Exogenous shock models: analytical characterization and probabilistic construction

Abstract

A new characterization for survival functions of multivariate failure-times arising in exogenous shock models with non-negative, continuous, and unbounded shocks is presented. These survival functions are the product of their ordered and individually transformed arguments. The involved transformations may depend on the specific order of the arguments and must fulfill a monotonicity condition. Conversely, every survival function of that form can be constructed using an exogenous shock model with independent and non-homogeneous shocks.

Appendices

Proof of Theorem 1

The theorem will be proven in four steps. Particularly, it is shown that $3. \Rightarrow 4. \Rightarrow 1. \Rightarrow 2. \Rightarrow 3.$ Proofs of used auxiliary results are deferred to “Appendix B”.

Remark 1

Under the assumptions of Theorem 1, particularly the representation of ${\bar{F}}$ in Eq. (6), the expression

$$\begin{aligned} g_{i}^{\pi } (t) = \frac{\prod _{j = 1}^{i} g_{j}^{\pi } (t) }{\prod _{j = 1}^{i - 1} g_{j}^{\pi } (t)} \end{aligned}$$

is invariant for different permutations with coinciding images of $[i-1]$ and i. If the first statement of the theorem is fulfilled, then $g_i^\pi$ has the interpretation of a conditional probability, i.e.

$$\begin{aligned} g_i^\pi (t) = {\mathbb {P}} \left( \tau _{\pi (i)}> t \mid \varvec{\tau }_{\pi ([i - 1])} > t \right) . \end{aligned}$$

Hence, the function $g_i^\pi$ only depends on $\pi ([i-1])$ and $\pi (i)$ and it is justified to work with ${\tilde{g}}^{\pi ([i]), \pi (i)}$.

Proof of$3. \Rightarrow 4.$ First observe that 4. is a special case of 3., hence $3. \Rightarrow 4.$ follows directly. $\square$

Proof of$4. \Rightarrow 1.$ Let 4. from Theorem 1 be fulfilled and define for independent random variables $Z_{I} \sim {\bar{S}}_{I}$, $\emptyset \not = I \subseteq [d]$, the random vector $\varvec{\tau }$ by

$$\begin{aligned} \tau _{i} :=\min \left\{ Z_{I} : i \in I \right\} , \quad \ i \in [d] . \end{aligned}$$

For $\varvec{t} \ge 0$ and $\pi \in {\mathcal {S}}_{d}$ with $t_{\pi (1)} \ge \cdots \ge t_{\pi (d)}$, using the independence of the shock variables and reordering the factors, it holds that

$$\begin{aligned} {\mathbb {P}}\left( \varvec{\tau }> \varvec{t} \right) = \prod _{\emptyset \not = I \subseteq [d]}{ {\mathbb {P}}\left( Z_{I}> \max _{i \in I}{t_{i}} \right) } = \prod _{i = 1}^{d}{ \left( \prod _{\begin{array}{c} I \subseteq \pi \left( \{i, i + 1, \ldots , d\}\right) \\ \pi (i) \in I \end{array}}{ {\mathbb {P}}\left( Z_{I} > t_{\pi (i)} \right) } \right) } . \end{aligned}$$

For $i \in [d]$ and $\pi (i) \in I \subseteq \pi (\{ i, \ldots , d\})$, by assumption, the survival function ${\bar{S}}_{I} \equiv {\bar{S}}_{I}^{\pi (i)}$ has a representation as in Eq. (10) with $m = \pi (i)$ and

$$\begin{aligned} \prod _{\begin{array}{c} I \subseteq \pi \left( \{i, i + 1, \ldots , d\}\right) \\ \pi (i) \in I \end{array}}{ {\mathbb {P}}\left( Z_{I} > t_{\pi (i)} \right) } = \prod _{\begin{array}{c} I \subseteq \pi \left( \{i, i + 1, \ldots , d\} \right) \\ \pi (i) \in I \end{array}}{ \left( \prod _{\begin{array}{c} J \subseteq I \\ \pi (i) \in J \end{array}}{ \left( {\tilde{g}}^{J \cup ([d] \backslash I), \pi (i)} \left( t_{\pi (i)}\right) \right) ^{(-1)^{|J |- 1}} } \right) } . \end{aligned}$$

Fix $K \subseteq [d]$ with $\pi \left( [i] \right) \subseteq K$; then $i \le |K |= k \le d$ and $1 \le j \le k - i + 1$. The expression ${\tilde{g}}^{K, \pi (i)}(t_{\pi (i)})$ with an exponent of $(-1)^{j - 1}$ appears $\left( {\begin{array}{c}k - i\\ j - 1\end{array}}\right)$ times, as there are exactly $\left( {\begin{array}{c}k - i\\ j - 1\end{array}}\right)$ possible choices for J with $\pi (i) \in J \subseteq K \backslash \pi ([i-1])$. Hence, the overall exponent of the expression ${\tilde{g}}^{K, \pi (i)}(t_{\pi (i)})$ is

$$\begin{aligned} \sum _{j = 1}^{k - i + 1}{ (-1)^{j - 1} \left( {\begin{array}{c}k - i\\ j - 1\end{array}}\right) } = \sum _{j = 0}^{k - i}{ (-1)^{j} \left( {\begin{array}{c}k - i\\ j\end{array}}\right) } = (1 - 1)^{k - i} = {\left\{ \begin{array}{ll} 1, &{} k = i \\ 0, &{} k > i , \end{array}\right. } \end{aligned}$$

where the latter expression follows with the binomial formula. Finally, it follows that

$$\begin{aligned} {\mathbb {P}}\left( \varvec{\tau } > \varvec{t} \right) = \prod _{i = 1}^{d}{ {\tilde{g}}^{\pi \left( [i] \right) , \pi (i)}\left( t_{\pi (i)}\right) } = \prod _{i = 1}^{d}{ g_{i}^{\pi }\left( t_{\pi (i)}\right) } . \end{aligned}$$

$\square$

In the following, $I_{1}$, $I_{2}$, $\{ \pi _{J} \}_{J \subseteq I_{2}}$, s and t (or a subset of these elements) fulfill the usual conditions if

• 1. $s > t \ge 0$,

• 2. $I_{1}, I_{2} \subseteq [d]$ with $I_{1} \cap I_{2} = \emptyset$ and $I_{2} \not = \emptyset$,

• 3. for $J \subseteq I_{2}$ one has

  • (a) $\pi _{J} \left( \{ 1, \ldots , |I_{1} |\} \right) = I_{1}$ (if $I_{1} \not = \emptyset$),

  • (b) $\pi _{J} \left( \{ |I_{1} |+ 1, \ldots , |I_{1} \cup J |\} \right) = J$,

  • (c) $\pi _{J} \left( \{ |I_{1} \cup J |+ 1, \ldots , |I_{2} |\} \right) = I_{2} \backslash J$.

If only a specific permutation $\pi$ is used, it is assumed that it fulfills this property for $J = I_{2}$.

Proof of$1. \Rightarrow 2.$ Let 1. in Theorem 1 be fulfilled and let $I_{1}$, $I_{2}$, $\{ \pi _{J} \}_{J \subseteq I_{2}}$, s and t fulfill the usual conditions. First assume that for arbitrary $\pi \in {\mathcal {S}}_{d}$ and $i \in [d]$ the functions $g_{i}^{\pi }$ are strictly positive on ${\mathbb {R}}_{+}$. Then

$$\begin{aligned} G_{I_{1}, I_{2}}^{\{ \pi _{J} \}_{J \subseteq I_{2}}} (s, t) = \frac{ \sum _{J \subseteq I_{2}}{ (-1)^{|J |} \prod _{j = 1}^{|I_{1} \cup J |}{ g_{j}^{\pi _{J}}(s) } \prod _{j = 1}^{|I_{2} \backslash J |}{ g_{|I_{1} \cup J |+ j}^{\pi _{J}}(t) } } }{ \prod _{j = 1}^{|I_{1} |}{ g_{j}^{\pi _{\emptyset }}(s) } } , \end{aligned}$$

(17)

where it is used that by 1. the diagonal of marginal survival functions of $\varvec{\tau }_{I_{1}}$ can be represented with every $\pi$ fulfilling $\pi (\{1, \ldots , |I_{1}|\}) = I_{1}$. Particularly, it holds that

$$\begin{aligned} {\mathbb {P}} \left( \tau _{i} > s, i \in I_{1} \right) = \prod _{j = 1}^{|I_{1} |}{ g_{j}^{\pi _{J_{1}}} (s) } = \prod _{j = 1}^{|I_{1} |}{ g_{j}^{\pi _{J_{2}}} (s) }, \ J_{1}, J_{2} \subseteq I_{2},\quad \ s \ge 0 . \end{aligned}$$

Subsequently, the numerator of Eq. (17) can be rewritten using the principle of inclusion and exclusion as

$$\begin{aligned}&\sum _{i = 0}^{|I_{2} |}{ (-1)^{i} \sum _{J \subseteq I_{2} : |J |= i}{ \prod _{j = 1}^{|J\cup I_{1} |}{ g_{j}^{\pi _{J}}(s) } \prod _{j = 1}^{|I_{2} \backslash J |}{ g_{|I_{1} \cup J |+ j}^{\pi _{J}}(t) } } } \\&\quad = {\mathbb {P}}\left( A_{\emptyset }^{I_{1}, I_{2}} \right) - \sum _{i = 1}^{|I_{2} |}{ (-1)^{i + 1} \sum _{J \subseteq I_{2} : |J |= i}{ {\mathbb {P}}\left( \bigcap _{j \in J}{ A_{j}^{I_{1}, I_{2}} } \right) } } \\&\quad = {\mathbb {P}}\left( A_{\emptyset }^{I_{1}, I_{2}} \right) - {\mathbb {P}}\left( \bigcup _{i \in I_{2}}{ A_{i}^{I_{1}, I_{2}} } \right) = {\mathbb {P}}\left( A^{I_{1}, I_{2}} \right) , \end{aligned}$$

where

$$\begin{aligned} A^{I_{1}, I_{2}}&:=\left\{ \tau _{i}> s \ \forall i \in I_{1}, \tau _{i} \in (t, s] \ \forall i \in I_{2} \right\} , \\ A_{\emptyset }^{I_{1}, I_{2}}&:=\left\{ \tau _{i}> s \ \forall i \in I_{1}, \tau _{i}> t \ \forall i \in I_{2} \right\} , \quad \text {and}\\ A_{i}^{I_{1}, I_{2}}&:=\left( \bigcap _{j \in I_{1} \cup \{ i \}}{ \{ \tau _{j}> s \} } \right) \cap \left( \bigcap _{j \in I_{2} \backslash \{ i \}}{ \{ \tau _{j} > t \} } \right) , \quad \ i \in I_{2} . \end{aligned}$$

It follows that

$$\begin{aligned} G_{I_{1}, I_{2}}^{\{ \pi _{J} \}_{J \subseteq I_{2}}} (s, t) = {\mathbb {P}} \left( \tau _{i} \in (t, s] \ \forall i \in I_{2} \mid \tau _{i} > s \ \forall i \in I_{1} \right) \end{aligned}$$

and subsequently that $G_{I_{1}, I_{2}}^{\{ \pi _{J} \}_{J \subseteq I_{2}}} (s, t)$ is non-negative and does not depend on the specific choice of $\{ \pi _{J} \}_{J \subseteq I_{2}}$.⁹

Now, by induction over i, the strict positivity, continuity, and non-increasingness of $g_{i}^{\pi }$ is proven for all $\pi \in {\mathcal {S}}_{d}$. This implies that $G_{I_{1}, I_{2}} (s, t)$ is continuous in s and t. For $i = 1$ and $\pi \in {\mathcal {S}}_{d}$, the assumptions of Theorem 1 imply that $g_{1}^{\pi }$ is strictly positive, continuous, and non-increasing. Let the claim be fulfilled for $j < i$, i.e. $g_{j}^{\pi }$ is strictly positive, continuous, and non-increasing for $j \le i -1$ and $\pi \in {\mathcal {S}}_{d}$.

Right-continuity and left-limits It is well known, see, e.g., Schweizer and Sklar (1983, Chp. 6), that copulae are Lipschitz-continuous with constant one. Hence, by exploiting the copula/survival function decomposition, it holds that

$$\begin{aligned} \left|{\bar{F}}(s_{1}, \ldots , s_{d}) - {\bar{F}} (t_{1}, \ldots , t_{d}) \right|\le \sum _{i = 1}^{d}{ \left|{\bar{F}}_{i} (s_{i}) - {\bar{F}}_{i} (t_{i}) \right|} , \quad \forall \varvec{t}, \varvec{s} \ge 0 \end{aligned}$$

and right-continuity as well as left-limits of ${\bar{F}}$ are inherited from the margins. For $\pi \in {\mathcal {S}}_{d}$ the survival function $t \mapsto {\mathbb {P}} \left( \min _{j\le i} \tau _{\pi (j)} > t \right)$ is right-continuous with left-limits and with

$$\begin{aligned} g_{i}^{\pi } (t) = \frac{\prod _{j = 1}^{i}{ g_{j}^{\pi } (t) }}{\prod _{j = 1}^{i - 1}{ g_{j}^{\pi } (t) }} = \frac{{\mathbb {P}} \left( \min _{j\le i} \tau _{\pi (j)} > t \right) }{\prod _{j = 1}^{i - 1}{ g_{j}^{\pi } (t) }} , \end{aligned}$$

right-continuity with left-limits for $g_{i}^{\pi }$ follows with the induction hypothesis.

Non-increasingness For $\pi \in {\mathcal {S}}_{d}$ and $s \ge t \ge 0$ define the vector $\varvec{u} (s, t)$ by

$$\begin{aligned} u_{\pi (j)} (s, t) :={\left\{ \begin{array}{ll} s ,&{} \quad \forall j < i , \\ t ,&{} \quad j = i , \\ 0 ,&{} \quad \forall j > i . \end{array}\right. } \end{aligned}$$

Then, by monotonicity of the measure ${\mathbb {P}}$, one has

$$\begin{aligned} {\mathbb {P}} \left( \varvec{\tau }> \varvec{u} (s,s) \right)&\le {\mathbb {P}} \left( \varvec{\tau } > \varvec{u} (s,t) \right) \\&\Leftrightarrow g_{i}^{\pi } (s) \prod _{j = 1}^{i - 1}{ g_{j}^{\pi } (s) } \le g_{i}^{\pi } (t) \prod _{j = 1}^{i - 1}{ g_{j}^{\pi } (s) } \\&\Leftrightarrow g_{i}^{\pi } (s) \le g_{i}^{\pi } (t) , \end{aligned}$$

where the induction hypothesis, i.e. $g_{j}^{\pi }$ is strictly positive for all $j < i$, is used.

Strict positivity Assume for $\pi \in {\mathcal {S}}_{d}$ that there exists a finite upper bound $s^{\star }$ for strict positivity of $g_{i}^{\pi }$, i.e. $s^{\star } :=\inf {\{ u > 0 : g_{i}^{\pi } (u) = 0 \}} < \infty$, and as $g_{i}^{\pi }$ is right-continuous and non-increasing we have that $g_{i}^{\pi } (s^{\star }) = 0$. For $t < s^{\star }$, let $I_{1} =\pi \left( \{ 1 , \ldots , i - 2\} \right)$ and $I_{2} = \pi (\{ i - 1, i \})$. Furthermore, let ${\tilde{\pi }}$ be the permutation which switches the positions of $i - 1$ and i in $\pi$, i.e. ${\tilde{\pi }} = \pi (i - 1, i)$. Assume w.l.o.g. that $s^{\star } \le u^{\star }$ for $u^{\star } :=\inf {\{ u > 0 : g_{i}^{{\tilde{\pi }}} (u) = 0 \}} \in \bar{{\mathbb {R}}}_{+}$ (else switch the roles of $\pi$ and ${\tilde{\pi }}$ and prove the contradiction for ${\tilde{\pi }}$ first). Then, with the induction hypothesis it holds that $g_{j}^{\pi }, g_{j}^{{\tilde{\pi }}} > 0 \ \forall j < i$ and, for $\pi _{\emptyset } \in \{ \pi , {\tilde{\pi }} \}$, that

$$\begin{aligned} 0 {\mathop {\le }\limits ^{\text {IH}}} G_{I_{1}, I_{2}}(s^{\star }, t)= & {} \prod _{j = i - 1}^{i}{ g_{j}^{\pi _{\emptyset }}(t) } - g_{i - 1}^{\pi }(s^{\star }) g_{i}^{\pi }(t) - g_{i - 1}^{{\tilde{\pi }}}(s^{\star }) g_{i}^{{\tilde{\pi }}}(t) + \prod _{j = i - 1}^{i}{ g_{j}^{\pi }(s^{\star }) } \nonumber \\= & {} g_{i - 1}^{\pi _{\emptyset }}(t) g_{i}^{\pi _{\emptyset }}(t) - g_{i - 1}^{\pi }(s^{\star }) g_{i}^{\pi }(t) - g_{i - 1}^{{\tilde{\pi }}}(s^{\star }) g_{i}^{{\tilde{\pi }}}(t) \nonumber \\= & {} {\left\{ \begin{array}{ll} \left( g_{i - 1}^{\pi }(t) - g_{i - 1}^{\pi } (s^{\star }) \right) g_{i}^{\pi }(t) - g_{i - 1}^{{\tilde{\pi }}}(s^{\star }) g_{i}^{{\tilde{\pi }}}(t) , &{} \pi _{\emptyset } = \pi \\ \left( g_{i - 1}^{{\tilde{\pi }}}(t) - g_{i - 1}^{{\tilde{\pi }}} (s^{\star }) \right) g_{i}^{{\tilde{\pi }}}(t) - g_{i - 1}^{\pi }(s^{\star }) g_{i }^{\pi }(t) , &{} \pi _{\emptyset } = {\tilde{\pi }}. \end{array}\right. } \end{aligned}$$

(18)

The last expression in Eq. (18) becomes negative if t is sufficiently close to $s^{\star }$:

• 1. If $u^{\star } > s^{\star }$, choose $\pi _{\emptyset } = \pi$. Then for $t \nearrow s^{\star }$ Eq. (18) approaches $-g_{i - 1}^{{\tilde{\pi }}}(s^{\star }) g_{i}^{{\tilde{\pi }}}(s^{\star }-)$.

  As $g_{i - 1}^{{\tilde{\pi }}} (s^{\star }) > 0$ by the induction hypothesis and $g_{i}^{{\tilde{\pi }}} (t) > 0 \ \forall t < u^{\star }$ with $s^{\star } < u^{\star }$ by the assumption made above it holds that

  $$\begin{aligned} 0 \le -g_{i - 1}^{{\tilde{\pi }}} (s^{\star }) g_{i}^{{\tilde{\pi }}} (s^{\star }-) < 0 . \end{aligned}$$

• 2. If $s^{\star } = u^{\star }$ and $g_{i}^{\pi _{\emptyset }} (s^{\star }-) > g_{i}^{\pi _{\emptyset }} (s^{\star }) = 0$ for at least one $\pi _{\emptyset } \in \{ \pi , {\tilde{\pi }} \}$, then for $t \nearrow s^{\star }$ Eq. (18) approaches $-g_{i - 1}^{\pi _{\emptyset }}(s^{\star }) g_{i}^{\pi _{\emptyset }}(s^{\star }-)$.

  As $g_{i - 1}^{\pi _{\emptyset }} (s^{\star }) > 0$ by the induction hypothesis and $g_{i}^{\pi _{\emptyset }} (s^{\star }-) > 0$ by the assumption made above it holds that

  $$\begin{aligned} 0 \le -g_{i - 1}^{\pi _{\emptyset }} (s^{\star }) g_{i}^{\pi _{\emptyset }} (s^{\star }-) < 0 . \end{aligned}$$

• 3. Otherwise, as $g_{j}^{\pi _{\emptyset }}$ for $j \in \{ i - 1, i\}$ have left-limits by the induction hypothesis, for every sequence $t_{k} \nearrow s^{\star }$ with $t_k \ne s^\star$, non-negative sequences $\{ a_{j, k}^{\pi _{\emptyset }} \}_{k \in {\mathbb {N}}}$ with $a_{j, k}^{\pi _{\emptyset }} (s^{\star } - t_{k}) \rightarrow 0$ for $k \rightarrow \infty$ can be found s.t.

  $$\begin{aligned} g_{j}^{\pi _{\emptyset }} (t_{k}) = g_{j}^{\pi _{\emptyset }} (s^{\star }-) + a_{j, k}^{\pi _{\emptyset }} (s^{\star } - t_{k}) ,\quad \ j\in \{i-1, i\} ,\quad \ k \in {\mathbb {N}} . \end{aligned}$$

  By the assumption on $s^\star$, it holds that $a_{i, k}^{\pi _\emptyset } > 0$ for all $k\in {\mathbb {N}}$ and $\pi _\emptyset \in \{\pi , {\tilde{\pi }}\}$. If $s^{\star } = u^{\star }$ and $g_{i}^{\pi _{\emptyset }} (s^{\star }-) = g_{i}^{\pi _{\emptyset }} (s^{\star }) = 0$ for all $\pi _{\emptyset } \in \{ \pi , {\tilde{\pi }} \}$, it follows from Eq. (18) and (left-)continuity of $g_{i - 1}^{\pi _{\emptyset }}$ that

  $$\begin{aligned} 0 \le {\left\{ \begin{array}{ll} a_{i - 1, k}^{\pi } a_{i, k}^{\pi } (s^{\star } - t_{k})^2 - g_{i - 1}^{{\tilde{\pi }}} (s^{\star }) a_{i, k}^{{\tilde{\pi }}} (s^{\star } - t_{k}) , {}&{} \pi _{\emptyset } = \pi \\ a_{i - 1, k}^{{\tilde{\pi }}} a_{i, k}^{{\tilde{\pi }}} (s^{\star } - t_{k})^2 - g_{i - 1}^{\pi } (s^{\star }) a_{i, k}^{\pi } (s^{\star } - t_{k}) , &{} \pi _{\emptyset } = {\tilde{\pi }} \end{array}\right. } \end{aligned}$$

  or equivalently

  $$\begin{aligned} 0 \le {\left\{ \begin{array}{ll} a_{i - 1, k}^{\pi } (s^{\star } - t_{k}) \frac{a_{i, k}^{\pi }}{a_{i, k}^{{\tilde{\pi }}}} - g_{i - 1}^{{\tilde{\pi }}} (s^{\star }) , {}&{} \pi _{\emptyset } = \pi \\ a_{i - 1, k}^{{\tilde{\pi }}} (s^{\star } - t_{k}) \frac{a_{i, k}^{{\tilde{\pi }}}}{a_{i, k}^{\pi }} - g_{i - 1}^{\pi } (s^{\star }) , &{} \pi _{\emptyset } = {\tilde{\pi }} . \end{array}\right. } \end{aligned}$$

  Now choose k sufficiently large and $\pi _{\emptyset }$ s.t. the fraction appearing in the upper equation is smaller or equal to 1, then

  $$\begin{aligned} \begin{aligned} 0 {}&\le {\left\{ \begin{array}{ll} a_{i - 1, k}^{\pi } (s^{\star } - t_{k}) - g_{i - 1}^{{\tilde{\pi }}} (s^{\star }) , {}&{} a_{i, k}^\pi \le a_{i, k}^{{\tilde{\pi }}} \\ a_{i - 1, k}^{{\tilde{\pi }}} (s^{\star } - t_{k}) - g_{i - 1}^{\pi } (s^{\star }) , &{} a_{i, k}^\pi > a_{i, k}^{{\tilde{\pi }}} \end{array}\right. } \\&< 0 , \end{aligned} \end{aligned}$$

where it is used that the respective first summand converges for $k \rightarrow \infty$ to 0 and the last summand is negative. Hence, a contradiction is found for each case and therefore $g_{i}^{\pi }(t) > 0 \ \forall t \in {\mathbb {R}}_{+}$.

Left-continuity Let $I_{1}$ and $I_{2}$ as well as $\pi$, ${\tilde{\pi }}$, and $\pi _{\emptyset }$ be as above. Then, for all $s > t \ge 0$ the function

$$\begin{aligned} {\mathbb {P}} \left( \tau _{i} \in (t, s], i \in I_{2} \mid \tau _{i} > s, i \in I_{1} \right) = G_{I_{1}, I_{2}} (s, t) \end{aligned}$$

has left-limits in t. Assume that there exists $s^{\dagger } \in {\mathbb {R}}_{+}^{\times }$ with $g_{i}^{\pi }(s^{\dagger }-) > g_{i}^{\pi }(s^{\dagger })$, then

$$\begin{aligned}&0 {\mathop {\le }\limits ^{\text {IH}}} \lim _{t \nearrow s^{\dagger }}{ G_{I_{1}, I_{2}}(s^{\dagger }, t) } \\&\quad = \lim _{t \nearrow s^{\dagger }}{} \left( \prod _{j = i - 1}^{i}{ g_{j}^{\pi _{\emptyset }}(t) } - g_{i - 1}^{\pi }(s^{\dagger }) g_{i}^{\pi }(t) - g_{i - 1}^{{\tilde{\pi }}}(s^{\dagger }) g_{i}^{{\tilde{\pi }}}(t) + \prod _{j = i - 1}^{i}{ g_{j}^{\pi }(s^{\dagger }) } \right) \\&\quad {\mathop {=}\limits ^{\pi _{\emptyset } = {\tilde{\pi }}, (\star )}} \left( g_{i}^{\pi }(s^{\dagger }) - g_{i}^{\pi }(s^{\dagger }-) \right) g_{i - 1}^{\pi }(s^{\dagger }) < 0 , \end{aligned}$$

where it is used in $(\star )$, that the first and third summand cancel out, when using that $g_{i - 1}^{{\tilde{\pi }}}$ is continuous under the induction hypothesis. This is a contradiction—hence $g_{i}^{\pi }$ is left-continuous. $\square$

Remark 2

The induction in the second part of the proof can be performed on the basis of statement 2. (instead of 1.) from Theorem 1 if the parts on right-continuity with left-limits and non-increasingness are replaced by the following lemma (as they rely on the survival function assumption of 1.). In particular, 2. implies $g_{i}^{\pi } \in \bar{{\mathcal {G}}}$ for all $i \in [d], \pi \in {\mathcal {S}}_{d}$.

Lemma 1

Let 2. from Theorem 1 be fulfilled and $g_{j}^{\pi }$ be right-continuous with left-limits, non-increasing, and strictly positive for all $j \le i - 1$ and $\pi \in {\mathcal {S}}_{d}$. Then $g_{i}^{\pi }$ is right-continuous with left-limits and non-increasing for all $\pi \in {\mathcal {S}}_{d}$.

Lemma 2

Assume that statement 2. of Theorem 1 is fulfilled and let $I_{1}$ and $I_{2}$ fulfill the usual conditions. Then for each $m \in I_{2}$, ${\bar{S}}_{I_{1}, I_{2}}^{m}$ is an ${\mathbb {R}}_{+}$-valued, positive, and continuous function on ${\mathbb {R}}_{+}$. Furthermore, ${\bar{S}}_{I_{1}, I_{2}}^{m}$ does not depend on $m\in I_2$, i.e.

$$\begin{aligned} {\bar{S}}_{I_{1}, I_{2}}^{m_{1}} (t) = {\bar{S}}_{I_{1}, I_{2}}^{m_{2}} (t) , \quad \forall t \ge 0, m_{1}, m_{2} \in I_{2} . \end{aligned}$$

(19)

Lemma 3

Let $I_{1}$ and $I_{2}$ fulfill the usual conditions and assume that ${\bar{S}}_{I_{1} \cup I_{2} \backslash J, J}^{m_{1}} = {\bar{S}}_{I_{1} \cup (I_{2} \backslash J), J}^{m_{2}} \in \bar{{\mathcal {G}}}$ for all $\emptyset \not = J \subseteq I_{2}$ and $m_{1}, m_{2} \in J$. Then for $s > t \ge 0$

$$\begin{aligned} G_{I_{1}, I_{2}} (s, t) = {\mathbb {P}} \left( {\check{\tau }}_{i} \in (t, s] \ \forall i \in I_{2} \right) , \end{aligned}$$

where

$$\begin{aligned} {\check{\tau }}_{i} :=\min {\left\{ \check{Z}_{J} : i \in J \subseteq I_{2} \right\} }, \quad \ i \in [d] \end{aligned}$$

with independent random shocks $\check{Z}_{J} \sim {\bar{S}}_{I_{1} \cup I_{2} \backslash J, J}$ for $\emptyset \not = J \subseteq I_{2}$.

The essence of the previous Lemma is the following: Let $I_1$ and $I_2$ fulfill the usual conditions, $Z_I\sim {\bar{S}}_I \in \bar{{\mathcal {G}}}$, $\emptyset \ne I\subseteq [d]$, $\varvec{\tau }$ be defined as in Eq. (1), and $\check{\varvec{\tau }} \in {\mathbb {R}}_+^{|I_2|}$ be defined by

$$\begin{aligned} {\check{\tau }}_i :=\min \left\{ \min \left\{ Z_J : J\cap (I_1\cup I_2) = I \right\} : i\in I\subseteq I_2 \right\} . \end{aligned}$$

Then

$$\begin{aligned} {\mathbb {P}} \left( \tau _i \in (t, s] \ \forall i \in I_2 \mid \tau _i> s \ \forall i \in I_1 \right) = {\mathbb {P}} \left( {\check{\tau }}_i \in (t, s] \ \forall i \in I_2 \right) , \quad \forall s > t \ge 0 . \end{aligned}$$

Lemma 4

Let $I_{1}$ and $I_{2}$ fulfill the usual conditions. Then, for a specific family of permutions, $\{ \pi _{J} \}_{J \subseteq I_{2}}$, the function $G_{I_{1}, I_{2}}^{\{ \pi _{J} \}_{J \subseteq I_{2}}}$ depends on $g_{i}^{\pi _{J}}, |I_{1} |+ 1 \le i \le |I_{1} \cup I_{2} |$, $J \subseteq I_{2}$. Therefore, write

$$\begin{aligned} G_{I_{1}, I_{2}}^{\{ \pi _{J} \}_{J \subseteq I_{2}}} \equiv G_{I_{1}, I_{2}}^{\left\{ g_{|I_{1} |+ 1}^{\pi _{J}}, g_{|I_{1} |+ 2}^{\pi _{J}}, \ldots , g_{|I_{1} \cup I_{2} |}^{\pi _{J}} \right\} _{J \subseteq I_{2}}} . \end{aligned}$$

Assume that $g_{i}^{\pi _{J}}, |I_{1} |+ 1 \le i \le |I_{1} \cup I_{2} |, J \subseteq I_{2}$ are positive. Then it holds for all $s \ge t \ge 0$ that

$$\begin{aligned} \begin{aligned} G_{I_{1}, I_{2}}^{\left\{ g_{|I_{1} |+ 1}^{\pi _{J}}, \ldots , g_{|I_{1} \cup I_{2} |}^{\pi _{J}} \right\} _{J \subseteq I_{2}}} (s, t)&= {\hat{g}}_{|I_{1} |+ 1}^{\pi _{\emptyset }}(t) \cdot g_{|I_{1} |+ 2}^{\pi _{\emptyset }}(t) \cdot \cdots \cdot g_{|I_{1} \cup I_{2} |}^{\pi _{\emptyset }}(t) \\&\quad \times \left( \frac{ g_{|I_{1} |+ 1}^{\pi _{\emptyset }}(t) }{ {\hat{g}}_{|I_{1} |+ 1}^{\pi _{\emptyset }}(t) } - \frac{ g_{|I_{1} |+ 1}^{\pi _{\emptyset }}(s) }{ {\hat{g}}_{|I_{1} |+ 1}^{\pi _{\emptyset }}(s) } \right) + \frac{g_{|I_{1} |+ 1}^{\pi _{\emptyset }} (s)}{{\hat{g}}_{|I_{1} |+ 1}^{\pi _{\emptyset }} (s)} \\&\quad \times G_{I_{1}, I_{2}}^{\left\{ {\hat{g}}_{|I_{1} |+ 1}^{\pi _{J}}, g_{|I_{1} |+ 2}^{\pi _{J}}, \ldots , g_{|I_{1} \cup I_{2} |}^{\pi _{J}} \right\} _{J \subseteq I_{2}}} (s, t) \end{aligned} \end{aligned}$$

(20)

for an arbitrary function ${\hat{g}}_{|I_{1} |+ 1}^{\pi _{\emptyset }}$ which is positive on ${\mathbb {R}}_{+}$, where

$$\begin{aligned} {\hat{g}}_{|I_{1} |+ 1}^{\pi _{J}}(s) :=\frac{ g_{|I_{1} |+ 1}^{\pi _{J}} (s) }{ g_{|I_{1} |+ 1}^{\pi _{\emptyset }} (s) } {\hat{g}}_{|I_{1} |+ 1}^{\pi _{\emptyset }} (s), \quad \ J \subseteq I_{2}, s \ge 0 , \end{aligned}$$

which are by definition positive functions on ${\mathbb {R}}_{+}$.

Lemma 5

For $k \in {\mathbb {N}}_{0}$, $j \ge 2$, let the functions ${\bar{F}}_{1, k}, \ldots , {\bar{F}}_{j, k}: [0, \infty ) \rightarrow (0, 1]$ as well as ${\bar{F}}_{1, k + 1}, \ldots , {\bar{F}}_{j - 1, k + 1}: [0, \infty ) \rightarrow (0, 1]$ be non-increasing with ${\bar{F}}_{l, k} = \frac{{\bar{F}}_{l - 1, k}}{{\bar{F}}_{l - 1, k + 1}}$ for $l \in \{ 2 , \ldots , j\}$. Then it holds that for $s\ge t \ge 0$

$$\begin{aligned} 0 \le {\bar{F}}_{j, k} (t) - {\bar{F}}_{j, k} (s) \le \left( \prod _{l = 1}^{j - 1}{ \frac{1}{ {\bar{F}}_{l, k + 1}(s) } } \right) \cdot \left( {\bar{F}}_{1, k} (t) - {\bar{F}}_{1, k} (s) \right) . \end{aligned}$$

Proof of$2. \Rightarrow 3.$ Let statement 2. in Theorem 1 be fulfilled, then due to Remark 2, Lemmas 1 and 2:

• For $i = 1, \ldots , d$ and $\pi \in {\mathcal {S}}_{d}$, it holds that $g_{i}^{\pi } \in \bar{{\mathcal {G}}}$.

• For $I_{1}$ and $I_{2}$ fulfilling the usual conditions and $m \in I_{2}$, the function ${\bar{S}}_{I_{1}, I_{2}}^{m}$ is well-defined as well as positive and continuous. Moreover, it does not depend on the specific $m \in I_{2}$ chosen, hence write ${\bar{S}}_{I_{1}, I_{2}}$.

It is left to prove that ${\bar{S}}_{I_{1}, I_{2}}$ is non-increasing for all $I_{1}, I_{2}$ fulfilling the usual conditions.

The claim is proven by induction over $|I_{2} |$. For $I_{2} = \{ m \}$, let $I_{1}$ and $I_{2}$ fulfill the usual conditions, then ${\bar{S}}_{I_{1}, I_{2}} = {\tilde{g}}^{I_{1} \cup I_{2}, m} \in \bar{{\mathcal {G}}}$. Now let $p > 1$ and assume that for all $I_{1}$ and $I_{2}$ fulfilling the usual conditions with $|I_{2} |< p$ it holds that ${\bar{S}}_{I_{1}, I_{2}} \in \bar{{\mathcal {G}}}$. Let $I_{1}, I_{2}$, $\{ \pi _{J} \}_{J \subseteq I_{2}}$, s, and t fulfill the usual conditions and $|I_{2} |= p$ and define the function ${\hat{g}}_{|I_{1} |+ 1}^{\pi _{\emptyset }} :=g_{|I_{1} |+ 1}^{\pi _{\emptyset }} / {\bar{S}}_{I_{1}, I_{2}}$, which is continuous and positive. With Lemma 4 it follows that

$$\begin{aligned} \begin{aligned} 0&\le G_{I_{1}, I_{2}}^{\left\{ g_{|I_{1} |+ 1}^{\pi _{J}}, g_{|I_{1} |+ 2}^{\pi _{J}}, \ldots , g_{|I_{1} \cup I_{2} |}^{\pi _{J}} \right\} _{J \in I_{2}}} (s, t) \\&= {\hat{g}}_{|I_{1} |+ 1}^{\pi _{\emptyset }} (t) g_{|I_{1} |+ 2}^{\pi _{\emptyset }} (t) \cdot \cdots \cdot g_{|I_{1} \cup I_{2} |}^{\pi _{\emptyset }} (t) \cdot \left( {\bar{S}}_{I_{1}, I_{2}} (t) - {\bar{S}}_{I_{1}, I_{2}} (s) \right) + {\bar{S}}_{I_{1}, I_{2}} (s) \\&\quad \times G_{I_{1}, I_{2}}^{\left\{ {\hat{g}}_{|I_{1} |+ 1}^{\pi _{J}}, g_{|I_{1} |+ 2}^{\pi _{J}}, \ldots , g_{|I_{1} \cup I_{2} |}^{\pi _{J}} \right\} _{J \in I_{2}}} (s, t) , \end{aligned} \end{aligned}$$

(21)

where ${\hat{g}}_{|I_{1} |+ 1}^{\pi _{J}} :=g_{|I_{1} |+ 1}^{\pi _{J}} / {\bar{S}}_{I_{1}, I_{2}}$ for $J \subseteq I_{2}$.

In light of Lemma 3, it makes sense to derive an exogenous shock model from

$$\begin{aligned} \left\{ {\hat{g}}_{|I_{1} |+ 1}^{\pi _{J}}, g_{|I_{1} |+ 2}^{\pi _{J}}, \ldots , g_{|I_{1} \cup I_{2} |}^{\pi _{J}} \right\} _{J \in I_{2}} . \end{aligned}$$

Hence one has to check, that for $\emptyset \not = J \subseteq I_{2}$ if $\bar{{\hat{S}}}_{I_{1} \cup I_{2} \backslash J, J} \in \bar{{\mathcal {G}}}$. Note that

$$\begin{aligned} \bar{{\hat{S}}}_{I_{1} \cup I_{2} \backslash J, J} = {\left\{ \begin{array}{ll} {\bar{S}}_{I_{1} \cup I_{2} \backslash J, J} , &{} \emptyset \not = J \subsetneq I_{2} \\ 1 , &{} J = I_{2} . \end{array}\right. } \end{aligned}$$

As ${\bar{S}}_{I_{1} \cup I_{2} \backslash J, J} \in \bar{{\mathcal {G}}}$ by the induction step for $\emptyset \not = J \subsetneq I_{2}$ and $\bar{{\hat{S}}}_{I_{1}, I_{2}} \equiv 1 \in \bar{{\mathcal {G}}}$, Lemma 3 can be used. Write for $s > t \ge 0$

$$\begin{aligned} G_{I_{1}, I_{2}}^{\left\{ {\hat{g}}_{|I_{1} |+ 1}^{\pi _{J}}, g_{|I_{1} |+ 2}^{\pi _{J}}, \ldots , g_{|I_{1} \cup I_{2} |}^{\pi _{J}} \right\} _{J \in I_{2}}} (s, t) = {\mathbb {P}}\left( {\hat{\tau }}_{i} \in (t, s] \ \forall i \in I_{2} \right) , \end{aligned}$$

where

$$\begin{aligned} {\hat{\tau }}_{i} :=\min {\left\{ {\hat{Z}}_{I} : i \in I \subseteq I_{2} \right\} }, \quad \ i \in I_{2} \end{aligned}$$

with independent ${\hat{Z}}_{I} \sim {\hat{S}}_{I_{1} \cup I_{2} \backslash I, I}$ for $\emptyset \not = I \subseteq I_{2}$. Let $s > t \ge 0$ and define

$$\begin{aligned} {\hat{A}}^{I_{1}, I_{2}} :=\left\{ {\hat{\tau }}_{i} \in (t, s] \ \forall i \in I_{2} \phantom {i_{1}^{2}} \right\} . \end{aligned}$$

Since ${\hat{Z}}_{I_{2}} = \infty$, there are at least two different sets $\emptyset \not = I, J \subsetneq I_{2}$ for which the respective shocks ${\hat{Z}}_{I}$, ${\hat{Z}}_{J}$ are minimal for one of their components. Moreover, this implies

$$\begin{aligned} {\hat{A}}^{I_{1}, I_{2}} \subseteq \bigcup _{\emptyset \not = I, J \subsetneq I_{2} : I \not = J}{ \left\{ t < {\hat{Z}}_{I}, {\hat{Z}}_{J} \le s \right\} } . \end{aligned}$$

From the sub-additivity of the probability measure ${\mathbb {P}}$, it follows that

$$\begin{aligned} {\mathbb {P}} ({\hat{A}}^{I_{1}, I_{2}})&= G_{I_{1}, I_{2}}^{\left\{ {\hat{g}}_{|I_{1} |+ 1}^{\pi _{J}}, g_{|I_{1} |+ 2}^{\pi _{J}}, \ldots , g_{|I_{1} \cup I_{2} |}^{\pi _{J}} \right\} _{J \in I_{2}}} (s, t) \\&\le \sum _{\begin{array}{c} \emptyset \not = I, J \subsetneq I_{2} \\ I \not = J \end{array}}{ {\mathbb {P}}\left( t < {\hat{Z}}_{I}, {\hat{Z}}_{J} \le s \right) } \\&\le \left( {\begin{array}{c}2^{|I_{2} |} - 2\\ 2\end{array}}\right) \max _{\emptyset \not = I \subsetneq I_{2}}{ \left( {\bar{S}}_{I_{1} \cup I_{2} \backslash I, I} (t) - {\bar{S}}_{I_{1} \cup I_{2} \backslash I, I} (s) \right) ^{2} } , \end{aligned}$$

where we used that for $\emptyset \not = I \subsetneq I_{2}$

$$\begin{aligned} {\mathbb {P}}\left( t < {\hat{Z}}_{I} \le s \right) = {\bar{S}}_{I_{1} \cup I_{2} \backslash I, I} (t) - {\bar{S}}_{I_{1} \cup I_{2} \backslash I, I} (s) . \end{aligned}$$

Note that for $\emptyset \not = J \subseteq I \subsetneq I_{2}$ and $m, n \in J$, $m \not = n$

$$\begin{aligned} {\bar{S}}_{I_{1} \cup (I_{2} \backslash I), J} (t)&= {\bar{S}}_{I_{1} \cup (I_{2} \backslash I), J}^{m} (t) \\&= \prod _{\begin{array}{c} \emptyset \not = L \subseteq J \\ m \in L \end{array}}{ \left( {\tilde{g}}^{L \cup I_{1} \cup (I_{2} \backslash I), m } (t) \right) ^{(-1)^{|L |- 1}} } \\&= \frac{ \prod _{\begin{array}{c} \emptyset \not = L \subseteq J \backslash \{ n \} \\ m \in L \end{array}}{ \left( {\tilde{g}}^{L \cup I_{1} \cup (I_{2} \backslash I), m } (t) \right) ^{(-1)^{|L |- 1}} } }{ \prod _{\begin{array}{c} \emptyset \not = K \subseteq J \backslash \{ n \} \\ m \in K \end{array}}{ \left( {\tilde{g}}^{K \cup \{ n \} \cup I_{1} \cup (I_{2} \backslash I), m } (t) \right) ^{(-1)^{|K |- 1}} } } \\&= \frac{ {\bar{S}}_{I_{1} \cup (I_{2} \backslash I), J \backslash \{ n \}}^{m} (t) }{ {\bar{S}}_{I_{1} \cup (I_{2} \backslash I) \cup \{ n \}, J \backslash \{ n \}}^{m} (t) } = \frac{ {\bar{S}}_{I_{1} \cup (I_{2} \backslash I), J \backslash \{ n \}} (t) }{ {\bar{S}}_{I_{1} \cup (I_{2} \backslash I) \cup \{ n \}, J \backslash \{ n \}} (t) } . \end{aligned}$$

Writing $b :=\left( {\begin{array}{c}2^{|I_{2} |} - 2\\ 2\end{array}}\right)$ and using Lemma 5 for ascending sequences $\emptyset \not = J_{1} \subsetneq \cdots \subsetneq J_{|I |} = I \subseteq I_{2}$ with $|J_I|= |I |$ as well as

• 1. ${\bar{F}}_{|J_{l} |, |I_{1} \cup (I_{2} \backslash I) |} \equiv {\bar{S}}_{I_{1} \cup (I_{2} \backslash I), J_{l}}$ for $l \in \left[ |I |\right]$ and

• 2. ${\bar{F}}_{|J_{l} |, |I_{1} \cup (I_{2} \backslash I) \cup (J_{l + 1} \backslash J_{l}) |} \equiv {\bar{S}}_{I_{1} \cup (I_{2} \backslash I) \cup (J_{l + 1} \backslash J_{l}), J_{l}}$ for $l \in \left[ |I |- 1\right]$

it follows that

$$\begin{aligned} {\mathbb {P}} ({\hat{A}}^{I_{1}, I_{2}})&\le b \max _{\begin{array}{c} \emptyset \not = I \subsetneq I_{2} \\ \emptyset \not = J_{1} \subsetneq \cdots \subsetneq J_{|I |} \\ J_{|I |} = I \end{array}}{ \left( \frac{ {\bar{S}}_{I_{1} \cup (I_{2} \backslash I), J_{1}} (t) - {\bar{S}}_{I_{1} \cup (I_{2} \backslash I), J_{1}} (s) }{ \prod _{l = 1}^{|I |- 1}{ {\bar{S}}_{I_{1} \cup (I_{2} \backslash I) \cup (J_{l + 1} \backslash J_{l}), J_{l}} (s) } } \right) ^{2} } \\&= b \max _{\begin{array}{c} \emptyset \not = I \subsetneq I_{2} \\ \emptyset \not = J_{1} \subsetneq \cdots \subsetneq J_{|I |} = I \\ J_{1} = \{ m \} \end{array}}{ \left( \frac{ {\tilde{g}}^{I_{1} \cup (I_{2} \backslash I) \cup J_{1}, m} (t) - {\tilde{g}}^{I_{1} \cup (I_{2} \backslash I) \cup J_{1}, m} (s) }{ \prod _{l = 1}^{|I |- 1}{ {\bar{S}}_{I_{1} \cup (I_{2} \backslash I) \cup (J_{l + 1} \backslash J_{l}), J_{l}} (s) } } \right) ^{2} } . \end{aligned}$$

Now let $\emptyset \not = I \subsetneq I_{2}$, $k = |I_{1} \cup (I_{2} \backslash I)|$, $J_{1} = \{ m\}$ and $\pi \in {\mathcal {S}}_{d}$ be a permutation fulfilling $\pi \left( \{ 1, \ldots , k \} \right) = I_{1} \cup (I_{2} \backslash I)$, $\pi (k + 1) = m$. Denote with ${\tilde{\pi }}$ the permutation, which switches the positions of m and $\pi (k)$, i.e. ${\tilde{\pi }} = \pi (k, k + 1)$. Then

$$\begin{aligned} 0&\le G_{I_{1} \cup (I_{2} \backslash I) \backslash \{ \pi (k) \}, \{ m, \pi (k) \}} (s, t) \\&= \prod _{j = 0}^{1}{ g_{k + j}^{{\tilde{\pi }}}(t) } - g_{k}^{\pi } (s) g_{k + 1}^{\pi } (t) - g_{k}^{{\tilde{\pi }}} (s) g_{k + 1}^{{\tilde{\pi }}} (t) + \prod _{j = 0}^{1}{ g_{k + j}^{\pi } (s) } \\&= g_{k + 1}^{{\tilde{\pi }}} (t) \left( g_{k}^{{\tilde{\pi }}} (t) - g_{k}^{{\tilde{\pi }}} (s) \right) - g_{k}^{\pi } (s) \left( g_{k + 1}^{\pi } (t) - g_{k + 1}^{\pi } (s) \right) , \end{aligned}$$

which is equivalent to

$$\begin{aligned} g_{k + 1}^{\pi } (t) - g_{k + 1}^{\pi } (s) \le \frac{g_{k + 1}^{{\tilde{\pi }}} (t)}{g_{k}^{\pi } (s)} \left( g_{k}^{{\tilde{\pi }}} (t) - g_{k}^{{\tilde{\pi }}} (s) \right) . \end{aligned}$$

This yields inductively the following inequality

$$\begin{aligned} g_{k + 1}^{\pi } (t) - g_{k + 1}^{\pi } (s) \le \prod _{l = 1}^{k}{ \frac{ {\tilde{g}}^{\pi (\{ 1, \ldots , l \}) \cup \{ m \}, m} (t) }{ {\tilde{g}}^{\pi (\{ 1, \ldots , l \}), \pi (l)} (s) } } \cdot \left( {\tilde{g}}^{\{ m \}, m} (t) - {\tilde{g}}^{\{ m \}, m} (s) \right) . \end{aligned}$$

Subsequently,

$$\begin{aligned} {\mathbb {P}} ({\hat{A}}^{I_{1}, I_{2}}) \le b p_{I_{1}, I_{2}} (s, t) q_{I_{2}} (s, t) \end{aligned}$$

with

$$\begin{aligned} p_{I_{1}, I_{2}} (s, t):= & {} \max _{\begin{array}{c} \emptyset \not = I \subsetneq I_{2} \\ \emptyset \not = J_{1} \subsetneq \cdots \subsetneq J_{|I |} = I \\ \pi \in \Pi _{I_{1}, I_{2}, I} \\ J_{1} = \{ m \} \end{array}}{ \left\{ \frac{ 1 }{ \prod _{l = 1}^{|I |- 1}{ {\bar{S}}_{I_{1} \cup (I_{2} \backslash I) \cup (J_{l + 1} \backslash J_{l}), J_{l}} (s) } } \right. } \\&\times \left. \prod _{l = 1}^{|I_{1} \cup (I_{2} \backslash I) |}{ \frac{ {\tilde{g}}^{\pi (\{ 1, \ldots , l\}) \cup \{ m \}, m} (t) }{ {\tilde{g}}^{\pi (\{ 1, \ldots , l \}), \pi (l) } (s) } } \right\} ^{2} , \end{aligned}$$

where $\Pi _{I_{1}, I_{2}, I}$ is the set of permutations fulfilling the conditions stated above and

$$\begin{aligned} q_{I_{2}} (s, t) :=\max _{m \in I_{2}}{\left\{ {\tilde{g}}^{\{ m \}, m} (t) - {\tilde{g}}^{\{ m \}, m} (s) \right\} ^{2}} . \end{aligned}$$

Let $s_{0} \ge s > t \ge t_{0} \ge 0$; the non-increasingness of the involved survival functions ${\bar{S}}_{I_{1} \cup (I_{2} \backslash I) \cup (J_{l + 1} \backslash J_{l}), J_{l}} (s)$, ${\tilde{g}}^{\pi (\{ [l] \}) \cup \{ m \}, m} (t)$, and ${\tilde{g}}^{\pi (\{ [l] \}), \pi (l)} (s)$ implies

$$\begin{aligned} p_{I_{1}, I_{2}} (s, t) \le p_{I_{1}, I_{2}} (s_{0}, t_{0}) , \quad \forall t < s \ \text {with} \ t, s \in [t_{0}, s_{0}] . \end{aligned}$$

Define for $s \ge t \ge 0$

$$\begin{aligned} \mu _{I_{2}} (s, t) = \sum _{m \in I_{2}}{ {\tilde{g}}^{\{ m\}, m} (t) - {\tilde{g}}^{\{ m\}, m} (s) } . \end{aligned}$$

As ${\tilde{g}}^{\{ m\}, m}, m \in I_{2}$ are non-negative and non-increasing and $q_{I_{2}} (s, t) \ge 0$ all summands are non-negative and

$$\begin{aligned} \mu _{I_{2}} (s, t) \ge \sqrt{q_{I_{2}} (s, t)} \ge 0, \ s \ge t \ge 0 . \end{aligned}$$

Hence

$$\begin{aligned} 0&\le G_{I_{1}, I_{2}}^{\left\{ {\hat{g}}_{|I_{1} |+ 1}^{\pi _{J}}, g_{|I_{1} |+ 2}^{\pi _{J}}, \ldots , g_{|I_{1} \cup I_{2} |}^{\pi _{J}} \right\} _{J \in I_{2}}} (s, t) \\&\le b p_{I_{1}, I_{2}} (s_{0}, t_{0}) q_{I_{2}} (s, t) \\&\le b p_{I_{1}, I_{2}} (s_{0}, t_{0}) \mu _{I_{2}} (s_{0}, t_{0})^2 , \quad \forall t < s \ \text {with}\ t, s \in [t_{0}, s_{0}] . \end{aligned}$$

Now, the proof proceeds analogously as for copulas in the exchangeable case (see Mai et al. 2016, 1296 sq.) or bivariate exchangeable case (see Durante et al. 2008, 67).

The function ${\bar{S}}_{I_{1}, I_{2}}$ splits in positive and negative powers in the product terms and

$$\begin{aligned} {\bar{S}}_{I_{1}, I_{2}} (t)&= \prod _{i = 1}^{|I_{2} |}{ \left( \prod _{\begin{array}{c} J \subseteq I_{2} \\ |J |= i, m \in J \end{array}}{ {\tilde{g}}^{J \cup I_{1}, m}(t) } \right) ^{(-1)^{i - 1}} } \\&= \frac{ \prod _{i = 0}^{\lfloor (|I_{2} |- 1) / 2 \rfloor }{ \left( \prod _{\begin{array}{c} J \subseteq I_{2} \\ |J |= 2 i + 1, m \in J \end{array}}{ {\tilde{g}}^{J \cup I_{1}, m}(t) } \right) } }{ \prod _{i = 1}^{\lfloor |I_{2} |/ 2\rfloor }{ \left( \prod _{\begin{array}{c} J \subseteq I_{2} \\ |J |= 2 i, m \in J \end{array}}{ {\tilde{g}}^{J \cup I_{1}, m}(t) } \right) } } \\&{\mathop {\le }\limits ^{(\star )}} \frac{ \prod _{i = 0}^{\lfloor (|I_{2} |- 1) / 2 \rfloor }{ \left( \prod _{\begin{array}{c} J \subseteq I_{2} \\ |J |= 2 i + 1, m \in J \end{array}}{ {\tilde{g}}^{J \cup I_{1}, m}(t_{0}) } \right) } }{ \prod _{i = 1}^{\lfloor |I_{2} |/ 2\rfloor }{ \left( \prod _{\begin{array}{c} J \subseteq I_{2} \\ |J |= 2 i, m \in J \end{array}}{ {\tilde{g}}^{J \cup I_{1}, m}(s_{0}) } \right) } } \\&=:p_{\text {max}}^{I_{1}, I_{2}} (s_{0}, t_{0}) , \end{aligned}$$

where the monotonicity of ${\tilde{g}}^{I, m}$ is used in $(\star )$. Assume that ${\bar{S}}_{I_{1}, I_{2}}$ is not non-increasing, i.e. there exists $s_{0} > t_{0} \ge 0$ s.t. ${\bar{S}}_{I_{1}, I_{2}} (s_{0}) > {\bar{S}}_{I_{1}, I_{2}} (t_{0})$.

Case$q_{I_{1}} (s_{0}, t_{0}) = 0$: From Eq. (21) we get

$$\begin{aligned} 0&\le G_{I_{1}, I_{2}}^{\left\{ g_{|I_{1} |+ 1}^{\pi _{J}}, g_{|I_{1} |+ 2}^{\pi _{J}}, \ldots , g_{|I_{1} \cup I_{2} |}^{\pi _{J}} \right\} _{J \in I_{2}}} (s_{0}, t_{0}) \\&= \underbrace{{\hat{g}}_{|I_{1} |+ 1}^{\pi _{\emptyset }} (t_{0}) g_{|I_{1} |+ 2}^{\pi _{\emptyset }} (t_{0}) \ldots g_{|I_{1} \cup I_{2} |}^{\pi _{\emptyset }} (t_{0})}_{> 0} \underbrace{\left( {\bar{S}}_{I_{1}, I_{2}} (t_{0}) - {\bar{S}}_{I_{1}, I_{2}} (s_{0}) \right) }_{< 0} \\&< 0 \end{aligned}$$

which is a contradiction.

Case$q_{I_{1}} (s_{0}, t_{0}) > 0$: Let

$$\begin{aligned} a (s_{0}, t_{0}) :=\frac{{\bar{S}}_{I_{1}, I_{2}} (s_{0}) - {\bar{S}}_{I_{1}, I_{2}} (t_{0})}{\mu _{I_{2}}(s_{0}, t_{0})} > 0 \end{aligned}$$

then we can write

$$\begin{aligned} {\bar{S}}_{I_{1}, I_{2}} (t_{0}) - {\bar{S}}_{I_{1}, I_{2}} (s_{0}) = -a(s_{0}, t_{0}) \mu _{I_{2}}(s_{0}, t_{0}) . \end{aligned}$$

For all $k \ge 1$, one can find $s_{k}, t_{k} \in [t_{0}, s_{0}]$ with $s_{k} > t_{k}$ and

$$\begin{aligned} \mu _{I_{2}} (s_{k}, t_{k}) = \frac{\mu _{I_{2}} (s_{0}, t_{0})}{k} \end{aligned}$$

(22)

as well as

$$\begin{aligned} {\bar{S}}_{I_{1}, I_{2}} (t_{k}) - {\bar{S}}_{I_{1}, I_{2}} (s_{k}) \le -a(s_{0}, t_{0}) \mu _{I_{2}} (s_{s}, t_{k}) . \end{aligned}$$

This can be seen by setting $t^{(0, k)} :=t_{0}$, $t^{(k, k)} :=s_{0}$, and

$$\begin{aligned} t^{(j, k)} :=\left( \sum _{m \in I_{2}}{ {\tilde{g}}^{\{ m \}, m} } \right) ^{\leftarrow } \left( x^{(j, k)} \right) , \quad \ j \in \{ 1, \ldots , k - 1\} , \end{aligned}$$

where ${\leftarrow }$ denotes the generalized inverse for non-increasing functions¹⁰ and for $k \in \{ 0, \ldots , k\}$

$$\begin{aligned} x^{(j, k)} :=\frac{k - j}{k} \sum _{m \in I_{2}}{ {\tilde{g}}^{\{ m \}, m } (t_{0}) } + \frac{j}{k} \sum _{m \in I_{2}}{ {\tilde{g}}^{\{ m \}, m} (s_{0}) } . \end{aligned}$$

As ${\tilde{g}}^{\{ m \}, m}$ are continuous and non-decreasing the generalized inverse is a right-inverse¹¹ and

$$\begin{aligned} \mu _{I_{2}} \left( t^{(j, k)}, t^{(j - 1, k)}\right) {}&= \underbrace{\sum _{m \in I_{2}}{ {\tilde{g}}^{\{ m \}, m} \left( t^{(j - 1, k)}\right) }}_{= x^{(j - 1, k)}} - \underbrace{\sum _{m \in I_{2}}{ {\tilde{g}}^{\{ m \}, m} \left( t^{(j, k)}\right) }}_{= x^{(j, k)}} \\&= \frac{1}{k} \mu _{I_{2}} (s_{0}, t_{0}) . \end{aligned}$$

Assume that for all $j \in \{ 1, \ldots , k\}$ the following inequality holds

$$\begin{aligned} {\bar{S}}_{I_{1}, I_{2}} (t^{(j - 1, k)}) - {\bar{S}}_{I_{1}, I_{2}} (t^{(j, k)}) > - a(s_{0}, t_{0}) \mu _{I_{2}} (t^{(j, k)}, t^{(j - 1, k)}) . \end{aligned}$$

Then,

$$\begin{aligned} {\bar{S}}_{I_{1}, I_{2}} (t_{0}) - {\bar{S}}_{I_{1}, I_{2}} (s_{0})&= \sum _{j = 1}^{k}{ {\bar{S}}_{I_{1}, I_{2}} (t^{(j - 1, k)}) - {\bar{S}}_{I_{1}, I_{2}} (t^{(j, k)}) } \\&> - a(s_{0}, t_{0}) \sum _{j = 1}^{k}{ \mu _{I_{2}} (t^{(j, k)}, t^{(j - 1, k)}) } = - a (s_{0}, t_{0}) \mu _{I_{2}} (s_{0}, t_{0}) , \end{aligned}$$

which is a contradiction. Hence, with $t_{k} = t^{(j - 1, k)}$, $s_{k} = t^{(j, k)}$ for some $j \in \{ 1, \ldots , k\}$, Eq. (22) is fulfilled and $s_k > t_k$.

Combining Eq. (21) with these results gives for feasible $t_{k}$, $s_{k}$ (chosen as above)

$$\begin{aligned} 0&\le G_{I_{1}, I_{2}}^{\left\{ g_{|I_{1} |+ 1}^{\pi _{J}}, g_{|I_{1} |+ 2}^{\pi _{J}}, \ldots , g_{|I_{1} \cup I_{2} |}^{\pi _{J}} \right\} _{J \in I_{2}}} (s_{k}, t_{k}) \\&= \underbrace{{\hat{g}}_{|I_{1} |+ 1}^{\pi _{\emptyset }} (t_{k})}_{= \frac{g_{|I_{1} |+ 1}^{\pi _{\emptyset }} (t_{k})}{{\bar{S}}_{I_{1}, I_{2}} (t_{k})}} g_{|I_{1} |+ 2}^{\pi _{\emptyset }} (t_{k}) \cdot \cdots \cdot g_{|I_{1} \cup I_{2} |}^{\pi _{\emptyset }} (t_{k}) \cdot \underbrace{\left( {\bar{S}}_{I_{1}, I_{2}} (t_{k}) - {\bar{S}}_{I_{1}, I_{2}} (s_{k}) \right) }_{\le - a (s_{0}, t_{0}) \frac{\mu _{I_{2}} (s_{0}, t_{0})}{k}} \\&\quad + {\bar{S}}_{I_{1}, I_{2}} (s_{k}) G_{I_{1}, I_{2}}^{\left\{ {\hat{g}}_{|I_{1} |+ 1}^{\pi _{J}}, g_{|I_{1} |+ 2}^{\pi _{J}}, \ldots , g_{|I_{1} \cup I_{2} |}^{\pi _{J}} \right\} _{J \in I_{2}}} (s_{k}, t_{k}) \\&\le \frac{g_{|I_{1}|+ 1}^{\pi _{\emptyset }} (s_{0})}{p_{\text {max}}^{I_{1},I_{2}} (s_{0}, t_{0})} g_{|I_{1}|+ 2}^{\pi _{\emptyset }} (s_{0}) \cdot \cdots \cdot g_{|I_{1} \cup I_{2}|}^{\pi _{\emptyset }} (s_{0}) \\&\quad \times \left( -a (s_{0}, t_{0}) \mu _{I_{2}} (s_{0}, t_{0}) \frac{1}{k} \right) \\&\quad + b p_{\text {max}}^{I_{1}, I_{2}} (s_{0}, t_{0}) p_{I_{1}, I_{2}} (s_{0}, t_{0}) \mu _{I_{2}} (s_{0}, t_{0})^2 \frac{1}{k^2} . \end{aligned}$$

In particular, if the latter inequality is multiplied by k and the limit $k \rightarrow \infty$ is taken, then

$$\begin{aligned} 0 \le - \underbrace{\frac{1}{p_{\text {max}}^{I_{1}, I_{2}} (s_{0}, t_{0})}}_{> 0} \underbrace{a (s_{0}, t_{0})}_{> 0} \underbrace{\mu _{I_{1}, I_{2}} (s_{0}, t_{0})}_{> 0} \underbrace{\prod _{j = 1}^{|I_{2} |} g_{|I_{1} |+ j}^{\pi _{\emptyset }} (s_{0})}_{> 0} < 0 , \end{aligned}$$

which leads to a contradiction. $\square$

Proofs of supporting lemmas

Proof of Lemma 1

Let $I_{1}$, $I_{2}$, and $\pi$ fulfill the usual conditions with $|I_{2} |= 2$ and $|I_{1} |= i - 2$ and define ${\tilde{\pi }} = \pi (i - 1, i)$.

Right-continuity Let $s + h> s > t \ge 0$. As $G_{I_{1}, I_{2}} (s, t)$ is right-continuous in s it holds that

$$\begin{aligned} 0 = \lim _{h \searrow 0}{ G_{I_{1}, I_{2}} (s + h, t) - G_{I_{1}, I_{2}} (s, t) } {\mathop {=}\limits ^{\text {IH}}} \underbrace{g_{i - 1}^{\pi } (s)}_{{\mathop {>}\limits ^{\text {IH}}} 0} \lim _{h \searrow 0}{ \left( g_{i}^{\pi } (s + h) - g_{i}^{\pi } (s) \right) } , \end{aligned}$$

where it is used that under the induction hypothesis all but two terms cancel out.

Left-limits Let $s> s - h > t \ge 0$. As $G_{I_{1}, I_{2}} (s, t)$ and $g_{i - 1}^{\pi _{\emptyset }} (s)$, $\pi _{\emptyset } \in \{ \pi , {\tilde{\pi }}\}$ have left-limits in s and $g_{i - 1}^{\pi }$ is positive by induction hypothesis it follows that $g_{i}^{\pi }$ has left-limits:

$$\begin{aligned}&\lim _{h \searrow 0}{ g_{i}^{\pi } (s - h) } = \lim _{h \searrow 0}{} \left( \frac{G_{I_{1}, I_{2}} (s - h, t) - g_{i - 1}^{{\tilde{\pi }}} (t) g_{i}^{{\tilde{\pi }}} (t)}{g_{i - 1}^{\pi } (s - h)} \right. \\&\quad \left. -\frac{- g_{i - 1}^{\pi } (s - h) g_{i}^{\pi } (t) - g_{i - 1}^{{\tilde{\pi }}} (s - h) g_{i}^{{\tilde{\pi }}} (t)}{g_{i - 1}^{\pi } (s - h)} \right) . \end{aligned}$$

Non-increasingness Now, let $I_{1}$, $I_{2}$, and $\pi$ fulfill the usual conditions with $I_{2} = \{ \pi (i) \}$ and $I_{1} = \pi ([i - 1])$. As $G_{I_{1}, I_{2}}$ is non-negative, it holds for all $s > t \ge 0$ that

$$\begin{aligned} 0 \le G_{I_{1}, I_{2}} (s, t) = g_{i}^{\pi } (t) - g_{i}^{\pi } (s) . \end{aligned}$$

$\square$

Proof of Lemma 2

For $\pi \in {\mathcal {S}}_{d}$, due to Remark 2 and Lemma 1, it follows that the functions $g_{i}^{\pi }$, $i = 1, \ldots , d$ are positive, continuous functions on ${\mathbb {R}}_{+}$. Hence ${\bar{S}}_{I_{1}, I_{2}}^{m}$ is an ${\mathbb {R}}_{+}$-valued, positive, and continuous function for every $I_{1}, I_{2}$ fulfilling the usual conditions with $m \in I_{2}$.

In the following, it is proven, by induction over $|I_{2} |$, that Eq. (19) holds and furthermore, that for all $I_{1}$ and $I_{2}$ fulfilling the usual conditions

$$\begin{aligned} \prod _{i = 1}^{|I_{2} |} g_{|I_{1} |+ i}^{{\tilde{\pi }}} (t) = \prod _{i = 1}^{|I_{2} |} g_{|I_{1} |+ i}^{{\hat{\pi }}} (t) , \quad \forall t \ge 0 \end{aligned}$$

(23)

for all ${\tilde{\pi }}, {\hat{\pi }} \in {\mathcal {S}}_{d}$ fulfilling $\pi ([|I_{1} |]) = I_{1}$ and $\pi ([|I_{1} \cup I_{2} |] \backslash [|I_{1} |]) = I_{2}$ for $\pi \in \{ {\tilde{\pi }}, {\hat{\pi }} \}$. For $|I_{2} |= 1$ both claims are naturally fulfilled. Let both claims be fulfilled for $|I_{2} |< p$ and let $I_{1}$, $I_{2}$ as well as $\pi$ fulfill the usual conditions with $|I_{2} |= p$, $m \in I_{2}$ as well as $\pi (|I_{1} |+ 1) = m$, then for $t \ge 0$

$$\begin{aligned} \prod _{\emptyset \not = J \subseteq I_{2}}{ {\bar{S}}_{I_{1} \cup (I_{2} \backslash J), J}^{\pi (\min _{j \in J}{ \pi ^{-1}(j)})} (t) } {}&{\mathop {=}\limits ^{(\star )}} \prod _{i = 1}^{|I_{2} |}{ \prod _{\begin{array}{c} J \subseteq \pi (\{ |I_{1} |+ i, \ldots , |I_{1} \cup I_{2} |\}) \\ \pi (|I_{1} |+ i) \in J \end{array}}{ {\bar{S}}_{I_{1} \cup (I_{2}\backslash J), J}^{\pi (|I_{1} |+ i)} (t) } } \\&= \prod _{i = 1}^{|I_{2} |}{ \prod _{\begin{array}{c} J \subseteq \pi (\{ |I_{1} |+ i, \ldots , |I_{1} \cup I_{2} |\}) \\ \pi (|I_{1} |+ i) \in J \end{array}}} \\&\quad \times { \prod _{\begin{array}{c} L \subseteq J \\ \pi ( |I_{1} |+ i) \in L \end{array}}{ \left( {\tilde{g}}^{ L \cup I_{1} \cup (I_{2} \backslash J), \pi (|I_{1} |+ i) } (t) \right) ^{(-1)^{|L |- 1}} } } , \end{aligned}$$

where the factors in $(\star )$ are regrouped in a similar sense as for the alternative representation for the GMO survival function.

Now for $i \in [d]$ fix $\pi (\{ 1, \ldots , |I_{1}|+ i\}) \subseteq K \subseteq I_{1} \cup I_{2}$ and define $k = |K |$ as well as $1 \le l \le k - |I_{1} |- i + 1$. The expression ${\tilde{g}}_{k}^{K, \pi (|I_{1}|+ i)} (t)$ with exponent $(-1)^{l - 1}$ appears $\left( {\begin{array}{c}k - |I_{1} |- i\\ l - 1\end{array}}\right)$ times and the overall exponent for ${\tilde{g}}_{k}^{K, \pi (|I_{1} |+ i)}$ is

$$\begin{aligned} \sum _{l = 1}^{k - i - |I_{1} |+ 1}{ (-1)^{l - 1} \left( {\begin{array}{c}k - i - |I_{1} |\\ l - 1\end{array}}\right) } = {\left\{ \begin{array}{ll} 1 , &{} k = |I_{1} |+ i \\ 0 , &{} \text {else} \, . \end{array}\right. } \end{aligned}$$

Hence, as it holds for $k = |I_{1} |+ i$ that $K = \pi (\{ 1, \ldots , |I_{1} |+ i\})$ and

$$\begin{aligned} \prod _{\emptyset \not = J \subseteq I_{2}}{ {\bar{S}}_{I_{1} \cup (I_{2} \backslash J), J}^{\pi (\min _{j \in J}{ \pi ^{-1}(j)})} (t) } = \prod _{i = 1}^{|I_{2} |}{ g_{|I_{1} |+ i}^{\pi }(t) } \end{aligned}$$

or equivalently,

$$\begin{aligned} {\bar{S}}_{I_{1}, I_{2}}^{m} (t) = \frac{\prod _{i = 1}^{|I_{2} |}{ g_{|I_{1} |+ i}^{\pi }(t) }}{\prod _{\emptyset \not = J \subsetneq I_{2}}{ {\bar{S}}_{I_{1} \cup (I_{2} \backslash J), J}^{\pi (\min _{j \in J}{ \pi ^{-1}(j)})} (t) }} . \end{aligned}$$

(24)

By induction, the factors of the denominator of the r.h.s. in Eq. (24), ${\bar{S}}_{I_{1} \cup (I_{2} \backslash J), J}^{\pi (\min _{j \in J}{\pi ^{-1} (j)})}$, are independent of $\pi (\min _{j \in J}{ \pi ^{-1} (j)})$ and subsequently also of m. Moreover, for arbitrary $I_{1}$, $I_{2}$ and $\{ \pi _{J} \}_{J \subseteq I_{2}}$ fulfilling the usual conditions and $s \ge 0$

$$\begin{aligned} \prod _{j = 1}^{|I_{2} |}{ g_{|I_{1} |+ j}^{\pi _{I_{2}}} (s) } = (-1)^{|I_{2} |} \left( G_{I_{1}, I_{2}}^{\{ \pi _{J} \}_{J \subseteq I_{2}}} (s, 0) -\sum _{J \subsetneq I_{2}}{ (-1)^{|J |} \prod _{j = 1}^{|J |}{ g_{|I_{1} |+ j}^{\pi _{J}} (s) } } \right) . \end{aligned}$$

By induction and assumption, the r.h.s. does not depend on the specific family of permutations, $\{ \pi _{J} \}_{J \subseteq I_{2}}$, chosen, therefore Eq. (23) holds for $|I_{2} |= p$. In conclusion, the nominator in Eq. (24) does not depend on the specific $\pi$, and subsequently m, chosen and Eq. (19) holds for $|I_{2} |= p$. $\square$

Proof of Lemma 3

As in the proof of 4. to 1. one can derive analogously for $\varvec{t} \ge 0$ and $\pi \in {\mathcal {S}}_{d}$ with $t_{\pi (1)} \ge \cdots \ge t_{\pi (d)}$ as well as $\pi \left( \{ 1, \ldots , |I_{1} |\} \right) = I_{1}$ and $\pi \left( \{ |I_{1} |+ 1, \ldots , |I_{1} \cup I_{2} |\} \right) = I_{2}$ that

$$\begin{aligned} {\mathbb {P}} \left( {\check{\tau }}_{j} > t_{j} \ \forall j \in I_{2} \right) = \prod _{j = |I_{1} |+ 1}^{|I_{1} \cup I_{2} |}{ g_{j}^{\pi } \left( t_{\pi (j)}\right) } = \prod _{j = 1}^{|\check{I}_{2} |}{ \check{g}_{j}^{{\check{\pi }}} \left( \check{t}_{{\check{\pi }} (j)}\right) } , \end{aligned}$$

where for $\check{I}_{2} = \{ 1, \ldots , |I_{2} |\}$, ${\check{\pi }} \in {\mathcal {S}}_{|I_{2} |}$ is defined by

$$\begin{aligned} \pi (|I_{1} |+ j) = i_{{\check{\pi }} (j)} , \quad \forall j \in \check{I}_{2}, I_{2} = \{ i_{1}, \ldots , i_{|I_{2} |} \} \end{aligned}$$

and $\check{g}_{j}^{{\check{\pi }}} :=g_{|I_{1} |+ j}^{\pi }$ as well as $\check{t}_{{\check{\pi }} (j)} :=t_{\pi (|I_{1} |+ j)}$. Then, it holds for all $0 \le t < s$ that

$$\begin{aligned} {\mathbb {P}} \left( {\check{\tau }}_{j} \in (t, s] \ \forall j \in I_{2} \right) = \check{G}_{\emptyset , \check{I}_{2}}(s, t) = G_{I_{1}, I_{2}}(s, t) , \end{aligned}$$

where $\check{G}_{\emptyset , \check{I}_{2}}$ corresponds to Eq. (8) w.r.t. $\{ \check{g}_{j}^{{\check{\pi }}} \}_{j \in \check{I}_{2}, {\check{\pi }} \in {\mathcal {S}}_{|I_{2}|}}$. $\square$

Proof of Lemma 4

Every summand corresponding to a non-empty interval $\emptyset \not = J \subseteq I_{2}$ contains a term $g_{|I_{1} |+ 1}^{\pi _{J}} (s)$. Therefore the result follows by multiplying $G_{I_{1}, I_{2}}$ with $\frac{g_{|I_{1} |+ 1}^{\pi _{\emptyset }} (s)}{{\hat{g}}_{|I_{1} |+ 1}^{\pi _{\emptyset }} (s)}$ and its reciprocal, whereas the first summand in Eq. (20) is a correction term for the summand belonging to $J = \emptyset$. $\square$

Proof of Lemma 5

This is a direct corollary of Mai et al. (2016, lem. B.2 on p. 1295). $\square$