Operational Risk Modeling Analytics phần 7 pdf

EXTREME VALUE COPULAS 259 from setting A(w) to its upper bound A(w) = 1. At the other extreme, if A(w) = max(w, 1 - w), then there is perfect correlation, and hence perfect dependency with C (u, u) = u. It is convenient to write the index of upper tail dependence in terms of the dependence function A(w). The result is that 1-2u+C(u,u) 1 - 2u + u2A(1/2) Xu = lim = lim u-1 1-u u-1 1-u = lim 2 - 2A( 1/2)~'~(~/')-~ u-1 = 2 - 2A(1/2). If a copula is specified through A(w), then the index of upper tail depen- dency is easily calculated. There are several well-known copulas in this class. Gumbel copula The Gumbel copula was discussed previously as an example of an Archimedean copula. It is also an extreme value copula with dependence function From this, by setting w = l/2, the Gumbel copula is seen to have index of upper tail dependence of 2 - 2'1'. Galambos copula The Galambos copula [42] has the dependence function Unlike the Gumbel copula, it is not Archimedean. It has index of upper tail dependence of 2Y1/'. The bivariate copula is of the form An asymmetric version of the Galambos copula with three parameters has dependence function 0 5 a,p 5 1. It has index of upper tail dependence of (a-' +p-')-'/'. The one-parameter version is obtained by setting a 5 ,B = 1.The bivariate asymmetric Galambos copula has the form 260 MULTIVARIATE MODELS 1 08 06 z 04 02 '0 02 04 06 08 1 > U U Fig. 8.23 Galambos copula density (6 = 2.5) fig. 8.24 Galambos copula pdf (6 = 2.5) EXTREME VALUE COPULAS 261 Figures 8.23 and 8.24 demonstrate the clear upper tail dependence. Hiisler and Reiss copula The Husler and Reiss copula [57] has dependence function where Q(x) is the cdf of the standard normal distribution. When w = 1/2, A(1/2) = @(l/O), resulting in an index of upper tail dependence of 2-2@(1/6). Tawn copula The Gumbel copula can be extended to a three-parameter asymmetric ver- sion by introducing two additional parameters, CY and /3, into the dependence function [114] A(w) = (1 - a)w+(l-p) (1 - W)+{(CYW)~+ [@(I - w)]”’} , 0 5 a,P 5 1. This is called the Tawn copula. Note that the one-parameter version of A(w) is obtained by setting Q = /3 = 1. The bivariate asymmetric Gumbel copula has the form BB5 copula The BB5 copula [62] is another extension of the Gumbel copula but with only two parameters. Its dependence function is The BB5 copula has the form 262 MULTIVARIATE MODELS 1 08 06 > , 04 02 '0 02 04 06 08 1 0 02 04 06 08 1 U U Fig. 8.25 BB4 copula density (6 = 3, 6 = 1.2) 8.8 ARCHIMAX COPULAS Archimedean and extreme value copulas can be combined into a single class of copulas called Archimax copulas. Archimax copulas are represented as where 4(u) is a valid Archimedean generator and A (w) is a valid dependence function. This general setup allows for a wide range of copulas and therefore shapes of dis- tributions. The BB4 copula is one such example. It can be shown [20] that that this is itself a valid copula. BB4 copula The BB4 copula [62] is an Archimax copula with 6 4(U) = u- - 1, K9 2 0, as with the Clayton copula and A(w) = 1 - {wP6 + (1 - w)-~}-"', K9 > 0, 6 > 0, leading to the copula of the form It is illustrated in Figures 8.25 and 8.26. EXERCISES 263 Fig. 8.26 BB4 copula pdf (6 = 2, 6 = 1.2) 8.9 EXERCISES 8.1 Prove that the Clayton, Frank, and Ali-Mikhail-Haq copulas have no upper tail dependence. 8.2 Prove that the Gumbel copula has index of upper tail dependence equal to 2 - 2-‘f@. 8.3 Prove that the Gaussian copula has no upper tail dependence. Hint: Begin by obtaining the conditional distribution of X given Y = y from the bivariate normal distribution. 8.4 Prove that the t copula has index of upper tail dependence xu = 2 t”+l(-/Y) l+P u+l) Hint: Begin by showing that if (X, Y) comes from a bivariate t distribution, each with v degrees of freedom, conditional on Y = y, the random variable v+l x-py J- V+Y2 diq? has a t distribution with v + 1 degrees of freedom. 8.5 For the EV copula, show that if A(w)=max(w, 1 - w) , the copula is the straight line C (u, u) = u. 264 MULTIVARIATE MODELS 8.6 For the bivariate EV copula, show that A (w) = - 1nC (e-", e-('-")) . 8.7 Prove that the index of upper tail dependence of the Gumbel copula is 2 - 2118. Part 111 Statistical methods for calibrating models of ope rational risk This Page Intentionally Left Blank Review of mathematical statistics Nothing is as easy as it looks. -Murphy 9.1 INTRODUCTION In this chapter, we review some key concepts from mathematical statistics. Mathematical statistics is a broad subject that includes many topics not cov- ered in this chapter. For those topics that are covered in this chapter, it is assumed that the reader has had at least some prior exposure. The topics of greatest importance for constructing models are estimation and hypothesis testing. Because the Bayesian approach to statistical inference is often either ignored or treated lightly in introductory mathematical statistics texts and courses, it receives a more in-depth coverage in this text in Section 10.5. We begin by assuming that we have some data; that is, we have a sam- ple. We also assume that we have a model (ie, a distribution) that we wish to calibrate by estimating the “true” values of the parameters of the model. This data will be used to estimate the parameter values. The formula form of an estimate is called the estimator. The estimator is itself a random vari- able because it is a function of random variables, sometimes called a random function. The numerical value of the estimator based on data is called the estimate. The estimate is a single number. Because the parameter estimates are based on a sample from the population and not the entire population, they will not be exactly the true values, but 267 268 REVIEW OF MATHEMATICAL STATISTICS only estimates of the true values. In applications, it is important to have an idea of how good the estimates are by understanding the potential error of the estimates. One way to express this is with an interval estimate. Rather than focusing on a particular value, a range of plausible values can be presented. 9.2 POINT ESTIMATION 9.2.1 Introduction Regardless of how a model is estimated, it is extremely unlikely that the estimated model will exactly match the true distribution. Ideally, we would like to be able to measure the error we will be making when using the estimated model. But this is clearly impossible! If we knew the amount of error we had made, we could adjust our estimate by that amount and then have no error at all. The best we can do is discover how much error is inherent in repeated use of the procedure, as opposed to how much error we actually make with our current estimate. Therefore, this section is about the quality of the answers produced from the procedure, not about the quality of a particular answer. When constructing models, there are a number of types of error. Several will not be covered here. Among these are model error (choosing the wrong model) and sampling frame error (trying to draw inferences about a popula- tion that differs from the one sampled). An example of model error is selecting a Pareto distribution when the true distribution is Weibull. An example of sampling frame error is using sampled losses from one process to estimate those of another. The type of error we can measure is the error that is due to the use of a sample from the population to make inferences about the entire population. Errors occur when the items sampled do not represent the population. As noted earlier, we cannot know whether the particular items sampled today do or do not represent the population. We can, however, estimate the extent to which estimators are affected by the possibility of a nonrepresentative sample. The approach taken in this section is to consider all the samples that might be taken from the population. Each such sample leads to an estimated quan- tity (for example, a probability, a parameter value, or a moment). We do not expect the estimated quantities to always match the true value. For a sensible estimation procedure we do expect that for some samples the quantity will match the true value, for many it will be close, and for only a few it will be quite different. If we can construct a measure of how well the set of potential estimates matches the true value, we have a good idea of the quality of our es- timation procedure. The approach outlined here is often called the classical or frequentist approach to estimation. [...]... Number of drivers 0 1 2 3 4 5 or more 81 ,71 4 11,306 1,618 250 40 7 Table 10.2 Data Set B $ 27 $4 57 $82 $680 $115 $855 $126 $ 877 $155 $ 974 $161 $1193 $243 $1340 $294 $1884 $340 $2558 $384 $15 ,74 3 Table 10.3 Data Set C Payment range 0- $75 00 $75 00-$ 17, 500 $ 17, 500-$32,500 $32,500-$ 67, 500 $ 67, 500-$125,000 $125,000-$300,000 Over $300,000 Number of losses 99 42 29 28 17 9 3 Data Set B These numbers are artificial... bank has been assuming that, for a particular type of operational risk, the average loss is $1200 You wish to put this assumption to a rigorous test The following data representing recent operational risk losses of the same type What are the hypotheses for this problem? 27 4 57 82 680 115 855 126 877 155 974 161 1193 243 1340 294 1884 340 2558 384 15 ,74 3 Let p be the population mean One possible hypothesis... 99 lnjF (75 00) - F(O)]+ 421n[F( 17, 500) - F (75 00)]+ + 3 ln[l - F(300,000)] - 99 in( 1 - e - 7 5 0 0 / @ ) + 42 ln(e -75 00/@- e- 17, 500/@) + + 3 ln e - 3 0 ~ , 0 ~ O / @ A numerical routine is needed t o produce % = 29 ,72 1, and the value of the loglikelihood function is -406.03 10.3.4 Truncated or censored data When data are censored, there is no additional complication As noted in Example 10 .7, right... is E(X) = [1(1)+ 2(2) + 3(3) + 4(2) + 5(3) + 6(2) + 7( 2) + 9(1)]/16 = 4.5 270 REVIEW OF MATHEMATICAL STATISTICS Table 9.1 The 16 possible outcomes in Example 9.4 Table 9.2 The 16 possible sample means in Example 9.4 1 3 2 4 5 7 3 5 2 5 6 9 4 3 6 7 Table 9.3 Distribution of sample mean in Example 9.4 X p,y(x) 2 2/16 1 1/16 3 3/16 4 2/16 5 3/16 6 2/16 7 2/16 9 1/16 and so the sample mean is an unbiased... moments to estimate parameters f o r the exponential, gamma, and Pareto distributions for Data Set B METHOD OF MOMENTS AND PERCENTILE MATCHING 2 87 The first two sample moments are b; = + 15 ,74 3) 1,424.4, + + 15 ,74 32) = 13,238,441.9 L ( 2 7 + 20 & = h( 27 For the exponential distribution the equation is 8 = 1424.4 with the obvious solution, 8 = 1,424.4 For the gamma distribution, the two equations... UMVUE within the class of estimators of this form 9.9 Two different estimators, 81 and 82, are being considered To test their performance, 75 trials have been simulated, each with the true value set at 0 = 2 The following totals were obtained: 75 75 75 j=1 j=l 75 j=1 where 8ij is the estimate based on the j t h simulation using estimator 8, Estimate the MSE for each estimator and determine the relative... distribution with B = 800 known Then use the model to estimate the losses in excess of thresholds of 0, $200, and $400 Using the shifting approach, the data set has 14 points ($43, $94, $140, $184, $2 57, $480, $655, $ 677 , $77 4, $993, $1140, $1684, $2358, and $15,543) The likelihood function is 14 l(a) = c [ l n a j=1 14 In a = 14 In a = + aln800 - ( a + 1)ln(zj + SOO)] + 93.5846a - 103.969(a + 1) - 103.969 - 10.384a,... enter the study at an advanced age lNTRODUCTlON Machine 1 2 3 4 5 6 7 8 9 10 11 12 13 14 First observed 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Last observed 0.1 0.5 0.8 0.8 1.8 1.8 2.1 2.5 2.8 2.9 2.9 3.9 4.0 4.0 Event w w w f w w w w w f f w f w Machine 16 17 18 19-30 31 32 33 34 35 36 37 38 39 40 First observed Last observed 0 0 0 0 0.3 0 .7 1.0 1.8 2.1 2.9 2.9 3.2 3.4 3.9 4.8 4.8 4.8 5.0 5.0 5.0 4.1 3.1 3.9... The equations to solve are 0.3 = F ( 185.6) = 1 - (185.: + 0)O' e 0.8 = F(1,310.6) =z 1 - Q + (1310.6 0 ) ' 6 In 0 .7 = -0.356 675 = a In (185.6 0 ) ' 6 ln0.2 = -1.609438 = a In (1,310.6 6) ' + + - 1.609438 -0.356 675 = 4.512338 = Numerical methods can be used to solve this equation for $ = 71 5.03 Then, from the first equation, 0.3 = 1 - (185.~?~5.03) a ' which yields & = 1.54559 The estimates are much... function It is easier to maximize the logarithm of the likelihood function Because it occurs so often, we denote the loglikelihood function as Z(O) = lnL(8) Then Z(8) = -71 nO - 4159 8-', Z'(8) = -78 -1 + 4159 O-' = 0, e= 4159 = 594.14 7 In this case, the calculus technique of setting the first derivative equal t o zero is easy to do Also, evaluating the second derivative at this solution produces 0 a . type. What are the hypotheses for this problem? 27 82 115 126 155 161 243 294 340 384 4 57 680 855 877 974 1193 1340 1884 2558 15 ,74 3 Let p be the population mean. One possible hypothesis. particular type of operational risk, the average loss is $1200. You wish to put this assumption to a rigorous test. The following data representing recent operational risk losses of the. means in Example 9.4 1 2 3 5 2 3 4 6 3 4 5 7 5 6 7 9 Table 9.3 Distribution of sample mean in Example 9.4 X 1 2 3 4 5 6 7 9 p,y(x) 1/16 2/16 3/16 2/16 3/16 2/16 2/16