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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 $1-p/(1-p)$.

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.

Write $D_k=M_k-M_{k-1}$ for the martingale difference sequence. Bounded increments mean that one deterministic $C<\infty$ satisfies $|D_k|\le C$ for every $k$, almost surely. Then $\mathbb{E}{D}_k=0$ and, for $j<k$,The increments are uncorrelated random variables, uniformly bounded in $L^2$. Applying the supplied strong law for uniformly L2-bounded uncorrelated random variables gives $n^{-1}{\sum }_{k=1}^n{D}_k\to 0$ almost surely. Also $M_0/n\to 0$, because $M_0$ is finite almost surely. Therefore a martingale with bounded increments has zero linear growth:This is the strong law for martingales with bounded increments.

For the submartingale, let $\Delta Y_k=Y_k-Y_{k-1}$ and define its predictable drift and compensator byThis is the Doob decomposition in discrete time. The process $N_n$ is a martingale. Since $|\Delta Y_k|\le C$, we have $0\le a_k\le C$, so $|N_k-N_{k-1}|\le 2C$. The result just proved gives $N_n/n\to 0$ almost surely, and $A_n\ge 0$. Consequently the strong law for submartingales with bounded increments isThe compensator may grow, so the conclusion is a lower bound rather than a claim of convergence to zero.

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 $1-r=(1-2p)/(1-p)$.

If a submartingale has uniformly bounded increments, then ${\mathrm{liminf}}_n{Y}_n/n\ge 0$ almost surely. In its Doob decomposition in discrete time, the predictable compensator is nondecreasing, while the martingale part has bounded increments. The strong law for martingales with bounded increments makes the latter negligible on the linear scale.