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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 $\mathrm{Bin}(n,p)$.

For Brownian motion and deterministic partitions of an interval $[0,t]$ whose mesh $|\pi |$ tends to zero, the displayed limit holds in L2 space. Split the error into $U_{\pi }={\sum }_j{B}_{t_j}((\Delta B_j{)}^2-\Delta t_j)$ and the Riemann sum error $V_{\pi }={\sum }_j{B}_{t_j}\Delta t_j-{\int }_0^t{B}_{s}ds$. The summands of $U_{\pi }$ are martingale differences, hence orthogonal in L2 space, and the Gaussian fourth moment givesThe Cauchy-Schwarz inequality and the Brownian increment variance give $\mathbb{E}{V}_{\pi }^2\le (t/2){\sum }_j(\Delta t_j{)}^2\le t^2|\pi |/2$. Both vanish. This proves the limit directly without a stochastic change-of-variables formula.

In a positive recurrent birth-death-immigration jump chain, let $I_n$ and $A_n$ count immigration events and upward steps before time $n$. The centered immigration indicators form bounded martingale differences; the strong law for martingales with bounded increments and the countable-state Markov chain ergodic theorem give $I_n/n\to {\sum }_k{\pi }_k\theta /(k+\theta )$ almost surely. The analogous argument for upward steps gives $A_n/n\to {\sum }_k{\pi }_k{u}_k$. Summing detailed balance gives ${\sum }_k{\pi }_k{u}_k={\sum }_k{\pi }_k{d}_k=1/2$, since every step is upward or downward. Therefore $I_n/A_n$ converges to the displayed fraction. For this chain the explicit stationary distribution reduces the fraction to $1-p/(1-p)$.

To establish the Brownian weighted quadratic variation in L2 for the deterministic partition of an interval, decomposewhere $U_{\pi }={\sum }_j{B}_j((\Delta B_j{)}^2-\Delta t_j)$ and $V_{\pi }={\sum }_j{B}_j\Delta t_j-{\int }_0^t{B}_{s}ds$.

The summands in $U_{\pi }$ are martingale differences: $B_j$ is ${\mathcal{F}}_{t_j}$-measurable, and the following Brownian increment is independent of ${\mathcal{F}}_{t_j}$. Their conditional means are zero, so distinct summands are orthogonal in L2 space. Since $\mathbb{E}{B}_j^2=t_j$ and $\mathbb{E}[((\Delta B_j{)}^2-\Delta t_j{)}^2]=2(\Delta t_j{)}^2$,For the Riemann sum error, the Cauchy-Schwarz inequality givesThus the triangle inequality in L2 space provesThe time integration variable is $s$.

For a martingale and a stopping time $T\le N$, expand the stopped value using its martingale differences:The event $\{T\ge k\}=\{T>k-1\}$ belongs to the filtration ${\mathcal{F}}_{k-1}$. Each summand is integrable and its conditional expectation given ${\mathcal{F}}_{k-1}$ is zero. Taking expectations of this finite sum therefore proves $\mathbb{E}{M}_T=\mathbb{E}{M}_0$ for every bounded stopping time. This proves the forward direction of the bounded-stopping-time characterization of a martingale without assuming an optional stopping theorem.

Use the uniform probability measure on the symmetric group $S_n$ and the unnormalized Hamming distanceLet $f:S_n\to \mathbb{R}$ have Lipschitz constant $K>0$ for this distance. We will prove the one-sided concentration inequalitiesConsequently $\mathbb{P}(|f-\mathbb{E}f|\ge t)\le 2e^{-t^2/(2n{K}^2)}$ for $t>0$. A zero Lipschitz constant makes $f$ constant, so that case is immediate.

Let $\sigma$ be a uniform permutation, reveal $\sigma (1),\dots ,\sigma (k)$ successively, and define the Doob exposure martingaleFix the first $k-1$ images. For an available value $a$, let $h(a)$ be the conditional mean when $\sigma (k)=a$. Given two available values $a,b$, left-composition with the transposition exchanging $a,b$ is a bijection from completions with next image $a$ to completions with next image $b$. It fixes the previously exposed images and changes exactly two coordinates of each complete permutation. Since conditional completions are uniform, averaging the Lipschitz condition over this pairing gives $|h(a)-h(b)|\le 2K$.

Thus, conditionally on the first $k-1$ images, $M_k-M_{k-1}$ has mean zero and takes values in an interval of length at most $2K$. We need the range-length version of the Hoeffding lemma: if a real random variable $Z$ has mean zero and lies in an interval of length $\ell$, thenHere is a proof. For $\psi (\lambda )=\log \mathbb{E}{e}^{\lambda Z}$, differentiation gives ${\psi }^{\prime \prime }(\lambda )={\mathrm{Var}}_{\lambda }(Z)$ under the exponentially tilted probability distribution. Any distribution on an interval of length $\ell$ has variance at most ${\ell }^2/4$, because its squared distance from the interval midpoint is at most ${\ell }^2/4$. Since $\psi (0)={\psi }^{\prime }(0)=0$, integrating this second-derivative bound gives the lemma for both signs of $\lambda$. The same argument holds for every conditional distribution.

Apply it conditionally to each martingale difference and use the tower property of conditional expectation repeatedly:The exponential Markov bound now gives, for $\lambda >0$,Choose $\lambda =t/(n{K}^2)$ to obtain the upper tail in the box, and apply the argument to $-f$ for the lower tail. This proves concentration of Lipschitz functions on the symmetric group. Using only the bound $|M_k-M_{k-1}|\le 2K$ in the absolute-increment Azuma-Hoeffding inequality would lose a factor four in the exponent; the conditional range length is what yields the constant needed in part (ii).

$Y_{j,k}$ is the number in group $k$ still observed and event-free immediately before event time $a_j$: its group-specific risk set size. Set $Y_j=Y_{j,0}+Y_{j,1}$. Under the common hazard function and conditional on exactly one event at that time, each at-risk individual has the same infinitesimal event probability. ThereforeThe indicator $g_{\pi (j)}$ records whether the event came from group 1, while $p(j,1)$ is its conditional expectation under equal hazards. Thus $z_j$ is the observed-minus-expected group-1 event count at that time. Under the null these are martingale differences, with conditional varianceThe log-rank statistic $z={\sum }_j{z}_j$ has null mean zero. Its predictable variance estimate is $V={\sum }_j{p}_j(1-p_j)$, so the equal-hazards test usesUse a two-sided large-sample log-rank test if no direction was prespecified; $V>0$ and adequate information are needed. The variance is the no-ties form. A positive $z$ indicates more group-1 events than expected. Independent censoring conditional on the model is required for these at-risk event intensities to describe the data.

Under the proportional hazards model, the total instantaneous group-$k$ rate is $Y_{j,k}{e}^{\beta k}{h}_0(a_j)$. Conditioning on the next event cancels the baseline hazard, givingKeep $z_j$ defined using its null probability $p(j,1)$, as in the question. Its conditional expectation under the alternative isIt is zero at $\beta =0$, as required, and has the sign of $\beta$ when both groups remain at risk. Replacing $p(j,1)$ inside the definition of $z_j$ with the alternative probability would give a different residual, whose mean would instead be zero under that alternative.

Let $I_n$ count immigration events during the first $n$ transitions, and let $A_n$ count upward steps during those transitions. Write $h(k)=\theta /(k+\theta )$ for the conditional immigration probability. If $J_m$ is the indicator that transition $m$ to $m+1$ is immigration, then $J_m-h(X_m)$ is a bounded martingale difference with respect to the history that records event types as well as population sizes. The strong law for martingales with bounded increments impliesSince $h$ is bounded, the countable-state Markov chain ergodic theorem givesApplying the same argument to upward-step indicators gives $A_n/n\to {\sum }_k{\pi }_k{u}_k$ almost surely. To evaluate this denominator, sum detailed balance over all neighboring edges:Since $u_k+d_k=1$ and ${\sum }_k{\pi }_k=1$, both sums equal $1/2$. The denominator limit is positive, so division is valid and the immigration fraction among upward steps isThis is the requested answer in terms of $\pi$. Substituting part (iii) additionally gives ${\sum }_k{\pi }_k\theta /(k+\theta )=(1-r)/2$, so the fraction simplifies to $1-r=(1-2p)/(1-p)$.

The adapted process is bounded by $1$, so every $X_n$ is an integrable random variable, indeed a square-integrable random variable. Independence of $U_{n+1}$ from ${\mathcal{F}}_n$ and its continuous uniform distribution giveThis calculation with a random threshold is justified by conditioning on the ${\mathcal{F}}_n$-measurable value of $X_n$: for a fixed $t\in [0,1]$, the Lebesgue integral ${\int }_0^1{1}_{\{u\le t\}}du$ equals $t$. More formally, independence integrates the bounded Borel measurable function $(t,u)\mapsto 1_{\{u\le t\}}$ over its second coordinate. Thereforewhich verifies the martingale property for this filtration.

Since $B_{n+1}^2=B_{n+1}$, the conditional expectation of the squared martingale difference isThe law of total expectation yields $\mathbb{E}[(X_{n+1}-X_n{)}^2]=\frac{1}{4}\mathbb{E}[X_n(1-X_n)]$. In particular, $\mathbb{E}{X}_n=x$ for every $n$.