Order and exit decisions under non-increasing price curves for products with short life cycles

Abstract

We consider a supplier selling a product with a relatively short life cycle and following a non-increasing price curve. Because of the short cycle, there is a single procurement opportunity at the beginning of the cycle. The objective of the supplier is to determine the initial order quantity and the time to remove the product from the market in order to maximize her profits. We study this problem in a continuous-time framework where the demand is modeled with a non-homogeneous Poisson process having a general intensity function and the pricing strategy is given by an arbitrary non-increasing function. We give a rigorous mathematical analysis for the problem and show how it can be solved in two stages. We also consider the special case with piecewise constant intensity and price functions. For this case, we show that the optimal exit time is included in the set of break points of these functions. This brings a fast method to obtain the optimal solution for this special case.

Appendices

Appendix A: Approximating the intensity and price functions

In this section, we give a proof of Lemma 10. In this proof we show how one can control the error associated with approximating the true intensity and price functions by piecewise constant functions selected according to (37). For this optimization problem we have shown after Lemma 14 that one can reduce the search for the optimal $\tau$ to a finite predetermined set. Below, we introduce some notation and give auxiliary results to pave the way for the proof of Lemma 10. To start, we first derive the following easy inequalities.

Let for any $a\in {\mathbb {N}}$ and $a_{*}\in {\mathbb {N}}$ the functions $\lambda _a$ and $p_{a_{*}}$ denote these approximating functions as given in (36–38) with the error

$$\begin{aligned} \Vert \lambda - \lambda _a \Vert \le \frac{1}{a} \quad \text {and} \quad \Vert p - p_{a_{*}} \Vert \le \frac{1}{a_{*}}. \end{aligned}$$

(65)

The integers a and $a_{*}$ can be the same or different.

If we define $\Lambda _a (t) := \int _0^t \lambda _a (s) \, ds$, then it is easy to see that

$$\begin{aligned} | \Lambda (t) - \Lambda _a(t) | = \Big | \int _0^t ( \lambda (s) - \lambda _a (s) ) \, ds \Big | \le \int _0^t | \lambda (s) - \lambda _a (s) | \, ds \le \frac{t}{a}. \end{aligned}$$

(66)

Also, by the triangular inequality, $p_{a_{*}}\le p$, and p being a non-increasing function, it follows that

$$\begin{aligned} | \lambda (t) p(t) -\lambda _a(t)p_{ a_{*}}(t) |\le & {} | \lambda (t)p(t)- \lambda (t)p_{ a_{*}}(t) | + | \lambda (t) p_{ a_{*}}(t) - \lambda _a(t)p_{ a_{*}}(t) | \nonumber \\\le & {} \frac{\lambda (t)}{a_{*}}+\frac{p(t)}{a}\le \frac{{\overline{\lambda }} }{a_{*}}+\frac{p(0)}{a} \end{aligned}$$

(67)

where ${\overline{\lambda }} := \Vert \lambda \Vert = \sup _{t \le {\overline{\tau }}} \lambda (t) < \infty \,$ (recall that $\lambda$ is assumed to be locally bounded and we have $\lim _{t \rightarrow \infty } \lambda (t) = 0$). Note, by using a different intensity function we implicitly consider a separate Poisson process $N_a$. Let $\sigma _{i,a}$ be the ith arrival moment of this new process $N_a$. Using this notation, one can show the following auxiliary result.

Corollary 18

For $t \ge 0$ and $i \ge 1$, the absolute difference of the tail probabilities of $\sigma _{i}$ and $\sigma _{i,a}$ is bounded as

$$\begin{aligned} \big | {\mathbb {P}}\, (\sigma _{i}> t) - {\mathbb {P}}( \sigma _{i,a}>t ) \big | \le \frac{\Lambda (t)^{i-1}t}{(i-1)!a}. \end{aligned}$$

(68)

Proof

Using a time transformation argument, the ith arrival time of the nonhomogeneous Poisson process N can be represented as $\sigma _i = \Lambda ^{\leftarrow } (S_i)$ with $S_i$ denoting the ith arrival time of a homogeneous Poisson process with unit arrival rate. This shows

$$\begin{aligned} {\mathbb {P}}\, (\sigma _{i}> t) = {\mathbb {P}}\, ( S_{i}>\Lambda ( t )) = \int _{\Lambda ( t )}^{\infty } e^{-u} \frac{ u^{i-1}}{(i-1)!} du. \end{aligned}$$

By a similar argument we obtain

$$\begin{aligned} {\mathbb {P}}\, ( \sigma _{i,a}>t) = \int _{\Lambda _a ( t )}^{\infty } e^{-u} \frac{ u^{i-1}}{(i-1)!} du , \end{aligned}$$

This yeilds using $\Lambda _a ( t )\le \Lambda (t)$ for every $t\ge 0$

$$\begin{aligned} \big | {\mathbb {P}}\, (\sigma _{i}> t ) - {\mathbb {P}}( \sigma _{i,a}>t ) \big | = \int _{\Lambda _{a} ( t ) }^{\Lambda ( t ) } e^{-u} \frac{ u^{i-1}}{(i-1)!} du . \end{aligned}$$

Applying now (66) and $e^{-u}\le 1$ for every $u\ge 0$, we obtain

$$\begin{aligned} \int _{\Lambda _{a}(t)}^{\Lambda (t)}e^{-u}\frac{u^{i-1}}{(i-1)!}du\le \frac{\Lambda (t)^{i-1}t}{(i-1)!a} , \end{aligned}$$

and this shows the desired result. $\square$

In the next corollary, we give an upperbound on the absolute difference between the revenues obtained when we use the product of the approximating price and arrival intensity functions, and the one obtained with the product of the original price and arrival intensity function.

Corollary 19

Introducing

$$\begin{aligned} \kappa _{\Lambda }(x,\tau ):=\tau \left( 1+\frac{\Lambda (\tau )^{x}}{x!}\right) \end{aligned}$$

the absolute difference in revenues (see (4)) is bounded as

$$\begin{aligned}&\left| {\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x}}p(t)\lambda (t)dt\right) -{\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x,a}}p_{a_{*}}(t) \lambda _{a}(t)dt\right) \right| \nonumber \\&\quad \le \frac{p(0)\kappa _{\Lambda } (x,\tau )}{a}+\frac{\Lambda (\tau )}{a_{*}} . \end{aligned}$$

(69)

Proof

We write the absolute difference in (69) as

$$\begin{aligned}&\left| {\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x}}p(t)\lambda (t)dt\right) -{\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x,a}}p_{a_{*}}(t)\lambda _{a}(t)dt\right) \right| \nonumber \\&\quad = \left| \int _{0}^{\tau }p(t)\lambda (t)\mathbb {P}(\sigma _{x}>t)dt-\int _{0}^{\tau }p_{a_{*}}(t)\lambda _{a}(t)\mathbb {P}{(\sigma }_{x,a}>t)dt\right| \nonumber \\&\quad \le \int _{0}^{\tau }\left| p(t)\lambda (t)\mathbb {P}{(\sigma }_{x}>t)-p_{a_{*}}(t)\lambda _{a}(t)\mathbb {P}{(\sigma }_{x,a}>t)\right| dt . \end{aligned}$$

(70)

By the triangle inequality it follows using (67) and (68) that

$$\begin{aligned}&\left| p(t)\lambda (t)\mathbb {P}{(\sigma }_{x}>t)-p_{a_{*}}(t)\lambda _{a}(t)\mathbb {P}{(\sigma }_{x,a}>t)\right| \\&\quad \le \left| p(t)\lambda (t)\mathbb {P}{(\sigma }_{x}>t)-p(t)\lambda (t)\mathbb {P}{(\sigma }_{x,a}>t)\right| \\&\qquad +\left| p(t)\lambda (t)\mathbb {P}{(\sigma }_{x,a}>t)-p_{a_{*}}(t)\lambda _{a}(t)\mathbb {P}{(\sigma } _{x,a}>t)\right| \\&\quad \le p(0)\lambda (t)\left| \mathbb {P}{(\sigma }_{x}>t)-\mathbb {P}{(\sigma }_{x,a}>t)\right| +\left| p(t)\lambda (t)-p_{a_{*}}(t)\lambda _{a}(t)\right| \\&\quad \le p(0)\lambda (t)\frac{\Lambda (t)^{x-1}t}{(x-1)!a}+\frac{\lambda (t)}{a_{*}}+\frac{p(t)}{a} . \end{aligned}$$

This implies by relation (70) that

$$\begin{aligned}&\left| {\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x}}p(t)\lambda (t)dt\right) -{\mathbb {E}}\left( \int _{0}^ {\tau \wedge \sigma _{x,a}}p_{a}(t)\lambda _{a}(t)dt\right) \right| \\&\quad \le \frac{p(0)\tau }{a}\left( 1+\frac{\Lambda (\tau )^{x} }{x!}\right) +\frac{\Lambda (\tau )}{a_{\star }}, \end{aligned}$$

and we have shown (69). $\square$

An easy consequence of Corollary 19 (take $p(t) = p_{a_{*}} (t) = 1$ for all $t \ge 0$ and hence $a_{*}=\infty$) is

$$\begin{aligned} \left| \mathbb {E}\left( \int _{0}^{\tau \wedge \sigma _{x}}\lambda (t)dt\right) -\mathbb {E}\left( \int _{0}^{\tau \wedge \sigma _{x,a}}\lambda _{a}(t)dt\right) \right| \le \frac{\kappa _{\Lambda }(x,\tau )}{a}. \end{aligned}$$

(71)

Finally we give in the next corollary an upperbound on the absolute difference in expected inventory costs under arrival processes N and $N_a$.

Corollary 20

The absolute difference of inventory holding costs (see (6)) under N and $N_a$ is bounded as

$$\begin{aligned}&\left| h{\mathbb {E}}\left( \int _{0}^{\tau }(x-N(t))^{+}dt\right) -h{\mathbb {E}}\left( \int _{0}^{\tau }(x-N_{a}(t))^{+}\right) dt\right| \nonumber \\&\quad \le \frac{h\tau ^{2}}{2a}\sum \limits _{i=0}^{x-1} \frac{\Lambda (\tau )^{i}}{i!} . \end{aligned}$$

(72)

Proof

In order to use our bounds on the difference of tail probabilities in Corollary 18 we rewrite this difference in terms of the arrival times as

$$\begin{aligned}&\left| h{\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x}}(x-N(t))^{+} dt\right) -h{\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x,a}} (x-N_{a}(t))^{+}dt\right) \right| \nonumber \\&\quad = \left| h{\mathbb {E}}\left( \sum \limits _{i=1}^{x}\sigma _{i}\wedge \tau \right) -h{\mathbb {E}}\left( \sum \limits _{i=1}^{x}\sigma _{i,a}\wedge \tau \right) \right| \nonumber \\&\quad \le h\sum \limits _{i=1}^{x}\left| {\mathbb {E}}\left( \sigma _{i} \wedge \tau \right) -{\mathbb {E}}\left( \sigma _{i,a}\wedge \tau \right) \right| . \end{aligned}$$

(73)

Observe that $\sigma _{i} \wedge \tau = \int _0^\tau 1_{ \{ s < \sigma _{i} \} } \, ds$, and so by Fubini’s Theorem, ${\mathbb {E}}\, (\sigma _{i} \wedge \tau ) = \int _0^\tau {\mathbb {P}}(s < \sigma _{i}) \, ds.$ Similarly, we also have ${\mathbb {E}}(\, \sigma _{i,a} \wedge \tau ) = \int _0^\tau {\mathbb {P}}( s < \sigma _{i,a} ) \, ds$. This yields

$$\begin{aligned}&\big | {\mathbb {E}}\, (\sigma _{i} \wedge \tau )- {\mathbb {E}}\, (\sigma _{i,a} \wedge \tau ) \big | = \Big | \int _0^\tau {\mathbb {P}}( s< \sigma _{i}) - {\mathbb {P}}( s< \sigma _{i,a}) ds \Big | \\&\quad \le \int _0^\tau \big | {\mathbb {P}}( s< \sigma _{i}) - {\mathbb {P}}( s < \sigma _{i,a} ) \big | ds. \end{aligned}$$

Applying Corollary 18 to the difference of probabilities for a given i above we obtain

$$\begin{aligned} \left| {\mathbb {E}}\left( \sigma _{i}\wedge \tau \right) -{\mathbb {E}}\left( \sigma _{i,a}\wedge \tau \right) \right| \le \int _{0}^{\tau }\frac{\Lambda (s)^{i-1}s}{(i-1)!a}ds\le \frac{\Lambda (\tau )^ {i-1}\tau ^{2}}{2(i-1)!a} . \end{aligned}$$

Finally using this upperbound for every $1\le i\le x$ in (73) yields

$$\begin{aligned}&\left| h{\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x}}(x-N(t))^{+} dt\right) -h{\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x,a}} (x-N_{a}(t))^{+}dt\right) \right| \\&\quad \le \frac{h\tau ^{2}}{2a}\sum \limits _{i=1}^{x}\frac{\Lambda (\tau )^{i-1}}{(i-1)!} , \end{aligned}$$

and this shows the desired inequality. $\square$

We can now give the proof of Lemma 10.

Proof of Lemma 10

Decomposing the cost and revenue components in (7), and writing the difference between the true model and that of the approximation we obtain by (71) and Corollary 19 and 20

$$\begin{aligned}&\left| \Psi _{a}(x,\tau )-\Psi (x,\tau )\right| \\&\quad \le \left\{ \begin{array} [c]{l} \left| {\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x}}(p(t)-\upsilon )\lambda (t)dt\right) -{\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x,a} }(p_{a}(t)-\upsilon )\lambda _{a}(t)dt\right) \right| \\ +\left| h{\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x}}(x-N(t))^{+} dt\right) -h{\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x,a}} (x-N_{a}(t))^{+}dt\right) \right| \end{array} \right. \\&\quad \le \left\{ \begin{array} [c]{l} \left| {\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x}}p(t)\lambda (t)dt\right) -{\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x,a}}p_{a_{*} }(t)\lambda _{a}(t)dt\right) \right| \\ +\left| \upsilon {\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x} }p(t)\lambda (t)dt\right) -\upsilon {\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x,a}}p_{a}(t)\lambda _{a}(t)dt\right) \right| \\ +\left| h{\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x}}(x-N(t))^{+} dt\right) -h{\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x,a}} (x-N_{a}(t))^{+}dt\right) \right| \end{array} \right. \\&\quad \le \frac{(p(0)+\upsilon )\kappa _{\Lambda }(x,\tau )}{a}+\frac{\Lambda (\tau )}{a}+\frac{h\tau ^{2}}{2a}\sum \limits _{\imath =0}^{x-1}\frac{\Lambda (\tau )}{i!}^{i}. \end{aligned}$$

This establishes the desired result. $\square$

Appendix B: Discretization of $\tau$-space when $\Lambda (\infty ) = \infty$

Here, we show how one can construct the finite discretization set ${\mathcal {D}}_\varepsilon$ on the $\tau$-space for Stage II of the optimization procedure when $\Lambda (\infty ) = \infty$; see the construction in (32) in Sect. 4. Note that when $\Lambda (\infty ) = \infty$,

$$\begin{aligned} {\mathbb {P}}(\sigma _i> \tau ) = {\mathbb {P}}( \Lambda (\sigma _i) > \Lambda (\tau ) ) = \int _{\Lambda (\tau )}^\infty \frac{t^{i-1} e^{-t}}{(i-1)!} \,dt \rightarrow 0 \quad \text {as }\tau \rightarrow \infty . \end{aligned}$$

In particular, we have an easy upper bound thanks to Markov inequality ${\mathbb {P}}[ \Lambda (\sigma _i) > \Lambda (\tau ) ] \le \frac{1}{\Lambda (\tau )} {\mathbb {E}}\Lambda (\sigma _i) = \frac{i}{\Lambda (\tau )}$. Alternatively, it also follows directly

$$\begin{aligned} {\mathbb {P}}(\sigma _i > \tau ) = {\mathbb {P}}( N(\tau ) \le i-1) = e^{-\Lambda (\tau )} \sum \limits _{j=0}^{i-1} \frac{[\Lambda (\tau )]^{j} }{j!} \le e^{-\Lambda (\tau )} \frac{ [\Lambda (\tau )]^{i}-1}{\Lambda (\tau )-1}. \end{aligned}$$

Lemma 21

For $x \le {\overline{x}}$ any $\tau \le p^{\leftarrow }(\upsilon ) \le \infty$ we have

$$\begin{aligned} \Psi (x,p^{\leftarrow }(\upsilon ))-\Psi (x,\tau ) \le (p(\tau )-\upsilon ) \sum \limits _{i=1}^{{\overline{x}}}\mathbb {P}{(\sigma }_{i} >\tau ). \end{aligned}$$

(74)

Proof

To show the above inequality it follows using the explicit form of $\Psi (x,\tau )$ in (7)

$$\begin{aligned}&\Psi (x,p^{\leftarrow }(\upsilon ))-\Psi (x,\tau ) \\&\quad = {\mathbb {E}}\left( \int _{\tau \wedge \sigma _{x}}^{p^{\leftarrow }(\upsilon )\wedge \sigma _{x} }(p(t)-\upsilon )\lambda (t)dt\right) \\&\qquad -\,h{\mathbb {E}}(\left( \int _{\tau \wedge \sigma _{x}}^{p^{\leftarrow }(\upsilon )\wedge \sigma _{x}}(x-N(t))^{+} dt\right) \\&\quad \le {\mathbb {E}}\left( \int _{\tau \wedge \sigma _{x}}^{p^{\leftarrow }(\upsilon )\wedge \sigma _{x}}(p(t)-\upsilon )\lambda (t)dt\right) \\&\quad = {\mathbb {E}}\left( \sum _{i=1}^{x}(p(\sigma _{i})-\upsilon )1_{\{\tau <\sigma _{i}\le p^{\leftarrow }(\upsilon )\}}\right) , \end{aligned}$$

where the last identity follows from an application of Doob’s stopping theorem. Since p is non-increasing, we obtain

$$\begin{aligned}&\Psi (x,p^{\leftarrow }(\upsilon ))-\Psi (x,\tau ) \le (p(\tau )-\upsilon ){\mathbb {E}}\left( \sum \limits _{i=1}^{x}1_{\{\tau <\sigma _{i}\le p^{\leftarrow }(\upsilon )\}}\right) \\&\quad \le (p(\tau )-\upsilon )\sum \limits _{i=1}^{{\overline{x}} }\mathbb {P}{(\sigma }_{i}>\tau ), \end{aligned}$$

and this shows the result. $\square$

An immediate corollary of Lemma 21 is the following result.

Corollary 22

We know by Lemma 4 that ${\overline{x}}$ in (13) is an upper bound for the optimal order quantity. This shows $x_Q(p^{\leftarrow }(\upsilon )) \le {\overline{x}}$ (see also (16)), and since $x_Q(p^{\leftarrow }(\upsilon ))$ is also feasible for ($Q_{\tau }$) with any $\tau \le p^{\leftarrow }(\upsilon )$, we obtain due to (74) that

$$\begin{aligned}&\Psi _Q(p^{\leftarrow }(\upsilon )) - \Psi _Q(\tau ) \le \Psi (x_Q(p^{\leftarrow }(\upsilon )) , p^{\leftarrow }(\upsilon )) \nonumber \\&\quad - \Psi (x_Q(p^{\leftarrow }(\upsilon )), \tau ) \le (p(\tau ) -\upsilon ) \, \sum \limits _{i= 1}^{{\overline{x}}} {\mathbb {P}}\{ \sigma _i > \tau \}. \end{aligned}$$

(75)

Therefore in problem (Q), we can set the right end point

$$\begin{aligned} \min \left\{ \tau \le p^{\leftarrow }(\upsilon ):(p(\tau )-\upsilon )\sum \limits _{i=1}^{{\overline{x}} }\mathbb {P}{(\sigma }_{i}>\tau )\le \epsilon \right\} < \infty , \end{aligned}$$

which is the smallest $\tau$ value for which the error terms in (75) are less than the given $\varepsilon > 0$. Then, we fill in the sets ${\mathcal {D}}_{\varepsilon }$ again starting from $\tau _{0} = 0$ and iterating relation (32) once again until this right end point is reached.

Appendix C: Supplementary proofs

Proof of Lemma 2

Since $t \mapsto p(t) - \upsilon$ is non-increasing it follows using the definition of the objective function $\Psi$ in (7), for $\tau \le \Lambda ^{\leftarrow } (\alpha x)$ that

$$\begin{aligned} \Psi (x,\tau )\le & {} {\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x} }(p(t)-\upsilon )dN(t)\right) +(\upsilon -c)x\\\le & {} (p(0)-\upsilon ){\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x} }dN(t)\right) +(\upsilon -c)x\\\le & {} (p(0)-\upsilon ){\mathbb {E}}(N(\tau ))+(\upsilon -c)x\\\le & {} (p(0)-\upsilon ){\mathbb {E}}(N(\Lambda ^{\leftarrow }(\alpha x))+(\upsilon -c)x\\= & {} (p(0)-\upsilon )\Lambda (\Lambda ^{\leftarrow }(\alpha x))+(\upsilon -c)x\\= & {} 0, \end{aligned}$$

and that concludes the proof. $\square$

Proof of Lemma 3

To show $\Psi (x, \tau ) \le 0$ we observe using Jensen’s inequality that

$$\begin{aligned} {\mathbb {E}}(x- N(t))^+ \ge (x- {\mathbb {E}}N(t))^+ = (x- \Lambda (t) )^+\ge x- \Lambda (t). \end{aligned}$$

(76)

Using the lower bound in (76) and ${\mathbb {E}}\int _{0}^{\sigma _{x}}dN(t)=x$, we obtain for $\Lambda ^{\leftarrow }(\alpha x)$ finite and $\tau \ge \Lambda ^{\leftarrow }(\alpha x)$ that

$$\begin{aligned} \Psi (x,\tau )\le & {} {\mathbb {E}}\left( \int _{0}^{\tau \wedge \sigma _{x} }(p(t)-\upsilon )dN(t)\right) +(\upsilon -c)x-h\int _{0}^{\tau }(x-\Lambda (t))dt\nonumber \\\le & {} (p(0)-\upsilon ){\mathbb {E}}\left( \int _{0}^{\sigma _{x}}dN(t)\right) +(\upsilon -c)x-h\int _{0}^{\Lambda ^{\leftarrow }(\alpha x)}(x-\Lambda (t))dt\nonumber \\\le & {} (p(0)-c)x-hx\Lambda ^{\leftarrow }(\alpha x)+h\Lambda (\Lambda ^{\leftarrow }(\alpha x))\Lambda ^{\leftarrow }(\alpha x)\nonumber \\= & {} [p(0)-c-h(1-\alpha )\Lambda ^{\leftarrow }(\alpha x)]x . \end{aligned}$$

(77)

Since $x\ge {\overline{x}}:=\left\lceil \frac{1}{\alpha }\Lambda \left( \frac{p(0)-\upsilon }{h}\right) \right\rceil$ and $\alpha :=\frac{c-\upsilon }{p(0)-\upsilon }$ it follows

$$\begin{aligned} \Lambda ^{\leftarrow }(\alpha x)\ge \Lambda ^{\leftarrow }(\alpha {\overline{x}} )\ge \frac{p(0)-\upsilon }{h}=\frac{p(0)-c}{h(1-\alpha )}, \end{aligned}$$

and substituting this in (77) gives a non-positive upper bound. For $\Lambda ^{\leftarrow }(\alpha x)$ infinite and hence $\Lambda (t)\le \alpha x$ we obtain $\int _{0}^{\tau }(x-\Lambda (t))dt\ge \tau (1-\alpha )x$, and this shows $\Psi (x,\tau )\le (p(0)-c)x-\tau (1-\alpha )x.$ Hence for $\tau =\infty =\Lambda ^{\leftarrow }(\alpha x)$ and $x\ge {\overline{x}}$ we again obtain $\Psi (x,\infty )=-\infty \le 0$. $\square$

Proof of Lemma 16

It follows for every $i=1,\ldots ,n$ that

$$\begin{aligned} \int _{0}^{a_{i+1}}p(t)[f_{\sigma _{k+2}}(t)-f_{\sigma _{k+1}}(t)]dt=\sum \limits _{j=1}^{i}p_{j}\int _{a_{j}}^{a_{j+1}}(f_{\sigma _{k+2}}(t)-f_{\sigma _{k+1}}(t))dt.\nonumber \\ \end{aligned}$$

(78)

By relation (23) we observe $f_{\sigma _{k+2}}(t)-f_{\sigma _{k+1}} (t)=-f_{k+1}^{^{\prime }}(t)$ with $f_{k+1}^{^{\prime }}(t)$ the derivative of the function $f_{k+1}:{\mathbb {R}}_{+}\rightarrow {\mathbb {R}}_{+}$ given by

$$\begin{aligned} f_{k+1}(t)=\mathbb {P}(N(t)=k+1)\mathbb {=}e^{-\Lambda (t)}\frac{\Lambda (t)^{k+1}}{(k+1)!}, \end{aligned}$$

and so we obtain using $N(a_{1})=0$ and (78)

$$\begin{aligned}&\int _{0}^{a_{i+1}}p(t)[f_{\sigma _{k+2}}(t)-f_{\sigma _{k+1}}(t)]dt \\&\quad = \sum \limits _{j=1}^{i}p_{j}\left[ \mathbb {P}(N(a_{j})=k+1)-\mathbb {P}( N(a_{j+1})=k+1)\right] \\&\quad = \sum \limits _{j=1}^{i-1}p_{j+1}[\mathbb {P}(N(a_{j+1})=k+1)-\sum \limits _{j=1}^{i}p_{j}\mathbb {P}(N(a_{j+1})=k+1)]\\&\quad = \sum \limits _{j=1}^{i-1}\Delta p_{j}\mathbb {P}(N(a_{j+1})=k+1)-p_{i} \mathbb {P}(N(a_{i+1})=k+1). \end{aligned}$$

Hence we have verified relation (58). $\square$