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In the slotted ALOHA model, a station retains one packet until it successfully transmits. Conditional on $(S_t,N_t)=(S,N)$, take fresh independent trials with a Bernoulli distribution of parameter $1/S$ for the $N$ backlogged stations. Let arrivals be independent across slots and independent of these trials, with Poisson distribution of mean $\nu$. New arrivals join the next slot's backlog. An idle slot serves nobody, exactly one attempt serves one packet, and a collision serves nobody. Consequently the backlog changes by arrivals minus the success indicator.

Under these independence assumptions, $(S_t,N_t)$ is a time-homogeneous Markov chain: the feedback probabilities and the independent arrival law depend only on its current state. Merely specifying Poisson arrival marginals would not suffice. For example, take $S_t=1$, $a=b=c=0$, $N_0=0$, and arrivals $Y_0=H$, $Y_1=K$, $Y_2=H$ for independent random variables $H,K$ with Poisson distributions. Histories with $(N_1,N_2)=(0,2)$ and $(2,2)$ both have current backlog two, but the next backlog is respectively two and four. The usual model therefore includes fresh arrivals as an assumption.

Write $p_0=(1-1/S{)}^N$ and $p_1=(N/S)(1-1/S{)}^{N-1}$. Set $p_1=0$ for $N=0$, and use their exact binomial distribution probabilities at $S=1$. Away from the clipping boundary, the conditional drifts areNear $S=1$, the latter must use $\max (1,S+d)-S$ for each update $d$, rather than $d$ itself.

When both coordinates are large with $N/S\to \kappa$, the Poisson limit theorem approximates the attempt count, which has a binomial distribution, by a Poisson distribution of mean $\kappa$. Thus $p_0\to e^{-\kappa }$ and $p_1\to \kappa e^{-\kappa }$. Rescaling state by $K$ and slot time by $K$ motivates the fluid approximation of slotted ALOHAThis is an interior fluid approximation of slotted ALOHA, not the exact conditional drift at a clipped boundary.

Here is one explicit set of sufficient conditions, independent of the arrival rate within the subcritical range:Put $c_0=2/(e-2)$. ThenHence $g$ is negative below one and positive above one: the controller decreases the attempt denominator when offered contention is too small, and increases it when contention is too large.

Use the ratio time change for a homogeneous fluid model, $du/dt=1/s$. The equations becomeFor $\kappa \ge 1$,since $g(\kappa )\ge 0$ and $A(1+c_0)>1$. Also $h(1)=\nu -e^{-1}<0$. Therefore $h$ is negative on $[1,\infty )$, and by continuity on $[\bar{\kappa },\infty )$ for some $\bar{\kappa }<1$. Its value at zero is $\nu \ge 0$, so nonnegative backlog is preserved.

The ratio remains bounded by $K=\max (\kappa (0),1)$ and eventually enters $[0,\bar{\kappa }]$: on the compact interval $[\bar{\kappa },K]$, its derivative is bounded above by a strictly negative number. Once inside it cannot cross upward. The bounded ratio and smooth coefficients make the transformed equations exist for all $u\ge 0$. In this region $g(\kappa )\le g(\bar{\kappa })<0$, so $s(u)$ decreases at least exponentially in $u$. The original time satisfies $t(u)={\int }_0^{u}s(v)dv$ and has a finite limit $T$; boundedness of $\kappa$ gives $n(u)=\kappa (u)s(u)\to 0$ as well. Thus every nonnegative interior fluid trajectory drains to the origin in finite fluid time. At the origin the ratio equation is undefined; the usual stopped fluid trajectory is held there afterward. A state with $s=0<n$ enters the interior under the continuous limiting boundary drift $\dot{s}=c>0$; the same argument then applies.

This proof supplies sufficient conditions, not a characterization of every stabilizing triplet. It also does not assert the positive recurrent Markov chain property of the original stochastic chain solely from the heuristic ODE.

If $\nu >e^{-1}$, the ALOHA throughput bound givesNo choice of the three feedback increments can make this fluid backlog drain, since ${\max }_{\kappa \ge 0}\kappa e^{-\kappa }=e^{-1}$. For the displayed controller, the supercritical fluid trajectory is global and grows instead of draining. At $\nu =e^{-1}$ the ray $n=s>0$ consists of stationary fluid states, explaining why the strict load inequality matters.

For fixed $k$ and sufficiently large $N$,The finite product and last factor tend to one, while $(1-\lambda /N{)}^N\to e^{-\lambda }$. Thus the Poisson limit theorem givesThese numbers form a probability mass function because they are nonnegative and their sum is $e^{-\lambda }{\sum }_{k\ge 0}{\lambda }^k/k!=1$, by the exponential series. They define the Poisson distribution of parameter $\lambda$. Only $N\ge \lambda$ is needed to make the binomial parameter valid, which is sufficient for the limit.

Use the slotted ALOHA model with one outstanding packet per active station and a collision channel: a slot with exactly one attempt succeeds, whereas an idle slot or a slot with two or more attempts delivers nothing. Let $Y_t$ be independent Poisson random variables of mean $\nu$, independent also of the transmission choices. Arrivals join after the current slot's attempts. Given $(S_t,N_t)=(S,N)$, the attempt count has a binomial distribution with parameters $N$ and $1/S$. Consequently, for $N\ge 1$,For $N=0$, set $p_0=1$ and $p_1=0$. The equation $N_{t+1}=N_t+Y_t-1_{\{Z_t=1\}}$ adds arrivals and removes precisely one packet following a successful slot. The estimate $S_t\ge 1$ controls the common transmission probability and updates using the idle, success, or collision feedback. The conditional distribution of the next state depends only on $(S_t,N_t)$, so this pair is a Markov chain.

Away from the reflecting floor in $S$, its one-step conditional expectations areWhen $N$ and $S$ are large with $N/S\approx \kappa$, the Poisson limit theorem gives $p_0\approx e^{-\kappa }$ and $p_1\approx \kappa e^{-\kappa }$. Treating these approximate drifts as derivatives motivates the fluid approximation of slotted ALOHA:The floor at $S=1$ is negligible in this interior scaling. This calculation motivates the ordinary differential equations; it is not a proof of a stochastic fluid limit.

For $(a,b,c)=(2-e,0,1)$, putThen $g^{\prime }(\kappa )=(e-2+\kappa )e^{-\kappa }>0$ and $g(1)=0$. On the domain $s>0$, introduce time $\tau$ by $d\tau /dt=1/s$. Differentiating $\kappa =n/s$ giveswhere $r(\kappa )=\kappa [1-(e-2+\kappa )e^{-\kappa }]$.

The function $r$ is strictly increasing from zero to infinity, and $r(1)=e^{-1}$. To check the monotonicity explicitly,Indeed, with $d=3-e\in (0,1)$,because its discriminant is $d^2-4d<0$; and $e^{\kappa }\ge 1+\kappa$. Thus the scalar ordinary differential equation for $\kappa$ points toward the unique value ${\kappa }_{\ast }$ satisfying $r({\kappa }_{\ast })=\nu$. Every trajectory with $s(0)>0$, $n(0)\ge 0$ has bounded $\kappa$ and $\kappa (\tau )\to {\kappa }_{\ast }$. Positivity of $s$ follows from $s(\tau )=s(0)\exp ({\int }_0^{\tau }g(\kappa (u))du)$.

If $0\le \nu <e^{-1}$, then ${\kappa }_{\ast }<1$ and $g({\kappa }_{\ast })<0$. Thus $s(\tau )$ decays exponentially for large $\tau$, andThis is finite-time draining of a fluid model. The unmodified equations have $n/s$ undefined at the origin, so the statement that trajectories converge to the origin needs this endpoint interpretation; continuing with an absorbing zero trajectory is an additional fluid-model convention, not a solution of the displayed equations at the origin.

If $\nu >e^{-1}$, then ${\kappa }_{\ast }>1$ and $g({\kappa }_{\ast })>0$. Now $s(\tau )$ grows exponentially, physical time tends to infinity, andIn particular the trajectory escapes to infinity. Even without the ratio analysis, $\dot{n}\ge \nu -e^{-1}>0$, since ${\max }_{\kappa \ge 0}\kappa e^{-\kappa }=e^{-1}$.

For the actual Markov chain, it is essential not to replace the finite-$N$ success probability by $e^{-1}$: a single backlogged packet can succeed with probability one. Instead, for $N\ge 2$, maximizing over $S\ge 1$ gives the ALOHA throughput boundIf $\nu >e^{-1}$, choose $K_0$ and $r_0<\nu$ so that $p_1(N,S)\le r_0$ for all $N>K_0$, uniformly in $S$. For sufficiently small $\theta >0$,because its derivative at zero is $r_0-\nu <0$. Independence of arrivals and attempts giveswhenever $N>K_0$. For any integer $K\ge K_0$, stop at the stopping time ${\sigma }_K=\inf \{t:N_t\le K\}$. The stopped process $e^{-\theta N_{t\wedge {\sigma }_K}}$ is a nonnegative supermartingale. The optional stopping theorem, first at bounded times and then by a limit, yields the exponential-supermartingale escape boundFrom every state with $N\le K$, a Poisson distribution arrival burst has a uniformly positive probability of sending the next backlog to at least $K+m$, for any fixed $m>0$. From there the probability of never revisiting $N\le K$ is at least $1-e^{-\theta m}$. By the Markov property at successive visits, the probability of infinitely many visits to this strip is zero: each visit has a uniform positive chance of being the last. This holds for every integer $K\ge K_0$, soTherefore the Markov chain is transient for every finite choice of $a,b,c$. The proof is uniform in $S$ and does not infer stochastic transience merely from the approximate ordinary differential equations.

Writing $Y_i$ for team1_goal and $x_i$ for team1_xg, the first hypothesis concerns only the expected value:The second specifies the entire conditional probability distribution:with independence between matches as assumed in the question. The Poisson limit theorem, often called the law of small numbers in this setting, suggests a Poisson distribution when a goal count is the sum of many approximately independent rare scoring opportunities.

Let ${\epsilon }_1,\dots ,{\epsilon }_n$ be independent $\mathrm{Bernoulli}(\lambda /n)$ variables and $S_n={\sum }_i{\epsilon }_i$. Apply the tensorization of entropy to $f(S_n)$ and then apply the stated Bernoulli log-Sobolev inequality in each coordinate. If $S_n^{(i)}=S_n-{\epsilon }_i$, this givesFor each fixed $i$, $(S_n^{(i)},{\epsilon }_i)$ converges in distribution to $(X,0)$, where $X\sim \mathrm{Poisson}(\lambda )$. The Poisson limit theorem in fact gives convergence in total variation. The assumptions $K_1\le f\le K_2$ and $|Df|\le K_3$ make all displayed integrands bounded, so expectations and entropy pass to the limit. Since $n(\lambda /n)(1-\lambda /n)\to \lambda$,

In a uniform balls-in-bins allocation with $m\to \infty$ and $n/m\to \lambda \in (0,\infty )$, a fixed bin count has convergence in distribution to a Poisson distribution variable of mean $\lambda$ by the Poisson limit theorem. If $N_j$ counts bins containing exactly $j$ balls, symmetry and the linearity of expectation give $\mathbb{E}[N_j]/m=\mathbb{P}(B_1=j)\to e^{-\lambda }{\lambda }^j/j!$ for each fixed $j$. In particular the expected empty fraction tends to $e^{-\lambda }$, and the expected fraction with at least two balls tends to $1-(1+\lambda )e^{-\lambda }$. These are statements about expectations; convergence of the random fractions requires an additional argument.