Operational Risk Modeling Analytics phần 4 ppt

THE (a, b, 1) CLASS 121 If the original values were ail available, then the zero-truncated probabilities could have all been obtained by multiplying the original values by 1/(1 - 0.362887) = 1.569580. For the zero-modified random variable, pf = 0.6 arbitrarily. From (5.4), pr = (1 - 0.6)(0.302406)/(1 - 0.362887) = 0.189860. Then p? = 0.189860 (5 + $+) = 0.110752, py = 0.110752 (5 + $4) = 0.055376. In this case, each original negative binomial probability has been multiplied by (1 - 0.6)/(1 - 0.362887) = 0.627832. Also note that, for j 2 1, py = 0.4~;. 0 Although we have only discussed the zero-modified distributions of the (a, b, 0) class, the (a, b, 1) class admits additional distributions. The (a, b) parameter space can be expanded to admit an extension of the negative bi- nomial distribution to include cases where -1 < T < 0. For the (a, b, 0) class, T > 0 is required. By adding the additional region to the sample space, the “extended” truncated negative binomial (ETNB) distribution has parameter restrictions ,B > 0, T > -1, T # 0. To show that the recursive equation pk=pk-l US- , k=2,3 , , (5.8) ( 3 with po = 0 defines a proper distribution, it is sufficient to show that for any value of pl , the successive values of pk obtained recursively are each positive and that C&pk < co. For the ETNB, this must be done for the parameter space (see Exercise 5.5). When T f 0, the limiting case of the ETNB is the logarithmic distribution with (see Exercise 5.6). The pgf of the logarithmic distribution is (5.10) (see Exercise 5.7). The zero-modified logarithmic distribution is created by assigning an arbitrary probability at zero and reducing the remaining proba- bilities. 122 MODELS FOR THE NUMBER OF LOSSES: COUNTlNG DISTRIBUTIONS It is also interesting that the special extreme case with -1 < r < 0 and p -+ 00 is a proper distribution, sometimes called the Sibuya distribution. It has pgf P(z) = 1 - (1 - z)-~, and no moments exist (see Exercise 5.8). Distributions with no moments are not particularly interesting for modeling loss numbers (unless the right tail is subsequently modified) because an infinite number of losses are expected. If this is the case, the risk manager should be fired! Example 5.7 Determine the probabilities for an ETNB distribution with r = -0.5 and /3 = 1. Do this both for the truncated version and for the modified version with pf = 0.6 set arbitrarily. We have a = 1/(1 + 1) = 0.5 and b = (-0.5 - 1)(1)/(1 + 1) = -0.75. We also have pr = -0.5(1)/[(1 + l)0.5 - (1 + l)] = 0.853553. Subsequent values are p; = 0.5 - - (0.853553) = 0.106694, ( O,,) p; = ( 0.5 - - O,,) (0.106694) = 0.026674. For the modified probabilities, the truncated probabilities need to be multi- plied by 0.4 to produce py = 0.341421, py = 0.042678, and py = 0.010670. Note: A reasonable question is to ask if there is a “natural” member of the ETNB distribution, that is, one for which the recursion would begin with pl rather than pa. For that to be the case, the natural value of po would have to satisfy pl = (0.5 - 0.75/l)p0 = -0.25~0. This would force one of the two probabilities to be negative and so there is no acceptable solution. It is easy 0 to show that this occurs for any r < 0. There are no other members of the (a, b, 1) class beyond those discussed above. A summary is given in Table 5.4. 5.7 COMPOUND FREQUENCY MODELS A larger class of distributions can be created by the processes of compounding any two discrete distributions. The term compounding reflects the idea that the pgf of the new distribution P(z) is written as P(z) = PrV LPM (z)] 7 (5.11) where PN(z) and PM (z) are called the primary and secondary distributions, respectively. The compound distributions arise naturally as follows. Let N be a count- ing random variable with pgf PN(z). Let MI, M2,. . . be identically and COMPOUND FREQUENCY MODELS 123 Table 5.4 Members of the (a, b, 1) class a b Parameter space Distributiona Po Poisson e-' 0 X X>O ZT Poisson 0 0 X X>O ZM Poisson Arbitrary 0 X X>O Binomial (1 -qy (m + 1)- O<q<l ZT binomial 0 (m + 1)- O<q<l ZM binomial Arbitrary (m + 1)- O<q<l 4 4 1-4 1'4 '74 1-4 _- 1-4 1-4 P (r - 1)- P r>O,P>o 1+P Negative binomial (1 + P)-' - ETNB 0 P r>-l,b P>o 1+P (r - 1)- P 1+P r > I,~ P > o P 1+P (r - 1)- P Arbitrary - 1+P l+P ZM ETNB Geometric (1 + PY 1-tp Po P>O ZT geometric 0 Lo P>O Arbitrary ~ Po P>O 1+P 1+D ZM geometric P>O P>O P 1+P P 1+P p 1+P P ZM logarithmic Arbitrary - l+B Logarithmic 0 aZT = zero truncated, ZM = zero modified. bExcluding T = 0, which is the logarithmic distribution. independently distributed counting random variables with pgf PM (2). As- suming that the Mjs do not depend on N, the pgf of the random sum S = MI + M2 +. . + MN (where N = 0 implies that S = 0) is Ps ( Z) = PN [PM (2)). This is shown as 124 MODELS FOR THE NUMBER OF LOSSES: COUNTlNG DlSTRlBUTlONS k=O n=O n=O k=O 33 n=O = piv [PM (.)I. In operational risk contexts, this distribution can arise naturally. If N repre- sents the number of loss-causing events and {Mk; k = 1,2,. . . , N} represents the number of losses (errors, injuries, failures, etc.) from the events, then S represents the total number of losses for all such events. This kind of in- terpretation is not necessary to justify the use of a compound distribution. If a compound distribution fits data well, that may be enough justification itself. Also, there are other motivations for these distributions, as presented in Section 5.9. Example 5.8 Demonstrate that any zero-modified distribution is a compound distribution. Consider a primary Bernoulli distribution. It has pgf PN(z) = 1 - q + 42. Then consider an arbitrary secondary distribution with pgf PM(z). Then, from formula (5.11) we obtain PS(z) = PN[PM(z)] = 1 - q + qpM(z). From formula (5.3), it is clear that this is the pgf of a ZM distribution with q = That is, the ZM distribution has assigned arbitrary probability py at zero, while PO is the probability assigned at zero by the secondary distribution. 0 M 1 - Po 1 -Po Example 5.9 Consider the case where both M and N have the Poisson dis- tribution. Determine the pgf of this distribution. This distribution is called the Poisson-Poisson or Neyrnan Type A distri- bution. Let PN(z) = e and Phf(z) = eA2('-'). Then COMPOUND FREQUENCY MODELS 125 When X2 is a lot larger than XI (for example, XI = 0.1 and Xz = 10) the 17 resulting distribution will have two local modes. Example 5.10 Demonstrate that the Poisson-logarithmic distribution is a negative binomia1,as compound Poisson-logarithmic distribution. The negative binomial distribution has pgf P(.) = [l - P(. - 1)]-r. Suppose PN(z) is Poisson(X) and PM(z) is logarithmic(P); then eV[P&f(.)l = .xp{"IM(.) - 11) -A/ ln(l+B) = [l - P(. - l)] = [l - P(. - 1)]-T, where r = A/ ln(l+P). This shows that the negative binomial distribution can be written as a compound Poisson distribution with a logarithmic secondary distribution. 0 The above example shows that the "Poisson-logarithmic" distribution does not create a new distribution beyond the (a, b, 0) and (a, b, 1) classes. As a result, this combination of distributions is not useful to us. Another com- bination that does not create a new distribution beyond the (a, b, 1) class is the compound geometric distribution where both the primary and secondary distributions are geometric. The resulting distribution is a zero-modified geo- metric distribution, as shown in Exercise 5.12. The following theorem shows that certain other combinations are also of no use in expanding the class of dis- tributions through compounding. Suppose Ps(z) = p~[Pj~(z); 01 as before. Now, PM(T) can always be written as hf(z) = fo + (1 - fo)Pif(.) (5.12) where P&(z) is the pgf of the conditional distribution over the positive range (in other words, the zero-truncated version). Theorem 5.11 Suppose the pgfPN(z; 0) satisfies PN(z; 0) = B[O(Z - l)] for some parameter 0 and some function B(z) that is independent of0. That is, the parameter 0 and the argument z only appear in the pgf as O(z - 1). 126 MODELS FOR THE NUMBER OF LOSSES: COUNTING DISTRIBUTIONS There may be other parameters as well, and they may appear anywhere in the pgf. Then Ps(z) = Plv[P~(z); 61 can be rewritten as Proof: This shows that adding, deleting, or modifying the probability at zero in the secondary distribution does not add a new distribution because it is equivalent to modifying the parameter 6 of the primary distribution. This means that, for example, a Poisson primary distribution with a Poisson, zero-truncated Poisson, or zero-modified Poisson secondary distribution will still lead to a Neyman Type A (Poisson-Poisson) distribution. 5.8 RECURSIVE CALCULATION OF COMPOUND PROBABILITIES The probability of exactly k losses can be written as 02 Pr(S = IC) = C Pr(S = kj~ = n) Pr(N = n) n=O 30 = C Pr(Ml+ . . . + M~ = IC~N = n) Pr(N = n) n=O 30 = C Pr(M1+ . . . + M~ = IC) Pr(N = n). (5.13) n=O Letting gn = Pr(S = n), p, = Pr(N = n), and f, = Pr(M = n), this is rewritten as gk xl)nfzn (5.14) where fin, k = O,l,. , is the n-fold convolution of the function fk, k = 0,1,. . ., that is, the probability that the sum of n random variables which are each independent and identically distributed (iid) with probability function fk will take on value k. oi) n=O RECURSIVE CALCULATION OF COMPOUND PROBABILITIES 127 When P~y(z) is chosen to be a member of the (a, b, 0) class, pk= U-k- Pk-1, k=1,2 , , (5.15) and a simple recursive formula can be used. This formula avoids the use of convolutions and thus reduces the computations considerably. Theorem 5.12 (Panjer recursion) If the primary distribution is a member of the (a, b, 0) class, the recursive formula is ( 3 (5.16) Proof: From formula (5.15), TZP~ = a(n - l)pn-I+ (U + b)pn-l. Multiplying each side by [P~(z)]~-lPh(z) and summing over n yields n=l n=l 33 Because Ps (z) = C,"==, p, [ Pm (z)]", the previous equation is 128 MODELS FOR THE NUMBER OF LOSSES: COUNTING DlSTRlBUTlONS Therefore, Rearrangement yields the recursive formula (5.16). 0 This recursion (5.16) has become known as the Panjer recursion after its introduction as a computational tool for aggregate losses by Panjer [88]. Its use here is numerically equivalent to its use for aggregate losses in Chapter 6. In order to use the recursive formula (5.16), the starting value go is required and is given in Theorem 5.15. Theorem 5.13 If the primary distribution is a member of the (a, b, 1) class, the recursive formula is [pl - (a + b)PO]fk + c:=, (a + bj/k) fjgk-j gk = , k = 1,2,3 , . (5.17) Proof: It is similar to the proof of Theorem 5.12 and is left to the reader. 0 Example 5.14 Develop the Panjer recursive formula for the case where the primary distribution is Poisson. 1 - afo In this case a = 0 and b = A, yielding the recursive form The starting value is, from (5.11), go = Pr(S = 0) = P(O) = evjPM(0)l = PN(f0) - - ,-W-fo). (5.18) Distributions of this type are called compound Poisson distributions’. When the secondary distribution is specified, the compound distribution is called 0 Poisson-X, where X is the name of the secondary distribution. The method used to obtain go applies to any compound distribution. Theorem 5.15 For any compound distribution, go = PN( fo), where PN(z) is the pgf of the primary distribution and fo is the probability that the secondary distribution takes on the value zero. ‘In some textbooks, the term conipound distribution, as in “compound Poisson,” refers to what are called in this book “mixed distributions.” RECURSIVE CALCULATION Of COMPOUND PROBABILITIES 129 Proof: See the second line of equation (5.18). 0 Note that the secondary distribution is not required to be in any special form. However, to keep the number of distributions manageable, secondary distributions will be selected from the (a, b, 0) or the (a, b, 1) class. Example 5.16 Calculate the probabilities for the Poisson-ETNB distribution where X = 3 for the Poisson distribution and the ETNB distribution has r = -0.5 and fl= 1. From Example 5.7 the secondary probabilities are fo = 0, f1 = 0.853553, f2 = 0.106694, and f3 = 0.026674. From equation (5.18), go = exp[-3(1 - O)] = 0.049787. For the Poisson primary distribution, a = 0 and b = 3. The recursive formula (5.16) becomes 3j k k c j = 1 (3j lk) fj gk- j - - c Tfjgk-3. j=1 1 - O(0) gk = Then, 30) 3(1) 3(2) 91 = -0.853553(0.049787) = 0.127488, 1 2 2 3 3 3 92 = -0.853553(0.127488) + -0.106694(0.049787) = 0.179163, g3 = "00.853553(0.179163) + -0.106694(0.127488) 3(2) +-0.026674(0.049787) 3(3) = 0.1841 14. r- U The following example uses the Panjer recursion to illustrate the equiva- lence between the Poisson-X and Poisson-zero-modified-X distributions, where X can be any distribution. Example 5.17 Determine the probabilities for a Poisson-zero-modified ETNB distribution where the parameters are X = 7.5, pf = 0.6, r = -0.5, and /3 = 1. From Example 5.7 the secondary probabilities are fo = 0.6, fl = 0.341421, f2 = 0.042678, and f3 = 0.010670. From equation (5.18), go = exp[-7.5(1 - 0.6)] = 0.049787. For the Poisson primary distribution, a = 0 and b = 7.5. The recursive formula (5.16) becomes 130 MODELS FOR THE NUMBER OF LOSSES: COUNTING DISTRIBUTIONS Then, 91 = __ 7'5(1) 0.341421 (0.049787) = 0.127487, 1 92 = - 7~5~(')0.341421(0.127487) + 7'5(2)0.042678(0.049787) = 0.179161, 93 = - 7~5(1)0.341421(0.179161) + 7'5(2) 0.042678(0.127487) 2 3 3 +- 7.5(3) 0.010670(0.049787) = 0.184112. Except for slight rounding differences, these probabilities are the same as those obtained in Example 5.16. 0 5.9 AN INVENTORY OF DISCRETE DISTRIBUTIONS In the previous sections of this chapter, we have introduced the simple (a, b, 0) class, generalized to the (a, b, 1) class, and then used compounding to create a larger class of distributions. In this section, we summarize the distributions introduced in those sections. There are relationships among the various distributions similar to those of Section 4.3.2. The specific relationships are given in Table 5.5. It is clear from earlier developments that members of the (a, b,O) class are special cases of members of the (a, b, 1) class and that zero-truncated distributions are special cases of zero-modified distributions. The limiting cases are best discovered through the probability generating function, as was done on page 113, where the Poisson distribution is shown to be a limiting case of the negative binomial distribution. We have not listed compound distributions where the primary distribution is one of the two parameter models such as the negative binomial or Poisson- inverse Gaussian. This was done because these distributions are often them- selves compound Poisson distributions and, as such, are generalizations of distributions already presented. This collection forms a particularly rich set of distributions in terms of shape. However, many other distributions are also possible. Many others are discussed in Johnson, Kotz, and Kemp [65], Douglas [24], and Panjer and Willmot [93]. 5.9.1 The distributions in this class have support on 0,1,. . . . For this class, a particular distribution is specified by setting PO and then using pk = (a + b/k)pk-l. Specific members are created by setting PO, a, and b. For any member, p(1) = (a+b)/(l-a), and for higherj, = (aj+b)p~(~-~)/(l-a). The variance is (a + b)/(l - u)~. The (a, b, 0) class [...]... Poisson distribution with A2 = 3 and secondary distribution 42 (2) = 0.25,q3(2) = 0.6, and q4(2) = 0.15 Determine the distribution of S = S + S 1 z FURTHER PROP€RJ/ES OF THE COMPOUND POISSON CLASS We have X = X I 141 + X2 = 2 + 3 = 5 Then 41 q 2 43 q4 + + + + = 0 .4( 0.2) 0.6(0) = 0.08, = 0 .4( 0.7) O.s(O.25) = 0 .43 , = 0 .4( 0.1) O.S(O.6) = 0 .40 , = 0 .4( 0) 0.6(0.15) = 0.09 Thus, S has a compound Poisson distribution... third central moment are 0.12 546 14, 0.1299599, and 0. 140 1737, respectively For these numbers, p3 + 2p = 7. 543 865 - 3a2 (a2- P I 2 / P From among the Poisson-binomial, negative binomial, Polya-Aeppli, Neyman Type A, and Poisson-ETNB distributions, only the latter is appropriate For this distribution, an estimate of r can be obtained from r+2 7. 543 865 = r f l resulting in r = -0. 847 1851 In Example 12.13 a... p = 0. 94, il = 0.11, and i 2 = 0.70 This means that about 94% of drivers were “good” with a risk of XI = 0.11 expected accidents per year and 6% were “bad” with a risk of A2 = 0.70 expected accidents per year 0 This example illustrates two important points about finite mixtures First, the model is probably oversimplified in the sense that risks (e.g., drivers) probably exhibit a continuum of risk levels... in operational risk modeling, we will not compute risk measures for the number of losses, although this is possible We are generally interested in the sum of the total losses measured in dollar terms A Tail-Value-at -Risk measure for the number of losses would give the expected number of losses conditional on the fact that a certain quantile was exceeded This may be useful for gaining insight into risk. .. compound Poisson class of probability distributions is the fact that it is closed under convolution We have the following theorem  140 MODELS FOR THE NUMBER OF LOSSES: COUNTING DISTRIBUTIONS Table 5.6 Hossack et al data No of cars No of losses 565,6 64 68,7 14 5,177 365 24 6 0 0 1 2 3 4 5 6+ Theorem 5.20 Suppose that S has a compound Poisson distribution with i Poisson parameter X i and secondary distribution... applies equally to discrete distributions Theorem 5 .40 Let X be a member of the AEDF subject to the above conditions Then the Tuil-Value- &Risk can be written as +h TVaR,(X) = p where o2 = 1 / X and h = & In (F (xp; A)) 8, 0 1 54 MODELS FOR THE NUMBER OF LOSSES: COUNTING DlSTRlBUTlONS Example 5 .41 Obtain the TVuRfor the Poisson distribution The Tail-Value-at -Risk is p +h where d h = - In (F (x,;8, , A)) a@... of this distribution is M ( t )= e x p { -P [ ( l - - 4 2pt)1/2 - 11) Hence, the inverse Gaussian distribution is infinitely divisible ( [ M ( t )l ]n is / the mgf of an inverse Gaussian distribution with p replaced by p / n ) From (formula 5. 24) with X = 1, the pgf of the mixed distribution is 41 1'2 - 1)) EFFECT OF EXPOSURE ON LOSS COUNTS 149 Table 5.7 Pairs of compound and mixed Poisson distributions... class for most problems in modeling count data 5. 14 EFFECT OF EXPOSURE ON LOSS COUNTS Assume that the current set of risks consists of n entities, each of which could produce losses Let Nj be the number of losses produced by the j t h entity Then N = N1 N, If we assume that the Nj are independent and identically distributed, then + + PrV(z)= [PN,(z)ln Now suppose the set of risks is expected to expand... two-point mixture Example 5.26 Suppose risks can be classified as "good risks" and "bad risks, " each group with its own Poisson distribution Determine the p f for this model and fit it t o the data from Example 11.5 This model and its application t o the data set are from Trobliger [118] in connection with automobile drivers From formula (5.23) the pf is POISSON MIXTURES 145 The maximum likelihood estimates3... O.s(O.25) = 0 .43 , = 0 .4( 0.1) O.S(O.6) = 0 .40 , = 0 .4( 0) 0.6(0.15) = 0.09 Thus, S has a compound Poisson distribution with Poisson parameter X = 5 and secondary distribution q1 = 0.08,qz = 0 4 3 , ~ 0 .40 , and 4 4 = 0.09 = Numerical values of the distribution of S may be obtained using the recursive formula beginning with P r ( S = 0) = e P 5 In various situations the convolution of negative binomial . 7'5(1) 0. 341 421 (0. 049 787) = 0.12 748 7, 1 92 = - 7~5~(')0. 341 421(0.12 748 7) + 7'5(2)0. 042 678(0. 049 787) = 0.179161, 93 = - 7~5(1)0. 341 421(0.179161) + 7'5(2) 0. 042 678(0.12 748 7). -0.853553(0. 049 787) = 0.12 748 8, 1 2 2 3 3 3 92 = -0.853553(0.12 748 8) + -0.1066 94( 0. 049 787) = 0.179163, g3 = "00.853553(0.179163) + -0.1066 94( 0.12 748 8) 3(2) +-0.0266 74( 0. 049 787). + 1)- O<q<l ZM binomial Arbitrary (m + 1)- O<q<l 4 4 1 -4 1&apos ;4 ' 74 1 -4 _- 1 -4 1 -4 P (r - 1)- P r>O,P>o 1+P Negative binomial (1 + P)-'