Matching tower information with piecewise Pareto

Abstract

Pricing tools for non-proportional reinsurance treaties often only provide layer prices, but no layer-independent collective risk model. There are, however, situations where such a layer-independent model is needed. Examples are large loss and catastrophe loss models for proportional reinsurance treaties. We show that the expected losses of a tower of reinsurance layers can always be matched using a piecewise Pareto distributed severity and provide an algorithm that can be used to convert layer information into a layer-independent collective risk model.

Appendices

Appendix 1: Technical details about the Pareto distribution

The purpose of this appendix is to define the Pareto alpha between two pieces of information (expected excess frequencies and/or expected losses of layers) more rigorously than in Sect. 3, and to prove existence and uniqueness. Moreover, we prove a technical lemma that is needed in Example 3.

Let $0<t\le t_1<t_2$ and let $f_1> f_2$ be the expected frequencies in excess of $t_1$ and $t_2$, respectively. Then we have

$$\begin{aligned} \frac{1-F_{t,\alpha }(t_2)}{1-F_{t,\alpha }(t_1)}=\frac{f_2}{f_1} \quad \Longleftrightarrow \quad \alpha =\frac{\log (f_2/f_1)}{\log (t_1/t_2)}. \end{aligned}$$

Definition 5

In this situation, we say that

$$\begin{aligned} \alpha =\frac{\log (f_2/f_1)}{\log (t_1/t_2)} \end{aligned}$$

is the Pareto alpha between $(t_1,f_1)$ and $(t_2,f_2)$.

Let $\alpha$ be the Pareto alpha between $(t_1,f_1)$ and $(t_2,f_2)$ and let $S=\sum _{n=1}^NX_n$ be a collective risk model with claim sizes $X_n\sim {\text {Pareto}}(t,\alpha )$. Then the expected frequency in excess of $t_i$ is given by ${\text {E}}(N)\cdot (1-F_{t,\alpha }(t_i))$, i.e. the model has the expected frequency $f_1$ in excess of $t_1$ if and only if the expected frequency in excess of $t_2$ equals $f_2$.

Lemma 4

Let $0<t\le a< b<+\infty$. Let $c:=b-a$ and let $f>0$ be the expected frequency in excess of t. Then the map

$$\begin{aligned} \Psi ^{(t,f)}_{c{\text {xs}}a}:(0,+\infty )\rightarrow (0,f\cdot c),\quad \alpha \mapsto f\cdot I_{t,\alpha }(a,b) \end{aligned}$$

is a strictly decreasing homeomorphism. In case of an unlimited layer $+\infty$ xs a we have a strictly decreasing homeomorphism

$$\begin{aligned} \Psi ^{(t,f)}_{+\infty {\text {xs}}a}:(1,+\infty )\rightarrow (0,+\infty ),\quad \alpha \mapsto f\cdot I_{t,\alpha }(a,+\infty ). \end{aligned}$$

Proof

We only consider the case of a limited layer. The proof for unlimited layers is similar. Since

$$\begin{aligned} (0,+\infty )\rightarrow [0,1],\quad \alpha \mapsto 1-F_{t,\alpha }(x)=\left( \frac{t}{x}\right) ^\alpha \end{aligned}$$

is strictly decreasing for all $x\in (a,b)$, the map

$$\begin{aligned} \Psi ^{(t,f)}_{c{\text {xs}}a}(\alpha )=f\cdot \int _a^b (1-F_{t,\alpha }(x)) \, dx \end{aligned}$$

is strictly decreasing, too. We have

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\Psi ^{(t,f)}_{c{\text {xs}}a}(\alpha ) = f\cdot (b-a) \quad \text { and }\quad \lim _{\alpha \rightarrow +\infty }\Psi ^{(t,f)}_{c{\text {xs}}a}(\alpha )=0, \end{aligned}$$

i.e. $\Psi ^{(t,f)}_{c{\text {xs}}a}$ is surjective. Moreover, $\Psi ^{(t,f)}_{c{\text {xs}}a}$ is continuous and consequently $\Psi ^{(t,f)}_{c{\text {xs}}a}$ is a homeomorphism. $\square$

In the proof of Lemma 4 we have used the fact that a bijective, strictly monotonic and continuous function $f:I\rightarrow J\subset \mathbb {R}$, where I is an open interval, is always a homeomorphism, i.e. the inverse $f^{-1}$ is continuous as well. We will also use this fact in the proofs below.

Definition 6

Let $0<t\le a<b\le +\infty$ and $c:=b-a$. Let e denote the expected loss of the layer c xs a and let f be the expected frequency in excess of t. If $0<e< f\cdot c$ then

$$\begin{aligned} \left( \Psi ^{(t,f)}_{c{\text {xs}}a}\right) ^{-1}(e) \end{aligned}$$

is called the Pareto alpha between (t, f) and c xs a.

Let $t \le t_1\le a$ and $f_1>e/c$. Let $\alpha$ be the Pareto alpha between $(t_1,f_1)$ and c xs a, and assume that $S=\sum _{n=1}^N X_n$ is a collective risk model with $X_n\sim {\text {Pareto}}(t,\alpha )$. Then the model has the expected frequency $f_1$ in excess of $t_1$, if and only if the expected loss of the layer c xs a equals e. Note that $f_1>e/c$ only excludes the extreme case of a total loss model (cf. Sect. 2).

Lemma 5

Let $0<t\le a<b\le +\infty$ and $c:=b-a$. Let $e>0$ denote the expected loss of the layer c xs a and let $f_1>f_2>e/c$. For $i\in \{1,2\}$ let $\alpha _i$ be the Pareto alpha between $(t,f_i)$ and c xs a. Then we have

$$\begin{aligned} f_1\cdot (1- F_{t,\alpha _1}(a)) >f_2\cdot (1- F_{t,\alpha _2}(a)) \end{aligned}$$

and if $b<+\infty$, then

$$\begin{aligned} f_1\cdot (1- F_{t,\alpha _1}(b)) <f_2\cdot (1- F_{t,\alpha _2}(b)). \end{aligned}$$

Let $\sum _{n=1}^{N_i}X_{i,n}$, $i\in \{1,2\}$ be collective risk models with ${\text {E}}(N_i)=f_i$ and $X_{i,n}\sim {\text {Pareto}}(t,\alpha _i)$, which both match the expected loss of a layer c xs a with $a\ge t$ and let $f_1>f_2$. Then Lemma 5 states that the model with $i=1$ has a greater expected layer entry frequency, whereas the model with $i=2$ has a greater expected layer exit frequency (if $b<+\infty$). Lemma 5 is used in Example 3.

Proof

We have $f_1\cdot I_{t,\alpha _2}(a,b)>f_2\cdot I_{t,\alpha _2}(a,b)=e=f_1\cdot I_{t,\alpha _1}(a,b)$. Applying Lemma 4 we conclude that $\alpha _1>\alpha _2$. Since

$$\begin{aligned} f_1\cdot \int _a^b(1-F_{t,\alpha _1}(x))\, dx =e=f_2\cdot \int _a^b(1-F_{t,\alpha _2}(x))\, dx, \end{aligned}$$

there exits an $x\in (a,b)$ such that $f_1\cdot (1- F_{t,\alpha _1}(x)) \ge f_2\cdot (1- F_{t,\alpha _2}(x))$. Since $\alpha _1>\alpha _2$, we have

$$\begin{aligned} f_1\cdot (1- F_{t,\alpha _1}(a))&=f_1\cdot (1- F_{t,\alpha _1}(x))\cdot \left( \frac{x}{a}\right) ^{\alpha _1}\\&> f_2\cdot (1- F_{t,\alpha _2}(x))\cdot \left( \frac{x}{a}\right) ^{\alpha _2} =f_2\cdot (1- F_{t,\alpha _2}(a)). \end{aligned}$$

For the case $b<+\infty$, the proof of $f_1\cdot (1- F_{t,\alpha _1}(b)) <f_2\cdot (1- F_{t,\alpha _2}(b))$ is similar. $\square$

Lemma 6

Let $0<t\le a_1<a_2$ and let $b_1\le b_2\le +\infty$ with $b_i>a_i$ and $c_i:=b_i-a_i$. If $b_2<+\infty$, then the map

$$\begin{aligned} \Phi _{c_2{\text {xs}}a_2}^{c_1{\text {xs}}a_1}:(0,+\infty )\rightarrow (c_1/c_2,+\infty ),\quad \alpha \mapsto \frac{I_{t,\alpha }(a_1,b_1)}{I_{t,\alpha }(a_2,b_2)} \end{aligned}$$

is well-defined (i.e. does not depend on t) and is a strictly increasing homeomorphism. If $b_1<b_2=+\infty$, then we have the (well-defined) strictly increasing homeomorphism

$$\begin{aligned} \Phi _{+\infty {\text {xs}}a_2}^{c_1{\text {xs}}a_1}:(1,+\infty )\rightarrow (0,+\infty ),\quad \alpha \mapsto \frac{I_{t,\alpha }(a_1,b_1)}{I_{t,\alpha }(a_2,+\infty )}. \end{aligned}$$

If $b_1=b_2=+\infty$, then we have the (well-defined) strictly increasing homeomorphism

$$\begin{aligned} \Phi _{+\infty {\text {xs}}a_2}^{+\infty {\text {xs}}a_1}:(1,+\infty )\rightarrow (1,+\infty ),\quad \alpha \mapsto \frac{I_{t,\alpha }(a_1,+\infty )}{I_{t,\alpha }(a_2,+\infty )}. \end{aligned}$$

Proof

The case $b_1<b_2=+\infty$ follows easily from

$$\begin{aligned} \frac{I_{t,\alpha }(a_1,b_1)}{I_{t,\alpha }(a_2,+\infty )}=-\left( \frac{a_2}{b_1}\right) ^{\alpha -1} + \left( \frac{a_2}{a_1}\right) ^{\alpha -1}, \end{aligned}$$

which is a consequence of Lemma 1 in Sect. 3. The case $b_1=b_2=+\infty$ follows directly from

$$\begin{aligned} \frac{I_{t,\alpha }(a_1,+\infty )}{I_{t,\alpha }(a_2,+\infty )}=\left( \frac{a_2}{a_1}\right) ^{\alpha -1}, \end{aligned}$$

which is also a consequence of Lemma 1. For the case $b_2<+\infty$, which is slightly more difficult to prove, see Riegel [11]. $\square$

Definition 7

Let $0<t\le a_1<a_2$ and let $b_1\le b_2\le +\infty$ with $b_i>a_i$ and $c_i:=b_i-a_i$. Moreover, let $e_1>0$ and $e_2>0$ be the expected losses of the layers $c_1$ xs $a_1$ and $c_2$ xs $a_2$, respectively. If $b_2<+\infty$, then we additionally assume that $e_1/e_2>c_1/c_2$ and if $b_1=b_2=+\infty$, then we require $e_1/e_2>1$. Then

$$\begin{aligned} \left( \Phi _{c_2{\text {xs}}a_2}^{c_1{\text {xs}}a_1}\right) ^{-1}\left(e_1/e_2\right) \end{aligned}$$

is called the Pareto alpha between the layers $c_1$ xs $a_1$ and $c_2$ xs $a_2$.

Let $\alpha$ be the Pareto alpha between the layers $c_1$ xs $a_1$ and $c_2$ xs $a_2$ (with $a_1<a_2$ and $a_1+c_1 \le a_2+c_2$) and assume that $S=\sum _{n=1}^N X_n$ is a collective risk model with $t\le a_1$ and $X_n\sim {\text {Pareto}}(t,\alpha )$. Then the model matches the expected loss $e_1$ for the layer $c_1$ xs $a_1$ if and only if it matches the expected loss $e_2$ of the layer $c_2$ xs $a_2$. Note that the condition $e_1/e_2>c_1/c_2$ in the case $c_2<+\infty$ simply means that the risk rate on line of the lower layer $c_1$ xs $a_1$ is greater than the risk rate on line of the higher layer $c_2$ xs $a_2$ (cf. Sect. 2).

Appendix 2: Proof of Proposition 1

Let L denote the Lévy metric, i.e.

$$\begin{aligned} L(F,G):=\inf \left\{ \varepsilon > 0\,|\, F(x-\varepsilon )-\varepsilon \le G(x) \le F(x+\varepsilon )+\varepsilon \text { for all }x\in \mathbb {R}\right\} \end{aligned}$$

for distribution functions F and G. Let G be a distribution function with $G(0)=0$ and let $\varepsilon > 0$. We show that there exist parameter vectors $\mathbf {t}$ and $\varvec{\alpha }$ such that $L(F_{\mathbf {t},\varvec{\alpha }}, G) \le \varepsilon$. Choose $0<\delta \le \varepsilon$ such that $G(\delta )<1-\delta$. This is possible since G is right-continuous. Let

$$\begin{aligned} n:=1+\min \{ k\in \mathbb {N}\,|\, G(k\cdot \delta ) \ge 1-\delta \}. \end{aligned}$$

We define $t_1:=\delta /2$, $s_1:=1$ and for $k=2,\ldots , n$

$$\begin{aligned} t_k:=(k-1)\cdot \delta ,\quad s_k:=\min (1-G(t_k), \delta /2). \end{aligned}$$

For $k=1,\ldots ,n-1$ let

$$\begin{aligned} \alpha _k:=\frac{\log (s_{k+1}/s_k)}{\log (t_k/t_{k+1})} \end{aligned}$$

and choose an arbitrary $\alpha _n>0$. Let $\mathbf {t}:=(t_1,\ldots ,t_{n})$, $\varvec{\alpha }:=(\alpha _1,\ldots ,\alpha _n)$. Since $\alpha _k$ is the Pareto alpha between $(t_k,s_k)$ and $(t_{k+1},s_{k+1})$ we then have

$$\begin{aligned} F_{\mathbf {t},\varvec{\alpha }}(k\cdot \delta )=G(k\cdot \delta ) \end{aligned}$$

for $k=0,\ldots ,n-2$. Moreover, we have $F_{\mathbf {t},\varvec{\alpha }}(x),G(x)\in [1-\delta ,1]$ for $x\ge (n-1)\cdot \delta$. This implies

$$\begin{aligned} L(F_{\mathbf {t},\varvec{\alpha }},G)\le \delta \le \varepsilon . \end{aligned}$$

For $x\in [k\delta ,(k+1)\delta ]$ with $k+1\le n-2$ we have

$$\begin{aligned} F_{\mathbf {t},\varvec{\alpha }}(x-\delta )-\delta<F_{\mathbf {t},\varvec{\alpha }}(k\delta )=G(k\delta )\le G(x)\le G((k+1)\delta )=F_{\mathbf {t},\varvec{\alpha }}((k+1)\delta )<F_{\mathbf {t},\varvec{\alpha }}(x+\delta )+\delta . \end{aligned}$$

For $x\in [(n-2)\delta ,(n-1)\delta ]$ we have

$$\begin{aligned} F_{\mathbf {t},\varvec{\alpha }}(x-\delta )-\delta <F_{\mathbf {t},\varvec{\alpha }}((n-2)\delta )=G((n-2)\delta )\le G(x)\le 1 \le F_{\mathbf {t},\varvec{\alpha }}((n-1)\delta )+\delta \le F_{\mathbf {t},\varvec{\alpha }}(x+\delta )+\delta . \end{aligned}$$

For $x\ge (n-1)\delta$ we have

$$\begin{aligned} F_{\mathbf {t},\varvec{\alpha }}(x-\delta )-\delta \le F_{\mathbf {t},\varvec{\alpha }}(x)-\delta \le 1-\delta \le G(x) \le 1\le F_{\mathbf {t},\varvec{\alpha }}((n-1)\delta )+\delta \le F_{\mathbf {t},\varvec{\alpha }}(x+\delta )+\delta . \end{aligned}$$

$\square$

Appendix 3: Proof of Lemma 3

We have to show that it is possible to find an $f_1>e_1/c_1$ such that Matching Algorithm 1 does not stop at an $i<k$ in Step 2.

The case $k=1$ is clear. For $k\in \{2,3\}$ we only sketch the proof and leave the (simple) technical details to the reader. Let $\alpha (f_i)$ denote the Pareto alpha between $(a_i,f_i)$ and the layer $c_i$ xs $a_i$. Then, for every $i<k$,

$$\begin{aligned} \phi _i:(e_i/c_i,+\infty ) \rightarrow (0,e_i/c_i),\quad f_i\mapsto \left( \frac{a_i}{a_{i+1}}\right) ^{\alpha (f_i)} \end{aligned}$$

is a strictly decreasing homeomorphism. If $k=3$, then we choose $f_2$, such that

$$\begin{aligned} \frac{e_1}{c_1}>f_2>\frac{e_2}{c_2} \quad \text {and}\quad \phi _2(f_2)>\frac{e_3}{c_3} \end{aligned}$$

(possible, since $e_1/c_1>e_2/c_2>e_3/c_3$). If $k=2$, then we only require $e_1/c_1>f_2>e_2/c_2$. Then Matching Algorithm 1 provides the requested result if we start with $f_1:=\phi _1^{-1}(f_2)$. $\square$

Appendix 4: Proof of Theorem 1

Either frequencies $f_i$ with $f_1 > e_1/c_1$ and $e_{i-1}/c_{i-1}>f_i>e_{i}/c_{i}$ for $i=2,\ldots ,k$ are given or they are calculated in Step 1 of Matching Algorithm 2. After Step 2 of the algorithm, where we define $s_i:=f_i/f_1$ and $l_i:=e_i/f_1$, the preconditions of the following proposition are fulfilled.

Proposition 2

Consider a sequence of attachment points $0<a_1<\cdots <a_{k}$, a sequence of excess probabilities $1=s_1>s_2>\cdots>s_{k}>0$ and a sequence of loss expectations $l_1,\ldots ,l_k>0$ for the layers $a_{i+1}-a_i$ xs $a_i$ (with $a_{k+1}:=+\infty$), such that

$$\begin{aligned} s_i>\frac{l_i}{a_{i+1}-a_i}>s_{i+1} \end{aligned}$$

for $i=1,\ldots ,k-1$. Let $n=2k-1$. Then there exist parameters $\mathbf {t}=(t_1,\ldots ,t_n)$ with $t_{2i-1}=a_i$ and $\varvec{\alpha }=(\alpha _1,\ldots ,\alpha _n)$ with $\alpha _i>0$ such that

$$\begin{aligned} 1-F_{\mathbf {t},\varvec{\alpha }}(a_i) =s_i \quad \text { and } \quad I_{\mathbf {t},\varvec{\alpha }}(a_i,a_{i+1}) =l_i \end{aligned}$$

for $i=1,\ldots ,k$.

In the following proof of Proposition 2, it is explicitly shown how the parameter vectors $\mathbf {t}$ and $\varvec{\alpha }$ can be calculated inductively. Matching Algorithm 2 uses exactly the same approach for the calculation of the parameter vectors of the piecewise Pareto distribution. Therefore, the algorithm always leads to the desired collective risk model.

Proof

We use induction to prove this statement. In the base case $k=1$ we can use $t_1:=a_1$ and $\alpha _1:=a_1/l_1+1$ (cf. Lemma 1). For the inductive step $k\rightarrow k+1$ we assume now that the statement is true for a $k\ge 1$. Then we have thresholds $\mathbf {t}^{(k)}=(t_1,\ldots ,t_{2k-1})$ with $t_{2i-1}=a_i$ and $\varvec{\alpha }^{(k)}=(\alpha _1,\ldots ,\alpha _{2k-2},\alpha _{2k-1}^{(k)})$ such that

$$\begin{aligned} 1-F_{\mathbf {t}^{(k)},\varvec{\alpha }^{(k)}}(a_i)=s_i \quad \text { and } \quad I_{\mathbf {t}^{(k)},\varvec{\alpha }^{(k)}}(a_i,a_{i+1})=l_i \end{aligned}$$

for $i=1,\ldots ,k$. Let $t_{2k+1}:=a_{k+1}$ and

$$\begin{aligned} \alpha _{2k+1}:=\frac{s_{k+1}a_{k+1}}{l_{k+1}}+1. \end{aligned}$$

We will show that there exist $t_{2k}\in (a_k,a_{k+1})$ and $\alpha _{2k-1},\alpha _{2k}> 0$ such that

$$\begin{aligned} 1-F_{\mathbf {t},\varvec{\alpha }}(a_{k+1})= & {} s_{k+1}\quad \text { and}\end{aligned}$$

(1)

$$\begin{aligned} I_{\mathbf {t},\varvec{\alpha }}(a_{k},a_{k+1})= & {} l_{k} \end{aligned}$$

(2)

for $\mathbf {t}=(t_1,\ldots ,t_{2k+1})$ and $\varvec{\alpha }=(\alpha _1,\ldots ,\alpha _{2k+1})$. Due to $1-F_{\mathbf {t},\varvec{\alpha }}(a_{k+1})=s_{k+1}$ and the definition of $\alpha _{2k+1}$ we then also have

$$\begin{aligned} I_{\mathbf {t},\varvec{\alpha }}(a_{k+1},+\infty )=l_{k+1} \end{aligned}$$

(cf. Lemma 1 in Sect. 3). For $\tau \in (a_k,a_{k+1})$ and $\alpha \ge 0$ we define

$$\begin{aligned} \sigma ^{(k)}(\tau ,\alpha ):=\frac{\ln (s_{k+1}/s_k)-\alpha \ln (a_k/\tau )}{\ln (\tau /a_{k+1})} \end{aligned}$$

and

$$\begin{aligned} \lambda ^{(k)}(\tau ,\alpha ):=s_k\cdot I_{(a_{k},\tau ), (\alpha ,\sigma (\tau ,\alpha ))}(a_k,a_{k+1}). \end{aligned}$$

For $\alpha _{2k-1}>0$ and $\alpha _{2k}>0$ we have

$$\begin{aligned} (1) \;&\Longleftrightarrow \; \left( \frac{a_k}{t_{2k}}\right) ^{\alpha _{2k-1}}\left( \frac{t_{2k}}{a_{k+1}}\right) ^{\alpha _{2k}}=\frac{s_{k+1}}{s_k} \\&\Longleftrightarrow \;\;\alpha _{2k-1}\cdot \ln (a_k/t_{2k})+\alpha _{2k}\cdot \ln (t_{2k}/a_{k+1})=\ln (s_{k+1}/s_k)\\&\Longleftrightarrow \;\;\alpha _{2k}=\frac{\ln (s_{k+1}/s_k)-\alpha _{2k-1}\ln (a_k/t_{2k})}{\ln (t_{2k}/a_{k+1})}, \end{aligned}$$

i.e. (1) is equivalent to

$$\begin{aligned} \alpha _{2k}=\sigma ^{(k)}(t_{2k},\alpha _{2k-1}). \end{aligned}$$

(3)

If (3) is fulfilled then (2) is equivalent to

$$\begin{aligned} \lambda ^{(k)}(t_{2k},\alpha _{2k-1})=l_k. \end{aligned}$$

(4)

For fixed $\tau \in (a_k,a_{k+1})$ the functions $\alpha \mapsto \sigma ^{(k)}(\tau ,\alpha )$ and $\alpha \mapsto \lambda ^{(k)}(\tau ,\alpha )$ are continuous and strictly decreasing and we have

$$\begin{aligned} \sigma ^{(k)}(\tau ,0)=\frac{\ln (s_{k+1}/s_k)}{\ln (\tau /a_{k+1})} \quad \text {and}\quad \sigma ^{(k)}\left( \tau ,\frac{\ln (s_{k+1}/s_k)}{\ln (a_{k}/\tau )}\right) =0. \end{aligned}$$

Therefore, for a given $t_{2k}\in (a_k,a_{k+1})$, the Eqs. (3) and (4) can be solved with $\alpha _{2k-1}>0$ and $\alpha _{2k}=\sigma ^{(k)}(t_{2k},\alpha _{2k-1})>0$ if and only if

$$\begin{aligned} \lambda ^{(k)} \left( t_{2k},\frac{\ln (s_{k+1}/s_k)}{\ln (a_{k}/t_{2k})}\right)< l_k< \lambda ^{(k)}(t_{2k},0). \end{aligned}$$

The functions

$$\begin{aligned} \tau \mapsto \lambda ^{(k)} \left( \tau ,\frac{\ln (s_{k+1}/s_k)}{\ln (a_{k}/\tau )}\right) \quad \text { and }\quad \tau \mapsto \lambda ^{(k)}(\tau ,0) \end{aligned}$$

are strictly increasing and we have

$$\begin{aligned} \lim _{\tau \searrow a_k} \lambda ^{(k)} \left( \tau ,\frac{\ln (s_{k+1}/s_k)}{\ln (a_{k}/\tau )}\right) =s_{k+1}\cdot (a_{k+1}-a_k) <l_k \end{aligned}$$

and

$$\begin{aligned} \lim _{\tau \nearrow a_{k+1}}\lambda ^{(k)} \left( \tau ,0\right) =s_k\cdot (a_{k+1}-a_k)>l_k. \end{aligned}$$

Let

$$\begin{aligned} \tau _l^{(k)}:=\inf \left\{ \tau \in (a_k,a_{k+1})\,\left| \,\lambda ^{(k)} (\tau ,0\right. )>l_k\right\} \end{aligned}$$

and

$$\begin{aligned} \tau _u^{(k)}:=\sup \left\{ \tau \in (a_k,a_{k+1})\,\left| \,\lambda ^{(k)} \left( \tau ,\frac{\ln (s_{k+1}/s_k)}{\ln (a_{k}/\tau )}\right. \right) <l_k\right\} . \end{aligned}$$

We have $\tau _u^{(k)}>a_k$, and if $\tau _l^{(k)}>a_k$, then we have

$$\begin{aligned} \lambda ^{(k)}\left( \tau _l^{(k)},\frac{\ln (s_{k+1}/s_k)}{\ln (a_{k}/\tau _l^{(k)})}\right) <\lambda ^{(k)}(\tau _l^{(k)},0)=l_k. \end{aligned}$$

Since $\tau _l^{(k)}<a_{k+1}$, it results that $\tau _l^{(k)}<\tau _u^{(k)}$. We select a $t_{2k}\in (\tau _l^{(k)},\tau _u^{(k)})$ and obtain

$$\begin{aligned} \lambda ^{(k)} \left( t_{2k},\frac{\ln (s_{k+1}/s_k)}{\ln (a_{k}/t_{2k})}\right)<\lim _{\tau \nearrow \tau _u^{(k)}}\lambda ^{(k)}\left( \tau ,\frac{\ln (s_{k+1}/s_k)}{\ln (a_{k}/\tau )}\right) \le l_k\le \lim _{\tau \searrow \tau _l^{(k)}} \lambda ^{(k)}(\tau ,0)<\lambda ^{(k)}(t_{2k},0), \end{aligned}$$

i.e. Eqs. (3) and (4) can be solved with $\alpha _{2k-1}>0$ such that $\alpha _{2k}:=\sigma ^{(k)}(t_{2k},\alpha _{2k-1})>0$. $\square$