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Suppose $\epsilon =o(1)$ and $\epsilon \ge n^{-1/6}$. The breadth-first exploration of a binomial random graph has deterministic drift $F(t)=\epsilon t-t^2/(2n)+O({\epsilon }^{3}n+\epsilon )$ up to $3\epsilon n$ steps. Its weighted martingale has variance $O(\epsilon n)$. For $h=\sqrt{\epsilon }+(n{\epsilon }^3{)}^{-1/8}$, the Doob L2 maximal inequality makes its maximum smaller than $h{\epsilon }^{2}n/6$ with high probability. Then $A_t>0$ throughout $[h\epsilon n,(2-h)\epsilon n]$, giving a graph component of order $(2-o(1))\epsilon n$.

For the upper expectation bound, let $K=\lceil {\epsilon }^{-3}\rceil$. The dominating binomial branching process has branching survival probability $(2+o(1))\epsilon$ by the binomial branching survival correction. Its branching process conditioned on extinction has mean $1-\epsilon +O({\epsilon }^2+\epsilon /n)$ and total-progeny expected value $O(1/\epsilon )$. Therefore $\mathbb{E}{N}_{\ge K}\le n\rho +O(n/(\epsilon K))$. Since $L_1\le K+N_{\ge K}$ and $K=o(\epsilon n)$, this yields the matching upper bound. Controlling rare large components is necessary to conclude an expected value asymptotic from a typical-size statement.

For $p=(1+\epsilon )/n$, the branching survival probability satisfies $1-\rho =(1-p\rho {)}^n$. If $(1+\epsilon {)}^2<n$, comparison with a Poisson branching process and expansion of the logarithm giveFor the upper bound use $\epsilon ={\sum }_{j\ge 1}(1-n{p}^{j+1}){\rho }^j/(j+1)$ and retain its first nonnegative term. For fixed $n>1$, the leading term is instead $2n\epsilon /(n-1)$. In particular $n=2$ gives $\rho =4\epsilon /(1+\epsilon {)}^2$, showing that the uncorrected exact bound $\rho \le 2\epsilon$ is false for small positive $\epsilon$.

For $0<\epsilon \le 1/4$ and $n\epsilon \ge 2$, the uncorrected exact upper bound does hold for the binomial branching process. Here $n{p}^2\le \epsilon$ and $n{p}^3\le 1/4$, so at $x=2\epsilon$ the first two terms of the nonnegative series give $\epsilon (1-n{p}^2)+(4/3){\epsilon }^2(1-n{p}^3)\ge \epsilon$. Monotonicity of that series yields $\rho \le 2\epsilon$. This recovers the intended large-$n$ bracket without claiming it for every finite reproduction law.

During breadth-first search of $G(n,p)$ let $U_t$ count unseen vertices, $A_t$ active vertices, and $t$ explored vertices. When $A_{t-1}=0$ start a new unseen root, recorded by $b_t=1$; otherwise $b_t=0$. The newly discovered count is conditionally $\mathrm{Bin}(U_{t-1}-b_t,p)$. Subtract its conditional mean to obtain a martingale difference $D_t$. With $q=1-p$, $U_0=n$ and $A_0=0$, iteration givesThe last sum is a martingale. Positive $A_t$ over an interval means that the graph component under exploration does not finish there. Independently adding phantom children couples each component exploration below a binomial branching process with offspring law $\mathrm{Bin}(n,p)$.

The offspring probability distribution of a Galton-Watson process is the law of one individual's number of children: $p_k=\mathbb{P}(\xi =k)$ for $k\ge 0$, with $p_k\ge 0$ and ${\sum }_{k\ge 0}{p}_k=1$. Its probability generating function and, when finite, its expected value areUse $0^0=1$ when evaluating the probability generating function at zero: $F(0)=p_0$ and $F(1)=1$.

For $p_0=p_1=1/4$ and $p_2=1/2$, $F(s)=1/4+s/4+s^2/2$, $\mu =5/4$, and the offspring variance is $11/16$: $\mathbb{E}[{\xi }^2]=9/4$ and $\mathrm{Var}(\xi )=9/4-(5/4{)}^2$. The generation probability mass functions pictured under Galton-Watson process come from repeatedly composing this probability generating function. They describe the whole generation, whereas $(p_k)$ describes one individual's reproduction. ^[ref]

The whole offspring distribution matters for extinction. Another law, $p_0=3/8$ and $p_2=5/8$, has the same offspring mean $5/4$ but branching extinction probability $3/5$, since $3/8+5s^2/8=s$ has roots of a polynomial $3/5$ and $1$. Consequently the two laws have the same expected generation size in a Galton-Watson process at each generation but different survival chances. A binomial branching process and a Poisson branching process likewise specify different offspring distributions; a common mean does not make their probability generating functions identical.

View the accessible rooms as the open cluster of bond percolation on a rooted ternary tree. Each accessible room has three forward corridors, independently open with probability $2/3$, so its number of accessible children has the binomial distribution $\mathrm{Bin}(3,2/3)$. The counts at successive levels form a binomial branching process with offspring probability generating functionLet $q_N$ be the probability that the root has no open path to level $N$. Then $q_0=0$, and independence of the three descendant subtrees givesThe sequence increases to the smallest fixed point $q=\varphi (q)$ in $[0,1]$, the branching extinction probability. Thus $q_N$ is close to that solution for large $N$.

Connectivity is the relevant escape event: if no window is reachable, escape is impossible; if one is reachable, ordinary random wandering reaches a window eventually with probability one. Before reaching level $N$ the guest moves within a finite connected graph, so an accessible exit cannot be avoided forever. This interpretation does not require the guest to move monotonically outward.

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.

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.