Laplace Transforms of Renewal Measures

Why Transform the Renewal Measure?

A renewal process records the successive times at which events occur. Its renewal measure keeps track of the expected number of those events falling in each time set. Laplace transforms are useful here because a renewal epoch is a sum of independent interarrival times: the transform of that sum is a power of the transform of one interarrival. Summing over epochs then produces a geometric series.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((X_n)_{n\ge1}\) be independent, identically distributed, finite random variables with \(\mathbb{P}(X_n>0)=1\). Define the renewal epochs by \(S_n=X_1+\cdots+X_n\), for \(n\ge1\). For a Borel set \(B\subseteq[0,\infty)\), define the renewal measure \(U\) by \(U(B)=\mathbb{E}\big[\sum_{n\ge1}\mathbf{1}_{\{S_n\in B\}}\big]\). This convention counts renewals at positive epochs only; there is no initial renewal at time zero. The measure may have atoms, for example when interarrival times are deterministic.

The main formula applies for \(s>0\). Positivity of the interarrivals is important: it gives \(e^{-sX_1}<1\) almost surely when \(s>0\), so \(0<\phi(s)<1\). The transform of the measure is therefore a convergent geometric sum. At \(s=0\), every renewal epoch contributes one to the transform, so the sum diverges. At negative \(s\), the exponential weight grows rather than decays, and the transform also diverges.

The Geometric-Series Formula

Proof. Fix \(a>0\). Since \(X_1>0\) almost surely, \(e^{-aX_1}<1\) almost surely, and consequently \(0<\phi(a)<1\). For \(t\ge0\), the event \(S_n\le t\) implies \(e^{a(t-S_n)}\ge1\). Hence $$ \mathbb{P}(S_n\le t)\le e^{at}\mathbb{E}[e^{-aS_n}]=e^{at}\phi(a)^n. $$ The equality uses independence of \(X_1,\ldots,X_n\). By summing these bounds, $$ U([0,t])=\sum_{n=1}^{\infty}\mathbb{P}(S_n\le t) \le e^{at}\sum_{n=1}^{\infty}\phi(a)^n =e^{at}\frac{\phi(a)}{1-\phi(a)}<\infty. $$ This proves local finiteness. Now fix \(s>0\). By the definition of \(U\) and Tonelli’s theorem for nonnegative functions, $$ \widehat U(s)=\sum_{n=1}^{\infty}\mathbb{E}[e^{-sS_n}]. $$ Independence gives \(\mathbb{E}[e^{-sS_n}]=\prod_{k=1}^{n}\mathbb{E}[e^{-sX_k}]=\phi(s)^n\). Since \(0<\phi(s)<1\), the geometric series converges to \(\phi(s)/(1-\phi(s))\). At \(s=0\), each term \(\mathbb{E}[e^{0}]\) is \(1\), so the sum is infinite. If \(s<0\), then \(e^{-sX_1}>1\) almost surely, so \(\phi(s)>1\) or \(\phi(s)=\infty\); in either case the nonnegative series of epoch transforms diverges. Tonelli’s theorem gives \(\widehat U(s)=\infty\) in both cases. \(\square\)

The formula uses the measure of renewal epochs, not the measure obtained by adding a renewal at zero. That distinction accounts for the numerator \(\phi(s)\): the sum begins at \(n=1\). Local finiteness also ensures that the renewal measure behaves as a measure on bounded time intervals, even if its total mass is infinite.

From the Measure to the Renewal Function

A related quantity is the renewal function \(m(t)=\mathbb{E}N(t)\), where \(N(t)=\sum_{n\ge1}\mathbf{1}_{\{S_n\le t\}}\) counts renewals by time \(t\), including a renewal exactly at \(t\). By the definition of \(U\), \(m(t)=U([0,t])\). Its Laplace transform, with respect to ordinary integration in \(t\), differs from the Laplace transform of the measure by a factor of \(1/s\).

Proof. Since \(m(t)=\int_{[0,\infty)}\mathbf{1}_{\{x\le t\}}\,U(dx)\), Tonelli’s theorem allows the order of integration to be exchanged: $$ \int_0^\infty e^{-st}m(t)\,dt =\int_{[0,\infty)}\left(\int_x^\infty e^{-st}\,dt\right)U(dx). $$ For \(s>0\), the inner integral is \(e^{-sx}/s\). Therefore the expression equals \(\frac1s\int_{[0,\infty)}e^{-sx}U(dx)=\widehat U(s)/s\). Substitution of the theorem’s formula gives the stated result. Whether the count includes an event at the endpoint \(t\) does not change this ordinary integral, since a single endpoint has zero length in the inner integration. \(\square\)

For example, with exponential interarrivals of rate \(2\), the measure transform is \(2/s\), while the renewal function transform is \(2/s^2\). These are transforms of different objects: the first integrates against the renewal measure, and the second integrates the expected count as a function of time.

Convergence, Interpretation, and a Common Error

For positive interarrivals, the real convergence domain is exact: the renewal-measure transform is finite precisely for \(s>0\). This conclusion does not require a finite mean interarrival time. The proof relies on positivity, independence, and the geometric sum; it does not use a long-run renewal rate or an asymptotic approximation.

A frequent mistake is to start the sum at \(n=0\), as if there were an initial renewal at zero, and then report \(1/(1-\phi(s))\) as the transform of \(U\). For the deterministic example with interarrival time \(2\) and \(s=\ln 2\), that incorrect expression gives \(1/(1-1/4)=4/3\). But the defined measure has atoms only at \(2,4,6,\ldots\), and its transform is \(1/3\). The extra \(1\) in \(4/3=1+1/3\) is exactly the transform of a unit atom at zero. If the initial renewal is intentionally included, the modified measure has transform \(1/(1-\phi(s))\); otherwise the correct formula is \(\phi(s)/(1-\phi(s))\).

Another useful check is the endpoint \(s=0\). Substituting formally into the geometric expression would give a denominator \(1-\phi(0)=0\), since \(\phi(0)=1\). This is not a removable algebraic issue: the total mass of \(U\) is \(\sum_{n\ge1}\mathbb{P}(S_n\in[0,\infty))=\sum_{n\ge1}1=\infty\). For negative \(s\), the weight \(e^{-st}\) increases with \(t\), so it cannot restore convergence. These checks align the algebra with the measure’s definition.