Laplace Transforms for Solving Renewal Equations

From a Renewal Equation to an Algebraic Equation

Renewal equations describe quantities whose future value depends on the first interarrival and the process that follows it. In the time domain, this dependence appears as a convolution. A Laplace transform turns that convolution into multiplication, so the equation can often be solved by ordinary algebra. The final step is inversion: translating the algebraic expression back into a function of time.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \(X_1,X_2,\ldots\) be independent, identically distributed, strictly positive interarrivals with distribution \(F\). Write \(S_n=X_1+\cdots+X_n\), \(N(t)=\max\{n:S_n\le t\}\), and \(m(t)=\mathbb{E}N(t)\), for \(t\ge0\). The renewal function satisfies \(m(t)=\sum_{n\ge1}\mathbb{P}(S_n\le t)\). The earlier material on renewal equations gives its first-renewal equation, \(m(t)=F(t)+\int_{[0,t]}m(t-x)\,dF(x)\).

Definition: For a measurable function \(h:[0,\infty)\to\mathbb{R}\), its Laplace transform at \(s>0\), when the integral is finite, is \(\widehat h(s)=\int_0^\infty e^{-st}h(t)\,dt\). For an interarrival distribution \(F\), write \(\phi(s)=\mathbb{E}[e^{-sX_1}]=\int_{[0,\infty)}e^{-sx}\,dF(x)\). Strict positivity of \(X_1\) implies \(0\le\phi(s)<1\) for every \(s>0\).

The transform of the distribution function \(F(t)=\mathbb{P}(X_1\le t)\) is not \(\phi(s)\): it is \(\phi(s)/s\). Indeed, Tonelli's theorem, applied to the nonnegative integrand, gives \(\int_0^\infty e^{-st}F(t)\,dt=\mathbb{E}[\int_{X_1}^\infty e^{-st}\,dt]=\phi(s)/s\). This distinction matters because renewal equations usually contain the distribution function, while their convolution kernel is the distribution measure \(dF\).

Theorem (Transform solution of a renewal equation): Suppose \(X_1>0\) almost surely, and \(g\) and \(u\) are nonnegative measurable functions satisfying \(u(t)=g(t)+\int_{[0,t]}u(t-x)\,dF(x)\) for \(t\ge0\). At any \(s>0\) for which \(\widehat g(s)\) and \(\widehat u(s)\) are finite, \(\widehat u(s)=\widehat g(s)+\phi(s)\widehat u(s)\), and hence \(\widehat u(s)=\widehat g(s)/(1-\phi(s))\).

Proof. Fix \(s>0\). Transform the convolution term and use Tonelli's theorem; nonnegativity permits changing the order of integration:

\begin{aligned} \int_0^\infty e^{-st}\int_{[0,t]}u(t-x)\,dF(x)\,dt &=\int_{[0,\infty)}\int_x^\infty e^{-st}u(t-x)\,dt\,dF(x)\\ &=\int_{[0,\infty)}e^{-sx}\,dF(x)\int_0^\infty e^{-sy}u(y)\,dy\\ &=\phi(s)\widehat u(s). \end{aligned}

In the second line, the substitution is \(y=t-x\). Transforming both sides of the given equation therefore gives \(\widehat u(s)=\widehat g(s)+\phi(s)\widehat u(s)\). Since \(X_1>0\) almost surely, \(\phi(s)<1\), so division by \(1-\phi(s)>0\) gives the stated formula. \(\square\)

For the renewal function, take \(g(t)=F(t)\). The formula for the transform of a distribution function then yields

\widehat m(s)=\frac{\phi(s)}{s(1-\phi(s))},\qquad s>0.

This identity also follows directly from the renewal sum. For each \(n\ge1\), the transform of \(t\mapsto\mathbb{P}(S_n\le t)\) is \(\mathbb{E}[\int_{S_n}^\infty e^{-st}\,dt]=\phi(s)^n/s\). Summing these nonnegative terms gives \(\widehat m(s)=\sum_{n\ge1}\phi(s)^n/s\), a convergent geometric series because \(\phi(s)<1\). This argument verifies that the transform is finite even when no separate moment assumption on \(X_1\) has been made.

Worked Inversions

Worked Example: Exponential Interarrivals

Suppose the interarrivals are independent exponential random variables with rate \(\lambda>0\). Find the renewal function from its transform. Here \(F(t)=1-e^{-\lambda t}\) for \(t\ge0\), and direct integration gives \(\phi(s)=\lambda/(\lambda+s)\) for \(s>0\). Substitution into the renewal-function formula gives

\widehat m(s) =\frac{\lambda/(\lambda+s)} {s\left(1-\lambda/(\lambda+s)\right)} =\frac{\lambda}{s^2}.

The standard transform pair \(\int_0^\infty e^{-st}t\,dt=1/s^2\), obtained by integration by parts, identifies \(m(t)=\lambda t\). The transform equation has not assumed that the renewal count has independent increments; it uses only the renewal equation and the exponential interarrival transform.

Check by substitution. The renewal equation requires \(m(t)=1-e^{-\lambda t}+\int_0^t m(t-x)\lambda e^{-\lambda x}\,dx\). With \(m(t)=\lambda t\), the integral is \(\lambda^2\int_0^t(t-x)e^{-\lambda x}\,dx=\lambda t-1+e^{-\lambda t}\). Thus the right side is \(1-e^{-\lambda t}+\lambda t-1+e^{-\lambda t}=\lambda t\), as required.

Worked Example: Deterministic Interarrivals

Suppose every interarrival equals \(2\), so \(X_k=2\) almost surely. Then \(\phi(s)=e^{-2s}\), and

\widehat m(s)=\frac{e^{-2s}}{s(1-e^{-2s})} =\frac{1}{s}\sum_{n=1}^\infty e^{-2ns}.

For each \(n\ge1\), the transform pair \(e^{-2ns}/s\longleftrightarrow\mathbf{1}_{[2n,\infty)}(t)\) follows by integrating \(e^{-st}\) from \(2n\) to infinity. Inverting term by term gives \(m(t)=\sum_{n\ge1}\mathbf{1}_{[2n,\infty)}(t)\). For any fixed \(t\), only finitely many terms are nonzero, and the number of integers \(n\ge1\) with \(2n\le t\) is \(\lfloor t/2\rfloor\). Consequently \(m(t)=\lfloor t/2\rfloor\), including at renewal times: the count uses \(S_n\le t\), not \(S_n<t\).

Check. At \(t=5\), the renewals occur at \(2\) and \(4\), so \(N(5)=2\) and the formula gives \(\lfloor5/2\rfloor=2\). At \(t=4\), both renewals at \(2\) and \(4\) are counted, so \(N(4)=2=\lfloor4/2\rfloor\). The transform series also checks algebraically: summing \(\sum_{n\ge1}e^{-2ns}/s\) gives \(e^{-2s}/[s(1-e^{-2s})]\).

Worked Example: A Forced Equation with a Deterministic Kernel

Consider the renewal equation \(u(t)=1+\int_{[0,t]}u(t-x)\,dF(x)\), again with \(X_1=2\) almost surely. The integral is \(u(t-2)\) when \(t\ge2\), and zero when \(t<2\). Thus the equation is \(u(t)=1+\mathbf{1}_{[2,\infty)}(t)u(t-2)\). Its transform, using \(\widehat g(s)=1/s\) and \(\phi(s)=e^{-2s}\), is

\widehat u(s)=\frac{1}{s(1-e^{-2s})} =\frac{1}{s}\sum_{n=0}^\infty e^{-2ns}.

Inversion gives \(u(t)=\sum_{n\ge0}\mathbf{1}_{[2n,\infty)}(t)\). On every bounded time interval this sum has finitely many nonzero terms, and \(u(t)=\lfloor t/2\rfloor+1\). For example, \(u(t)=1\) on \(0\le t<2\), and \(u(t)=2\) on \(2\le t<4\). At \(t=2\), the equation gives \(u(2)=1+u(0)=2\), confirming that the endpoint is included.

Check by substitution. If \(0\le t<2\), the integral is zero and \(u(t)=1\), so the equation holds. If \(t\ge2\), then \(u(t-2)=\lfloor(t-2)/2\rfloor+1=\lfloor t/2\rfloor\). Hence \(1+u(t-2)=\lfloor t/2\rfloor+1=u(t)\), including when \(t\) is an even integer.

What Inversion Does—and Does Not—Do

The transform equation is useful only when the algebraic expression can be inverted. In the examples, inversion used a standard transform pair or a geometric series of shifted step functions. For more complicated expressions, one may use partial fractions, transform tables, or a contour-integral inversion formula. The Bromwich inversion theorem is a standard result: under suitable exponential-order and piecewise-smoothness conditions, a function can be recovered at its continuity points from its transform by integrating \(e^{st}\widehat u(s)\) along a vertical line to the right of the transform's singularities. Its proof relies on Fourier inversion and is not given here.

There is also a standard uniqueness theorem for Laplace transforms: if two locally integrable functions have finite transforms for all sufficiently large real \(s\) and those transforms agree there, the functions agree almost everywhere. This theorem ensures that a valid inversion identifies the solution up to almost-everywhere equality. When the equation or model requires pointwise values—as it does at deterministic renewal times—check those values directly in the original equation. Transform equality alone does not settle a chosen value at a discontinuity.

Common pitfall: confusing the two transforms. For exponential interarrivals, using \(\phi(s)=\lambda/(\lambda+s)\) as though it were \(\widehat F(s)\) would give the incorrect calculation \(\widehat m(s)=\phi(s)/(1-\phi(s))=\lambda/s\), and hence the incorrect answer \(m(t)=\lambda\). The renewal equation contains \(F(t)\), whose transform is \(\phi(s)/s\), not \(\phi(s)\). Including the missing factor gives \(\widehat m(s)=\phi(s)/[s(1-\phi(s))]=\lambda/s^2\), and inversion gives \(m(t)=\lambda t\). Substitution into the renewal equation confirms the corrected answer.

To apply the method reliably, identify whether the kernel is a distribution measure \(dF\) or a function, check that the transforms are finite at the chosen \(s\), solve only where \(1-\phi(s)\ne0\), and then verify the proposed inverse in the original equation. The method gives an algebraic simplification, not permission to skip the conditions that make transformation and inversion valid.

Check Your Understanding

Use the transform identities above, and pay particular attention to whether endpoints are included.

For exponential interarrivals with rate \(\lambda\), derive \(\widehat m(s)\) from \(\phi(s)\) and identify its inverse.
If \(X_k=3\) almost surely, write \(\widehat m(s)\) as a geometric series and state \(m(6)\) under the convention \(S_n\le t\).
For the forced equation \(u(t)=1+\int_{[0,t]}u(t-x)\,dF(x)\) with \(X_k=2\), compute \(u(4)\) and verify the equation at that time.
Explain why the transform of the distribution function is \(\phi(s)/s\), rather than \(\phi(s)\).
What conditions justify dividing by \(1-\phi(s)\), and why should an inverse be checked at discontinuity points?

Laplace Transforms for Solving Renewal Equations

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