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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.

For the unit-rate exponential distribution, direct integration givesFor $x>0$, maximizing $\theta x+\log (1-\theta )$ gives $\theta =1-1/x$, and therefore $I(x)=x-1-\log x$. For $x\le 0$ the supremum is infinite, as seen by taking $\theta \to -\infty$.

For $a>1$, $I$ is increasing on $[a,\infty )$, and its continuity gives ${\inf }_{[a,\infty )}I={\inf }_{(a,\infty )}I=I(a)$. The closed-set upper bound and open-set lower bound of the Cramér theorem squeeze the desired limit to $-I(a)$. For $a=1$, both infima are $0$, because $I(1)=0$ and $I(x)\to 0$ as $x\downarrow 1$. For $0\le a<1$, the strong law of large numbers gives $S_n/n\to 1$ almost surely, so the tail probability tends to $1$; at $a=0$ it is identically $1$.

Consequently the exponential sample-mean upper-tail rate isThe threshold case uses the rate-function lower bound on the open half-line, rather than applying the law of large numbers at equality.

The strong law of large numbers gives $S_n/n\to -1$ almost surely, so $S_n\to -\infty$. Therefore the infinite-horizon maximum $Z$ is finite and attained almost surely; also $Z\ge 0$ because $S_0=0$.

Apply the Doob maximal inequality to $M_k=e^{2S_k}$ through time $n$. Since $\mathbb{E}{M}_n=1$,These finite-horizon events increase to $\{Z>t\}$. Continuity of probability under increasing unions givesFor $t<0$ the same bound is trivial, because its right side exceeds $1$. Only bounded-horizon maximal inequalities were used before passing to the limit; no unjustified stopping at an infinite-horizon crossing time is needed.

The Skorokhod embedding of a centered random walk states that a random walk with independent identically distributed centered steps of finite variance ${\sigma }^2$ can be realized on an appropriate probability space aswhere $B$ is a standard Brownian motion and the $T_n$ are finite stopping times. More precisely, the stopped positions have the same joint law as the given random walk, and the pairsmay be chosen independent and identically distributed. Their spatial component has the step law, and $\mathbb{E}(T_n-T_{n-1})={\sigma }^2$. Repeating the one-step Skorokhod embedding theorem with the Strong Markov property gives this formulation. In the present normalization, the mean time increment is one, and the strong law of large numbers gives $T_n/n\to 1$ almost surely.

The Donsker invariance principle states that the linearly interpolated diffusively rescaled random walkconverges weakly as a random element of $C[0,1]$, equipped with the uniform norm, to standard Brownian motion restricted to $[0,1]$. At $t=1$ the fractional term is zero. The only step assumptions needed here are zero mean, unit variance, and independent identical distributions; a higher moment or bounded support is not required. This is a functional central limit theorem, concerning the entire interpolated path rather than only its endpoint.