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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 a planar simple symmetric random walk write its coordinates as $(S_n,T_n)$ and rotate to $(S_n+T_n,S_n-T_n)$. Each step now has four equally likely values $(1,1),(1,-1),(-1,1),(-1,-1)$. Thus the two rotated coordinates are independent one-dimensional simple symmetric random walks.

A return is impossible at odd times, and at time $2n$ its probability isFor $n\ge 2$, the supplied estimate gives $1/(4n)<p_{2n}(0,0)<1/n$. At $n=1$ the lower bound is actually equality, $p_2(0,0)=1/4$; this harmless strict-endpoint typo does not affect the argument. The harmonic series diverges, so the criterion proved above shows that the origin is recurrent, in the terminology of a recurrent Markov chain. Translation invariance gives the same conclusion for every state. The Markov chain is irreducible, so from any starting point it hits any fixed state almost surely and revisits it infinitely often. One can see the hitting assertion directly: repeated returns to the starting point give repeated positive-probability opportunities to follow a finite path to the target.

For a Markov chain invariant measure, $xP=x$, and hence $x{P}^n=x$. The irreducible Markov chain property supplies, for each $j$, an $n$ with $(P^n{)}_{kj}>0$. Nonnegativity gives $x_j\ge x_k(P^n{)}_{kj}>0$. Every coordinate of a nonzero invariant measure is positive.

Start the Markov chain at $k$, and let $T_k=\inf \{n\ge 1:X_n=k\}$ be its first return time. Each product in the series is the probability of a path with no intermediate visit to $k$. ThusThis is the expected number of visits to $j$ in an excursion, including the return endpoint and excluding the starting point. Because the chain is recurrent Markov chain, $T_k<\infty$ almost surely, and exactly one visit to $k$ is counted: ${\gamma }_k^k=1$. Counting the starting point instead of the return endpoint gives the equivalent usual excursion occupation measure.

For the lower bound, let $J=I\setminus \{k\}$, $Q=(p_{ij}{)}_{i,j\in J}$, and $r=(p_{kj}{)}_{j\in J}$. The stationarity equations on $J$, with $x_k=1$, are $x_J=r+x_{J}Q$. Iterating and using nonnegativity yieldsThe $j$th coordinate on the right is precisely the partial excursion series for $j\ne k$. Taking its increasing limit gives $x_j\ge {\gamma }_j^k$; for $j=k$ equality follows from $x_k={\gamma }_k^k=1$. The iteration is valid also on a countable state space because all summands are nonnegative.