Estimation of the marginal expected shortfall under asymptotic independence

We study the asymptotic behavior of the marginal expected shortfall when the two random variables are asymptotic independent but positive associated, which is modeled by the so-called tail dependent coefficient. We construct an estimator of the marginal expected shortfall which is shown to be asymptotically normal. The finite sample performance of the estimator is investigated in a small simulation study. The method is also applied to estimate the expected amount of rainfall at a weather station given that there is a once every 100 years rainfall at another weather station nearby.


Introduction
Let X and Y denote two risk factors.The marginal expected shortfall (MES) is defined as where Q Y is the quantile function of Y and p is a small probability.The name of MES originates from its application in finance as an important ingredient for constructing a systemic risk measure; see for instance [1] and [8].In actuarial science, this quantity is known as the multivariate extensions of tail conditional expectation (or, conditional tail expectation); see for instance [5] and [11].
Under the assumption that X is in the Fréchet domain of attraction, [7] has established the following asymptotic limit; see Proposition 1 in that paper.With Q X the quantile function of X, where a > 0 if X and Y are asymptotic dependent and a = 0 if they are asymptotic independent.Based on this result, an estimator for MES is established in [7], which is not applicable for asymptotic independent data.It is the goal of this paper to study the asymptotic behavior of MES and to develop an estimation of MES for asymptotic independent data.Under the framework of multivariate extreme value theory, there are various ways to describe asymptotic dependence, for instance by means of exponent measure, spectral measure or Pickands dependence functions, cf.Chapter 6 of [12] and Chapter 8 of [3].However, these measures don't distinguish the relative strength of the extremal dependence for an asymptotic independent pair.The so-called coefficient of tail dependence introduced by [14] is mostly used to measure the strength of the extremal dependence for an asymptotic independent pair.In this paper, we make use of the coefficient of tail dependence, denoted as η to model asymptotic independence.Namely, we assume that there exists an η ∈ (0, 1] such that the following limit exists and is positive: We are interested in the scenario that η ∈ (1/2, 1), which corresponds to asymptotic independence but positive association of X and Y .For this type of distributions, one has that is, the joint extremes of (X, Y ) happen much more often than those of a distribution with independent components of X and Y .This gives an intuitive explanation that even if the pair are asymptotic independent, the extremal dependence can still be strong and thus needs to be accounted for.We also assume that X is in the Fréchet domain of attraction, so it has a heavy right tail.As our result shall reveal, the risk represented by MES can also be very big under the combination of positive association and X being heavy tailed, cf.Proposition 2.1.Thus from the application point of view, it is very important to consider MES for such a model assumption.This paper is organised as follows.Section 2 contains the main theoretical results on the limit behaviour of M ES and the asymptotic normality of the proposed estimator of MES.The performance of the estimation method is illustrated by a simulation study in Section 3 and by an application to precipitation data in Section 4. The proofs of the main theorems are provided in Section 5.

Main results
We first derive the asymptotic limit for MES as p → 0, based on which we shall then construct an estimator for MES.Let F 1 and F 2 denote the marginal distribution functions of X and Y , respectively.As usual in extreme value analysis, it is more convenient to work with, in stead of the quantile function, the tail quantile defined as U j = 1 1−Fj ← , j = 1, 2, where ← denotes the left continuous inverse.Then MES can be written as We now present our model, namely, assumptions on the tail distribution of X and the extremal dependence of X and Y .First, we assume that X has a heavy right tail, that is, there exists a Second, we assume the positive association of X and Y .Precisely, there exists an η ∈ (1/2, 1] such that for all (x, y) ∈ (0, ∞) 2 , the following limit exists As a consequence, c is a homogeneous function of order 1 η .The condition of (2.2) is also assumed in [13] for estimating η.Note that if η = 1, it corresponds to X and Y being asymptotic dependent.For η < 1, this condition is linked to the so-called hidden regular variation (cf.[15]) in the following way: where ν * is defined in (3) of [15].In order to obtain the limit result on θ p for p → 0, we need a second order strengthening condition of (2.1).

A(1)
We also need some technical conditions on the extremal dependence of X and Y .For t > 0, define ) is a continuous function, then we have , then the result of (2.4) coincide with the result of Proposition 1 in [7].
Provided with a random sample (X 1 , Y 1 ), . . ., (X n , Y n ), we now construct an estimation of θ p , where p = p(n) → 0, as n → ∞.Propositions 2.1 suggests the following approximation.With t sufficiently large, From this extrapolation relation, the remaining task is to estimate η, γ 1 and θ k n .There are wellknown existing methods for estimating γ 1 and η; see Chapters 3 and 7 of [12].For θ k n , we propose a nonparametric estimator given by θk Let γ1 and η denote estimators of γ 1 and η, respectively.We construct the following estimator for θ p : θp = θk/n k np . (2.6) Next we prove the asymptotic normality of θp .The following conditions will be needed.

Simulation Study
In this section, we study the finite sample performance of our method.We apply our estimator given by (2.6) to data simulated from the following two distributions.We consider sample size n = 5000, and p = 10/n, 1/n and 1/(10n).Let Z 1 , Z 2 , and Z 3 be independent Pareto random variables with parameters 0.3, 0.4, and 0.4 , where B is a Bernoulli (1/2) random variable independent of Z 1 , Z 2 , and Z 3 .
For both distributions, γ 1 = 0.4, η = 3 4 and c(x, y) = d(x ∧ y) 1/η , 0 < x, y < ∞, for some constant d > 0. And assumptions B(1)-B (7) are satisfied by both models with properly chosen δ, β 1 , β 2 , τ and α.To complete our estimator given by (2.6), we use the Hill estimator for γ 1 and an estimator for η proposed in [13].Let k 1 and k 2 be two intermediate sequences.Define where i and R Y i denoting the ranks of X i and Y i in their respective samples.
For p = 10/n, we compare our estimator with the nonparametric estimator, namely, θemp = 1 10 which is obtained by letting k/n = p in (2.5).
For each estimator, we compute the relative error defined as bias p = 1 m m i=1 θp,i θp − 1, where θp,i is an estimate based on the i-th sample.A relative error for θemp is computed in the same way, imsart-generic ver.2014/10/16 file: MES_CAI_MUSTA.tex date: September 2, 2018 q q q q q q q q q q q q q q q q q q q q q q q q q q −1.0 −0.5 0.0 0.5 1.0 1.5

Example1
bias emp bias p1 bias p2 bias p3 q q q q q q q q q q q q q q q q q q q q q q −1.0 denoted as bias emp .Figure 1 shows the relative errors obtained by generating m = 500 samples for each scenario.From the boxplots, for the situation where the empirical estimator is applicable, that is p = 10/p, our estimator has a slightly larger bias but a smaller variance.As p becomes smaller, the empirical estimator is not applicable, yet our estimator still has decent performance with growing variance.

Application
We apply our estimation to daily precipitation data from two weather stations in the Netherlands, namely Cabauw and Rotterdam.The distance between these two stations is about 32 km.The station Cabauw is close to the Lek river while the station Rotterdam is close to the river Nieuwe Maas, which is the continuation of Lek.Heavy rainfall at both stations might lead to a severe flood in this region.Thus, the expected amount of rainfall in Cabauw given a heavy rainfall in Rotterdam is an important risk measure for the hydrology safety control.We estimate this quantity based on the data from August 1st, 1990 to December 31, 2016.After removing the missing values, there are in total 9605 observations.There is open access to the data at http: //projects.knmi.nl/klimatologie/uurgegevens/selectie.cgi.
Let X be the daily rainfall at Cabauw and Y be the daily rainfall at Rotterdam.Before applying our estimation, we shall look at the sign of the extreme value index of X and the extremal dependence of X and Y .From the Hill estimates of γ as shown in right panel of Figure 2, we conclude that γ > 0, which is in line with the existing literature.For instance, [4] obtains 0.1082 as the estimate of γ for the daily rainfall in the Netherlands and [10] reports 0.066 as the estimate of γ for the daily rainfall in the southwest of England.
Next, we compute the Hill estimator of η given by (3.1).The estimates are above 0.5 as shown in the right panel of Figure 2.
Finally, we apply our estimator to answer the following question.Provided that the amount of rainfall in Rotterdam exceeds the M -year return level, what is the expected amount of rainfall in Cabauw, respectively, for M = 50 and 100?Let R M denote the M -year return level.[9] gives the definition of R M as the level expected to be exceeded once every M years.As we consider the daily precipitation, R M = U 2 (365M ).Choosing k 1 = k 2 = 200, we obtain the following estimates of γ and η: γ1 = 0.326 and η = 0.835.Figure 3 plots the estimates of θ p against k, from which we conclude k = 50 lying in the interval where the estimates are stable.We thus report the following estimates of θp : 41.6 mm for M = 50 and 45.5 mm for M = 100.

Proofs
Proof of Proposition 2.1.We recall that for any positive random variable Z, the expectation can be written as Then by definition of θ p and a change of variable, we have (5.1) For any fixed x, by (2.2) and the continuity of the function x → c(x, 1), we have We shall apply the generalized dominated convergence theorem to validate that lim t→∞ By assumption A(1), for any > 0, there exists t 0 such that Hence, for c 1 = (d + )/(d − ) and x > c 1 (t 0 /t) γ1 , we get Consequently, for x > c 1 (t 0 /t) γ1 , On the other hand, for 0 Then f t (x) ≤ g t (x).By generalized dominated convergence theorem, it is then sufficient to prove that x −β1/γ1 dx → 0, by Assumptions A(2) and A(3).
Through out the proof section, we denote the convergence speed in Theorem 2.1 by From Assumption B(5), T n → ∞, as n → ∞.By construction, the asymptotic normality of θp depends on the asymptotic behaviour of θk/n , which is given in Proposition 5.1.
Proposition 5.1.Under the assumptions of Theorem 2.1, it holds We shall show that these four terms all converges to unity at certain rates.First, from the assumption that √ k( γ1 − γ 1 ) = O P (1), it follows that In the last equality, we used the assumption that lim n→∞ (n/k) −1/2η+1/2 log d n = 0. Recall that T n is defined in (5.2).
In the same way, we get Tn .Combining Propositions 5.1 and 2.1, we derive that That is, T n (I 3 − 1) where Γ 1 is a normal distribution with mean zero and variance, , which is the limit distribution in Theorem 2.1.Then we deal with the last term, I 4 .Here we need a rate for the convergence in Proposition 2.1.Continuing with (5.1), By the regular variation of 1 − F 1 , we have lim n→∞ s n (x) = x, for any x > 0. By Lemma 6.1 (iii) and (v) in the Appendix, we have that It follows from Assumptions B(4) and B(5) that Combing this result with (5.3) and (5.5) leads to Thus, we obtain θp The proof is completed.
6. Appendix: Proof of Proposition 5.1 In this section, all the limit processes involved are defined in the same probability space via the Skorohod construction, i.e. they are only equal in distribution to the original processes.If we define Note that e n ) is the (k + 1)th order statistics of a random sample from the standard uniform distribution.
We first investigate the asymptotic behavior of Let W (y) denote a mean zero Gaussian process on [1/2, 2] with covariance structure The convergence of the process holds in distribution in the Skorokhod space D([1/2, 2]).
imsart-generic ver.2014/10/16 file: MES_CAI_MUSTA.tex date: September 2, 2018 Before proving Proposition 6.1, we first show two lemmas.The first lemma states some some properties of the functions c t (x, y) and c(x, y), that will be used frequently in the proof.The second lemma is established to compute the covariance of the limit process in Proposition 6.1.
Proof.The first statement follows from simple transformations of the integral.Indeed we have By the homogeneity property of c(x, y): c(kx, ky) = k First we deal with the first term in the right hand side of (6.7).By the homogeneity property of c(x, y), we have Note that, for any 0 > 0, for sufficiently large n and x < l n , (see [6], page 85) This implies that by a Taylor expansion, we obtain Furthermore, using the triangular inequality and Cauchy-Schwartz, we get Going back to (6.6), we obtain sup T n ln 0 → 0, because of assumption B( 5).
Next, we deal with the second term in the right hand side of (6.7).By Cauchy-Schwartz and assumption B(1), we obtain for some constant C > 0.Moreover, since Furthermore, the triangular inequality yields Note that, by assumption B(3), imsart-generic ver.2014/10/16 file: MES_CAI_MUSTA.tex date: September 2, 2018 Then, using the definition of s n , a change of variable and Jensen inequality, we obtain Hence, assumption B(5 On the other hand, using the definition of s n , we get because of assumption B( 5).
(iv) We write sup The first term in the right hand side of the inequality converges to zero by assumptions B(2)-B(3).Moreover, by assumption B(1), we have imsart-generic ver.2014/10/16 file: MES_CAI_MUSTA.tex date: September 2, 2018 (v) We first give the proof for (6.4).By Assumptions B(2) and B(3), we have sup Next we obtain an upper bound for the integral in the last equality.Since s n (x) is monotone and s n (1) = 1, we get the following bound for the integral from zero to one, which is shown to be O(1) in [7] (page 438).Moreover, using the definition of s n , a change of variable and Jensen inequality, we obtain By Assumption B(5), Thus, (6.4) is proved.The proof for (6.3) can be obtained in a similar way.We use the triangular inequality as in (6.10) to get sup A 1 converges to zero by (6.11).Moreover, as in (6.7), by assumption B(5).Finally, (see (6.9)).
As a result, by ( 5) and a change of variable, we obtain The statement follows from (6.2).
Proof of Proposition 6.1.For i = 1, . . ., n, let where s n (x) is defined in (5.4).Similarly, we have , and it enables us to write the left hand side of (6.1) as where Recall that we have lim n→∞ s n (x) = x by the regular variation of 1 − F 1 .We shall study a simpler process obtained by replacing s n (x) with x in (6.13): P − → 0 Step 1: Proof of (6.15)Using the definitions and the triangular inequality we write All three terms in the left hand side converge to zero by Lemma 6.1 (iii) and (v).
Step 2: Proof of (6.16)We aim to apply Theorem 2.11.9 in [17].We will prove that the four conditions of this theorem are satisfied.Here (F, ρ , we get, with δ as defined in Assumption B (1).
= n The last convergence follows from that T n → ∞ and Lemma 6.2.b) Take a sequence δ n → 0.Then, by the triangular inequality and that y 1 , y 2 ≥ 1/2, it follows that where A n (y 1 , 2) is defined as in Lemma 6.2.Thus, the second summand converges to zero lim n→∞ E[A n (y 1 , 2)] < ∞ and δ n → 0.Moreover, by the triangular inequality and Lemma 6.1 (ii, iv), Consider the partition given by I n,j = [1/2 + (j − 1)∆ n , 1/2 + j∆ n ].Then N = 3/2∆ n .We aim at finding ∆ n such that for every sequence δ n → 0 it holds By the same reasoning for (6.18), we obtain For the first term we have B n ≤ K 1 ∆ n for some constant K 1 > 0 by Lemma 6. by Holder inequality, we obtain for some constant K 2 .The last equality is obtained by applying Lemma 6.2 and choosing q = (2 + δ)/2 and 1/p + 1/q = 1.Second, by the same reasoning for (6.19), the triangular inequality and Lemma 6.1 (ii), (iv), we get a second bound on C n , for some τ * ∈ (τ, 0) we use the first bound on C n , i.e.
, for sufficiently large n, 2 and we use the second bound on C n with Hence, in this case,  converges a normal distribution, where N n,i = M j=1 a j Z n,i (y j ).This will follow from the Lindeberg-Feller central limit theorem (see e.g.[16]), once we show that for each > 0, which converges to zero by (6.17).For (6.21), we write We have now verified the four conditions required by Theorem 2.11.9 in [17], which leads to the conclusion that n i=1 (Z n,i − E[Z n,i ]) converges in distribution to a Gaussian process.Finally, we compute the covariance structure of the limit process.

Fi g 2 :
The Hill estimates of γ 1 for the daily precipitation at Cabauw (left panel) and the estimates of η for the daily precipitations at Cabauw and Rotterdam (right panel).
x, 1) dx −2γ1 .The proof of Proposition 5.1 is postponed to the Appendix.Proof of Theorem 2.1.Recall that d n = k np .By the definition of θp , we make the following decomposition θp , y) dx −γ1 → 0.