Renewal Equations with an Initial Delay

How an Initial Delay Changes the Count

An ordinary renewal process starts with an interarrival time having the same law as every later interarrival. A delayed renewal process separates the first wait from the subsequent waits. This distinction changes the renewal-count equation: after conditioning on the initial delay, the count includes the first renewal if it has occurred, followed by an ordinary renewal count over the remaining time.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $D$ be a finite, nonnegative random variable, and let $X_1,X_2,\ldots$ be independent and identically distributed, strictly positive random variables with common distribution $F$. Assume $D$ is independent of the entire sequence $(X_k)_{k\ge1}$. Define $S_0=0$ and $S_n=X_1+\cdots+X_n$ for $n\ge1$. The delayed renewal times are $D+S_0,D+S_1,\ldots$. Counts below include a renewal at the endpoint $t$, and, if $D=0$, include a renewal at time zero.

Definition: For $t\ge0$, the ordinary renewal count and its mean are $N(t)=\sum_{n\ge1}\mathbf{1}_{\{S_n\le t\}}$ and $m(t)=\mathbb{E}N(t)$. The delayed renewal count and its mean are $N_D(t)=\sum_{n\ge0}\mathbf{1}_{\{D+S_n\le t\}}$ and $m_D(t)=\mathbb{E}N_D(t)$. Write $G(B)=\mathbb{P}(D\in B)$ for the probability distribution of $D$, and $G(t)=\mathbb{P}(D\le t)$.

Because the interarrivals are strictly positive, only finitely many renewals occur by any fixed time almost surely. The expected counts are finite as well: there is some $\varepsilon>0$ with $\mathbb{P}(X_1\ge\varepsilon)>0$, and accumulating more than $t/\varepsilon$ such interarrivals forces the sum past $t$. The number of trials needed to obtain that many interarrivals of size at least $\varepsilon$ has finite expectation. Also, $N(0)=0$ and $m(0)=0$.

Theorem (Delayed renewal equation): Under the assumptions above, for every $t\ge0$, $$ m_D(t)=G(t)+\int_{[0,t]}m(t-x)\,G(dx) =\int_{[0,t]}\bigl(1+m(t-x)\bigr)\,G(dx). $$ Here the integral is with respect to the probability distribution of $D$, including any atom at zero or at $t$.

Proof. Fix $t\ge0$. On the event $\{D>t\}$, no delayed renewal has occurred by time $t$, so $N_D(t)=0$. On $\{D\le t\}$, the renewal at time $D$ has occurred, and the later renewals by time $t$ are exactly those ordinary renewals whose elapsed time from $D$ is at most $t-D$. Thus $N_D(t)=1+N(t-D)$ on $\{D\le t\}$. The sequence of later interarrivals is independent of $D$, so conditioning on $D$ gives $$ m_D(t)=\mathbb{E}\!\left[\mathbf{1}_{\{D\le t\}}\bigl(1+m(t-D)\bigr)\right] =\int_{[0,t]}\bigl(1+m(t-x)\bigr)\,G(dx). $$ Separating the integral of $1$, which is $G(t)$, gives the first form. At $D=t$, the residual time is zero and $m(0)=0$; the formula therefore counts the renewal at $t$ exactly once. $\square$

The equation is a conditioning formula, not a claim that the delayed process has the same renewal function as an ordinary process. It uses the ordinary mean $m$ for the renewals after the first one. The initial delay enters through its full distribution $G$, not just its mean.

Computing the Delayed Mean

Worked Example: A Two-Point Delay and Two-Point Interarrivals

Suppose $X_k=1$ or $2$, each with probability $\tfrac12$, and independently $D=0$ or $2$, each with probability $\tfrac12$. Find $m_D(4)$. First check the ordinary mean values needed in the delayed equation.

The ordinary first-step renewal equation is $m(t)=F(t)+\mathbb{E}[\mathbf{1}_{\{X_1\le t\}}m(t-X_1)]$, where $F(t)=\mathbb{P}(X_1\le t)$. Since $m(0)=0$, for the times required here it gives

\begin{aligned} m(1)&=\tfrac12,\\ m(2)&=1+\tfrac12m(1)+\tfrac12m(0) =1+\tfrac14=\tfrac54,\\ m(3)&=1+\tfrac12m(2)+\tfrac12m(1) =1+\tfrac58+\tfrac14=\tfrac{15}{8},\\ m(4)&=1+\tfrac12m(3)+\tfrac12m(2) =1+\tfrac{15}{16}+\tfrac{10}{16} =\tfrac{41}{16}. \end{aligned}

For example, $F(3)=F(4)=1$, since both possible interarrivals are at most those times. In the calculation of $m(4)$, the numerators add to $16+15+10=41$, so $m(4)=\tfrac{41}{16}$. Conditioning on the two possible delays now gives

\begin{aligned} m_D(4) &=\tfrac12\bigl(1+m(4)\bigr)+\tfrac12\bigl(1+m(2)\bigr)\\ &=\tfrac12\left(1+\tfrac{41}{16}\right) +\tfrac12\left(1+\tfrac54\right)\\ &=\tfrac12\cdot\tfrac{57}{16}+\tfrac12\cdot\tfrac{36}{16} =\tfrac{93}{32}. \end{aligned}

Check by conditioning on each delay: if $D=0$, the count has mean $1+m(4)=\tfrac{57}{16}$; if $D=2$, it has mean $1+m(2)=\tfrac94=\tfrac{36}{16}$. Their equally weighted average is $\tfrac{93}{32}$, as above.

When $D$ has a density $g$, the same equation becomes $m_D(t)=\int_0^t(1+m(t-x))g(x)\,dx$. The endpoint convention remains $\le t$; for a continuous delay law the endpoints have probability zero, but the integral still accounts for all delays in the interval.

Worked Example: Uniform Delay with Deterministic Interarrivals

Let each interarrival equal $2$ with probability $1$, and let $D$ be uniform on $[0,1]$, independently. Find $m_D(\tfrac52)$. In the ordinary process, renewal times are $2,4,6,\ldots$, so $m(u)=\lfloor u/2\rfloor$ for $u\ge0$, with a renewal counted at equality.

Here the density of $D$ is $1$ on $[0,1]$. In the delayed equation, $t-x=\tfrac52-x$ is at least $2$ exactly when $x\le\tfrac12$. Thus $m(\tfrac52-x)=1$ for $0\le x\le\tfrac12$, and it is $0$ for $\tfrac12<x\le1$. The single boundary point $x=\tfrac12$ is included in the first range; it has probability zero under the uniform law. Therefore

\begin{aligned} m_D\!\left(\tfrac52\right) &=\int_0^{1/2}(1+1)\,dx+\int_{1/2}^1(1+0)\,dx\\ &=2\cdot\tfrac12+1\cdot\tfrac12 =\tfrac32. \end{aligned}

Check by counting renewals directly: the first renewal at $D$ always occurs by time $\tfrac52$. A second renewal at $D+2$ occurs by that time exactly when $D\le\tfrac12$, an event of probability $\tfrac12$. A third renewal cannot occur because $D+4>\tfrac52$. The expected count is therefore $1+\tfrac12=\tfrac32$, agreeing with the integral.

A Transform Formula for Delayed Counts

The delayed equation can also be summarized through Laplace transforms. For $s>0$, define $\widetilde F(s)=\mathbb{E}[e^{-sX_1}]$, $\widetilde G(s)=\mathbb{E}[e^{-sD}]$, and $\widehat m_D(s)=\int_0^\infty e^{-st}m_D(t)\,dt$. Since $X_1>0$ almost surely, $0\le\widetilde F(s)<1$.

Theorem (Laplace transform of the delayed renewal mean): For every $s>0$, $$ \widehat m_D(s)=\frac{\widetilde G(s)}{s\bigl(1-\widetilde F(s)\bigr)}. $$ The formula requires that $D$ be independent of the ordinary interarrivals.

Proof. By definition, $m_D(t)=\sum_{n\ge0}\mathbb{P}(D+S_n\le t)$. All summands are nonnegative, so Tonelli's theorem permits interchanging the sum, expectation, and integral. For any nonnegative random variable $Y$, $$ \int_0^\infty e^{-st}\mathbf{1}_{\{Y\le t\}}\,dt =\int_Y^\infty e^{-st}\,dt =\frac{e^{-sY}}{s}. $$ Apply this identity with $Y=D+S_n$, take expectations, and use independence to obtain $$ \widehat m_D(s) =\frac1s\sum_{n\ge0}\mathbb{E}[e^{-s(D+S_n)}] =\frac{\widetilde G(s)}s\sum_{n\ge0}\widetilde F(s)^n =\frac{\widetilde G(s)}{s(1-\widetilde F(s))}. $$ The geometric series converges because $\widetilde F(s)<1$. $\square$

Worked Example: Exponential Delay and Exponential Interarrivals

Let the interarrivals be exponential with rate $1$, and let $D$ be an independent exponential random variable with rate $2$. The ordinary renewal count is a rate-$1$ Poisson count, so $m(u)=u$. Use the delayed equation to find $m_D(t)$.

The delay has density $2e^{-2x}$ for $x\ge0$. Substitution in the equation gives

\begin{aligned} m_D(t) &=\int_0^t(1+t-x)2e^{-2x}\,dx\\ &=(1-e^{-2t})+\int_0^t(t-x)2e^{-2x}\,dx. \end{aligned}

For the remaining integral, $\int_0^t2e^{-2x}\,dx=1-e^{-2t}$, and integration by parts gives $\int_0^t2xe^{-2x}\,dx=\tfrac12-(t+\tfrac12)e^{-2t}$. Hence

\begin{aligned} \int_0^t(t-x)2e^{-2x}\,dx &=t(1-e^{-2t})-\left(\tfrac12-(t+\tfrac12)e^{-2t}\right)\\ &=t-\tfrac12+\tfrac12e^{-2t}, \end{aligned}

and therefore $m_D(t)=t+\tfrac12-\tfrac12e^{-2t}$. Check at $t=0$: the formula gives $0+\tfrac12-\tfrac12=0$, as required because $D>0$ almost surely and no renewal has occurred at time zero. As $t$ increases, the formula includes the first delayed renewal and the expected ordinary count after it.

Assumptions and Common Pitfalls

Independence is essential to the conditioning step: given $D=x$, the later interarrivals must still have their original common law. If the interarrivals depend on $D$, the conditional future count need not equal $m(t-x)$, and the displayed equation need not hold. Also, $m_D$ is not generally obtained by substituting the delay distribution for the interarrival distribution in an ordinary renewal equation; the subsequent renewals still use $F$.

Common mistake: omitting the first delayed renewal. In the first example, using $\tfrac12m(4)+\tfrac12m(2)$ gives $\tfrac12(\tfrac{41}{16}+\tfrac{20}{16})=\tfrac{61}{32}$, but this omits the renewal at $D$ in each case. The correct conditional means are $1+m(4)$ and $1+m(2)$, so the answer is $\tfrac{93}{32}$. In particular, the term $1$ cannot be dropped even when $D=0$: the delayed count includes the renewal at time zero by definition.

Endpoint conventions matter as well. A renewal at exactly $t$ is counted because the event in the definition is $\{D+S_n\le t\}$. When the delay has an atom at $t$, the equation includes that mass and uses $m(0)=0$ for the remaining ordinary count. When the delay has a continuous distribution, a single endpoint has probability zero, but the same equation and convention remain valid.

Check Your Understanding

For each question, specify whether the initial renewal at time zero is counted under the definition used here.

Suppose $D$ is equally likely to be $1$ or $3$. Write $m_D(t)$ in terms of the ordinary renewal mean $m$, making the endpoint condition explicit.
Why does the delayed renewal equation use $m(t-x)$, rather than $m_D(t-x)$, after conditioning on $D=x$?
For deterministic interarrivals of length $2$ and $D$ uniform on $[0,1]$, compute $m_D(\tfrac32)$ and explain why no second renewal can occur by that time.
State the independence assumption needed for the Laplace-transform formula and identify where it is used in the proof.
If the distribution of $D$ has an atom at $t$, which term in the delayed renewal equation accounts for the renewal at exactly $t$?

Renewal Equations with an Initial Delay

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