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Independent marking is the hypothesis that guarantees the Poisson thinning theorem. Equal marginal retention probabilities do not guarantee it. Consider one Bernoulli random variable $B$ with success probability $p\in (0,1)$, independent of a Poisson process, and retain every arrival when $B=1$ and none when $B=0$. Each arrival is retained with probability $p$, but the marks share the same coin. In an interval whose original count has Poisson distribution of mean $m>0$, the retained count is $R=BN$ and the rejected count is $D=(1-B)N$, soThe extra variance shows that $R$ is not Poisson distributed; the negative covariance proves dependence. Its no-arrival probability is $1-p+p{e}^{-m}$, rather than $e^{-pm}$. For the toy values $p=1/2,m=2$, both rules have retained mean $1$, but the common-coin rule has variance $2$ and no-arrival probability $(1+e^{-2})/2\approx 0.567668$, compared with variance $1$ and no-arrival probability $e^{-1}\approx 0.367879$ for independent marking. These formulas follow by conditioning on $B$ and using the law of total variance.

An arrival-dependent or adversarial retention rule can change the law just as strongly. Keeping every second arrival gives $R_t=\lfloor N_t/2\rfloor$. Its successive retained gaps are sums of two independent exponential random variables, hence have the Erlang distribution of shape two and rate $\lambda$, not an exponential distribution. The resulting renewal process and its count variance are analyzed in Every second Poisson arrival has underdispersed counts, and the two dependent streams in alternating split of a Poisson process. As an even more restrictive rule, keeping only the first arrival ever gives $R_t=1_{\{N_t\ge 1\}}$, a Bernoulli distribution at each $t>0$, so it cannot have a positive-mean Poisson distribution. A thinning rule must specify how decisions depend on arrivals and on one another; the independent-marking conclusion cannot be applied automatically to arbitrary deletion.

A renewal process counts arrivals $N(t)=\max \{j:A_j\le t\}$, where $A_0=0,A_j=S_1+\cdots +S_j$ and the holding times are independent, identically distributed positive random variables. Their sum tends to infinity almost surely, so the count is finite on bounded time intervals. Use $N(s)=X_s$ in the question's notation. The index $J=N(s)+1=\min \{j:A_j>s\}$ is a stopping index for the holding-time sequence. Conditional on $J=j$ and the first $j$ holding times, every later $S_{j+k}$ retains its original probability distribution. Summing over $j$ provesThe straddling time $T_1=S_J$ is different. Let $\overline{F}(t)=P(S_1>t)$ and let $U$ be the renewal measure $U(B)={\sum }_{j\ge 0}P(A_j\in B)$. Independence of the next holding time from its start givesFor two numbers in $[0,1]$, their minimum is at least their product. Since $\overline{F}(\max (t,a))=\min (\overline{F}(t),\overline{F}(a))$, subtracting $\overline{F}(t)$ times the second integral provesThese formulae handle holding-time atoms with the stated renewal convention $A_j\le s<A_{j+1}$; sums may be interchanged by nonnegativity.

For rate-$\lambda$ exponentials, the renewal process is Poisson. The elapsed age $B_s=s-A_{N(s)}$ has probability distribution $\min (E,s)$ with $E\sim \mathrm{Exp}(\lambda )$: for $0\le b<s$, $P(B_s>b)$ is the probability of no arrivals in $(s-b,s]$, namely $e^{-\lambda b}$, with an atom $e^{-\lambda s}$ at $s$. The forward residual $R_s$ is an independent rate-$\lambda$ exponential by independent future increments. Thus $T_1=B_s+R_s$. Conditioning on the age gives the exact survival lawFor $s,t>0$ the extra term is strictly positive, proving strict inequality. For fixed $t$ and $s\to \infty$, the limit is $e^{-\lambda t}(1+\lambda t)$, which equals the survival probability of the sum of two independent rate-$\lambda$ exponentials. This is the finite-time form of the stochastic length bias of a renewal interval.

The equilibrium-delay theorem for a renewal process with positive holding time $S$, finite mean $\mu$, and distribution function $F$ says that the first future renewal must have the equilibrium residual-life distribution,independently of subsequent ordinary holding times. This is the residual lifetime seen at a stationary observation time, rather than an ordinary inter-renewal lifetime.

Here $\mu =3/\lambda$ and $1-F(t)=e^{-\lambda t}(1+\lambda t+{\lambda }^2{t}^2/2)$. Thus the required initial-delay density isEquivalently, choose with equal probability a sum of one, two, or three independent rate-$\lambda$ exponential random variables. The three Erlang distributions have densities that average to the displayed density. A concrete stationary construction chooses the phase modulo three of a stationary Poisson clock uniformly; the number of exponential stages until its next third-stage renewal is then uniform on $\{1,2,3\}$.

Assume the first-fork choices are independent Bernoulli routings, as in Poisson thinning. Put $r=2\lambda /3$. Cars assigned to A form a Poisson process of rate $\lambda /3$, also an ordinary renewal process with independent exponential holding times of rate $\lambda /3$. The stream to the second fork is an independent Poisson process of rate $r$.

B receives its odd-numbered arrivals. Its initial delay is exponential of rate $r$, followed by independent Erlang distributions of shape two and rate $r$ between successive B arrivals. Thus B is a delayed renewal process, not Poisson. C receives the even-numbered arrivals: its first and all subsequent holding times are independent Erlang$(2,r)$, so C is an ordinary renewal process, not Poisson. Their holding-time density and survivor are $r^{2}t{e}^{-rt}$ and $e^{-rt}(1+rt)$. Both have asymptotic rate $r/2=\lambda /3$; equality of these mean rates with A does not make the processes Poisson.

Let $S_k={\sum }_{\ell =1}^k{Y}_{\ell }$, $S_0=0$, be the epochs of the renewal process. The definition uses the next epoch at or after $n$, so $X_n=0$ at an epoch. If $X_n=i\ge 1$, no renewal occurs before $n+i$ and necessarily $X_{n+1}=i-1$.

If $X_n=0$, time $n$ is a renewal epoch, and the next unused interarrival time is independent of the entire observed past with the original distribution. To justify the random index, condition on each possible number $k$ of renewals up to $n$: the event identifying $k$ depends only on $Y_1,\dots ,Y_k$, whereas $Y_{k+1}$ is independent of those variables. Summing over $k$ preserves that distribution and independence. Therefore the residual lifetime Markov chain has transition probabilitiesIts reachable state space is $E=\{j\ge 0:\mathbb{P}(Y_1>j)>0\}$, either an initial finite segment or all nonnegative integers. The conditional next-step law depends only on $X_n$, proving the Markov property. Initially $X_0=0$. The reset is $Y-1$, rather than $Y$, because one unit of time has elapsed by the next step.

A renewal process has independent identically distributed, nonnegative, almost surely finite inter-arrival times $X_i$, epochs $S_0=0,S_n={\sum }_{i=1}^n{X}_i$, and count $N(t)=\max \{n:S_n\le t\}$. Allowing zero inter-arrivals is harmless provided $\mathbb{P}(X_1>0)>0$. The hypothesis $\mathbb{E}{X}_1>0$ ensures there is $\epsilon >0$ with $p=\mathbb{P}(X_1\ge \epsilon )>0$.

The strong law of large numbers for the indicators gives $S_n\ge \epsilon {\sum }_{i=1}^n{1}_{X_i\ge \epsilon }\to \infty$, so $N(t)$ is finite for finite $t$. Conversely every fixed $S_n$ is finite, so eventually $N(t)\ge n$. On their countable common probability-one event, this proves $\boxed{N(t)\to \infty \text{ as }t\to \infty \text{ almost surely}}$.

For fixed $t$, put $L=\lfloor t/\epsilon \rfloor$. The hinted comparison process has epochs $S_n^{\prime }=\epsilon {\sum }_{i=1}^n{1}_{X_i\ge \epsilon }\le S_n$, hence $N(t)\le N^{\prime }(t)$. Equivalently,If $p<1$, for $n>L$ this is bounded by $C_{t,p}{n}^L(1-p{)}^n$, by summing the first $L+1$ binomial distribution. Thus ${\sum }_n{e}^{\theta n}\mathbb{P}(N(t)=n)<\infty$ for $0<\theta <-\log (1-p)$. If $p=1$, $N(t)\le L$ deterministically. Therefore a strictly positive exponential moment exists for every fixed finite $t$, with the same allowed $\theta$ for all such $t$. Almost-sure finiteness of the inter-arrivals is implicit in a proper renewal process; allowing an infinite inter-arrival with positive probability would invalidate the divergence claim.

Let ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$ be the natural filtration. An integer-valued random variable $M$ is a stopping time with respect to $(X_i)$ whenThus whether one has stopped by time $n$ can be decided from the first $n$ observations, without seeing future variables. Equivalently, $\{M\ge i\}\in {\mathcal{F}}_{i-1}$.

Write the stopped sum asThe event $\{M\ge i\}$ depends only on $X_1,\dots ,X_{i-1}$, so its indicator and $X_i$ are independent random variables. Since $\mathbb{E}|X_1|<\infty$ and $\mathbb{E}M<\infty$, Tonelli theorem givesWe may consequently exchange expectation and summation. Using the tail-sum formula for $M$ proves Wald's equation:

For the renewal process, put $S_n=X_1+\cdots +X_n$ and $N_t=\max \{n:S_n\le t\}$. The random variable $N_t+1$ is a stopping time. Since $S_{N_t+1}>t$, Wald's equation givesand hence

For $c>0$, set $Y_i=X_i\wedge c$ and ${\mu }_c=\mathbb{E}{Y}_i$. Wald's equation still applies to the stopping time $N_t+1$ and the independent variables $Y_i$. Because $S_{N_t}\le t$,ThereforeBy the monotone convergence theorem, ${\mu }_c\uparrow \mu$ as $c\to \infty$. Combining the bounds proves the elementary renewal theorem

For a fixed word over $q$ equally likely symbols, the waiting time for a word in independent uniform symbols has mean $\sum q^k$, where $k$ runs through the lengths of the borders of a word, including the full word. The word lava' has no proper nonempty border, so $$ \boxed{\mathbb E\tau_{\mathtt{lava}}=100^4=100\,000\,000}. $$ The word aa' has borders of lengths $1$ and $2$, soFor a direct check, let $E_0$ be the expected remaining time with no trailing a' and $E_1$ with one trailing a'. Thenwhich gives $E_0=10100$.

For independent identically distributed nonnegative random variables ${\xi }_1,{\xi }_2,\dots$, define the renewal epochsThe associated renewal process isthe number of renewals by time $t$.

PutThis is the renewal process generated by the interarrival times. Since the variables have a density and positive finite mean, $S_n\to \infty$ almost surely, so $N_t$ is finite and tends to infinity with $t$. The strong law givesOn this probability-one event,Dividing by $N_t$ and using $N_t\to \infty$ givesBoth outside expressions tend to $1/\lambda$, soThis is the renewal counting law of large numbers.

First identify the recurrence type, rather than infer an occupation limit from a one-time probability. At time $k(a+b)$, return to zero requires exactly $kb$ upward steps. The binomial distribution giveswhere the last step is Stirling's approximation. At all other positive times the return probability is zero. Hence ${\sum }_{n\ge 0}{\mathbb{P}}_0(S_n=0)=\infty$. The recurrence criterion by return probabilities proves the recurrent property; by the irreducible Markov chain property established in (i), every state is recurrent. The equivalence of positive recurrence and existence of a stationary distribution for an irreducible Markov chain on a countable state space, together with (iii), makes this chain null recurrent.

For completeness, the zero occupation proportion follows directly by treating returns as a renewal process. Let $L_1,L_2,\dots$ be successive strictly positive return-cycle lengths to zero. The Strong Markov property makes them independent and identically distributed, the recurrent property makes them finite almost surely, and the null recurrent property gives $\mathbb{E}{L}_1=\infty$. For each positive integer $K$, the strong law of large numbers applied to $\min (L_j,K)$ givesTake a common event of probability one for these countably many $K$. By the monotone convergence theorem, the right side increases to infinity. Thus the number of completed return cycles by time $n$, divided by $n$, tends to zero. Counting the initial visit changes this ratio by at most $1/n$. Consequently,The walk visits zero infinitely often, but the asymptotic fraction of time spent there is zero.

Let $d=\gcd (a,b)>1$ and put $a^{\prime }=a/d$, $b^{\prime }=b/d$. Then $a^{\prime },b^{\prime }$ are coprime integers, and $p=b^{\prime }/(a^{\prime }+b^{\prime })$. The six requested revisions are as follows.

(b)(i): communicating classes. Every increment is divisible by $d$, so the residue class modulo $d$ is preserved. Conversely, if $j-i$ is divisible by $d$, Bézout's identity applied to $a^{\prime },b^{\prime }$ gives a path just as in (b)(i). Thus the original Markov chain is reducible on $\mathbb{Z}$, with exactly the $d$ closed communicating classesThe specified walk starting at zero stays in $d\mathbb{Z}$.

(b)(ii): period of a state in a Markov chain. The return equation $au=bv$ reduces to $a^{\prime }u=b^{\prime }v$. Hence $u=b^{\prime }k$, $v=a^{\prime }k$. For every state, the period of a state in a Markov chain is thereforeThis is still at least two, so no state is aperiodic.

(b)(iii): stationary distribution. The conclusion is unchanged: there is no stationary distribution on $\mathbb{Z}$. The maximum-mass proof in (b)(iii) applies verbatim, without using coprimality. Equivalently, each closed communicating class, after translating and dividing its state labels by $d$, is the same centered random walk with step sizes $a^{\prime },-b^{\prime }$ and has no stationary distribution.

(b)(iv): occupation of zero. On $d\mathbb{Z}$, the rescaled walk $S_n/d$ has exactly the transition probabilities of the coprime walk with parameters $a^{\prime },b^{\prime }$. The return-probability argument in (b)(iv) therefore makes this class null recurrent. The renewal process proof of the occupation limit applies unchanged:Starting in another residue class would make every visit to zero impossible, also giving occupation proportion zero.

(c)(i): modified communicating classes. These are unchanged: $C=\{-c-b,\dots ,c+a\}$ is one finite closed communicating class, and all other states are transient singleton classes. The original proof used consecutive reset destinations and never used coprimality. Resetting to the $b$ consecutive integers $c-b+1,\dots ,c$ connects different residue classes, so it would be incorrect to split $C$ according to residue classes modulo $d$.

(c)(ii): limiting distribution. The column-sum calculation, the irreducible Markov chain property on a finite class, positive self-transition probabilities, and geometric entrance argument also required no coprimality. Thus for every initial state the same convergence in distribution holds:In particular, the denominator $2c+a+b+1$ is unchanged; it is not divided by $d$.

If every interspike interval is at least $t_{\mathrm{abs}}>0$, a sustained spike train has mean firing rate at most $1/t_{\mathrm{abs}}$. Across a window of duration $T$, the event count satisfies $N(T)\le 1+T/t_{\mathrm{abs}}$; dividing by $T$ and taking the long-time limit gives the bound. The extra one accounts for a possible first event. The bound permits much slower firing and says nothing about whether stronger drive can maintain the limiting rate in a real neuron.

For the chosen phenomenological paired-pulse recovery, the hold alone gives $500\mathrm{Hz}$. Add the further assumption that every successful event resets precisely the same recovery curve, with no accumulation from earlier events. Then $0.35\mathrm{nA}$ onset-timed pulses can sustain at most $1000/4.942488\approx 202.327\mathrm{Hz}$ in that model. This repeatable reset is an extra assumption: a two-pulse result alone cannot establish it. A real neuron's sustained rate also depends on neural adaptation, input delivery and evolving ion-channel inactivation. The existing leaky integrate-and-fire model similarly adds integration time to its imposed hold.

A homogeneous Poisson process of rate $\lambda$ has interspike intervals with an exponential distribution, soWith the purely illustrative choices $\lambda =100\mathrm{Hz}$ and $t_{\mathrm{abs}}=0.002\mathrm{s}$, this is $1-e^{-0.2}\approx 0.181269$. Thus about $18.13\%$ of intervals predicted by that model violate this chosen hold. Matching a mean spike train count does not make a stimulus-only Poisson process respect recent firing.

A minimal alternative is a renewal process with independent and identically distributed intervals $S=t_{\mathrm{abs}}+X$, where $X$ has an exponential distribution of rate $\lambda$. Its hazard function is zero during the hold and $\lambda$ afterwards. Its mean interval is $t_{\mathrm{abs}}+1/\lambda$ and long-run firing rate isThe same $100{\mathrm{s}}^{-1}$ rate after a $2\mathrm{ms}$ hold gives $83.333\mathrm{Hz}$, not $100\mathrm{Hz}$. This separate stochastic example has no gradual relative recovery; it illustrates why an available-state rate differs from an observed mean firing rate. Smooth dependence on time since the last event can extend the hazard function, while longer histories require more than this simple renewal process. Time-rescaled spike-train diagnostics can test the resulting history-dependent predictions against recorded spike trains.