A sufficient condition for the subexponential asymptotics of GI/G/1-type Markov chains with queueing applications

Abstract

The main contribution of this paper is to present a new sufficient condition for the subexponential asymptotics of the stationary distribution of a GI/G/1-type Markov chain with the stochastic phase transition matrix in non-boundary levels, which implies no possibility of jumps from level “infinity” to level zero. For simplicity, we call such Markov chains GI/G/1-type Markov chains without disasters because they are used to analyze semi-Markovian queues without “disasters”, which are negative customers who remove all the customers in the system (including themselves) on their arrivals. We first demonstrate the application of our main result to the stationary queue length distribution in the standard BMAP/GI/1 queue. Thereby we present new asymptotic formulas and derive the existing formulas under weaker conditions than those in the literature. We also apply our main result to the stationary queue length distributions in two queues: One is a MAP/$\mathrm{GI}$/1 queue with the $(a,b)$-bulk-service rule (i.e., MAP/$\mathrm{GI}^{(a,b)}$/1 queue); and the other is a MAP/$\mathrm{GI}$/1 retrial queue with constant retrial rate.

Appendices

Appendix 1: Proof of Theorem 3.1

We present three lemmas for the proof of Theorem 3.1.

Lemma 8.1

Suppose that Assumption 2.1 is satisfied. If Assumption 3.1 holds for some $Y \in {\mathcal {L}}$, then

$$\begin{aligned} \lim _{k \rightarrow \infty } \sum _{m = 1}^{\infty } {\overline{\varvec{A}}(k+m) \varvec{L}(m) \over {\mathsf {P}}(Y > k) }= & {} {\varvec{c}_A\varvec{\pi }(\varvec{I} - \varvec{R} ) (\varvec{I} - \varvec{\varPhi }(0)) \over -\sigma }, \end{aligned}$$

(8.1)

$$\begin{aligned} \lim _{k \rightarrow \infty } \sum _{m = 1}^{\infty } {\overline{\varvec{B}}(k+m) \varvec{L}(m) \over {\mathsf {P}}(Y > k) }= & {} {\varvec{c}_B\varvec{\pi }(\varvec{I} - \varvec{R} ) (\varvec{I} - \varvec{\varPhi }(0)) \over -\sigma }. \end{aligned}$$

(8.2)

Proof

We prove (8.1) only. As for (8.2), we can prove it in a similar way to the proof of (8.1). It follows from Lemma 3.1.3 in Kimura et al. (2013) that

$$\begin{aligned} \lim _{n\rightarrow \infty } \sum _{l=0}^{\tau -1}\varvec{L}(n\tau +l) = \tau \varvec{e} \varvec{\psi }, \end{aligned}$$

(8.3)

where

$$\begin{aligned} \varvec{\psi } = \varvec{\pi }(\varvec{I} - \varvec{R} ) (\varvec{I} - \varvec{\varPhi }(0))/(-\sigma ), \end{aligned}$$

(8.4)

and $\tau$ denotes the period of an Markov additive process with kernel $\{\varvec{A}(k);k\in {\mathbb {Z}}\}$ (see Kimura et al. 2010, Appendix B). According to (8.3), we fix $\varepsilon >0$ arbitrarily and $m_{*}:= m_{*} (\varepsilon )$ such that for all $m \ge m_{*}$ and $l = 0,1,\ldots ,\tau -1$,

$$\begin{aligned}&\varvec{e}( \varvec{\psi } - \varepsilon \varvec{e}^{\mathrm{t}} /\tau ) \le {1 \over \tau } \sum _{l=0}^{\tau -1} \varvec{L} (\lfloor m / \tau \rfloor \tau + l) \le \varvec{e}( \varvec{\psi } + \varepsilon \varvec{e}^{\mathrm{t}} /\tau ). \end{aligned}$$

(8.5)

Further since $\varvec{L}(m) \le \varvec{e}\varvec{e}^{\mathrm{t}}$ for all $m \in {\mathbb {N}}$, it follows from (3.1) and $Y \in {\mathcal {L}}$ that

$$\begin{aligned} \limsup _{k \rightarrow \infty } \sum _{m=1}^{m_{*}-1} { \overline{\varvec{A}}(k+m) \varvec{L}(m) \over {\mathsf {P}}(Y > k) }\le & {} \sum _{m=1}^{m_{*}-1} \limsup _{k \rightarrow \infty } { \overline{\overline{\varvec{A}}}(k+m-1)\varvec{e}\varvec{e}^{\mathrm{t}} - \overline{\overline{\varvec{A}}}(k+m)\varvec{e}\varvec{e}^{\mathrm{t}} \over {\mathsf {P}}(Y > k) }\\= & {} \varvec{O}, \end{aligned}$$

and thus

$$\begin{aligned} \lim _{k\rightarrow \infty } \sum _{m=1}^{\infty } { \overline{\varvec{A}} (k+m) \varvec{L}(m) \over {\mathsf {P}}(Y > k) } = \lim _{k \rightarrow \infty } \sum _{m=m_{*}}^{\infty } { \overline{\varvec{A}} (k+m) \varvec{L}(m) \over {\mathsf {P}}(Y > k) }. \end{aligned}$$

(8.6)

As a result, to prove (8.1) it suffices to show that for any fixed $\varepsilon > 0$,

$$\begin{aligned} \limsup _{k \rightarrow \infty } \sum _{m=m_{*}}^{\infty } {\overline{\varvec{A}}(k+m) \varvec{L}(m) \over {\mathsf {P}}(Y > k) }\le & {} \varvec{c}{}_A(\varvec{\psi } + \varepsilon \varvec{e}^{\mathrm{t}}/\tau ), \end{aligned}$$

(8.7)

$$\begin{aligned} \liminf _{k \rightarrow \infty } \sum _{m=m_{*}}^{\infty } {\overline{\varvec{A}}(k+m) \varvec{L}(m) \over {\mathsf {P}}(Y > k) }\ge & {} \varvec{c}{}_A(\varvec{\psi } - \varepsilon \varvec{e}^{\mathrm{t}}/\tau ). \end{aligned}$$

(8.8)

Indeed, letting $\varepsilon \downarrow 0$ in (8.7) and (8.8) and using (8.4), we obtain

$$\begin{aligned} \lim _{k \rightarrow \infty } \sum _{m=m_{*}}^{\infty } { \overline{\varvec{A}} (k+m) \varvec{L}(m) \over {\mathsf {P}}(Y > k) } = {\varvec{c}_A\varvec{\pi }(\varvec{I} - \varvec{R} ) (\varvec{I} - \varvec{\varPhi }(0)) \over -\sigma }. \end{aligned}$$

Substituting the above equation into (8.6), we have (8.1).

We first show that (8.7) holds. By definition, $\{\overline{\varvec{A}}(k);k\in {\mathbb {Z}}_+\}$ is nonincreasing. We thus obtain

$$\begin{aligned} \sum _{m=m_{*}}^{\infty } \overline{\varvec{A}} (k+m) \varvec{L}(m)\le & {} \sum _{n= \lfloor m_{*}/\tau \rfloor }^{\infty } \sum _{l=0}^{\tau -1} \overline{\varvec{A}}(k + n\tau + l) \varvec{L}(n\tau + l) \\\le & {} \sum _{n=\lfloor m_{*}/\tau \rfloor }^{\infty } \overline{\varvec{A}}(k + n\tau ) \sum _{l=0}^{\tau -1} \varvec{L}(n\tau + l)\\\le & {} \sum _{n=\lfloor m_{*}/\tau \rfloor }^{\infty } {1 \over \tau } \sum _{i=0}^{\tau -1} \overline{\varvec{A}}(k + n\tau - i) \cdot \sum _{l=0}^{\tau -1} \varvec{L}(n\tau + l). \end{aligned}$$

Substituting (8.5) into the above inequality yields

$$\begin{aligned} \sum _{m=m_{*}}^{\infty } { \overline{\varvec{A}} (k+m) \varvec{L}(m) \over {\mathsf {P}}(Y > k) }\le & {} \sum _{n=\lfloor m_{*}/\tau \rfloor }^{\infty } \sum _{i=0}^{\tau -1} { \overline{\varvec{A}}(k + n\tau - i) \varvec{e} \over {\mathsf {P}}(Y > k) } (\varvec{\psi } + \varepsilon \varvec{e}^{\mathrm{t}}/\tau ) \nonumber \\= & {} { \overline{\overline{\varvec{A}}}(k + \lfloor m_{*}/\tau \rfloor \tau - \tau ) \varvec{e} \over {\mathsf {P}}(Y > k) } (\varvec{\psi } + \varepsilon \varvec{e}^{\mathrm{t}}/\tau ). \end{aligned}$$

(8.9)

From (8.9), (3.1) and $Y \in {\mathcal {L}}$, we have (8.7).

Next we show that (8.8) holds. Since $\{\overline{\varvec{A}}(k)\}$ is nonincreasing, we have

$$\begin{aligned} \sum _{m=m_{*}}^{\infty } \overline{\varvec{A}} (k+m) \varvec{L}(m)\ge & {} \sum _{n=\lceil m_{*}/\tau \rceil }^{\infty } \sum _{l=0}^{\tau -1} \overline{\varvec{A}}(k + n\tau + l) \varvec{L}(n\tau + l) \\\ge & {} \sum _{n=\lceil m_{*}/\tau \rceil }^{\infty } \overline{\varvec{A}}(k + n\tau + \tau + 1) \sum _{l=0}^{\tau -1} \varvec{L}(n\tau + l) \\\ge & {} \sum _{n=\lceil m_{*}/\tau \rceil }^{\infty } {1 \over \tau } \sum _{i=1}^{\tau } \overline{\varvec{A}}(k + n\tau + \tau + i) \cdot \sum _{l=0}^{\tau -1} \varvec{L}(n\tau + l). \end{aligned}$$

Combining this with (8.5) yields

$$\begin{aligned} \sum _{m=m_{*}}^{\infty } { \overline{\varvec{A}} (k+m) \varvec{L}(m) \over {\mathsf {P}}(Y > k) }\ge & {} \sum _{n=\lceil m_{*}/\tau \rceil +1}^{\infty } \sum _{i=1}^{\tau } { \overline{\varvec{A}}(k + n\tau + i) \varvec{e} \over {\mathsf {P}}(Y > k) } (\varvec{\psi } - \varepsilon \varvec{e}^{\mathrm{t}}/\tau ). \\= & {} { \overline{\overline{\varvec{A}}}(k + \lceil m_{*}/\tau \rceil \tau + \tau ) \varvec{e} \over {\mathsf {P}}(Y > k) } (\varvec{\psi } - \varepsilon \varvec{e}^{\mathrm{t}}/\tau ). \end{aligned}$$

Therefore similarly to (8.7), we can obtain (8.8). $\square$

Lemma 8.2

Suppose that Assumption 2.1 is satisfied. If Assumption 3.1 holds for some $Y \in {\mathcal {L}}$, then

$$\begin{aligned} \lim _{k \rightarrow \infty } {\overline{\varvec{R}}(k) \over {\mathsf {P}}(Y > k) }= & {} {\varvec{c}_A\varvec{\pi } ( \varvec{I} - \varvec{R} ) \over -\sigma }, \end{aligned}$$

(8.10)

$$\begin{aligned} \lim _{k \rightarrow \infty } {\overline{\varvec{R}_0}(k) \over {\mathsf {P}}(Y > k) }= & {} {\varvec{c}_B\varvec{\pi } ( \varvec{I} - \varvec{R} ) \over -\sigma }. \end{aligned}$$

(8.11)

Proof

From (2.6), we have

$$\begin{aligned} \overline{\varvec{R}}(k) = \left[ \overline{\varvec{A}}(k) + \sum _{m = 1}^{\infty } \overline{\varvec{A}}(k+m) \varvec{L}(m) \right] (\varvec{I} - \varvec{\varPhi }(0))^{-1}. \end{aligned}$$

(8.12)

Further it follows from (3.1) and $Y \in {\mathcal {L}}$ that

$$\begin{aligned} \lim _{k \rightarrow \infty } {\overline{\varvec{A}}(k) \over {\mathsf {P}}(Y > k) } \le \lim _{k \rightarrow \infty } {\overline{\overline{\varvec{A}}}(k-1)\varvec{e}\varvec{e}^{\mathrm{t}} - \overline{\overline{\varvec{A}}}(k)\varvec{e}\varvec{e}^{\mathrm{t}} \over {\mathsf {P}}(Y > k) } = \varvec{O}. \end{aligned}$$

Thus (8.12) yields

$$\begin{aligned} \lim _{k \rightarrow \infty } {\overline{\varvec{R}}(k) \over {\mathsf {P}}(Y > k) } = \lim _{k \rightarrow \infty } \sum _{m = 1}^{\infty } {\overline{\varvec{A}}(k+m) \varvec{L}(m) \over {\mathsf {P}}(Y > k) } ( \varvec{I} - \varvec{\varPhi }(0) )^{-1} . \end{aligned}$$

(8.13)

Substituting (8.1) into (8.13), we obtain (8.10). Similarly, using (2.5), we can prove (8.11). $\square$

Lemma 8.3

Suppose that Assumption 2.1 is satisfied. If Assumption 3.1 holds for some $Y \in {\mathcal {S}}$, then

$$\begin{aligned} \lim _{k \rightarrow \infty } {\overline{\varvec{F}}(k) \over {\mathsf {P}}(Y > k)} = {(\varvec{I} - \varvec{R} )^{-1} \varvec{c}_A\varvec{\pi } \over -\sigma }. \end{aligned}$$

(8.14)

Proof

It follows from (2.2) that

$$\begin{aligned} \sum _{k=0}^{\infty }\varvec{F}(k) = (\varvec{I} - \varvec{R})^{-1}. \end{aligned}$$

(8.15)

Applying Lemma 6 in Jelenković and Lazar (1998) to (2.2) and using (8.15) yield

$$\begin{aligned} \lim _{k \rightarrow \infty } {\overline{\varvec{F}}(k) \over {\mathsf {P}}(Y > k)}= & {} (\varvec{I} - \varvec{R} )^{-1} \lim _{k \rightarrow \infty } {\overline{\varvec{R}}(k) \over {\mathsf {P}}(Y > k) }(\varvec{I} - \varvec{R})^{-1}. \end{aligned}$$

From this and (8.10), we have (8.14). $\square$

Finally we are ready to prove Theorem 3.1. Applying Proposition 9.1 to (2.3) and using (8.11), (8.14) and (8.15), we obtain

$$\begin{aligned} \lim _{k \rightarrow \infty } {\overline{\varvec{x}}(k) \over {\mathsf {P}}(Y > k) } = {\varvec{x}(0) \over -\sigma } \left[ \varvec{c}_B \varvec{\pi } + \varvec{R}_0(\varvec{I} - \varvec{R})^{-1} \varvec{c}_A \varvec{\pi } \right] . \end{aligned}$$

Substituting (2.4) into the above equation yields (3.2). The proof of Theorem 3.1 is completed.

Appendix 2: Auxiliary results for the proof and applications of the main result

This section summarizes some known results, which are used for the proof and applications of the main result. We first present an asymptotic result on the convolution of matrix sequences with subexponential tails, which is utilized several times in Sects. 4–6 and Appendix 1.

Proposition 9.1

(Masuyama 2011, Proposition A.3) Let $\{\varvec{M}(k);k\in {\mathbb {Z}}_+\}$ and $\{\varvec{N}(k);k\in {\mathbb {Z}}_+\}$ denote finite-dimensional nonnegative matrix sequences such that their convolution $\{\varvec{M}*\varvec{N}(k);k\in {\mathbb {Z}}_+\}$ is well-defined and $\varvec{M}:= \sum _{k=0}^{\infty }\varvec{M}(k)$ and $\varvec{N}:= \sum _{k=0}^{\infty }\varvec{N}(k)$ are finite. Suppose that for some random variable $U \in {\mathcal {S}}$,

$$\begin{aligned} \lim _{k\rightarrow \infty }{\overline{\varvec{M}}(k) \over {\mathsf {P}}(U > k)} = \widetilde{\varvec{M}} \ge \varvec{O}, \qquad \lim _{k\rightarrow \infty }{\overline{\varvec{N}}(k) \over {\mathsf {P}}(U > k)} = \widetilde{\varvec{N}} \ge \varvec{O}, \end{aligned}$$

where $\widetilde{\varvec{M}} = \widetilde{\varvec{N}} = \varvec{O}$ is allowed. We then have

$$\begin{aligned} \lim _{k\rightarrow \infty }{ \overline{\varvec{M} *\varvec{N}}(k) \over {\mathsf {P}}(U > k) } = \widetilde{\varvec{M}} \varvec{N} + \varvec{M} \widetilde{\varvec{N}}. \end{aligned}$$

Next we describe some of the asymptotic results on a (regenerative) cumulative process sampled at heavy-tailed random times (for details, see Masuyama 2013b), which are used in Sects. 4, 5 and 6.

Let $\{B(t);t\in {\mathbb {R}}_+\}$ denote a stochastic process on $(-\infty ,\infty )$, where $|B(0)| < \infty$ with probability one. We assume that there exist regenerative points $0 \le \tau _0 < \tau _1 < \tau _2 < \cdots$ such that for any $n \in {\mathbb {Z}}_+$, $\{B(t+\tau _n)-B(\tau _n); t \ge 0\}$ is stochastically equivalent to $\{B(t+\tau _0)-B(\tau _0); t \ge 0\}$ and is independent of $\{B(u);0 \le u < \tau _n\}$. The process $\{B(t);t\in {\mathbb {R}}_+\}$ is called (regenerative) cumulative process, which is introduced by Smith (1955).

Let $\varDelta \tau _0 = \tau _0$ and $\varDelta \tau _n = \tau _n-\tau _{n-1}$ for $n\in {\mathbb {N}}$. Let

$$\begin{aligned} \varDelta B_n= & {} \left\{ \begin{array}{ll} B(\tau _0), &{} n = 0,\\ B(\tau _n)-B(\tau _{n-1}), &{} n \in {\mathbb {N}}, \end{array} \right. \quad \varDelta B_n^{*} = \left\{ \begin{array}{ll} \displaystyle \sup _{0 \le t \le \tau _0} \max (B(t), 0), &{} n = 0,\\ \displaystyle \sup _{\tau _{n-1} \le t \le \tau _n} B(t) - B(\tau _{n-1}), &{} n \in {\mathbb {N}}. \end{array} \right. \end{aligned}$$

It then follows that $\varDelta B_n^{*} \ge \varDelta B_n$ for $n\in {\mathbb {Z}}_+$ and that $\{\varDelta \tau _n;n\in {\mathbb {N}}\}$ (resp. $\{\varDelta B_n;n\in {\mathbb {N}}\}$ and $\{\varDelta B_n^{*};n\in {\mathbb {N}}\}$) is a sequence of i.i.d. random variables, which is independent of $\varDelta \tau _0$ (resp. $\varDelta B_0$ and $\varDelta B_0^{*}$).

Remark 9.1

The counting process $\{N(t);t \in {\mathbb {R}}_+\}$ of BMAP $\{\varvec{C},\varvec{D}(1),\varvec{D}(2),\ldots \}$ is a (regenerative) cumulative process such that regenerative points are hitting times to any fixed background state and the regenerative cycle follows a phase-type distribution (see Eqs. (3.3)–(3.5) in Masuyama 2013b).

We now assume that

$$\begin{aligned}&{\mathsf {P}}(\varDelta \tau _n \in {\mathbb {R}}_+) = {\mathsf {P}}(\varDelta B_n^{*} \in {\mathbb {R}}_+) = 1~~(n=0,1), \\&{\mathsf {E}}[|\varDelta B_1|] <\infty ,~~~ 0 < {\mathsf {E}}[\varDelta \tau _1] < \infty ,~~~ b := {{\mathsf {E}}[\varDelta B_1] \big / {\mathsf {E}}[\varDelta \tau _1] } > 0. \end{aligned}$$

We then have the following results.

Proposition 9.2

(Masuyama 2013b, Theorem 3.3) Suppose that $T$ is a nonnegative random variable independent of $\{B(t);t\in {\mathbb {R}}_+\}$. Further suppose that (i) $T \in {\mathcal {L}}^{\mu }$ for some $\mu \ge 2$; (ii) ${\mathsf {E}}[(\varDelta \tau _1)^2] < \infty$ and ${\mathsf {E}}[(\varDelta B_1)^2] < \infty$; and (iii) ${\mathsf {E}}[\exp \{Q(\varDelta B_n^{*})\}] < \infty$ ($n=0,1$) for some cumulative hazard function $Q \in {\mathcal {SC}}$ such that $x^{1/\mu } = O(Q(x))$. We then have ${\mathsf {P}}(B(T) > bx) \mathop {\sim }\limits ^{x} {\mathsf {P}}(T > x)$.

Corollary 9.1

Suppose that $T$ is a nonnegative random variable independent of $\{(N(t),J(t));t\in {\mathbb {R}}_+\}$, where $\{N(t)\}$ and $\{J(t)\}$ denote the counting process and the background Markov chain, respectively, of BMAP $\{\varvec{C},\varvec{D}(1),\varvec{D}(2),\ldots \}$ introduced in Sect. 4.1. Suppose that (i) $T \in {\mathcal {L}}^{\mu }$ for some $\mu \ge 2$; and (ii) $\sum _{k=1}^{\infty } \exp \{Q(k)\} \varvec{D}(k) < \infty$ ($n=0,1$) for some cumulative hazard function $Q \in {\mathcal {SC}}$ such that $x^{1/\mu } = O(Q(x))$. We then have ${\mathsf {P}}(N(T) > k \mid J(0) = i) \mathop {\sim }\limits ^{k} {\mathsf {P}}(T > k/\lambda )$ for $i \in {\mathbb {M}}$.

Proof

It suffices to prove that conditions (i)–(iii) of Proposition 9.2 are satisfied. For this purpose, fix $B(t) = N(t)$ for $t\in {\mathbb {R}}_+$ and $J(0) = i \in {\mathbb {M}}$. Let $\tau _0 = 0$ and $\tau _n$ ($n \in {\mathbb {N}}$) denote the $n$th hitting time of state $i$. Clearly, $\varDelta B_0^{*} = 0$ and thus ${\mathsf {E}}[\exp \{Q(\varDelta B_0^{*})\}] < \infty$.

Since the regenerative cycle follows a phase-type distribution (see Remark 9.1), we have ${\mathsf {E}}[(\varDelta \tau _1)^2] < \infty$. Further since $\{B(t)=N(t);t\in {\mathbb {R}}_+\}$ is nondecreasing, we have $\varDelta B_n^{*} = \varDelta B_n$ for all $n \in {\mathbb {Z}}_+$. Therefore it follows from the renewal reward theorem (see, e.g., Wolff 1989, Chapter 2, Theorem 2) that

$$\begin{aligned} {{\mathsf {E}}[\varDelta B_1^{*}] \over {\mathsf {E}}[\varDelta \tau _1]} = \lambda \in (0,\infty ), \qquad {{\mathsf {E}}[\exp \{Q(\varDelta B_1^{*})\}] \over {\mathsf {E}}[\varDelta \tau _1]} = \varvec{\varpi }\sum _{k=1}^{\infty } \exp \{Q(k)\} \varvec{D}(k) \varvec{e} < \infty , \end{aligned}$$

which lead to ${\mathsf {E}}[\exp \{Q(\varDelta B_1^{*})\}] < \infty$ and thus ${\mathsf {E}}[(\varDelta B_1)^2] < \infty$ (see Remark 2.4 in Masuyama 2013b). As a result, conditions (i)–(iii) of Proposition 9.2 are satisfied. $\square$

We also have a result similar to Corollary 9.1:

Proposition 9.3

(Masuyama (2013b), Corollary 3.1) Suppose that $T$ is a nonnegative random variable independent of $\{(N(t),J(t));t\in {\mathbb {R}}_+\}$, where $\{N(t)\}$ and $\{J(t)\}$ denote the counting process and the background Markov chain, respectively, of BMAP $\{\varvec{C},\varvec{D}(1),\varvec{D}(2),\ldots \}$ introduced in Sect. 4.1. Suppose that (i) $T \in {\mathcal {C}}$; (ii) ${\mathsf {E}}[T] < \infty$; and (iii) $\overline{\varvec{D}}(k)= o( {\mathsf {P}}(T > k) )$. We then have ${\mathsf {P}}(N(T) > k \mid J(0) = i) \mathop {\sim }\limits ^{k} {\mathsf {P}}(T > k/\lambda )$ for $i \in {\mathbb {M}}$.

Remark 9.2

According to the original statement of Corollary 3.1 in Masuyama (2013b), we have ${\mathsf {P}}(N(T) > k) \mathop {\sim }\limits ^{k} {\mathsf {P}}(T > k/\lambda )$. In fact, the statement of this corollary holds for any initial distribution of the background Markov chain, which follows from Theorem 3.5 in Masuyama (2013b). Thus we can obtain the asymptotic formula conditioned on the initial background state, as stated in Proposition 9.3.