Asymptotic capital allocation based on the higher moment risk measure

Abstract

We investigate capital allocation based on the higher moment risk measure at a confidence level $q\in (0,1)$. To reflect the excessive prudence of today’s regulatory frameworks in banking and insurance, we consider the extreme case with $q\uparrow 1$ and study the asymptotic behavior of capital allocation for heavy-tailed and asymptotically independent/dependent risks. Some explicit asymptotic formulas are derived, demonstrating that the capital allocated to a specific line is asymptotically proportional to the Value at Risk of the corresponding individual risk. In addition, some numerical studies are conducted to examine their accuracy.

Appendix

1.1 Lemmas

The following first lemma is a restatement of Proposition 2.2.3 of [2]:

Lemma A.1

If $f\in \textrm{RV}_{-\alpha }$ for some $\alpha \ge 0$, then for arbitrary $0<\varepsilon <1$ there is some $x_{0}>0$ such that the Potter’s bounds

$$\begin{aligned} (1-\varepsilon )\left( y^{-\alpha -\varepsilon }\wedge y^{-\alpha +\varepsilon }\right) \le \frac{f(xy)}{f(x)}\le (1+\varepsilon )\left( y^{-\alpha -\varepsilon }\vee y^{-\alpha +\varepsilon }\right) \end{aligned}$$

hold whenever $x\ge x_{0}$ and $xy\ge x_{0}$.

The next lemma is closely related to Theorem 3.1 of [4], but neither of them covers the other:

Lemma A.2

Under the conditions of Theorem 3.1, we have

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{P\left( S>x\right) }{{\overline{F}}(x)} =\Theta \qquad \text {and} \qquad \lim _{q\uparrow 1}\frac{F_{S}^{\leftarrow }(q)}{F^{\leftarrow }(q)}=\Theta ^{1/\alpha }. \end{aligned}$$

Proof

We only need to prove the first relation, which immediately implies the second one as ${\overline{F}}\in \textrm{RV}_{-\alpha }$. Denote by ${\mathcal {I}}$ the subset of $\{1,\ldots ,d\}$ such that $\theta _{i}>0$ for $i\in {\mathcal {I}}$ and $\theta _{i}=0$ for $i\in {\mathcal {I}}^{c}=\{1,\ldots ,d\}\backslash {\mathcal {I}}$, and denote by $d_{1}$ and $d_{2}=d-d_{1}$ the cardinalities of the sets ${\mathcal {I}}$ and ${\mathcal {I}}^{c}$, respectively. Write $S_{{\mathcal {I}}}=\sum _{i\in {\mathcal {I}}}X_{i}$ and $S_{{\mathcal {I}} ^{c}}=\sum _{i\in {\mathcal {I}}^{c}}X_{i}$. Note that ${\mathcal {I}}\ne \emptyset$ since $\Theta >0$. By Theorem 3.1 of [4] and the conditions in (3.1),

$$\begin{aligned} P\left( S_{{\mathcal {I}}}>x\right) \sim \sum _{i\in {\mathcal {I}}}P\left( X_{i}>x\right) \sim \left( \sum _{i\in {\mathcal {I}}}\theta _{i}\right) {\overline{F}}(x). \end{aligned}$$

(A.1)

Thus, we will have proved the lemma if ${\mathcal {I}}^{c}=\emptyset$. Now assume ${\mathcal {I}}^{c}\ne \emptyset$. Then for every small $\varepsilon \in (0,1)$, applying (A.1) we obtain

$$\begin{aligned} P\left( S>x\right)\le & {} P\left( S_{{\mathcal {I}}}>(1-\varepsilon )x\right) +P\left( S_{{\mathcal {I}}^{c}}>\varepsilon x\right) \\\le & {} \left( \sum _{i\in {\mathcal {I}}}\theta _{i}\right) {\overline{F}} ((1-\varepsilon )x)+\sum _{i\in {\mathcal {I}}^{c}}P\left( X_{i}>\frac{ \varepsilon }{d_{2}}x\right) \\= & {} \left( \sum _{i\in {\mathcal {I}}}\theta _{i}\right) {\overline{F}} ((1-\varepsilon )x)+o\left( {\overline{F}}(x)\right) \\\sim & {} (1-\varepsilon )^{-\alpha }\left( \sum _{i\in {\mathcal {I}}}\theta _{i}\right) {\overline{F}}(x). \end{aligned}$$

On the other hand,

$$\begin{aligned} P\left( S>x\right)\ge & {} P\left( S_{{\mathcal {I}}}>(1+\varepsilon )x,S_{ {\mathcal {I}}^{c}}>-\varepsilon x\right) \\= & {} P\left( S_{{\mathcal {I}}}>(1+\varepsilon )x\right) -P\left( S_{{\mathcal {I}} }>(1+\varepsilon )x,S_{{\mathcal {I}}^{c}}\le -\varepsilon x\right) \\\ge & {} P\left( S_{{\mathcal {I}}}>(1+\varepsilon )x\right) -\sum _{i\in {\mathcal {I}}}\sum _{j\in {\mathcal {I}}^{c}}P\left( X_{i}>\frac{1+\varepsilon }{d_{1}} x,X_{j}\le -\frac{\varepsilon }{d_{2}}x\right) . \end{aligned}$$

By (A.1), the first term on the right-hand side is asymptotic to $(1+\varepsilon )^{-\alpha }\left( \sum _{i\in {\mathcal {I}}}\theta _{i}\right) {\overline{F}}(x)$. With $c=\frac{1+\varepsilon }{d_{1}}\wedge \frac{ \varepsilon }{d_{2}}$, each summand in the second term is no more than

$$\begin{aligned} P\left( X_{i}>cx,X_{j}\le -cx\right) =o\left( P\left( X_{i}>cx\right) \right) =o\left( {\overline{F}}(x)\right) , \end{aligned}$$

since $X_{i}$ and $X_{j}$ are QAI. In summary,

$$\begin{aligned} (1+\varepsilon )^{-\alpha }\left( \sum _{i\in {\mathcal {I}}}\theta _{i}\right) {\overline{F}}(x)\lesssim P\left( S>x\right) \lesssim (1-\varepsilon )^{-\alpha }\left( \sum _{i\in {\mathcal {I}}}\theta _{i}\right) {\overline{F}}(x). \end{aligned}$$

By the arbitrariness of $\varepsilon$ and the fact that $\Theta =\sum _{i\in {\mathcal {I}}}\theta _{i}$ since $\sum _{i\in {\mathcal {I}}^{c}}\theta _{i}=0$, we complete the proof. $\square$

In the following lemma, we establish a novel inequality in the spirit of Potter’s upper bound in Lemma A.1 but in a multidimensional scenario. Notably, it does not require any dependence information among the individual random variables.

Lemma A.3

Consider d real-valued random variables $X_{1}$, ..., $X_{d}$ distributed by $F_{1}$, ..., $F_{d}$, respectively, and denote by S their sum (1.1). Assume that there is some reference distribution F on ${\mathbb {R}}_{+}$, with a tail ${\overline{F}}\in \textrm{RV }_{-\alpha }$ for some $\alpha \ge 0$, such that the limit

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{\overline{F_{i}}(x)}{{\overline{F}}(x)} =\theta _{i}\in [0,\infty ) \end{aligned}$$

(A.2)

exists for each $i=1,\ldots ,d$. Then for every small $0<\varepsilon <1$, there are some large $K>0$ and $x_{1}>0$ dependent on $\varepsilon$ such that the inequality

$$\begin{aligned} P\left( X_{k}>ux,S>vx\right) \le K\left( \left( u\vee v\right) ^{-\alpha -\varepsilon }\vee \left( u\vee v\right) ^{-\alpha +\varepsilon }\right) {\overline{F}}(x) \end{aligned}$$

(A.3)

holds whenever $ux\ge x_{1}$ and $vx\ge x_{1}$.

Proof

In this proof, for ease of exposition we use K to denote a large positive constant, which depends on $\varepsilon$ but not on the working variables u, v, and x, and which may vary from place to place. By (A.2),

$$\begin{aligned} \overline{F_{i}}(x)\le K{\overline{F}}(x),\qquad x\ge 0,\ i=1,\ldots ,d. \end{aligned}$$

(A.4)

For every small $\varepsilon \in (0,1)$, we derive

$$\begin{aligned}{} & {} P\left( X_{k}>ux,S>vx\right) \nonumber \\{} & {} \quad \le P\left( X_{k}>ux,X_{k}>(1-\varepsilon )vx\right) +P\left( X_{k}>ux,\sum _{i=1,i\ne k}^{d}X_{i}>\varepsilon vx\right) \nonumber \\{} & {} \quad \le P\left( X_{k}>\left( u\vee (1-\varepsilon )v\right) x\right) +\sum _{i=1,i\ne k}^{d}P\left( X_{k}>ux,X_{i}>\frac{\varepsilon }{d-1} vx\right) \nonumber \\{} & {} \quad =I_{k}+\sum _{i=1,i\ne k}^{d}I_{ki}. \end{aligned}$$

(A.5)

Choose $x_{1}=\frac{d-1}{\varepsilon }x_{0}\ge x_{0}$, where $x_{0}$ is specified in Lemma A.1, so that whenever $ux\ge x_{1}$ and $vx\ge x_{1}$ we have $ux\ge x_{0}$ and $\frac{\varepsilon }{d-1} vx\ge x_{0}$. For the first term $I_{k}$ in (A.5), by (A.4 ) and Lemma A.1,

$$\begin{aligned} I_{k}\le & {} K{\overline{F}}(\left( u\vee (1-\varepsilon )v\right) x) \nonumber \\\le & {} K\left( \left( u\vee (1-\varepsilon )v\right) ^{-\alpha -\varepsilon }\vee \left( u\vee (1-\varepsilon )v\right) ^{-\alpha +\varepsilon }\right) {\overline{F}}(x) \nonumber \\\le & {} K\left( \left( u\vee v\right) ^{-\alpha -\varepsilon }\vee \left( u\vee v\right) ^{-\alpha +\varepsilon }\right) {\overline{F}}(x). \end{aligned}$$

(A.6)

For the second term in (A.5), if $u\ge v$, we have

$$\begin{aligned} I_{ki}\le & {} P\left( X_{k}>ux\right) \\\le & {} K\left( u^{-\alpha -\varepsilon }\vee u^{-\alpha +\varepsilon }\right) {\overline{F}}(x) \\= & {} K\left( \left( u\vee v\right) ^{-\alpha -\varepsilon }\vee \left( u\vee v\right) ^{-\alpha +\varepsilon }\right) {\overline{F}}(x), \end{aligned}$$

while if $u<v$, we have

$$\begin{aligned} I_{ki}\le & {} P\left( X_{i}>\frac{\varepsilon }{d-1}vx\right) \\\le & {} K\left( \left( \frac{\varepsilon }{d-1}v\right) ^{-\alpha -\varepsilon }\vee \left( \frac{\varepsilon }{d-1}v\right) ^{-\alpha +\varepsilon }\right) \\\le & {} K\left( v^{-\alpha -\varepsilon }\vee v^{-\alpha +\varepsilon }\right) \\= & {} K\left( \left( u\vee v\right) ^{-\alpha -\varepsilon }\vee \left( u\vee v\right) ^{-\alpha +\varepsilon }\right) {\overline{F}}(x). \end{aligned}$$

Thus, in any case,

$$\begin{aligned} \sum _{i=1,i\ne k}^{d}I_{ki}\le K\left( \left( u\vee v\right) ^{-\alpha -\varepsilon }\vee \left( u\vee v\right) ^{-\alpha +\varepsilon }\right) {\overline{F}}(x). \end{aligned}$$

(A.7)

Plugging (A.6)–(A.7) into (A.5) yields (A.3). $\square$

The following last lemma bears a resemblance to Lemma A.3 of [3], but we come up with a much shorter proof in the current setting:

Lemma A.4

Under the conditions of Theorem 3.1(ii), it holds for every $u,v>0$ that

$$\begin{aligned} P\left( X_{k}>ux,S>vx\right) \sim \theta _{k}\left( u\vee v\right) ^{-\alpha }{\overline{F}}(x). \end{aligned}$$

(A.8)

Proof

To obtain a suitable upper bound, we start from (A.5) in which $\varepsilon \in (0,1)$ is arbitrarily fixed. For the first term $I_{k}$, by (3.1) for $F_{k}$ for which $\theta _{k}>0$, and by ${\overline{F}}\in \textrm{RV}_{-\alpha }$,

$$\begin{aligned} I_{k}\sim \theta _{k}\left( u\vee (1-\varepsilon )v\right) ^{-\alpha } {\overline{F}}(x). \end{aligned}$$

For the second term, since $X_{k}$ and $X_{i}$ are QAI, we have

$$\begin{aligned} I_{ki}=o\left( {\overline{F}}(x)\right) \end{aligned}$$

as in the second half of the proof of Lemma A.2. Substituting these results into (A.5) yields

$$\begin{aligned} P\left( X_{k}>ux,S>vx\right) \lesssim \theta _{k}\left( u\vee (1-\varepsilon )v\right) ^{-\alpha }{\overline{F}}(x). \end{aligned}$$

The corresponding lower bound can be obtained in a similar way. We derive

$$\begin{aligned}{} & {} P\left( X_{k}>ux,S>vx\right) \\{} & {} \quad \ge P\left( X_{k}>ux,X_{k}>(1+\varepsilon )vx,\sum _{i=1,i\ne k}^{d}X_{i}>-\varepsilon vx\right) \\{} & {} \quad \ge P\left( X_{k}>ux,X_{k}>(1+\varepsilon )vx\right) -P\left( X_{k}>ux,\sum _{i=1,i\ne k}^{d}X_{i}\le -\varepsilon vx\right) \\{} & {} \quad \ge P\left( X_{k}>\left( u\vee (1+\varepsilon )v\right) x\right) -\sum _{i=1,i\ne k}^{d}P\left( X_{k}>ux,X_{i}<-\frac{\varepsilon }{d-1} vx\right) \\{} & {} \quad \sim \theta _{k}\left( u\vee (1+\varepsilon )v\right) ^{-\alpha }{\overline{F}}(x)+o\left( {\overline{F}}(x)\right) . \end{aligned}$$

Based on these upper and lower bounds and by the arbitrariness of $\varepsilon$, we obtain (A.8). $\square$

1.2 Proof of Theorem 3.1

(i) By Lemma A.2, $\overline{F_{S}}\in \textrm{RV}_{-\alpha }$. Thus, for $p\ge 1$, by Theorem 4.1 of [19],

$$\begin{aligned} \rho (S)\sim & {} \frac{\alpha }{p^{(p-1)/\alpha }(\alpha -p)^{1-p/\alpha }} \textrm{B}(\alpha -p,p)^{1/\alpha }F_{S}^{\leftarrow }({\tilde{q}}) \\\sim & {} \frac{\alpha }{p^{(p-1)/\alpha }(\alpha -p)^{1-p/\alpha }}\textrm{B} (\alpha -p,p)^{1/\alpha }\Theta ^{1/\alpha }F^{\leftarrow }({\tilde{q}}), \end{aligned}$$

where in the last step we have applied $F_{S}^{\leftarrow }({\tilde{q}})\sim \Theta ^{1/\alpha }F^{\leftarrow }({\tilde{q}})$ stated in Lemma A.2. This proves (3.2).

(ii) We need to formulate the proof into two parts: $p=1$ and $p>1$.

$\underline{{\hbox {Step 1:\ Assume}} p=1.}$

For this case, ${\tilde{q}}=q$. We start from (2.5). For brevity, denote $x=F_{S}^{\leftarrow }(q)$, which tends to $\infty$ as $q\uparrow 1$. We expand

$$\begin{aligned} \Lambda \left( X_{k},S\right)= & {} \frac{1}{1-q}\int _{0}^{\infty } \left( P\left( X_{k}>y,S>x\right) -P\left( X_{k}<-y,S>x\right) \right) \textrm{d}y \nonumber \\= & {} \frac{x}{1-q}\int _{0}^{\infty }\left( P\left( X_{k}>ux,S>x\right) -P\left( X_{k}<-ux,S>x\right) \right) \textrm{d}u \nonumber \\= & {} \frac{x}{1-q}\left( I_{1}-I_{2}+I_{3}\right) \end{aligned}$$

(A.9)

with

$$\begin{aligned} \left\{ \begin{array}{ll} I_{1}=\int _{\delta }^{\infty }P\left( X_{k}>ux,S>x\right) \textrm{d}u, \\ I_{2}=\int _{\delta }^{\infty }P\left( X_{k}<-ux,S>x\right) \textrm{d}u, \\ I_{3}=\int _{0}^{\delta }\left( P\left( X_{k}>ux,S>x\right) -P\left( X_{k}<-ux,S>x\right) \right) \textrm{d}u, \end{array} \right. \end{aligned}$$

with $\delta >0$ introduced to be arbitrarily fixed and small.

For $I_{1}$, by Lemma A.3, for every small $0<\varepsilon <(\alpha -1)\wedge 1$, there is some large $K>0$ such that, for all large x and all $u\ge \delta$,

$$\begin{aligned} \frac{P\left( X_{k}>ux,S>x\right) }{{\overline{F}}(x)}\le K\left( u\vee 1\right) ^{-\alpha +\varepsilon }. \end{aligned}$$

Note that $\int _{\delta }^{\infty }\left( u\vee 1\right) ^{-\alpha +\varepsilon }\textrm{d}u<\infty$. Thus, by the dominated convergence theorem

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{I_{1}}{{\overline{F}}(x)}=\int _{\delta }^{\infty }\lim _{x\rightarrow \infty }\frac{P\left( X_{k}>ux,S>x\right) }{ {\overline{F}}(x)}\textrm{d}u=\theta _{k}\int _{\delta }^{\infty }\left( u\vee 1\right) ^{-\alpha }\textrm{d}u, \nonumber \\ \end{aligned}$$

(A.10)

where the last step is due to Lemma A.4.

For $I_{2}$, we examine

$$\begin{aligned} \frac{P\left( X_{k}<-ux,S>x\right) }{{\overline{F}}(x)}\le & {} \frac{P\left( X_{k}<-ux,\sum _{i=1,i\ne k}^{d}X_{i}>(1+u)x\right) }{{\overline{F}}(x)} \\\le & {} \sum _{i=1,i\ne k}^{d}\frac{P\left( X_{k}<-ux,X_{i}>\frac{1+u}{d-1} x\right) }{{\overline{F}}(x)} \\\le & {} \sum _{i=1,i\ne k}^{d}P\left( \left. X_{k}<-{\tilde{u}}x\right| X_{i}>{\tilde{u}}x\right) \frac{P\left( X_{i}>ux\right) }{{\overline{F}}(x)}, \end{aligned}$$

where ${\tilde{u}}=u\wedge \frac{1+u}{d-1}$. As $x\rightarrow \infty$, each conditional probability $P\left( \left. X_{k}<-{\tilde{u}}x\right| X_{i}> {\tilde{u}}x\right)$ tends to 0 uniformly for all ${\tilde{u}}\ge \delta \wedge \frac{1+\delta }{d-1}$ since $X_{k}$ and $X_{i}$ are QAI. Moreover,

$$\begin{aligned} \frac{P\left( X_{i}>ux\right) }{{\overline{F}}(x)}\le \frac{P\left( X_{i}>ux\right) }{{\overline{F}}(ux)}\times \frac{{\overline{F}}(ux)}{{\overline{F}}(x)}. \end{aligned}$$

By (A.4), $\frac{P\left( X_{i}>ux\right) }{{\overline{F}}(ux)}$ is uniformly bounded for all $u>\delta$ and $x>0$. By Lemma A.1, for arbitrarily fixed and small $0<\varepsilon <\left( \alpha -1\right) \wedge 1$ and for all large $x>0$,

$$\begin{aligned} \frac{{\overline{F}}(ux)}{{\overline{F}}(x)}\le (1+\varepsilon )\left( u^{-\alpha -\varepsilon }\vee u^{-\alpha +\varepsilon }\right) , \end{aligned}$$

while $\int _{\delta }^{\infty }\left( u^{-\alpha -\varepsilon }\vee u^{-\alpha +\varepsilon }\right) \textrm{d}u<\infty$. Thus, by the dominated convergence theorem,

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{I_{2}}{{\overline{F}}(x)}=0. \end{aligned}$$

(A.11)

For $I_{3}$, we derive

$$\begin{aligned} \left| I_{3}\right| \le \int _{0}^{\delta }P\left( \left| X_{k}\right|>ux,S>x\right) \textrm{d}u\le \delta P\left( S>x\right) \sim \delta \Theta {\overline{F}}(x), \end{aligned}$$

(A.12)

where the last step is due to Lemma A.2.

Finally, substituting (A.10)–(A.12) into (A.9) and keeping in mind the arbitrariness of $\delta$, we obtain

$$\begin{aligned} \Lambda \left( X_{k},S\right)\sim & {} \frac{x}{1-q}{\overline{F}}\left( x\right) \theta _{k}\int _{0}^{\infty }\left( u\vee 1\right) ^{-\alpha } \textrm{d}u \\= & {} \frac{\alpha }{\alpha -1}\theta _{k}\frac{F_{S}^{\leftarrow }(q)}{1-q} {\overline{F}}\left( F_{S}^{\leftarrow }(q)\right) \\\sim & {} \frac{\alpha }{\alpha -1}\frac{\theta _{k}}{\Theta ^{1-1/\alpha }} F^{\leftarrow }(q), \end{aligned}$$

where the last step follows from $F_{S}^{\leftarrow }(q)\sim \Theta ^{1/\alpha }F^{\leftarrow }(q)$, as stated in Lemma A.2. This proves (3.3) for $p=1$.

$\underline{{\hbox {Step 2:\ Assume}} p>1.}$

We start from (2.6) in which $s_{*}$ is the unique solution to equation (2.4). By Lemma 2.2 of [19], as $q\uparrow 1$ we have $s_{*}\uparrow \infty$, and by relation (4.4) of [19], we have

$$\begin{aligned} s_{*}\sim \frac{(\alpha -p)^{p/\alpha }}{p^{(p-1)/\alpha }}\textrm{B} (\alpha -p,p)^{1/\alpha }\Theta ^{1/\alpha }F^{\leftarrow }({\tilde{q}}). \end{aligned}$$

(A.13)

We need to explicitize the asymptotics for both the numerator and the denominator of (2.6). For the numerator, we expand

$$\begin{aligned}{} & {} E\left[ X_{k}\left( S-s_{*}\right) _{+}^{p-1}\right] \\{} & {} \quad =\iint \limits _{(y,z)\in (0,\infty )^{2}}\left( P\left( X_{k}>y,S>s_{*}+z\right) -P\left( X_{k}<-y,S>s_{*}+z\right) \right) \textrm{d}y\textrm{ d}z^{p-1} \\{} & {} \quad =s_{*}^{p}\iint \limits _{(u,v)\in (0,\infty )^{2}}\left( P\left( X_{k}>s_{*}u,S>s_{*}(1+v)\right) -P\left( X_{k}<-s_{*}u,S>s_{*}(1+v)\right) \right) \textrm{d}u\textrm{d}v^{p-1}. \end{aligned}$$

Similar to the proof in step 1, we split the double integral above into three parts as

$$\begin{aligned} E\left[ X_{k}\left( S-s_{*}\right) _{+}^{p-1}\right] =s_{*}^{p}\left( I_{1}-I_{2}+I_{3}\right) \end{aligned}$$

(A.14)

with

$$\begin{aligned} \left\{ \begin{array}{ll} I_{1}=\iint \limits _{(u,v)\in (\delta ,\infty )\times (0,\infty )}P\left( X_{k}>s_{*}u,S>s_{*}(1+v)\right) \textrm{d}u\textrm{d}v^{p-1}, \\ I_{2}=\iint \limits _{(u,v)\in (\delta ,\infty )\times (0,\infty )}P\left( X_{k}<-s_{*}u,S>s_{*}(1+v)\right) \textrm{d}u\textrm{d}v^{p-1}, \\ I_{3}=\iint \limits _{(u,v)\in (0,\delta )\times (0,\infty )}\left( P\left( X_{k}>s_{*}u,S>s_{*}(1+v)\right) -P\left( X_{k}<-s_{*}u,S>s_{*}(1+v)\right) \right) \textrm{d}u\textrm{d}v^{p-1}, \end{array} \right. \end{aligned}$$

with $\delta >0$ introduced to be arbitrarily fixed and small. As in step 1, by virtue of Lemmas A.3–A.4, we can verify the applicability of the dominated convergence theorem in the following derivations, but we skip details:

$$\begin{aligned} \lim _{s_{*}\rightarrow \infty }\frac{I_{1}}{{\overline{F}}(s_{*})}= & {} \iint \limits _{(u,v)\in (\delta ,\infty )\times (0,\infty )}\lim _{s_{*}\rightarrow \infty }\frac{P\left( X_{k}>s_{*}u,S>s_{*}(1+v)\right) }{{\overline{F}}(s_{*})}\textrm{d}u\textrm{d}v^{p-1} \nonumber \\= & {} \theta _{k}\iint \limits _{(u,v)\in (\delta ,\infty )\times (0,\infty )}\left( u\vee (1+v)\right) ^{-\alpha }\textrm{d}u\textrm{d}v^{p-1} \end{aligned}$$

(A.15)

and

$$\begin{aligned} \lim _{s_{*}\rightarrow \infty }\frac{I_{2}}{{\overline{F}}(s_{*})} =\iint \limits _{(u,v)\in (\delta ,\infty )\times (0,\infty )}\lim _{s_{*}\rightarrow \infty }\frac{P\left( X_{k}<-s_{*}u,S>s_{*}(1+v)\right) }{{\overline{F}}(s_{*})}\textrm{d}u\textrm{d}v^{p-1}=0. \nonumber \\ \end{aligned}$$

(A.16)

We deal with $I_{3}$ as follows:

$$\begin{aligned} \frac{\left| I_{3}\right| }{{\overline{F}}(s_{*})}\le & {} \iint \limits _{(u,v)\in (0,\delta )\times (0,\infty )}\frac{P\left( \left| X_{k}\right|>s_{*}u,S>s_{*}(1+v)\right) }{\overline{F }(s_{*})}\textrm{d}u\textrm{d}v^{p-1} \nonumber \\\le & {} \delta \int _{0}^{\infty }\frac{P\left( S>s_{*}(1+v)\right) }{ {\overline{F}}(s_{*})}\textrm{d}v^{p-1} \nonumber \\\rightarrow & {} \delta \Theta \int _{0}^{\infty }(1+v)^{-\alpha }\textrm{d} v^{p-1}, \end{aligned}$$

(A.17)

where the last step is due to Lemma A.2 and the condition ${\overline{F}}\in \textrm{RV}_{-\alpha }$. Substituting (A.15)–(A.17) into (A.14) and keeping in mind the arbitrariness of $\delta$, we obtain

$$\begin{aligned} \lim _{s_{*}\rightarrow \infty }\frac{E\left[ X_{k}(S-s_{*})_{+}^{p-1} \right] }{s_{*}^{p}{\overline{F}}(s_{*})}= & {} \theta _{k}\iint \limits _{(u,v)\in (0,\infty )^{2}}\left( u\vee (1+v)\right) ^{-\alpha }\textrm{d}u\textrm{d}v^{p-1} \\= & {} \theta _{k}\frac{\alpha }{\alpha -1}(p-1)\textrm{B}(\alpha -p,p-1). \end{aligned}$$

For the denominator of (2.6), we derive

$$\begin{aligned} E\left[ \left( S-s_{*}\right) _{+}^{p-1}\right]= & {} \int _{0}^{\infty }P\left( S>s_{*}+z\right) \textrm{d}z^{p-1} \\= & {} s_{*}^{p-1}\int _{0}^{\infty }P\left( S>s_{*}(1+v)\right) \textrm{d}v^{p-1} \\\sim & {} s_{*}^{p-1}\Theta {\overline{F}}(s_{*})\int _{0}^{\infty }(1+v)^{-\alpha }\textrm{d}v^{p-1} \\= & {} s_{*}^{p-1}\Theta {\overline{F}}(s_{*})(p-1)\textrm{B}(\alpha -p+1,p-1), \end{aligned}$$

where in the second last step we have applied the dominated convergence theorem justified by Lemmas A.1–A.2.

Finally, we conclude that

$$\begin{aligned} \Lambda (X_{k},S)= & {} \frac{E\left[ X_{k}\left( S-s_{*}\right) _{+}^{p-1} \right] }{E\left[ \left( S-s_{*}\right) _{+}^{p-1}\right] } \\\sim & {} \frac{s_{*}^{p}{\overline{F}}(s_{*})\theta _{k}\frac{\alpha }{ \alpha -1}(p-1)\textrm{B}(\alpha -p,p-1)}{s_{*}^{p-1}\Theta {\overline{F}} (s_{*})(p-1)\textrm{B}(\alpha -p+1,p-1)} \\= & {} \frac{\theta _{k}\alpha }{\Theta \left( \alpha -p\right) }s_{*} \\\sim & {} \frac{\theta _{k}\alpha (\alpha -p)^{p/\alpha -1}}{\Theta ^{1-1/\alpha }p^{(p-1)/\alpha }}\textrm{B}(\alpha -p,p)^{1/\alpha }F^{\leftarrow }({\tilde{q}}), \end{aligned}$$

where the last step is due to (A.13). This proves (3.3) for $p>1$.

1.3 Proof of Theorem 3.2

(i) First of all, as pointed out in Remark 2.1, under the conditions of Theorem 3.2, in particular (3.5), when considering the limit behavior of the probability of $\frac{{\textbf{X}}}{ x}$ over a Borel set $B\subset [-\varvec{\infty },\varvec{ \infty }]\backslash \{{\textbf{0}}\}$, we can always restrict $\frac{{\textbf{X}} }{x}$ to $B\cap [{\textbf{0}},\varvec{\infty }]\backslash \{\textbf{ 0}\}$. Thus, it holds for every $u\ge -\infty$ and $v>0$ that

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{P\left( X_{k}>ux,S>vx\right) }{{\overline{F}} (x)}=\mu \left( {\textbf{x}}\in [{\textbf{0}},\varvec{\infty } ]:x_{k}>u_{+},\sum _{i=1}^{d}x_{i}>v\right) . \nonumber \\ \end{aligned}$$

(A.18)

In fact, by Lemma A.1 of [18], one can easily verify that $\mu$ assigns mass zero to the boundary of the set $\left( {\textbf{x}}\in [{\textbf{0}},\varvec{\infty }]:x_{k}>u_{+},\sum _{i=1}^{d}x_{i}>1+v\right)$. Relation (A.18) will be essentially used in the proof of (ii).

To complete the proof of (i), we take $u=-\infty$ and $v=1$ in (A.18) and obtain

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{P\left( S>x\right) }{{\overline{F}}(x)}=\mu \left( {\textbf{x}}\in [{\textbf{0}},\varvec{\infty } ]:\sum _{i=1}^{d}x_{i}>1\right) =\Xi . \end{aligned}$$

(A.19)

The limit result (A.19) implies that $\overline{F_{S}}\in \textrm{RV} _{-\alpha }$ and further that

$$\begin{aligned} F_{S}^{\leftarrow }({\tilde{q}})\sim \Xi ^{1/\alpha }F^{\leftarrow }({\tilde{q}} ), \end{aligned}$$

(A.20)

similar to Lemma A.2. Then by Theorem 4.1 of [19],

$$\begin{aligned} \rho (S)\sim \frac{\alpha }{p^{(p-1)/\alpha }(\alpha -p)^{1-p/\alpha }} \textrm{B}(\alpha -p,p)^{1/\alpha }\Xi ^{1/\alpha }F^{\leftarrow }({\tilde{q}} ). \end{aligned}$$

This proves (3.6).

(ii) The same as the proof of Theorem 3.1, we formulate the proof into two parts: $p=1$ and $p>1$.

$\underline{{\hbox {Step 1:\ Assume}} p=1.}$

For this case, ${\tilde{q}}=q$. We start from (2.5) and denote $s=F_{S}^{\leftarrow }(q)$, which tends to $\infty$ as $q\uparrow 1$. We expand

$$\begin{aligned}{} & {} \Lambda \left( X_{k},S\right) \nonumber \\{} & {} \quad =\frac{s}{1-q}\left( \left( \int _{0}^{\delta }+\int _{\delta }^{\infty }\right) P\left( X_{k}>ys,S>s\right) \textrm{d}y-\int _{0}^{\infty }P\left( X_{k}<-ys,S>s\right) \textrm{d}y\right) \nonumber \\{} & {} \quad =\frac{s}{1-q}\left( I_{1}+I_{2}-I_{3}\right) . \end{aligned}$$

(A.21)

with $\delta >0$ introduced to be arbitrarily fixed and small. For $I_{1}$, by (A.19),

$$\begin{aligned} I_{1}\le \delta P\left( S>s\right) \sim \delta \Xi {\overline{F}}(s). \end{aligned}$$

(A.22)

For $I_{2}$, by (A.18) with $v=1$, subject to an dominated convergence argument based on Lemma A.3, we obtain

$$\begin{aligned} \lim _{s\rightarrow \infty }\frac{I_{2}}{{\overline{F}}(s)}= & {} \int _{\delta }^{\infty }\lim _{s\rightarrow \infty }\frac{P\left( X_{k}>ys,S>s\right) }{ {\overline{F}}(s)}\textrm{d}y \nonumber \\= & {} \int _{\delta }^{\infty }\mu \left( {\textbf{x}}\in [{\textbf{0}}, \varvec{\infty }]:s_{k}>y,\sum _{i=1}^{d}s_{i}>1\right) \textrm{d}y. \end{aligned}$$

(A.23)

For $I_{3}$, following Remark 2.1, we have

$$\begin{aligned} \lim _{s\rightarrow \infty }\frac{I_{3}}{{\overline{F}}(s)}=0. \end{aligned}$$

(A.24)

For a direct verification, we derive

$$\begin{aligned} I_{3}\le & {} \int _{0}^{\delta }P\left( S>s\right) \textrm{d}y+\int _{\delta }^{\infty }P\left( X_{k}<-ys\right) \textrm{d}y \\= & {} \left( 1+o(1)\right) \delta \Xi {\overline{F}}(s)+o(1)\int _{\delta }^{\infty }P\left( X_{k}>ys\right) \textrm{d}y, \end{aligned}$$

where the last step is due to (A.19) and (3.5). For the integral $\int _{\delta }^{\infty }P\left( X_{k}>ys\right) \textrm{d}y$, by the dominated convergence theorem justified by Lemma A.1, we have

$$\begin{aligned} \lim _{s\rightarrow \infty }\frac{1}{{\overline{F}}(s)}\int _{\delta }^{\infty }P\left( X_{k}>ys\right) \textrm{d}y=\int _{\delta }^{\infty }\lim _{s\rightarrow \infty }\frac{P\left( X_{k}>ys\right) }{{\overline{F}}(s)} \textrm{d}y=\theta _{k}\int _{\delta }^{\infty }y^{-\alpha }\textrm{d}y. \end{aligned}$$

This verifies (A.24). Substituting (A.22)–(A.24) into (A.21) and bearing in mind the arbitrariness of $\delta$, we arrive at

$$\begin{aligned} \Lambda (X_{k},S)\sim & {} \frac{s{\overline{F}}(s)}{1-q}\int _{0}^{\infty }\mu \left( {\textbf{x}}\in [{\textbf{0}},\varvec{\infty } ]:s_{k}>y,\sum _{i=1}^{d}s_{i}>1\right) \textrm{d}y \\\sim & {} {F}^{\leftarrow }(q)\Xi ^{1/\alpha -1}\int _{0}^{\infty }\mu \left( {\textbf{x}}\in [{\textbf{0}},\varvec{\infty }]:s_{k}>y, \sum _{i=1}^{d}s_{i}>1\right) \textrm{d}y, \end{aligned}$$

where the last step is due to (A.20). This proves (3.7) for $p=1$.

$\underline{{\hbox {Step 2:\ Assume}} p>1.}$

We start from (2.6) in which $s_{*}$ is the unique solution to equation (2.4). By Lemma 2.2 of [19], as $q\uparrow 1$ we have $s_{*}\uparrow \infty$, and by relation (4.4) of [19],

$$\begin{aligned} s_{*}\sim \frac{(\alpha -p)^{p/\alpha }}{p^{(p-1)/\alpha }}\textrm{B} (\alpha -p,p)^{1/\alpha }\Xi ^{1/\alpha }F^{\leftarrow }({\tilde{q}}). \end{aligned}$$

As in the corresponding proof of Theorem 3.1, we need to explicitize the asymptotics for both the numerator and the denominator of (2.6). For the numerator, we still follow the expansion (A.14). As in step 1, by virtue of Lemma A.3 and relation (A.18), we can verify the applicability of the dominated convergence theorem in the following derivations, but we skip details:

$$\begin{aligned} \lim _{s*\rightarrow \infty }\frac{I_{1}}{{\overline{F}}(s_{*})}= & {} \iint \limits _{(y,z)\in (\delta ,\infty )\times (0,\infty )}\lim _{s*\rightarrow \infty }\frac{P\left( X_{k}>s_{*}y,S>s_{*}(1+z)\right) }{ {\overline{F}}(s_{*})}\textrm{d}y\textrm{d}z^{p-1} \\= & {} \iint \limits _{(y,z)\in (\delta ,\infty )\times (0,\infty )}\mu \left( {\textbf{x}}\in [{\textbf{0}},\varvec{\infty }]:x_{k}>y, \sum _{i=1}^{d}x_{i}>1+z\right) \textrm{d}y\textrm{d}z^{p-1} \end{aligned}$$

and

$$\begin{aligned} \lim _{s*\rightarrow \infty }\frac{I_{2}}{{\overline{F}}(s_{*})} =\iint \limits _{(y,z)\in (\delta ,\infty )\times (0,\infty )}\frac{P\left( X_{k}<-s_{*}y,S>s_{*}(1+z)\right) }{{\overline{F}}(s_{*})}\textrm{d }y\textrm{d}z^{p-1}=0. \end{aligned}$$

We deal with $I_{3}$ as follows:

$$\begin{aligned} \frac{\left| I_{3}\right| }{{\overline{F}}(s_{*})}\le & {} \iint \limits _{(y,z)\in (0,\delta )\times (0,\infty )}\frac{P\left( \left| X_{k}\right|>s_{*}y,S>s_{*}(1+z)\right) }{\overline{F }(s_{*})}\textrm{d}y\textrm{d}z^{p-1} \\\le & {} \delta \int _{0}^{\infty }\frac{P\left( S>s_{*}(1+z)\right) }{ {\overline{F}}(s_{*})}\textrm{d}z^{p-1} \\\rightarrow & {} \delta \Xi \int _{0}^{\infty }(1+z)^{-\alpha }\textrm{d}z^{p-1}. \end{aligned}$$

Since $\delta >0$ can be arbitrary small, it follows that

$$\begin{aligned}{} & {} E\left[ X_{k}\left( S-s_{*}\right) _{+}^{p-1}\right] \nonumber \\{} & {} \quad \sim s_{*}^{p}{\overline{F}}(s_{*})\iint \limits _{(y,z)\in (0,\infty )^{2}}\mu \left( {\textbf{x}}\in [{\textbf{0}},\varvec{\infty } ]:x_{k}>y,\sum _{i=1}^{d}x_{i}>1+z\right) \textrm{d}y\textrm{d}z^{p-1}. \nonumber \\ \end{aligned}$$

(A.25)

For the denominator of (2.6), we derive

$$\begin{aligned} E\left[ \left( S-s_{*}\right) _{+}^{p-1}\right]= & {} s_{*}^{p-1}\int _{0}^{\infty }P\left( S>s_{*}(1+z)\right) \textrm{d}z^{p-1} \nonumber \\\sim & {} s_{*}^{p-1}{\overline{F}}(s_{*})\Xi \int _{0}^{\infty }(1+z)^{-\alpha }\textrm{d}z^{p-1} \nonumber \\= & {} s_{*}^{p-1}{\overline{F}}(s_{*})\Xi (p-1)\textrm{B}(\alpha -p+1,p-1), \end{aligned}$$

(A.26)

where the second last step is subject to a dominated convergence argument justified by Lemma A.1 and relation (A.19). Substituting (A.25)–(A.26) into (2.6), we obtain (3.7) for $p>1$.