again contradicting that upper bound for sufficiently small positive $\epsilon$.

Here is the corrected binomial branching survival correction. The Poisson branching process of mean $1+\epsilon$ has branching survival probability $r$ satisfying $-\log (1-r)=(1+\epsilon )r$. Expanding the logarithm givesThus $2\epsilon /(1+2\epsilon )\le r\le 2\epsilon$, which implies the requested lower estimate $r\ge 2\epsilon -4{\epsilon }^2$. Since $(1-p+ps{)}^n\le e^{np(s-1)}$ on $[0,1]$, iteration of the two offspring probability generating functions gives $\rho \ge r$.

For the actual binomial branching process, expansion of its exact fixed-point equation givesIf $(1+\epsilon {)}^2<n$, all coefficients are nonnegative. Keeping the first term provesIn particular, for $n\to \infty$ and $\epsilon =o(1)$ this gives $\rho =2\epsilon +O({\epsilon }^2+\epsilon /n)$. The printed bounds also hold for the binomial branching process in an explicit large-$n$ regime: $0<\epsilon \le 1/4$ and $n\epsilon \ge 2$. Indeed these conditions give $n{p}^2\le \epsilon$ and $n{p}^3\le 1/4$. In the preceding nonnegative series, evaluation at $x=2\epsilon$ givesThe series is increasing, and its value at the actual branching survival probability $\rho$ is $\epsilon$, so $\rho \le 2\epsilon$. Together with the lower bound already proved, this recovers $2\epsilon -4{\epsilon }^2\le \rho \le 2\epsilon$ in that regime. In particular it applies eventually to part (ii). The counterexample shows why a regime condition is needed for a literal finite-$n$ statement.

For the probability generating function $G(t)=\mathbb{E}[t^X]$, differentiation givesThe second derivative is a factorial moment, not the raw second moment.

In the Galton-Watson process, condition on $X_n=m$. The next generation is a sum of $m$ independent offspring counts, and so its conditional generating function is $G(t{)}^m$. Averaging over $X_n$ givesThe conditional mean is $m\mu$ and the conditional variance is $m{\sigma }^2$. Thus iterated expectation gives $\mathbb{E}[X_n]={\mu }^n$ from one ancestor, and the law of total variance gives $v_{n+1}={\mu }^2{v}_n+{\sigma }^2{\mu }^n$, $v_0=0$. Iterating yields the branching-process generation momentsFor the critical case $\mu =1$, the correct limiting value is $v_n=n{\sigma }^2$. If $\mu =0$, nonnegative offspring are zero almost surely, so every later generation has zero mean and variance; handle this directly rather than use an ambiguous zero power. At generation zero the mean is one and variance zero.

The branching-process extinction criterion states that extinction probability is the smallest fixed point of $G$ in $[0,1]$: probabilities of extinction by successive generations are the iterates of $G$ from zero, increasing to that fixed point. Here the equation is $4q^3-7q+3=0$, factoring as $(q-1)(2q-1)(2q+3)=0$. The two admissible roots are $1/2$ and one, so

Let $G(z)={\sum }_{n\ge 0}{p}_n{z}^n$ be the probability generating function. For $|z|<1$ the series can be differentiated term by term: $\sum n{p}_n|z{|}^{n-1}$ is finite, even without a finite claim-count expected value. Multiply the Panjer claim-count class recurrence by $n$ and put $m=n-1$. ThenHenceThis argument is valid throughout the open unit disk. Values at boundary points may be obtained by a limit when the required derivatives exist; one need not assume $\mathbb{E}N<\infty$ to establish the identity.

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.

Condition on the count and use the independence of random variables of the claim sizes. For $n\ge 1$ their joint exponential transform factors, while for $n=0$ the empty sum contributes $1$. Consequently the law of total expectation givesThis random sum of independent claims transform uses both independence assumptions: the count must be independent of the entire claim-size sequence, and the sizes must be mutually independent with the same probability distribution. The probability generating function is interpreted through its defining nonnegative series. The identity holds as a finite moment-generating function wherever that series is finite; outside that domain the expectation and series can agree at $+\infty$. In particular a positive argument may take $M_X(r)$ beyond $1$, so finiteness does not follow merely from the usual unit-disk domain of a probability generating function.

Conditioning on the intensity in the Poisson mixture givesThus $N$ has the negative binomial distribution with two successes and success probability $p$, counting failures; explicitly $\mathbb{P}(N=j)=(j+1)p^2{q}^j$ for $j\ge 0$. The probability generating function is finite for real $z<1/q$.

The claim-size moment-generating function is $M_X(t)=(1-\mu t{)}^{-1}$. Substituting into the aggregate moment-generating function and using $p+q=1$ givesPut $L(t)=p/(p-\mu t)$, the moment-generating function of an exponential distribution with rate $p/\mu$. The identitythen yieldsThe gamma-mixed Poisson aggregate with exponential claims has three nonnegative mixture weights summing to one. By uniqueness of the moment-generating function near zero, the aggregate distribution isHere ${\delta }_0$ is the Dirac measure at zero. In particular $\mathbb{P}(S=0)=p^2$, consistently with $\mathbb{P}(N=0)$. The positive components have respective expected values $\mu /p$ and $2\mu /p$; their mixture distribution accounts for the possibility of no aggregate payout.

Write $m_X=\mathbb{E}{X}_1$ and $v_X=\mathrm{Var}(X_1)$. For the random sum of independent claims, conditioning on $N$ givesThe law of total expectation and the law of total variance therefore give the aggregate momentsThe first term in the variance measures variation of the individual claims at a fixed count; the second measures variation of the count itself. These formulas require the indicated moments to be finite.

For the moment-generating function, independent random variables givewhere $G_N(z)=\mathbb{E}[z^N]$ is the probability generating function. This identity holds wherever the expectations are finite; in particular a moment-generating function need not exist for positive $t$ for an arbitrary positive claim distribution. The empty sum for $N=0$ is zero.

Assume individual incubation delays are independent of the Inhomogeneous Poisson process of infections and of one another. This marking assumption is needed in addition to specifying a marginal incubation density.

In the discrete cohort model let $N_i\sim \mathrm{Poisson}(h_i)$ be the independent infection counts. Mark each infection by its eventual onset interval. For cohort $i$, the categories have probabilities $q_{ik}$ for observed intervals, with the remaining probability assigned to onsets outside the window. Poisson thinning makes the counts $N_{ik}$ in these categories mutually independent Poisson random variables with means $h_i{q}_{ik}$. One direct proof is their probability generating function:Different cohorts are independent. Summing their counts therefore gives independent Poisson onset counts in disjoint intervals:The same argument applies exactly in continuous time by the Independent marking theorem for Poisson point processes and mapping each marked infection to its onset time. The endpoint model approximates its means; its independent-Poisson conclusion is exact within that discrete model. A fixed cohort size would instead induce negatively correlated onset-bin counts, so the Poisson process infection assumption matters.

A rate-$\lambda$ Poisson process is a counting process starting at zero, with right-continuous nondecreasing integer-valued paths, independent stationary increments, and $N_t-N_s\sim \mathrm{Pois}(\lambda (t-s))$ for $0\le s<t$. Its jumps have size one almost surely.

For independent Poisson processes $N,M$, increments of $N+M$ on disjoint intervals are independent, because all the corresponding increments of the two processes are jointly independent. On an interval of length $t$, multiplying their probability generating functions givesThus the increment is Poisson with mean $(\lambda +\mu )t$. The path requirements hold, and independent processes almost surely have no simultaneous jumps. Therefore $N+M$ is a Poisson process of rate $\lambda +\mu$.

In an interval of length $t$, condition on the Poisson process count $X_t=k$. Independently marking its points gives $Y_t\mid X_t=k\sim \mathrm{Bin}(k,p)$. The joint probability generating function of kept and discarded counts isIt factors into the generating functions of independent Poisson variables of means $\lambda pt$ and $\lambda (1-p)t$. The same calculation applies to any interval, and counts and marks in disjoint intervals are independent. Hence $Y$ is a Poisson process of intensity $\boxed{\lambda p}$, and $Y_t$ and $X_t-Y_t$ are independent. This is the Poisson thinning theorem.

Let $u_n={\mathbb{P}}_0(X_n=0)$ and $f_n={\mathbb{P}}_0(T=n)$. Decomposing at the first return and using the Markov property gives the renewal equationFor $0<s<1$, define the probability generating functions $U(s)={\sum }_{n\ge 0}{u}_n{s}^n$ and $F(s)={\sum }_{n\ge 1}{f}_n{s}^n={\mathbb{E}}_0[s^T]$. Absolute convergence, or nonnegative summation, justifies the convolution identity $U=1+FU$. The binomial series givesSolving the renewal identity gives the first-return generating function of the simple symmetric random walkAs consistency checks, $F(s)\to 1$ as $s\uparrow 1$, confirming recurrence, while $F^{\prime }(s)=s/\sqrt{1-s^2}\to \infty$. By monotone convergence of ${\sum }_{n}n{f}_n{s}^{n-1}$, this last limit is ${\mathbb{E}}_{0}T=\infty$, confirming null recurrent states.

Since $\mathbb{P}(X_1=0)=e^{-\lambda }$, the indicator $Y$ is an unbiased estimator of $\theta$. The joint probability mass function isThe Fisher-Neyman factorization theorem therefore makes $T={\sum }_i{X}_i$ a sufficient statistic. More explicitly, conditioning on $T=t$ gives a multinomial distribution with $t$ trials and cell probabilities $1/n$, so $X_1\mid T=t$ has the binomial distribution $\mathrm{Bin}(t,1/n)$. Consequently conditioning by the Rao-Blackwell theorem givesFor $n=1$, interpret the value at $T=0$ as $1$ and at $T>0$ as $0$; thus the same expression uses $0^0=1$ in this probability convention. For $\lambda =0$ all observations and $T$ are zero almost surely. For $n>1$ and $\lambda >0$, the reduction in variance is strict: given $T=1$, the event $X_1=0$ still has a nondegenerate conditional probability. Directly, using the probability generating function of $T\sim \mathrm{Poisson}(n\lambda )$, the new variance is $e^{-2\lambda }(e^{\lambda /n}-1)$.