Aggregation-robustness and model uncertainty of regulatory risk measures

Abstract

Research related to aggregation, robustness and model uncertainty of regulatory risk measures, for instance, value-at-risk (VaR) and expected shortfall (ES), is of fundamental importance within quantitative risk management. In risk aggregation, marginal risks and their dependence structure are often modelled separately, leading to uncertainty arising at the level of a joint model. In this paper, we introduce a notion of qualitative robustness for risk measures, concerning the sensitivity of a risk measure to the uncertainty of dependence in risk aggregation. It turns out that coherent risk measures, such as ES, are more robust than VaR according to the new notion of robustness. We also give approximations and inequalities for aggregation and diversification of VaR under dependence uncertainty, and derive an asymptotic equivalence for worst-case VaR and ES under general conditions. We obtain that for a portfolio of a large number of risks, VaR generally has a larger uncertainty spread compared to ES. The results warn that unjustified diversification arguments for VaR used in risk management need to be taken with much care, and they potentially support the use of ES in risk aggregation. This in particular reflects on the discussions in the recent consultative documents by the Basel Committee on Banking Supervision.

Appendix: Proofs

1.1 A.1 A useful lemma

Before presenting the main proofs, we first state a lemma that is essential in proving the main results in Sects. 3 and 4 in this paper. Recall the definitions of the essential supremum and the essential infimum of random variables: for any random variable $S$,

$$\begin{aligned} \mathrm {ess}\ \mathrm {sup}\mbox{ }S =&\sup\{t: \mathbb {P}[S\leqslant t]< 1\},\\ \mathrm {ess}\ \mathrm {inf}\mbox{ }S =&\inf\{t:\mathbb {P}[S\leqslant t]>0\}. \end{aligned}$$

We denote $S_{n}=X_{1}+\cdots+X_{n}$ in the following. We remind the reader that this $S_{n}$ is different from the symbolic one in the notation of $\overline {\mathrm {VaR}}_{p}(S_{n})$. We hope this will not lead to notational confusion.

Lemma A.1

Suppose that $F_{i},i\in \mathbb {N}$, is a sequence of distributions on $[0,1]$. Then there exist $X_{i}\sim F_{i}$, $i\in \mathbb {N}$, such that for each $n\in \mathbb {N}$,

$$ \mathrm {ess}\ \mathrm {sup}\mbox{ }S_{n}-\mathrm {ess}\ \mathrm {inf}\mbox{ }S_{n}\leqslant1.$$

Proof

We first show that if $X$ and $Y$ are countermonotonic and both take values in $[0,1]$, then $\mathrm {ess}\ \mathrm {sup}\mbox{ }(X+Y)-\mathrm {ess}\ \mathrm {inf}\mbox{ }(X+Y)\leqslant1$. Since $X$ and $Y$ are countermonotonic, there exists $U\sim\mathrm{U}[0,1]$ such that $X=F^{-1}(U)$ and $Y=G^{-1}(1-U)$, where $F$ and $G$ are the distributions of $X$ and $Y$, respectively. For $u,v \in(0,1)$, one of $F^{-1}(u)-F^{-1}(v)$ and $G^{-1}(1-u)-G^{-1}(1-v)$ is nonpositive. Hence,

$$\begin{aligned} &F^{-1}(u)+G^{-1}(1-u)-\big(F^{-1}(v)+G^{-1}(1-v)\big)\\ &=\big(F^{-1}(u)-F^{-1}(v)\big)+\big(G^{-1}(1-u)-G^{-1}(1-v)\big)\\ &\leqslant \max\{F^{-1}(u)-F^{-1}(v),G^{-1}(1-u)-G^{-1}(1-v)\}\\&\leqslant1. \end{aligned}$$

Thus,

$$\begin{aligned} & \mathrm {ess}\ \mathrm {sup}\mbox{ }(X+Y)-\mathrm {ess}\ \mathrm {inf}\mbox{ }(X+Y)\\ &=\sup_{u\in(0,1)}\{F^{-1}(u)+G^{-1}(1-u)\}-\inf_{v\in(0,1)}\{ F^{-1}(v)+G^{-1}(1-v)\}\leqslant1. \end{aligned}$$

Let $X_{1} \sim F_{1}$. For $k\geqslant 2$, choose $X_{k}$ to be countermonotonic with $S_{k-1}$. Since $\mathrm {ess}\ \mathrm {sup}\mbox{ }(X_{1})-\mathrm {ess}\ \mathrm {inf}\mbox{ }(X_{1})\leqslant1$, by induction we get that

$$ \mathrm {ess}\ \mathrm {sup}\mbox{ }(S_{k})-\mathrm {ess}\ \mathrm {inf}\mbox{ }(S_{k})=\mathrm {ess}\ \mathrm {sup}\mbox{ }(S_{k-1}+X_{k})-\mathrm {ess}\ \mathrm {inf}\mbox{ }(S_{k-1}+X_{k})\leqslant1 $$

for all $k\geqslant 2$. □

Remark A.2

Lemma A.1 is of independent interest in the theory of negative dependence. Indeed, it shows that an extremely negatively dependent sequence always exists for uniformly bounded marginal distributions. The definition of and details on extremely negative dependence can be found in Wang and Wang [38]. In the latter paper, it was shown that an extremely negatively dependent sequence always exists for identical marginal $L_{1}$-distributions. Lemma A.1, as a new contribution, confirms that the same statement holds for inhomogeneous marginal distributions if we assume uniform boundedness.

The following useful corollary is directly implied by Lemma A.1.

Corollary A.3

Suppose that $F_{i},i\in \mathbb {N}$, is a sequence of distributions with bounded support. Then there exist $X_{i}\sim F_{i}$, $i\in \mathbb {N}$, such that for each $n\in \mathbb {N}$,

$$|S_{n}-\mathbb {E}[S_{n}]|\leqslant L_{n}.$$

where $L_{n}$ is the largest length of the support of $F_{i}, i=1,\dots,n$, that is,

$$ L_{n}=\max\{\mathrm {ess}\ \mathrm {sup}\mbox{ }X_{i}-\mathrm {ess}\ \mathrm {inf}\mbox{ }X_{i}:X_{i}\sim F_{i}, i=1,\dots,n\}. $$

1.2 A.2 Proof of Theorem 2.3

Proof

Suppose $\rho$ is a coherent distortion risk measure with distortion function $h$. Since $h$ is increasing and convex on (0,1), its has a left derivative on $(0,1)$, denoted as

$$ \delta(t):=\lim_{x\rightarrow0+} \frac{h(t)-h(t-x)}{x}, \quad t\in(0,1). $$

It follows from (2.3) that $\rho(X)=\int_{0}^{1} \mathrm {VaR}_{t}(X) \mathrm{d} h(t)= \int_{0}^{1} \mathrm {VaR}_{t}(X)\delta(t)\mathrm{d}t$. Note that since $\mathfrak{S}_{n}$ is compatible with a coherent risk measure $\rho$, we have that $\mathbb {E}[|X_{i}|]<\infty$, $X_{i}\sim F_{i}$, $i=1,\dots,n$. For $q \in(1/2,1)$, define

$$ \tilde{\rho}_{q}(X)=\frac{1}{1-h(q)}\int_{q}^{1} \mathrm {VaR}_{t}(X)\delta (t)\mathrm{d} t,\quad X \in \mathcal {X}_{0}. $$

We can easily check that $\tilde{\rho}_{q}$ is also a coherent distortion risk measure.

For any $S\in\mathfrak{S}_{n}(F_{1},\dots,F_{n})$, write $S=X_{1}+\cdots+X_{n}$, where $X_{i}\sim F_{i}$, $i=1,\dots,n$. For $q\in(1/2,1)$, we have that

$$\begin{aligned} \left|\rho(S)-\int_{1-q}^{q} \mathrm {VaR}_{t}(S)\delta(t)\mathrm{d}t\right |&=\left|\int _{0}^{1-q} \mathrm {VaR}_{t}(S)\delta(t)\mathrm{d}t+\int_{q}^{1} \mathrm {VaR}_{t}(S)\delta (t)\mathrm{d}t\right |\\ &\leqslant\left|\int_{0}^{1-q} \mathrm {VaR}_{t}(S)\delta(t)\mathrm{d}t\right |+\left|\big(1-h(q)\big)\tilde{\rho}_{q}(S)\right|\\ &\leqslant\delta(1-q)\int_{0}^{1-q} |\mathrm {VaR}_{t}(S)|\mathrm{d}t+\left |\big(1-h(q)\big)\tilde{\rho}_{q}(S)\right|. \end{aligned}$$

Note that

$$\begin{aligned} \left|\big(1-h(q)\big)\tilde{\rho}_{q}(S)\right |&\leqslant \bigg|\big(1-h(q)\big)\sum_{i=1}^{n} \tilde{\rho}_{q}(X_{i}) \bigg|=\bigg|\sum _{i=1}^{n} \int_{q}^{1} \mathrm {VaR}_{t}(X_{i})\delta(t) \mathrm{d}t\bigg|. \end{aligned}$$

On the other hand, by the comonotonic additivity of $\mathrm {VaR}_{t}$, $t\in (0,1)$, we have that

$$\begin{aligned} \int_{0}^{1-q} |\mathrm {VaR}_{t}(S)|\mathrm{d}t&=\int_{0}^{1-q} |\mathrm {VaR}_{t}(S\mathrm {I}_{\{S\geqslant 0\}})+\mathrm {VaR}_{t}(S\mathrm {I}_{\{S< 0\}})|\mathrm{d}t\\ &\leqslant\int_{0}^{1-q} \mathrm {VaR}_{t}(S\mathrm {I}_{\{S\geqslant 0\}})\mathrm{d}t +\int _{0}^{1-q}\mathrm {VaR}_{1-t}(-S\mathrm {I}_{\{S< 0\}})\mathrm{d}t\\ &\leqslant\int_{0}^{1-q} \mathrm {VaR}_{t}(|S|)\mathrm{d}t +\int_{0}^{1-q}\mathrm {VaR}_{1-t}(|S|)\mathrm{d}t\\ &\leqslant2(1-q)\mathrm {ES}_{q}(|S|)\\ &\leqslant2 (1-q)\sum_{i=1}^{n} \mathrm {ES}_{q}(|X_{i}|)\\ &= 2\sum_{i=1}^{n} \int_{q}^{1} \mathrm {VaR}_{t}(|X_{i}|)\mathrm{d}t. \end{aligned}$$

Note that for $i=1,\dots,n$, $\rho(X_{i})<\infty$ implies that $\int _{q}^{1} \mathrm {VaR}_{t}(X_{i})\delta(t) \mathrm{d}t\rightarrow0$ as $q\rightarrow1$, and that $\mathbb {E}[|X_{i}|]<\infty$ implies that $\int_{q}^{1} \mathrm {VaR}_{t}(|X_{i}|)\mathrm{d} t\rightarrow0$ as $q\rightarrow1$. As a consequence, as $q\rightarrow1$,

$$ \eta(q):=\left|\rho(S)-\int_{1-q}^{q} \mathrm {VaR}_{t}(S)\delta(t)\mathrm {d}t\right |\longrightarrow0 $$

uniformly in $S\in\mathfrak{S}_{n}$. Therefore, for each $\varepsilon >0$, there exists $1/2< q<1$ such that $\eta(q)<\varepsilon /3$. By Theorem 1 of Cont et al. [10], the distortion risk measure

$$ \hat{\rho}_{q}(X):=\frac{1}{2q-1}\int_{1-q}^{q} \mathrm {VaR}_{t}(X)\delta (t)\mathrm{d} t,\quad X\in \mathcal {X}_{0} $$

is continuous at all distributions with respect to weak convergence. As a consequence, for fixed $q\in(1/2,1)$ and $S, S_{1}, S_{2},{\dots} \in \mathfrak{S}_{n}$ with $S_{k}\to S$ weakly as $k \to\infty$, we have that there exists $K_{0}\in \mathbb {N}$ such that for $k\geqslant K_{0}$, $|\hat{\rho}_{q}(S_{k})-\hat{\rho}_{q}(S)|<\varepsilon /3$. Therefore, as $k\to\infty$,

$$ |\rho(S_{k})-\rho(S)|\leqslant(2q-1)|\hat{\rho}_{q}(S_{k})-\hat{\rho}_{q}(S)|+2\eta (q)< \varepsilon . $$

Since $\varepsilon$ is arbitrary, we conclude that $\rho$ is aggregation-robust. □

1.3 A.3 Proof of Theorem 2.5

Proof

We first show that distortion risk measures with a continuous distortion function on $[0,1]$ are aggregation-robust. Since $\mathcal {X}=L^{\infty}$, we suppose for some $M>0$ that $|X_{i}|\leqslant M$ a.s., $X_{i}\sim F_{i}$, for all $i=1,2,\dots$ For $q\in(1/2,1)$, we have that

$$\begin{aligned} \eta(q):=&\,\left|\rho(S)-\int_{1-q}^{q} \mathrm {VaR}_{t}(S) \mathrm {d}h(t)\right|\\=&\,\left |\int_{0}^{1-q} \mathrm {VaR}_{t}(S) \mathrm{d}h(t)+\int_{q}^{1} \mathrm {VaR}_{t}(S) \mathrm{d}h(t)\right|\\ \leqslant&\, nMh(1-q)+nM\big(h(1)-h(q)\big)\longrightarrow0, \end{aligned}$$

uniformly in $S\in\mathfrak{S}_{n}$. The rest of the proof is similar to the proof of Theorem 2.3.

Now suppose that $h$ is discontinuous at $p\in(0,1)$. Using the same argument as in Example 2.2, we can see that $\rho$ is not aggregation-robust. The case where $h$ is discontinuous at $p=0$ or $p=1$ can be obtained with similar counterexamples. □

1.4 A.4 Proof of Theorem 3.1

We use the following lemma, where an alternative definition of VaR is used, namely

$$ \mathrm {VaR}^{*}_{p}(X)=\inf\{x\in \mathbb {R}:\mathbb {P}[X\leqslant x]>p \}, \quad p\in(0,1). $$

The following lemma is analogous to Lemma 4.3 of [7], with the continuity condition on the marginal distributions removed. In the following, we set $\mathfrak{S}_{n}=\mathfrak{S}_{n}(F_{1},\dots,F_{n})$.

Lemma A.4

For $p\in(0,1)$,

$$\begin{aligned} \sup_{S\in\mathfrak{S}_{n}}\mathrm {VaR}^{*}_{p}(S) =&\sup\{\mathrm {ess}\ \mathrm {inf}\mbox{ }S: S\in \mathfrak{S}_{n}(F_{p,1},\dots,F_{p,n})\}, \end{aligned}$$

(A.1)

$$\begin{aligned} \inf_{S\in\mathfrak{S}_{n}}\mathrm {VaR}_{p}(S) =&\inf\{\mathrm {ess}\ \mathrm {sup}\mbox{ }S: S\in\mathfrak{S}_{n}(F_{1}^{p},\dots,F_{n}^{p})\}, \end{aligned}$$

(A.2)

where $F_{p,i}$ is the distribution of $F_{i}^{-1}(p+(1-p)U)$, and $F_{i}^{p}$ is the distribution of $F^{-1}_{i}(pU)$, $i=1,\dots,n$, for a random variable $U$ uniformly distributed on $[0,1]$.

Proof

We only need to show (A.1), as (A.2) is symmetric to (A.1). First, we show that

$$ \sup_{S\in\mathfrak{S}_{n}}\mathrm {VaR}^{*}_{p}(S)\leqslant\sup\{\mathrm {ess}\ \mathrm {inf}\mbox{ }S: S\in \mathfrak{S}_{n}(F_{p,1},\dots,F_{p,n})\}=:a_{0}. $$

For any $T\in\mathfrak{S}_{n}$, denote its distribution by $F_{T}$. Let $U\sim\mathrm{U}[0,1]$ be such that $T=F^{-1}_{T}(U)$, and define $A_{0}=\{U\geqslant p\}$. Write $T=X_{1}+\cdots+X_{n}$ where $X_{i}\sim F_{i}$, $i=1,\dots,n$. Clearly, the conditional random variable $T|A_{0}=X_{1}|A_{0}+\cdots+X_{n}|A_{0}$ is dominated (in stochastic order) by some $S_{0}\in\mathfrak{S}_{n}(F_{p,1},\dots,F_{p,n})$ since each $X_{i}|A_{0}$ is dominated by some $\hat{X}_{i}\sim F_{p,i}$. This implies that $\mathrm {ess}\ \mathrm {sup}\mbox{ }T_{0}\leqslant a_{0}$, where $T_{0}$ is distributed as $T|A_{0}$. Therefore, $\mathrm {VaR}_{p}^{*}(T)=\mathrm {ess}\ \mathrm {sup}\mbox{ }T_{0}\leqslant a_{0}$.

Next we show that

$$ \sup_{S\in\mathfrak{S}_{n}}\mathrm {VaR}^{*}_{p}(S)\geqslant a_{0}. $$

Note that by Lemma 4.2 of Bernard et al. [7], there exists $S_{0}\in\mathfrak{S}_{n}(F_{p,1},\dots,F_{p,n})$ such that $\mathrm {ess}\ \mathrm {inf}\mbox{ }S_{0}=a_{0}$. Let $U_{0}$ be a $\mathrm{U}[0,1]$ random variable, independent of $S_{0}$. Write

$$ T_{1}=\sum_{i=1}^{n} F_{i}^{-1}(U_{0})\mathrm {I}_{\{U_{0}< p\}}+ S_{0}\mathrm {I}_{\{U_{0}\geqslant p\}}. $$

It is easy to check that $T_{1}\in\mathfrak{S}_{n}$. As a consequence, $\mathrm {VaR}_{p}^{*}(T_{1})\geqslant \mathrm {ess}\ \mathrm {sup}\mbox{ }S_{0}=a_{0}$. □

Proof of Theorem 3.1

We first show that for $p \in(0,1)$ and $q\in(p,1]$,

$$\begin{aligned} &\sup\{\mathrm {ess}\ \mathrm {inf}\mbox{ }S: S\in\mathfrak{S}_{n}(F_{p,1},\dots,F_{p,n})\} \\ &\geqslant \sum_{i=1}^{n} \mu_{p,q}^{(i)}-\max_{i=1,\dots,n}\big(F_{i}^{-1}(q)-F_{i}^{-1}(p)\big). \end{aligned}$$

(A.3)

Since the case when $F_{i}^{-1}(q)=\infty$ for some $i$ is trivial, we suppose that $F_{i}^{-1}(q)<\infty$ for all $i=1,\dots,n$.

Let $F^{(i)}_{p,q}$ be the distribution of $W_{i}=F_{i}^{-1}(p+(q-p)U)$ for $0< p< q\leqslant1$. By Corollary A.3, there exist random variables $X_{i}\sim F^{(i)}_{p,q}$, $i=1,\dots,n$, such that

$$ X_{1}+ \cdots+X_{n} \geqslant \sum_{i=1}^{n} \mu_{p,q}^{(i)}-\max_{i=1,\dots ,n}\big(F^{-1}(q)-F^{-1}(p)\big). $$

Let $Z_{i}$, $i=1,\dots,n$, be any random variables with distribution $F_{q,i}$, and let $C$ be a set independent of $X_{1},\dots ,X_{n},Z_{1},\dots ,Z_{n}$, for which $\mathbb {P}[C]=(q-p)/(1-p)$. Define $Y_{i}=X_{i}\mathrm {I}_{C}+Z_{i}(1-\mathrm {I}_{C})$ for $i=1,\dots,n$. It is straightforward to check that $Y_{i}$ has distribution $F_{p,i}$ and

$$ Y_{1}+\cdots+Y_{n}\geqslant X_{1}+\cdots+X_{n} \geqslant \sum_{i=1}^{n} \mu _{p,q}^{(i)}-\max _{i=1,\dots,n}\big(F_{i}^{-1}(q)-F_{i}^{-1}(p)\big). $$

Thus

$$ \mathrm {ess}\ \mathrm {inf}\mbox{ }(Y_{1}+\cdots+Y_{n})\geqslant \sum_{i=1}^{n} \mu_{p,q}^{(i)}-\max _{i=1,\dots ,n}\big(F_{i}^{-1}(q)-F_{i}^{-1}(p)\big), $$

and we obtain (A.3). Since $\mathrm {VaR}_{p}(X)\geqslant \mathrm {VaR}^{*}_{r}(X)$ for any $r< p$ and any random variable $X$, we have that

$$\begin{aligned} \overline {\mathrm {VaR}}_{p}(S_{n}) &\geqslant \lim_{r\rightarrow p-}\sup_{S\in\mathfrak{S}_{n}}\mathrm {VaR}^{*}_{r}(S)\\ &\geqslant \lim_{r\rightarrow p-}\bigg(\sum_{i=1}^{n} \mu_{r,q}^{(i)}-\max _{i=1,\dots,n}\big(F_{i}^{-1}(q)-F_{i}^{-1}(r)\big)\bigg)\\ &=\sum_{i=1}^{n} \mu_{p,q}^{(i)}-\max_{i=1,\dots,n}\big(F_{i}^{-1}(q)-F_{i}^{-1}(p)\big). \end{aligned}$$

Note that here we use the fact that $F_{i}^{-1}$ is left-continuous for each $i$. Now, we have (A.3), and with Lemma A.4, we obtain the first inequality in (3.5). On the other hand,

$$ \overline {\mathrm {VaR}}_{p}(S_{n})\leqslant\sup_{S\in\mathfrak{S}_{n}}\mathrm {VaR}^{*}_{p}(S)=\sup\{ \mathrm {ess}\ \mathrm {inf}\mbox{ }S: S\in\mathfrak{S}_{n}(F_{p,1},\ldots,F_{p,n})\}\leqslant\sum_{i=1}^{n}\mu _{p,1}^{(i)} $$

always holds. Thus we obtain (3.5). We can show (3.6) similarly. □

1.5 A.5 Proof of Theorem 3.3

Proof

First, let us assume that $\mathbb {E}[X_{i}]=0$ for all $i\in \mathbb {N}$. Note that

$$ \overline {\mathrm {ES}}_{p}(S_{n})=\sum_{i=1}^{n} \mathrm {ES}_{p}(X_{i})=\sum_{i=1}^{n}\mu^{(i)}_{p,1} $$

for $X_{i}\sim F_{i}$. We use (3.5) and take $q_{n}=1-n^{-1}$ for $n$ large enough such that $q_{n}>p$. By (3.10), we have $\sum_{i=1}^{n}\mu^{(i)}_{p,1}>0$ for large $n$.

Note that by (3.9), $\mathbb {E}[|X_{i}|^{k}]\leqslant M$ uniformly. Therefore, $[F_{i}^{-1}(t)]^{k}(1-t)\leqslant M$ for $t\in(0,1)$, and we have

$$ F^{-1}_{i}(t)\leqslant\left(\frac{M}{1-t}\right)^{1/k}, \quad t\in(0,1), i\in \mathbb {N}. $$

Note that for $X_{i}\sim F_{i}$,

$$\begin{aligned} \mu^{(i)}_{p,1}-\mu^{(i)}_{p,q_{n}}&=\frac{1}{1-p}\mathbb {E}\big[X_{i}\mathrm {I}_{\{X_{i}\geqslant F_{i}^{-1}(p)\}}\big]-\frac{1}{q_{n}-p}\mathbb {E}\big[X_{i}\mathrm {I}_{\{ F_{i}^{-1}(q_{n})\geqslant X_{i}\geqslant F_{i}^{-1}(p)\}}\big] \\ &\leqslant\frac{1}{1-p}\mathbb {E}\big[X_{i}\mathrm {I}_{\{X_{i}\geqslant F^{-1}(q_{n})\}}\big] \\&=\frac{1}{1-p}\int_{q_{n}}^{1} F_{i}^{-1}(t) \mathrm{d}t \\&\leqslant\frac{1}{1-p}\int_{q_{n}}^{1} \left(\frac{M}{1-t}\right)^{1/k} \mathrm{d}t \\&=\frac{1}{1-p}\frac{1}{1-1/k} M^{1/k} (1-q_{n})^{1-1/k}. \end{aligned}$$

As a consequence, we have

$$\begin{aligned} \sup_{S\in\mathfrak{S}_{n}}\mathrm {VaR}_{p}(S)&\geqslant \sum_{i=1}^{n}\mu ^{(i)}_{p,q_{n}}- \max_{i=1,\dots ,n}(F^{-1}_{i}(q_{n})-F^{-1}_{i}(p)) \\ &\geqslant \sum_{i=1}^{n}\mu^{(i)}_{p,1}-\sum_{i=1}^{n}(\mu ^{(i)}_{p,1}-\mu^{(i)}_{p,q_{n}})-\max_{i=1,\dots,n }F^{-1}_{i}(q_{n}) \\ &\geqslant \sum_{i=1}^{n}\mu^{(i)}_{p,1}-\sum_{i=1}^{n}\frac {1}{1-p}\frac{1}{1-1/k} M^{1/k} (1-q_{n})^{1-1/k} \\ &\phantom{=}-\bigg(\frac{M}{1-q_{n}}\bigg)^{1/k} \\ &= \sum_{i=1}^{n}\mu^{(i)}_{p,1}- \frac{1}{1-p}\frac {1}{1-1/k} M^{1/k} n^{1/k}-M^{1/k}n^{1/k} \\ &=\sum_{i=1}^{n}\mu^{(i)}_{p,1}-O(n^{1/k}). \end{aligned}$$

(A.4)

By (3.10), it follows that

$$\begin{aligned} 1\geqslant \frac{\overline {\mathrm {VaR}}_{p}(S_{n})}{\sum_{i=1}^{n}\mu^{(i)}_{p,1}}\geqslant 1-\frac {O(n^{1/k})}{\sum_{i=1}^{n}\mu^{(i)}_{p,1}} \longrightarrow 1\quad \mbox{as }n\rightarrow\infty, \end{aligned}$$

hence we obtain (3.13).

Now for the case that $\mathbb {E}[X_{i}]\ne0$ for some $i\in \mathbb {N}$, we denote by $F_{i}^{*}$ the distribution of $X_{i}-\mathbb {E}[X_{i}]$ and set

$$ \mathfrak{S}_{n}^{*}=\{Y_{1}+\cdots+Y_{n}: Y_{i}\sim F_{i}^{*}, i=1,\dots,n\}. $$

Then by (A.4), with $\mathfrak{S}_{n}$ replaced by $\mathfrak{S}_{n}^{*}$, we have

$$\begin{aligned} \sup_{S\in\mathfrak{S}_{n} } \mathrm {VaR}_{p}(S)&=\sup_{S\in\mathfrak{S}^{*}_{n} } \mathrm {VaR}_{p}(S)+\sum_{i=1}^{n} \mathbb {E}[X_{i}]\\&=\sum_{i=1}^{n}(\mu^{(i)}_{p,1}-\mathbb {E}[X_{i}])-O(n^{1/k})+\sum_{i=1}^{n} \mathbb {E}[X_{i}]\\ &=\sum_{i=1}^{n}\mu ^{(i)}_{p,1}-O(n^{1/k}). \end{aligned}$$

Thus, (A.4) still holds for $\mathfrak{S}_{n}$ in the case $\mathbb {E}[X_{i}]\ne0$ for some $i$.

When (3.11) holds, by (A.4), we have that

$$\begin{aligned} 1\geqslant \frac{\overline {\mathrm {VaR}}_{p}(S_{n})}{\sum_{i=1}^{n}\mu^{(i)}_{p,1}}\geqslant 1-\frac {(\frac {1}{1-p}\frac{k}{k-1}+1) M^{1/k}(n^{1/k})}{\sum_{i=1}^{n}\mu ^{(i)}_{p,1}}\geqslant 1-Cn^{-1+1/k}, \end{aligned}$$

for $n$ sufficiently large. This leads to (3.14) and completes the proof of the theorem. □

1.6 A.6 Proof of Theorem 4.1

Proof

(i) Let us introduce $a_{n}=\overline {\mathrm {VaR}}_{q}(S_{n})$, $b_{n}=\underline {\mathrm {VaR}}_{q}(S_{n})$, $c_{n}=\overline {\mathrm {ES}}_{q}(S_{n})$ and $d_{n}=\underline {\mathrm {LES}}_{q}(S_{n})$. We have that

$$ \liminf_{n\rightarrow\infty}\frac{a_{n}-b_{n}}{c_{n}-d_{n}}=\liminf _{n\rightarrow\infty}\frac{a_{n}/c_{n}-b_{n}/c_{n}}{1-d_{n}/c_{n}}=\liminf _{n\rightarrow\infty}\frac{a_{n}/c_{n}-(b_{n}/d_{n})(d_{n}/c_{n})}{1-d_{n}/c_{n}}. $$

Note that by (4.1), we have $\limsup_{n\rightarrow\infty }d_{n}/c_{n}<1$. Further, by Theorem 3.3 and Corollary 3.4, we have $a_{n}/c_{n}\to1$ and $b_{n}/d_{n}\to1$. As a consequence,

$$ \liminf_{n\rightarrow\infty}\frac{a_{n}-b_{n}}{c_{n}-d_{n}}\geqslant 1. $$

Since $c_{n}\geqslant a_{n}\geqslant b_{n}\geqslant d_{n}$, we have that

$$ \frac{a_{n}-b_{n}}{c_{n}-d_{n}}\leqslant1\quad \Longrightarrow \quad \lim _{n\rightarrow\infty}\frac{a_{n}-b_{n}}{c_{n}-d_{n}}=1. $$

Write

$$ \frac{\overline {\mathrm {ES}}_{q}(S_{n})-\underline {\mathrm {LES}}_{q}(S_{n})}{\overline{\mathrm {ES}}_{p}(S_{n})-\underline {\mathrm {ES}}_{p}(S_{n})} =\frac {\overline{\mathrm {VaR}}_{q}(S_{n})-\underline{\mathrm {VaR}}_{q}(S_{n})}{\overline{\mathrm {ES}}_{p}(S_{n})-\underline {\mathrm {ES}}_{p}(S_{n})} \frac{a_{n}-b_{n}}{c_{n}-d_{n}}, $$

and we obtain the first equality in (4.2). The rest of (4.2) follows by noting that $\mathrm {ES}_{q}(X)\geqslant \mathrm {ES}_{p}(X)\geqslant \mathbb {E}[X]\geqslant \mathrm {LES}_{q}(X)$ for any random variable $X$ and any $0< p\leqslant q<1$.

(ii) This can be obtained from part (i) by noting that (3.9), (3.15) and (4.1) are all satisfied by the distribution of $X+c$, where $c$ is some constant. □