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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.

The useful autoregressive model is for the squares, not for signed observations. Put $Y_t=X_t^2$ andThenThe errors form a martingale difference sequence relative to the noise history, because the current standardized noise is independent of the past. When the fourth moment is finite they have finite variance and are uncorrelated across distinct times, although their conditional variance depends on the regressor. In centered form, $Y_t-m_2={\alpha }_2(Y_{t-2}-m_2)+u_t$.

For ordinary least squares, use the $n-2$ response-regressor pairs $y_t=X_t^2$, $r_t=X_{t-2}^2$, $t=3,\dots ,n$. With their separate means $\bar{y}$ and $\bar{r}$, minimize ${\sum }_{t=3}^n(y_t-a-b{r}_t{)}^2$. If the regressor sum of squares is positive, the estimators areThe two means use the matched pairs; replacing them indiscriminately by a single full-sample mean is not the exact least-squares formula. The noise-series representation supplies an ergodic stationary process. If $3{\alpha }_2^2<1$, finite regressor second moments and the error's zero conditional expectation justify the usual population regression and statistical consistency argument. The observations still define a finite-sample least-squares fit outside that moment range, but the ordinary finite-variance justification must not be claimed there. If parameter constraints are required, minimize the same criterion subject to $a>0$ and $0<b<1$, rather than assert that unconstrained estimates automatically satisfy them.

This covariance is also zero, but symmetry is essential to the proof. Write ${\epsilon }_t={\eta }_t|{\epsilon }_t|$, where ${\eta }_t$ is an independent fair sign. The normal distribution makes that sign independent of its magnitude and of every other driving variable. Conditional on the magnitudes and all noise except this sign, changing ${\eta }_t$ flips $X_t$ and leaves $X_t^2$ unchanged. Every future conditional scale uses only squared past observations, so $X_{t+h}$ is unchanged for $h>0$.

Thus $f(X_{t+h})$ is independent of the remaining fair sign in $X_t$. Averaging that sign gives $\mathbb{E}[X_{t}f(X_{t+h})]=0$, and henceThe Cauchy-Schwarz inequality again guarantees integrability. For odd $h$, the parity decomposition of a lag-two ARCH process also gives independence directly. For even $h$, the sign symmetry of an ARCH process supplies the argument; the martingale difference sequence property alone would not justify this reversed covariance.

The covariance is zero. Let ${\mathcal{F}}_t=\sigma ({\epsilon }_s:s\le t)$. The stationary noise-series representation makes $f(X_t)$ measurable with respect to ${\mathcal{F}}_t$, whereas the martingale difference sequence property gives $\mathbb{E}[X_{t+h}\mid {\mathcal{F}}_{t+h-1}]=0$. The law of total expectation therefore givesSince $\mathbb{E}{X}_{t+h}=0$, this is the required covariance. All products are integrable by the Cauchy-Schwarz inequality, using finite second moments of $X$ and the stipulated square integrability of $f(X_t)$.

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.

With symmetric driving noise whose sign is independent of its magnitude, changing the sign of one driving variable changes the corresponding $X_t$ but leaves future conditional scales unchanged. Consequently $\mathrm{Cov}(X_t,f(X_{t+h}))=0$ for $h>0$ and square-integrable $f(X_{t+h})$. This conclusion uses symmetry, beyond the martingale difference sequence property.

If a martingale has increments bounded by a single deterministic constant, then $M_n/n\to 0$ almost surely. Its increments form a uniformly $L^2$-bounded martingale difference sequence; apply the strong law for uniformly L2-bounded uncorrelated random variables to their partial sums and note $M_0/n\to 0$. No square-integrability of the initial value is needed beyond the usual martingale integrability.

If ${\sup }_n\mathbb{E}|X_n{|}^2<\infty$ and the $X_n$ are pairwise uncorrelated random variables, then $n^{-1}{\sum }_{k=1}^n(X_k-\mathbb{E}{X}_k)\to 0$ almost surely. Independence is unnecessary for this version of the strong law of large numbers. A uniformly bounded martingale difference sequence meets its hypotheses, since the differences have mean zero and are orthogonal in $L^2$.