# Estimating the Conditional Tail Expectation in the Case of Heavy-Tailed Losses

Authors: Abdelhakim Necir; Abdelaziz Rassoul; Ričardas Zitikis
Journal: Journal of Probability and Statistics
 (2010)

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## Abstract

The conditional tail expectation (CTE) is an important actuarial risk measure and a useful tool in financial risk assessment. Under the classical assumption that the second moment of the loss variable is finite, the asymptotic normality of the nonparametric CTE estimator has already been established in the literature. The noted result, however, is not applicable when the loss variable follows any distribution with infinite second moment, which is a frequent situation in practice. With a help of extreme-value methodology, in this paper, we offer a solution to the problem by suggesting a new CTE estimator, which is applicable when losses have finite means but infinite variances.


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## Body

## 1. Introduction

One of the most important actuarial risk measures is the conditional tail expectation (CTE) (see, e.g., [1]), which is the average amount of loss given that the loss exceeds a specified quantile. Hence, the CTE provides a measure of the capital needed due to the exposure to the loss, and thus serves as a risk measure. Not surprisingly, therefore, the CTE continues to receive increased attention in the actuarial and financial literature, where we also find its numerous extensions and generalizations (see, e.g., [2–8], and references therein). We next present basic notation and definitions.LetX be a loss random variable with cumulative distribution function (cdf) F. Usually, the cdf F is assumed to be continuous and defined on the entire real line, with negative loss interpreted as gain. We also assume the continuity of F throughout the present paper. The CTE of the risk or loss X is then defined, for every t∈(0,1), by(1.1)CTEF(t)=E[X∣X>Q(t)],
where Q(t)=inf{x:F(x)≥t} is the quantile function corresponding to the cdf F. Since the cdf F is continuous, we easily check that(1.2)CTEF(t)=11-t∫t1Q(s)ds.Naturally, the CTE is unknown since the cdfF is unknown. Hence, it is desirable to establish statistical inferential results such as confidence intervals for CTEF(t) with specified confidence levels and margins of error. We shall next show how to accomplish this task, initially assuming the classical moment assumption E[X2]<∞. Namely, suppose that we have independent random variables X1, X2, … , each with the cdf F, and let X1:n<⋯<Xn:n denote the order statistics of X1,…,Xn. It is natural to define an empirical estimator of CTEF(t) by the formula(1.3)CTÊn(t)=11-t∫t1Qn(s)ds,
where Qn(s) is the empirical quantile function, which is equal to the ith order statistic Xi:n for all s∈((i-1)/n,i/n], and for all i=1,…,n. The asymptotic behavior of the estimator CTÊn(t) has been studied by Brazauskas et al. [9], and we next formulate their most relevant result for our paper as a theorem.Theorem 1.1.
Assume thatE[X2]<∞. Then for every t∈(0,1), we have the asymptotic normality statement
(1.4)n(CTÊn(t)-CTEF(t))(1-t)→d𝒩(0,σ2(t)),
when n→∞, where the asymptotic variance σ2(t) is given by the formula
(1.5)σ2(t)=∫t1∫t1(min(x,y)-xy)dQ(x)dQ(y).The assumptionE[X2]<∞ is, however, quite restrictive as the following example shows. Suppose that F is the Pareto cdf with index γ>0, that is, 1-F(x)=x-1/γ for all x≥1. Let us focus on the case γ<1, because when γ≥1, then CTEF(t)=+∞ for every t∈(0,1). Theorem 1.1 covers only the values γ∈(0,1/2) in view of the assumption E[X2]<∞. When γ∈[1/2,1), we have E[X2]=∞ but, nevertheless, CTEF(t) is well defined and finite since E[X]<∞. Analogous remarks hold for other distributions with Pareto-like tails, an we shall indeed work with such general distributions in this paper.Namely, recall that the cdfF is regularly varying at infinity with index (-1/γ)<0 if(1.6)limt→∞1-F(tx)1-F(t)=x-1/γ
for every x>0. This class includes a number of popular distributions such as Pareto, generalized Pareto, Burr, Fréchet, Student, and so forth, which are known to be appropriate models for fitting large insurance claims, fluctuations of prices, log-returns, and so forth (see, e.g., [10]). In the remainder of this paper, therefore, we restrict ourselves to this class of distributions. For more information on the topic and, generally, on extreme value models and their manifold applications, we refer to the monographs by Beirlant et al. [11], Castillo et al. [12], de Haan and Ferreira [13], Resnick [14].The rest of the paper is organized as follows. In Section2 we construct an alternative, called “new”, CTE estimator by utilizing an extreme value approach. In Section 3 we establish the asymptotic normality of the new CTE estimator and illustrate its performance with a little simulation study. The main result, which is Theorem 3.1 stated in Section 3, is proved in Section 4.


## 2. Construction of a New CTE Estimator

We have already noted that the “old” estimatorCTÊn(t) does not yield the asymptotic normality (in the classical sense) beyond the condition E[X2]<∞. Indeed, this follows by setting t=0, in which case CTÊn(t) becomes the sample mean of X1,…,Xn, and thus the asymptotic normality of CTÊn(0) is equivalent to the classical Central Limit Theorem (CLT). Similar arguments show that the finite second moment is necessary for having the asymptotic normality (in the classical sense) of CTÊn(t) at any fixed “level” t∈(0,1). Indeed, note that the asymptotic variance σ2(t) in Theorem 1.1 is finite only if E[X2]<∞.For this reason, we next construct an alternative CTE estimator, which takes into account different asymptotic properties of moderate and high quantiles in the case of heavy-tailed distributions. Hence, from now on we assume thatγ∈(1/2,1). Before indulging ourselves into construction details, we first formulate the new CTE estimator:(2.1)CTẼn(t)=11-t∫t1-k/nQn(s)ds+kXn-k,nn(1-t)(1-γ̂),
where we use the simplest yet useful and powerful Hill's [15] estimator(2.2)γ̂n=1k∑i=1klogXn-i+1:n-logXn-k:n
of the tail index γ∈(1/2,1). Integers k=kn∈{1,…,n} are such that k→∞ and k/n→0 when n→∞, and we note at the outset that their choices present a challenging task. In Figures 1 and 2, we illustrate the performance of the new estimator CTẼn(t) with respect to the sample size n≥1, with the integers k=kn chosen according to the method proposed by Cheng and Peng [16]. Note that when t increases through the values 0.25, 0.50, 0.75, and 0.90 (panels (a)–(d), resp.), the vertical axes of the panels also increase, which reflects the fact that the larger the t gets, the more erratic the “new” and “old” estimators become. Note also that the empirical (i.e., “old”) estimator underestimates the theoretical CTEF(t), which is a well known phenomenon (see [17]).Values of the CTE estimatorCTẼn(t) (vertical axis) versus sample sizes n (horizontal axis) evaluated at the levels t=0.25, t=0.50, t=0.75, and t=0.90 (panels (a)–(d), resp.) in the Pareto case with the tail index γ=2/3.
(a)(b)(c)(d)Values of the CTE estimatorCTẼn(t) (vertical axis) versus sample sizes n (horizontal axis) evaluated at the levels t=0.25, t=0.50, t=0.75, and t=0.90 (panels (a)–(d), resp.) in the Pareto case with the tail index γ=3/4.
(a)(b)(c)(d)We have based the construction ofCTẼn(t) on the recognition that one should estimate moderate and high quantiles differently when the underlying distribution is heavy-tailed. For this, we first recall that the high quantile qs is, by definition, equal to Q(1-s) for sufficiently small s. For an estimation theory of high quantiles in the case of heavy-tailed distributions we refer to, for example, Weissman [18], Dekkers and de Haan [19], Matthys and Beirlant [20], Gomes et al. [21], and references therein. We shall use the Weissman estimator(2.3)q̃s=(kn)γ̂Xn-k:ns-γ̂,0<s<kn,
of the high quantile qs. Then we write CTEF(t) as the sum CTE1,n(t)+CTE2,n(t) with the two summands defined together with their respective empirical estimators CTẼ1,n(t) and CTẼ2,n(t) as follows:(2.4)CTE1,n(t)=11-t∫t1-k/nQ(s)ds≈11-t∫t1-k/nQn(s)ds=CTẼ1,n(t),CTE2,n(t)=11-t∫1-k/n1Q(s)ds≈11-t∫1-k/n1q̃1-sds=CTẼ2,n(t).
Simple integration gives the formula(2.5)CTẼ2,n(t)=kXn-k,nn(1-γ̂)(1-t).
Consequently, the sum CTẼ1,n(t)+CTẼ2,n(t) is an estimator of CTEF(t), and this is exactly the estimator CTẼn(t) introduced above. We shall investigate asymptotic normality of the new estimator in the next section, accompanied with an illustrative simulation study.


## 3. Main Theorem and Its Practical Implementation

We start this section by noting that Hill's estimatorγ̂n has been thoroughly studied, improved, and generalized in the literature. For example, weak consistency of γ̂n has been established by Mason [22] assuming only that the underlying distribution is regularly varying at infinity. Asymptotic normality of γ̂ has been investigated under various conditions by a number of researchers, including Csörgő and Mason [23], Beirlant and Teugels [24], Dekkers et al. [25], see also references therein.The main theoretical result of this paper, which is Theorem3.1 below, establishes asymptotic normality of the new CTE estimator CTẼn(t). To formulate the theorem, we need to introduce an assumption that ensures the asymptotic normality of Hill's estimator γ̂n. Namely, the cdf F satisfies the generalized second-order regular variation condition with second-order parameter ρ≤0 (see [26, 27]) if there exists a function a(t) which does not change its sign in a neighbourhood of infinity and is such that, for every x>0,(3.1)limt→∞1a(t)(1-F(tx)1-F(t)-x-1/γ)=x-1/γxρ/γ-1ρ/γ.
When ρ=0, then the ratio on the right-hand side of (3.1) is interpreted as logx. For statistical inference concerning the second-order parameter ρ, we refer, for example, to Peng and Qi [28], Gomes et al.[21], Gomes and Pestana [29]. Furthermore, in the formulation of Theorem 3.1, we shall also use the function A(z)=γ2a(𝕌(z)), where 𝕌(z)=Q(1-1/z).Theorem 3.1.
Assume that the cdfF satisfies condition (3.1) with γ∈(1/2,1). Then for any sequence of integers k=kn→∞ such that k/n→0 and k1/2A(n/k)→0 when n→∞, we have that for any fixed t∈(0,1),
(3.2)n(CTẼn(t)-CTEF(t))(1-t)(k/n)1/2Xn-k:n→d𝒩(0,σγ2),
where the asymptotic variance σγ2 is given by the formula
(3.3)σγ2=γ4(1-γ)4(2γ-1).The asymptotic varianceσγ2 does not depend on t, unlike the variance in Theorem 1.1. This is not surprising because the heaviness of the right-most tail of F makes the asymptotic behaviour of ∫t1(Qn(s)-Q(s))ds “heavier” than the classical CLT-type behaviour of ∫0t(Qn(s)-Q(s))ds, for any fixed t. This in turn implies that under the conditions of Theorem 3.1, statement (3.2) is equivalent to the same statement in the case t=0. The latter statement concerns estimating the mean E[X] of a heavy-tailed distribution. Therefore, we can view Theorem 3.1 as a consequence of Peng [30], and at the same time we can view results of Peng [30] as a consequence of Theorem 3.1 by setting t=0 in it. Despite this equivalence, in Section 4 we give a proof of Theorem 3.1 for the sake of completeness. Our proof, however, is crucially based on a powerful technique called the Vervaat process (see [31–33], for details and references).To discuss practical implementation of Theorem3.1, we first fix a significance level ς∈(0,1) and use the classical notation zς/2 for the (1-ς/2)-quantile of the standard normal distribution 𝒩(0,1). Given a realization of the random variables X1,…,Xn (e.g., claim amounts), which follow a cdf F satisfying the conditions of Theorem 3.1, we construct a level 1-ς confidence interval for CTEF(t) as follows. First, we choose an appropriate number k of extreme values. Since Hill's estimator has in general a substantial variance for small k and a considerable bias for large k, we search for a k that balances between the two shortcomings, which is indeed a well-known hurdle when estimating the tail index. To resolve this issue, several procedures have been suggested in the literature, and we refer to, for example, Dekkers and de Haan [34], Drees and Kaufmann [35], Danielsson et al. [36], Cheng and Peng [16], Neves and Fraga Alves [37], Gomes et al. [38], and references therein. In our current study, we employ the method of Cheng and Peng [16] for an appropriate value k* of the “parameter” k. Having computed Hill's estimator and consequently determined Xn-k*:n, we then compute the corresponding values of CTẼn(t) and σγ̂n2, and denote them by CTẼn*(t) and σγ̂n2*, respectively. Finally, using Theorem 3.1 we arrive at the following (1-ς)-confidence interval for CTEF(t):(3.4)CTẼn*(t)±zς/2(k*/n)1/2Xn-k*:nσγ̂n2*(1-t)n.To illustrate the performance of this confidence interval, we have carried out a small-scale simulation study based on the Pareto cdfF(x)=1-x-1/γ, x≥1, with the tail index γ set to 2/3 and 3/4, and the level t set to 0.75 and 0.90. We have generated 200 independent replicates of three samples of sizes n=1000, 2000, and 5000. For every simulated sample, we have obtained estimates CTẼn(t). Then we have calculated the arithmetic averages over the values from the 200 repetitions, with the absolute error (error) and root mean squared error (rmse) of the new estimator C̃n(t) reported in Table 1 (γ=2/3) and Table 2 (γ=3/4). In the tables, we have also reported 95%-confidence intervals (3.4) with their lower and upper bounds, coverage probabilities, and lengths.Table 1
Point estimatesCTẼn(t) and 95% confidence intervals for CTEF(t) when γ=2/3.
t=0.75CTEF(t)=7.005nk*CTẼn(t)errorrmseloweruppercoverlength10000546.8760.0450.3036.3567.3970.8391.04120001006.8310.0250.2316.4637.1990.8820.73650002197.1190.0160.1946.8817.3570.8950.476t=0.90CTEF(t)=12.533nk*CTẼn(t)errorrmseloweruppercoverlength100005412.7530.0170.53412.24113.2690.8471.028200010012.4870.0030.29412.13712.8380.8410.701500021912.4610.0050.23612.24612.6760.8870.430Table 2
Point estimatesCTẼn(t) and 95% confidence intervals for CTEF(t) when γ=3/4.
t=0.75CTEF(t)=9.719nk*CTẼn(t)errorrmseloweruppercoverlength10000519.5430.0180.5828.5899.5430.8540.95420001049.8080.0090.4669.15010.4660.8881.31650002229.7890.0070.4109.36310.2150.9150.852t=0.90CTEF(t)=18.494nk*CTẼn(t)errorrmseloweruppercoverlength100005118.1990.0150.98917.43718.9600.8741.523200010418.6960.0110.85818.05219.3400.8951.288500022218.5410.0020.79818.09218.9900.9250.898We note emphatically that the above coverage probabilities and lengths of confidence intervals can be improved by employing more precise but, naturally, considerably more complex estimators of the tail index. Such estimators are described in the monographs by Beirlant et al. [11], Castillo et al. [12], de Haan and Ferreira [13], and Resnick [14]. Since the publication of these monographs, numerous journal articles have appeared on the topic. Our aim in this paper, however, is to present a simple yet useful result that highlights how much Actuarial Science and developments in Mathematical Statistics, Probability, and Stochastic Processes are interrelated, and thus benefit from each other.


## 4. Proof of Theorem3.1

We start the proof of Theorem3.1 with the decomposition(4.1)(CTẼn(t)-CTEF(t))(1-t)=An,1(t)+An,2,
where(4.2)An,1(t)=∫t1-k/n(Qn(s)-Q(s))ds,An,2=k/n1-γ̂nXn-k:n-∫1-k/n1Q(s)ds.
We shall show below that there are Brownian bridges Bn such that(4.3)nAn,1(t)(k/n)1/2Q(1-k/n)=-∫01-k/nBn(s)dQ(s)(k/n)1/2Q(1-k/n)+oP(1),(4.4)nAn,2(k/n)1/2Q(1-k/n)=γ2(1-γ)2nkBn(1-kn)-γ(1-γ)2nk∫1-k/n1Bn(s)1-sds+oP(1).
Assuming for the time being that statements (4.3) and (4.4) hold, we next complete the proof of Theorem 3.1. To simplify the presentation, we use the following notation:(4.5)W1,n=-∫01-k/nBn(s)dQ(s)(k/n)1/2Q(1-k/n),W2,n=γ2(1-γ)2nkBn(1-kn),W3,n=-γ(1-γ)2nk∫1-k/n1Bn(s)1-sds.
Hence, we have the asymptotic representation(4.6)n(CTẼn(t)-CTEF(t))(1-t)(k/n)1/2Q(1-k/n)=W1,n+W2,n+W3,n+oP(1).
The sum W1,n+W2,n+W3,n is a centered Gaussian random variable. To calculate its asymptotic variance, we establish the following limits:(4.7)E[W1,n2]→2γ2γ-1,E[W2,n2]→γ4(1-γ)4,E[W3,n2]→2γ2(1-γ)4,2E[W1,nW2,n]→-2γ2(1-γ)2,2E[W1,nW3,n]→2γ(1-γ)2,2E[W2,nW3,n]→-2γ3(1-γ)4.
Summing up the right-hand sides of the above six limits, we obtain σγ2, whose expression in terms of the parameter γ is given in Theorem 3.1. Finally, since Xn-k:n/Q(1-k/n) converges in probability to 1 (see, e.g., the proof of Corollary in [39]), the classical Sultsky's lemma completes the proof of Theorem 3.1. Of course, we are still left to verify statements (4.3) and (4.4), which make the contents of the following two subsections.


### 4.1. Proof of Statement (4.3)

IfQ were continuously differentiable, then statement (4.3) would follow easily from the proof of Theorem 2 in [39]. We do not assume differentiability of Q and thus a new proof is required, which is crucially based on the Vervaat process (see [31–33], and references therein)(4.8)Vn(t)=∫0t(Qn(s)-Q(s))ds+∫-∞Q(t)(Fn(x)-F(x))dx.
Hence, for every t such that 0<t<1-k/n, which is satisfied for all sufficiently large n since t is fixed, we have that(4.9)An,1(t)=∫01-k/n(Qn(s)-Q(s))ds-∫0t(Qn(s)-Q(s))ds=-∫Q(t)Q(1-k/n)(Fn(x)-F(x))dx+Vn(1-kn)-Vn(t).
It is well known (see [31–33]) that Vn(t) is nonnegative and does not exceed -(Fn(Q(t))-t)(Qn(t)-Q(t)). Since the cdf F is continuous by assumption, we therefore have that(4.10)nVn(t)≤|en(t)||Qn(t)-Q(t)|,
where en(t) is the uniform empirical process n(Fn(Q(t))-F(Q(t)), which for large n looks like the Brownian bridge Bn(t). Note also that with the just introduced notation en, the integral on the right-hand side of (4.9) is equal to ∫Q(t)Q(1-k/n)en(F(x))dx. Hence,(4.11)nAn,1(t)(k/n)1/2Q(1-k/n)=-∫Q(t)Q(1-k/n)en(F(x))dx(k/n)1/2Q(1-k/n)+OP(1)|en(1-k/n)||Qn(1-k/n)-Q(1-k/n)|(k/n)1/2Q(1-k/n)+OP(1)|en(t)||Qn(t)-Q(t)|(k/n)1/2Q(1-k/n).
We shall next replace the empirical process en by an appropriate Brownian bridge Bn in the first integral on the right-hand side of (4.11) with an error term of magnitude oP(1), and we shall also show that the second and third summands on the right-hand side of (4.11) are of the order oP(1). The replacement of en by Bn can be accomplished using, for example, Corollary 2.1 on page 48 of Csörgő et al. [40], which states that on an appropriately constructed probability space and for any 0≤ν<1/4, we have that(4.12)sup1/n≤s≤1-1/n|en(s)-Bn(s)|s1/2-ν(1-s)1/2-ν=OP(n-ν).
This result is applicable in the current situation since we can always place our original problem into the required probability space, because our main results are “in probability”. Furthermore, since Q(t)≤x≤Q(1-k/n), we have that t≤F(x)≤1-k/n. Hence, statement (4.12) implies that(4.13)-∫Q(t)Q(1-k/n)en(F(x))dx(k/n)1/2Q(1-k/n)=-∫Q(t)Q(1-k/n)Bn(F(x))dx(k/n)1/2Q(1-k/n)+OP(1)∫Q(t)Q(1-k/n)(1-F(x))1/2-νdxnν(k/n)1/2Q(1-k/n).
Changing the variables of integration and using the property (k/n)1/2Q(1-k/n)→∞ when n→∞, we obtain that(4.14)-∫Q(t)Q(1-k/n)Bn(F(x))dx(k/n)1/2Q(1-k/n)=-∫01-k/nBn(s)dQ(s)(k/n)1/2Q(1-k/n)+oP(1).
The main term on the right-hand side of (4.14) is W1,n. We shall next show that the right-most summand of (4.13) converges to 0 when n→∞.Changing the variable of integration and then integrating by parts, we obtain the bound(4.15)∫Q(t)Q(1-k/n)(1-F(x))1/2-νdxnν(k/n)1/2Q(1-k/n)≤(1-s)1/2-νQ(s)|t1-k/nnν(k/n)1/2Q(1-k/n)+O(1)∫t1-k/n(1-s)-1/2-νQ(s)dsnν(k/n)1/2Q(1-k/n).
We want to show that the right-hand side of bound (4.15) converges to 0 when n→∞. For this, we first note that(4.16)(1-s)1/2-νQ(s)|t1-k/nnν(k/n)1/2Q(1-k/n)=1kν-(1-t)1/2-νQ(t)nν(k/n)1/2Q(1-k/n)→0.
Next, with the notation ϕ(u)=Q(1-u)/u1/2+ν, we have that(4.17)∫t1-k/n(1-s)-1/2-νQ(s)dsnν(k/n)1/2Q(1-k/n)=1kν∫k/n1-tϕ(s)ds(k/n)ϕ(k/n)→0
when n→∞, where the convergence to 0 follows from Result 1 in the Appendix of Necir and Meraghni [39]. Taking statements (4.15)–(4.17) together, we have that the right-most summand of (4.13) converges to 0 when n→∞.Consequently, in order to complete the proof of statement (4.3), we are left to show that the second and third summands on the right-hand side of (4.11) are of the order oP(1). The third summand is of the order oP(1) because |en(t)∥Qn(t)-Q(t)|=OP(1) and (k/n)1/2Q(1-k/n)→∞. Hence, we are only left to show that the second summand on the right-hand side of equation (4.11) is of the order oP(1), for which we shall show that(4.18)|en(1-k/n)|(k/n)1/2|Qn(1-k/n)Q(1-k/n)-1|=oP(1).
To prove statement (4.18), we first note that(4.19)|en(1-k/n)|(k/n)1/2≤|en(1-k/n)-Bn(1-k/n)|(k/n)1/2+|Bn(1-k/n)|(k/n)1/2.
The first summand on the right-hand side of bound (4.19) is of the order OP(1) due to statement (4.12) with ν=0. The second summand on the right-hand side of bound (4.19) is of the order OP(1) due to a statement on page 49 of Csörgő et al. [40] (see the displayed bound just below statement (2.39) therein). Hence, to complete the proof of statement (4.18), we need to check that(4.20)Qn(1-k/n)Q(1-k/n)=1+oP(1).
Observe that, for each n, the distribution of Qn(1-k/n) is the same as that of Q(En-1(1-k/n)), where En-1 is the uniform empirical quantile function. Furthermore, the processes {1-En-1(1-s),0≤s≤1} and {En-1(s),0≤s≤1} are equal in distribution. Hence, statement (4.20) is equivalent to(4.21)Q(1-En-1(k/n))Q(1-k/n)=1+oP(1).
From the Glivenko-Cantelli theorem we have that En-1(k/n)-k/n→0 almost surely, which also implies that En-1(k/n)→0 since k/n→0 by our choice of k. Moreover, we know from Theorem 0 and Remark 1 of Wellner [41] that(4.22)sup1/n≤s≤1s-1|En-1(s)-s|=oP(1),
from which we conclude that(4.23)nEn-1(k/n)k=1+oP(1).
Since the function s↦Q(1-s) is slowly varying at zero, using Potter's inequality (see the 5th assertion of Proposition B.1.9 on page 367 of de Haan and Ferreira [13],we obtain that(4.24)Q(1-En-1(k/n))Q(1-k/n)=(1+oP(1))(nEn-1(k/n)k)-γ±θ
for any θ∈(0,γ). In view of (4.23), the right-hand side of (4.24) is equal to 1+oP(1), which implies statement (4.21) and thus finishes the proof of statement (4.3).


### 4.2. Proof of Statement (4.4)

The proof of statement (4.4) is similar to that of Theorem 2 in Necir et al. [42], though some adjustments are needed since we are now concerned with the CTE risk measure. We therefore present main blocks of the proof together with pinpointed references to Necir et al.  [42] for specific technical details.We start the proof with the function𝕌(z)=Q(1-1/z) that was already used in the formulation of Theorem 3.1. Hence, if Y is a random variable with the distribution function G(z)=1-1/z, z≥1, then 𝕌(Y)=Q(G(Y))=dX because G(Y) is a uniform on the interval [0,1] random variable. Hence,(4.25)An,2=k/n1-γ̂n𝕌(Yn-k:n)-∫0k/n𝕌(1s)ds,
and so we have(4.26)nAn,2(k/n)1/2Q(1-k/n)=k(11-γ̂n𝕌(Yn-k:n)𝕌(n/k)-(n/k)∫0k/n𝕌(1/s)ds𝕌(n/k))=k(11-γ̂n𝕌(Yn-k:n)𝕌(n/k)-11-γ)+k(11-γ-∫1∞s-2𝕌(ns/k)ds𝕌(n/k)).
We next show that the right-most term in (4.26) converges to 0 when n→∞. For this reason, we first rewrite the term as follows:(4.27)k(11-γ-∫1∞s-2𝕌(ns/k)ds𝕌(n/k))=-k∫1∞1s2(𝕌(ns/k)𝕌(n/k)-sγ)ds.
The right-hand side of (4.27) converges to 0 (see notes on page 149 of Necir et al. [42]) due to the second-order condition (3.1), which can equivalently be rewritten as(4.28)limz→∞1A(z)(𝕌(zs)𝕌(z)-sγ)=sγsρ-1ρ
for every s>0, where A(z)=γ2a(𝕌(z)). Note that kA(n/k)→0 when n→∞. Hence, in order to complete the proof of statement (4.4), we need to check that(4.29)k(11-γ̂n𝕌(Yn-k:n)𝕌(n/k)-11-γ)=γ2(1-γ)2nkBn(1-kn)-γ(1-γ)2nk∫1-k/n1Bn(s)1-sds+oP(1).
With Hill's estimator written in the form(4.30)γ̂n=1k∑i=1klog(𝕌(Yn-i+1:n)𝕌(Yn-k:n)),
we proceed with the proof of statement (4.29) as follows:
(4.31)k(1-γ1-γ̂n𝕌(Yn-k:n)𝕌(n/k)-1)=k1-γ1-γ̂n(𝕌(Yn-k:n)𝕌(n/k)-(Yn-k:nn/k)γ)+k1-γ1-γ̂n((Yn-k:nn/k)γ-1)+kγ̂n-γ1-γ̂n.
Furthermore, we have that(4.32)kγ̂n-γ1-γ̂n=11-γ̂n1k∑i=1k(log(𝕌(Yn-i+1:n)𝕌(Yn-k:n))-γlog(Yn-i+1:nYn-k:n))+γ1-γ̂n1k∑i=1k(log(Yn-i+1:nYn-k:n)-1).
Arguments on page 156 of Necir et al. [42] imply that the first term on the right-hand side of (4.32) is of the order OP(kA(Yn-k:n)), and a note on page 157 of Necir et al. [42] says that kA(Yn-k:n)=oP(1). Hence, the first term on the right-hand side of (4.32) is of the order oP(1). Analogous considerations using bound (2.5) instead of (2.4) on page 156 of Necir et al. [42] imply that the first term on the right-hand side of (4.31) is of the order oP(1). Hence, in summary, we have that(4.33)k(1-γ1-γ̂n𝕌(Yn-k:n)𝕌(n/k)-1)=1-γ1-γ̂nk((Yn-k:nn/k)γ-1)+γ1-γ̂n1k∑i=1k(log(Yn-i+1:nYn-k:n)-1)+oP(1).
We now need to connect the right-hand side of (4.33) with Brownian bridges Bn. To this end, we first convert the Y-based order statistics into U-based (i.e., uniform on [0,1]) order statistics. For this we recall that the cdf of Y is G, and thus Y is equal in distribution to G-1(U), which is 1/(1-U). Consequently,(4.34)k(1-γ1-γ̂n𝕌(Yn-k:n)𝕌(n/k)-1)=1-γ1-γ̂nk((1(n/k)(1-Un-k:n))γ-1)+γ1-γ̂n1k∑i=1k(log((1-Un-k:n)(1-Un-i+1:n))-1)+oP(1).
Next we choose a sequence of Brownian bridges Bn (see pages 158-159 in [42] and references therein) such that the following two asymptotic representations hold:(4.35)k((1(n/k)(1-Un-k:n))γ-1)=-γnkBn(1-kn)+oP(1),1k∑i=1k(log((1-Un-k:n)(1-Un-i+1:n))-1)=nkBn(1-kn)-nk∫1-k/n1Bn(s)1-sds+oP(1).
Using these two statements on the right-hand side of (4.34) and also keeping in mind that γ̂n is a consistent estimator of γ (see [22]), we have that(4.36)k(1-γ1-γ̂n𝕌(Yn-k:n)𝕌(n/k)-1)=γ21-γnkBn(1-kn)-γ1-γnk∫1-k/n1Bn(s)1-sds+oP(1).
Dividing both sides of equation (4.36) by 1-γ, we arrive at (4.29). This completes the proof of statement (4.4) and of Theorem 3.1 as well.


## 4.1. Proof of Statement (4.3)

IfQ were continuously differentiable, then statement (4.3) would follow easily from the proof of Theorem 2 in [39]. We do not assume differentiability of Q and thus a new proof is required, which is crucially based on the Vervaat process (see [31–33], and references therein)(4.8)Vn(t)=∫0t(Qn(s)-Q(s))ds+∫-∞Q(t)(Fn(x)-F(x))dx.
Hence, for every t such that 0<t<1-k/n, which is satisfied for all sufficiently large n since t is fixed, we have that(4.9)An,1(t)=∫01-k/n(Qn(s)-Q(s))ds-∫0t(Qn(s)-Q(s))ds=-∫Q(t)Q(1-k/n)(Fn(x)-F(x))dx+Vn(1-kn)-Vn(t).
It is well known (see [31–33]) that Vn(t) is nonnegative and does not exceed -(Fn(Q(t))-t)(Qn(t)-Q(t)). Since the cdf F is continuous by assumption, we therefore have that(4.10)nVn(t)≤|en(t)||Qn(t)-Q(t)|,
where en(t) is the uniform empirical process n(Fn(Q(t))-F(Q(t)), which for large n looks like the Brownian bridge Bn(t). Note also that with the just introduced notation en, the integral on the right-hand side of (4.9) is equal to ∫Q(t)Q(1-k/n)en(F(x))dx. Hence,(4.11)nAn,1(t)(k/n)1/2Q(1-k/n)=-∫Q(t)Q(1-k/n)en(F(x))dx(k/n)1/2Q(1-k/n)+OP(1)|en(1-k/n)||Qn(1-k/n)-Q(1-k/n)|(k/n)1/2Q(1-k/n)+OP(1)|en(t)||Qn(t)-Q(t)|(k/n)1/2Q(1-k/n).
We shall next replace the empirical process en by an appropriate Brownian bridge Bn in the first integral on the right-hand side of (4.11) with an error term of magnitude oP(1), and we shall also show that the second and third summands on the right-hand side of (4.11) are of the order oP(1). The replacement of en by Bn can be accomplished using, for example, Corollary 2.1 on page 48 of Csörgő et al. [40], which states that on an appropriately constructed probability space and for any 0≤ν<1/4, we have that(4.12)sup1/n≤s≤1-1/n|en(s)-Bn(s)|s1/2-ν(1-s)1/2-ν=OP(n-ν).
This result is applicable in the current situation since we can always place our original problem into the required probability space, because our main results are “in probability”. Furthermore, since Q(t)≤x≤Q(1-k/n), we have that t≤F(x)≤1-k/n. Hence, statement (4.12) implies that(4.13)-∫Q(t)Q(1-k/n)en(F(x))dx(k/n)1/2Q(1-k/n)=-∫Q(t)Q(1-k/n)Bn(F(x))dx(k/n)1/2Q(1-k/n)+OP(1)∫Q(t)Q(1-k/n)(1-F(x))1/2-νdxnν(k/n)1/2Q(1-k/n).
Changing the variables of integration and using the property (k/n)1/2Q(1-k/n)→∞ when n→∞, we obtain that(4.14)-∫Q(t)Q(1-k/n)Bn(F(x))dx(k/n)1/2Q(1-k/n)=-∫01-k/nBn(s)dQ(s)(k/n)1/2Q(1-k/n)+oP(1).
The main term on the right-hand side of (4.14) is W1,n. We shall next show that the right-most summand of (4.13) converges to 0 when n→∞.Changing the variable of integration and then integrating by parts, we obtain the bound(4.15)∫Q(t)Q(1-k/n)(1-F(x))1/2-νdxnν(k/n)1/2Q(1-k/n)≤(1-s)1/2-νQ(s)|t1-k/nnν(k/n)1/2Q(1-k/n)+O(1)∫t1-k/n(1-s)-1/2-νQ(s)dsnν(k/n)1/2Q(1-k/n).
We want to show that the right-hand side of bound (4.15) converges to 0 when n→∞. For this, we first note that(4.16)(1-s)1/2-νQ(s)|t1-k/nnν(k/n)1/2Q(1-k/n)=1kν-(1-t)1/2-νQ(t)nν(k/n)1/2Q(1-k/n)→0.
Next, with the notation ϕ(u)=Q(1-u)/u1/2+ν, we have that(4.17)∫t1-k/n(1-s)-1/2-νQ(s)dsnν(k/n)1/2Q(1-k/n)=1kν∫k/n1-tϕ(s)ds(k/n)ϕ(k/n)→0
when n→∞, where the convergence to 0 follows from Result 1 in the Appendix of Necir and Meraghni [39]. Taking statements (4.15)–(4.17) together, we have that the right-most summand of (4.13) converges to 0 when n→∞.Consequently, in order to complete the proof of statement (4.3), we are left to show that the second and third summands on the right-hand side of (4.11) are of the order oP(1). The third summand is of the order oP(1) because |en(t)∥Qn(t)-Q(t)|=OP(1) and (k/n)1/2Q(1-k/n)→∞. Hence, we are only left to show that the second summand on the right-hand side of equation (4.11) is of the order oP(1), for which we shall show that(4.18)|en(1-k/n)|(k/n)1/2|Qn(1-k/n)Q(1-k/n)-1|=oP(1).
To prove statement (4.18), we first note that(4.19)|en(1-k/n)|(k/n)1/2≤|en(1-k/n)-Bn(1-k/n)|(k/n)1/2+|Bn(1-k/n)|(k/n)1/2.
The first summand on the right-hand side of bound (4.19) is of the order OP(1) due to statement (4.12) with ν=0. The second summand on the right-hand side of bound (4.19) is of the order OP(1) due to a statement on page 49 of Csörgő et al. [40] (see the displayed bound just below statement (2.39) therein). Hence, to complete the proof of statement (4.18), we need to check that(4.20)Qn(1-k/n)Q(1-k/n)=1+oP(1).
Observe that, for each n, the distribution of Qn(1-k/n) is the same as that of Q(En-1(1-k/n)), where En-1 is the uniform empirical quantile function. Furthermore, the processes {1-En-1(1-s),0≤s≤1} and {En-1(s),0≤s≤1} are equal in distribution. Hence, statement (4.20) is equivalent to(4.21)Q(1-En-1(k/n))Q(1-k/n)=1+oP(1).
From the Glivenko-Cantelli theorem we have that En-1(k/n)-k/n→0 almost surely, which also implies that En-1(k/n)→0 since k/n→0 by our choice of k. Moreover, we know from Theorem 0 and Remark 1 of Wellner [41] that(4.22)sup1/n≤s≤1s-1|En-1(s)-s|=oP(1),
from which we conclude that(4.23)nEn-1(k/n)k=1+oP(1).
Since the function s↦Q(1-s) is slowly varying at zero, using Potter's inequality (see the 5th assertion of Proposition B.1.9 on page 367 of de Haan and Ferreira [13],we obtain that(4.24)Q(1-En-1(k/n))Q(1-k/n)=(1+oP(1))(nEn-1(k/n)k)-γ±θ
for any θ∈(0,γ). In view of (4.23), the right-hand side of (4.24) is equal to 1+oP(1), which implies statement (4.21) and thus finishes the proof of statement (4.3).


## 4.2. Proof of Statement (4.4)

The proof of statement (4.4) is similar to that of Theorem 2 in Necir et al. [42], though some adjustments are needed since we are now concerned with the CTE risk measure. We therefore present main blocks of the proof together with pinpointed references to Necir et al.  [42] for specific technical details.We start the proof with the function𝕌(z)=Q(1-1/z) that was already used in the formulation of Theorem 3.1. Hence, if Y is a random variable with the distribution function G(z)=1-1/z, z≥1, then 𝕌(Y)=Q(G(Y))=dX because G(Y) is a uniform on the interval [0,1] random variable. Hence,(4.25)An,2=k/n1-γ̂n𝕌(Yn-k:n)-∫0k/n𝕌(1s)ds,
and so we have(4.26)nAn,2(k/n)1/2Q(1-k/n)=k(11-γ̂n𝕌(Yn-k:n)𝕌(n/k)-(n/k)∫0k/n𝕌(1/s)ds𝕌(n/k))=k(11-γ̂n𝕌(Yn-k:n)𝕌(n/k)-11-γ)+k(11-γ-∫1∞s-2𝕌(ns/k)ds𝕌(n/k)).
We next show that the right-most term in (4.26) converges to 0 when n→∞. For this reason, we first rewrite the term as follows:(4.27)k(11-γ-∫1∞s-2𝕌(ns/k)ds𝕌(n/k))=-k∫1∞1s2(𝕌(ns/k)𝕌(n/k)-sγ)ds.
The right-hand side of (4.27) converges to 0 (see notes on page 149 of Necir et al. [42]) due to the second-order condition (3.1), which can equivalently be rewritten as(4.28)limz→∞1A(z)(𝕌(zs)𝕌(z)-sγ)=sγsρ-1ρ
for every s>0, where A(z)=γ2a(𝕌(z)). Note that kA(n/k)→0 when n→∞. Hence, in order to complete the proof of statement (4.4), we need to check that(4.29)k(11-γ̂n𝕌(Yn-k:n)𝕌(n/k)-11-γ)=γ2(1-γ)2nkBn(1-kn)-γ(1-γ)2nk∫1-k/n1Bn(s)1-sds+oP(1).
With Hill's estimator written in the form(4.30)γ̂n=1k∑i=1klog(𝕌(Yn-i+1:n)𝕌(Yn-k:n)),
we proceed with the proof of statement (4.29) as follows:
(4.31)k(1-γ1-γ̂n𝕌(Yn-k:n)𝕌(n/k)-1)=k1-γ1-γ̂n(𝕌(Yn-k:n)𝕌(n/k)-(Yn-k:nn/k)γ)+k1-γ1-γ̂n((Yn-k:nn/k)γ-1)+kγ̂n-γ1-γ̂n.
Furthermore, we have that(4.32)kγ̂n-γ1-γ̂n=11-γ̂n1k∑i=1k(log(𝕌(Yn-i+1:n)𝕌(Yn-k:n))-γlog(Yn-i+1:nYn-k:n))+γ1-γ̂n1k∑i=1k(log(Yn-i+1:nYn-k:n)-1).
Arguments on page 156 of Necir et al. [42] imply that the first term on the right-hand side of (4.32) is of the order OP(kA(Yn-k:n)), and a note on page 157 of Necir et al. [42] says that kA(Yn-k:n)=oP(1). Hence, the first term on the right-hand side of (4.32) is of the order oP(1). Analogous considerations using bound (2.5) instead of (2.4) on page 156 of Necir et al. [42] imply that the first term on the right-hand side of (4.31) is of the order oP(1). Hence, in summary, we have that(4.33)k(1-γ1-γ̂n𝕌(Yn-k:n)𝕌(n/k)-1)=1-γ1-γ̂nk((Yn-k:nn/k)γ-1)+γ1-γ̂n1k∑i=1k(log(Yn-i+1:nYn-k:n)-1)+oP(1).
We now need to connect the right-hand side of (4.33) with Brownian bridges Bn. To this end, we first convert the Y-based order statistics into U-based (i.e., uniform on [0,1]) order statistics. For this we recall that the cdf of Y is G, and thus Y is equal in distribution to G-1(U), which is 1/(1-U). Consequently,(4.34)k(1-γ1-γ̂n𝕌(Yn-k:n)𝕌(n/k)-1)=1-γ1-γ̂nk((1(n/k)(1-Un-k:n))γ-1)+γ1-γ̂n1k∑i=1k(log((1-Un-k:n)(1-Un-i+1:n))-1)+oP(1).
Next we choose a sequence of Brownian bridges Bn (see pages 158-159 in [42] and references therein) such that the following two asymptotic representations hold:(4.35)k((1(n/k)(1-Un-k:n))γ-1)=-γnkBn(1-kn)+oP(1),1k∑i=1k(log((1-Un-k:n)(1-Un-i+1:n))-1)=nkBn(1-kn)-nk∫1-k/n1Bn(s)1-sds+oP(1).
Using these two statements on the right-hand side of (4.34) and also keeping in mind that γ̂n is a consistent estimator of γ (see [22]), we have that(4.36)k(1-γ1-γ̂n𝕌(Yn-k:n)𝕌(n/k)-1)=γ21-γnkBn(1-kn)-γ1-γnk∫1-k/n1Bn(s)1-sds+oP(1).
Dividing both sides of equation (4.36) by 1-γ, we arrive at (4.29). This completes the proof of statement (4.4) and of Theorem 3.1 as well.



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*Source: 596839-2010-04-12.xml*