Renewal process

A renewal process with independent identically distributed nonnegative interarrival times ${\xi }_1,{\xi }_2,\dots$ has renewal epochsand counting process $N_t=\max \{n:S_n\le t\}$.

Alternately routing a rate-$r$ Poisson process produces non-Poisson streams with common interarrival law Erlang$(2,r)$. The odd-numbered stream has an initial exponential delay, while the even-numbered stream is an ordinary renewal process. Their equilibrium delay survivor is $e^{-rt}(1+rt/2)$. The probability of no odd-stream point in $(s,s+t]$ is $e^{-rt}[1+rt(1-e^{-2rs})/2]$; for the even stream replace the minus by a plus. Conditioning on the parity of $N(s)$ proves these formulas.

For independent uniform distribution increments on $(0,1)$ and $0\le x\le 1$, let $N(x)$ be the first index at which their partial sum exceeds $x$. First-step analysis gives $M(x)=\mathbb{E}N(x)=1+{\int }_0^{x}M(s)ds$, hence $M=e^x$. Equivalently, $\mathbb{P}(N(x)>n)=x^n/n!$ is a simplex-volume probability and the tail-sum formula sums to $e^x$. The strict inequality gives $N(0)=1$ almost surely.

For a positive interarrival law $K$, the renewal measure $U={\sum }_{j\ge 0}{K}^{\ast j}$ counts the expected number of renewal epochs in a set, including the epoch at zero through $K^{\ast 0}={\delta }_0$. Its cumulative function is one plus the usual renewal function. Convolution against $U$ solves a renewal equation.

For positive nonarithmetic interarrival times with law $K$ and mean $m\in (0,\infty ]$, let $U={\sum }_{j\ge 0}{K}^{\ast j}$ be their renewal measure. If $g$ is directly Riemann integrable, then ${\int }_{[0,t]}g(t-x)U(dx)\to m^{-1}{\int }_0^{\infty }g(x)dx$, with $1/\infty =0$. Thus a locally bounded solution of $z=g+z\ast K$ has this limit through its renewal representation. Arithmetic laws require the corresponding lattice version.

A renewal equation decomposes a process at its first renewal. For returns of a Markov chain to a fixed state, $f_k$ is the first-return probability and $u_n$ the probability of being at that state at time $n$. Their probability generating functions satisfy $U=1+FU$.

For positive interarrivals with law $K$, the unique locally bounded solution of $z=g+z\ast K$ is $z=g\ast U={\sum }_{j\ge 0}g\ast K^{\ast j}$, where $U$ is the renewal measure. Iteration proves the identity: the residual is bounded on each compact interval by a constant times the probability that the sum of many positive interarrivals stays in that interval, which tends to zero.

A defective renewal equation is $z(u)=g(u)+{\int }_{[0,u]}z(u-x)Q(dx)$ where the nonnegative kernel measure has total mass less than one. Its renewal representation is a convergent sum of convolutions. Exponential tilting can turn the kernel into a probability measure, enabling the key renewal theorem.

The ultimate ruin probability satisfies a tail-forced defective renewal equation. Multiplying it by $e^{Ru}$ at an adjustment coefficient turns its kernel into the probability density $k_R$. Normalization is exactly $\lambda (M(R)-1)=cR$. The density is absolutely continuous even when the original claim law has atoms; thus it has a nonarithmetic distribution.

For nonnegative finite independent and identically distributed random variables as inter-arrivals with positive probability of exceeding $\epsilon >0$, write $p=\mathbb{P}(X_1\ge \epsilon )>0$. The count tail is bounded by $\mathbb{P}(\mathrm{Bin}(n,p)\le \lfloor t/\epsilon \rfloor )$, which has geometric decay times a polynomial in $n$. Hence $\mathbb{E}{e}^{\theta N(t)}<\infty$ for every fixed finite $t$ and sufficiently small $\theta >0$.

The renewal function is the expected number of renewals by time $t$:

For strictly positive independent identically distributed interarrival times of finite mean $\mu$,

A renewal-reward process attaches an integrable reward $R_n$ to each i.i.d. renewal cycle of length ${\xi }_n$. Its cumulative reward sums completed-cycle rewards and may include reward accrued in the current unfinished cycle.

For independent identically distributed regenerative cycles with $0<\mathbb{E}{\xi }_1<\infty$ and $\mathbb{E}|R_1|<\infty$, the long-run reward rate isalmost surely, provided the unfinished-cycle reward is negligible after division by $t$.

If a repaired machine has independent exponential lifetimes of rate $\lambda$ and is inspected and repaired every $m$ time units after each repair, then its cycle length is $m\lceil X/m\rceil$ and its working reward is $X$. Its long-run working proportion is

If $0<\mathbb{E}{\xi }_1=1/\lambda <\infty$, then the strong law $S_n/n\to 1/\lambda$ and the inequalities $S_{N_t}\le t<S_{N_t+1}$ imply $N_t/t\to \lambda$ almost surely.

The renewal interval containing $t$ has lengthSelection by a fixed observation time favours longer interarrival times.

For a nonnegative random variable $X$ with $0<\mathbb{E}X<\infty$, its size-biased version $\hat{X}$ is defined byfor bounded measurable $h$. If $X$ has density $f$, then $\hat{X}$ has density $xf(x)/\mathbb{E}X$.

Reweight probabilities proportionally to the nonnegative value being observed, then divide by its positive finite expected value. This produces a size-biased distribution. Values twice as large receive twice the original relative weight; this is a change in the sampling law rather than a deterministic multiplication of the sampled value.

For the positive-support geometric distribution with success probability $p$, size bias gives $\mathbb{P}(\tilde{Y}=k)=k{p}^2(1-p{)}^{k-1}$. The sum of two independent copies has $k$ positive-integer decompositions at value $k+1$, so its shifted probability is exactly this expression. Its probability generating function is $p^{2}s/(1-(1-p)s{)}^2$, also obtained by the size-biased probability generating function identity.

For a positive integer-valued random variable with finite mean $\mu$, the size-biased distribution has probabilities $k\mathbb{P}(Y=k)/\mu$. They sum to one by the definition of expected value. Multiplying the differentiated probability generating function by $s/\mu$ gives the displayed identity for $0\le s<1$, with the endpoint interpreted by its limit.

For exponential interarrival times of rate $\lambda$, the age and residual life at time $t$ converge jointly to independent rate-$\lambda$ exponential variables. Their sum therefore converges to the gamma distribution with density ${\lambda }^{2}x{e}^{-\lambda x}$, the size-biased exponential law.

At every deterministic $t>0$, the containing interval stochastically dominates an unselected interarrival time:Conditioning on the first interarrival gives a renewal equation; subtracting the constant tail probability leaves a renewal equation with a nonnegative forcing term.

The residual lifetime is the time remaining until the next renewal.

For non-arithmetic inter-renewal times with finite positive mean $\mu$, the excess $E(t)=S_{N(t)+1}-t$ converges in distribution to $E_{\infty }$ with

If an independent superposition of a Poisson process and a non-arithmetic finite-mean renewal process is itself a renewal process, then its first interarrival is exponential. Equating the limiting excess law computed as a renewal excess with the minimum of the Poisson and second-process excesses yields an integral equation whose survival-function solutions are exponential.

With bounded integer lifetimes, residual lifetime decreases deterministically and resets randomly at zero.

The invariant residual-life mass at i is proportional to the probability that a lifetime exceeds i.

For inter-renewal times with an Erlang distribution of integer shape $r$ and rate $\lambda$, the mean is $r/\lambda$ and survival function is $e^{-\lambda t}{\sum }_{j=0}^{r-1}(\lambda t{)}^j/j!$. The equilibrium first-delay density is survival divided by the mean,This is the equal mixture of Erlang distributions of shapes $1,\dots ,r$. It can be constructed by observing a Poisson clock with uniformly randomized phase modulo $r$, whose every $r$th event is a renewal.