Operational Risk Modeling Analytics phần 2 ppsx

... PROBABILITY CONCEPTS 20 0 150 2. 100 50 0 -50 1 I 0 50 100 1 50 20 0 25 0 300 X Fig. 2. 10 Excess loss variable -50 I I 0 50 100 150 20 0 25 0 300 X f;g. 2. 11 Left censored ... E(Xf) = 1 x2(0.01)dx = 3,333.33, 100 dx = 1,000, (. + 2, 000)4 dx = 4,000,000, O0 3 (2, 000)3 (x + 2, 000)4 E(X;) -1 x2 E(X3) = O(0.5) + l(0 .25 ) +...

Ngày tải lên: 09/08/2014, 19:22

... 191 191 1 92 1 94 198 198 20 5 20 5 20 7 20 8 20 8 21 0 21 3 21 4 21 7 21 9 22 1 22 1 22 2 22 6 22 7 22 8 22 9 23 0 23 0 vi CONTENTS 2. 4 Quantiles of a distribution 2. 5 Generating ... ’ theorem 10.5 .2 Inference and prediction 10.5.3 Computational issues 10.6 Exercises 26 7 26 7 26 8 26 8 26 9 2 75 27 7 28 0 28 3 28 3 28 6 28 9 28 9...

Ngày tải lên: 09/08/2014, 19:22

... Percentile matching, 26 5 Plot denszty Junction, 324 difference, 926 dzstribution funtion, 323 Poznt estimation, 24 6 Posterior distribution, 28 2 Predictzve distribution, 28 2 Prior distribution ... 137, 1 92, 304-305, 31 3 Polya- Aeppli, 125 Polya- Eggenberger, 130 posterior, 28 2 predictive, 28 2 prior, 28 1 scale, 67 Sibuya, 110 spliced, 83 subexponential, 1...

Ngày tải lên: 09/08/2014, 19:22

... first three recursions yield pi = 0.3 628 87 (5 + it) = 0.3 024 06, p2 = 0.3 024 06 (5 + f f) = 0.176404, p3 = 0.176404 (5 + ii) = 0.08 820 2. For the zero-truncated random variable, ... be reversed), For the variance, Var(X) = E(X2) - [E(X)I2 = EIE(X21A>l - ~ElE(Xl~)l )2 = E(Var(X1A) + [E(X/A) ]2} - {E[E(XlA)] }2 = E[Var(X/A)] + Var[E(X/A)]....

Ngày tải lên: 09/08/2014, 19:22

... = 3a2 - 2p + 2 - 'I2 P 3 (a2 - p )2 Polya-Aeppli: p3 = 3a2 - 2p + P Neyman Type A: p3 = 3g2 - T + 2 ( 02 - p )2 r+l p Poisson-ETNB: p3 = 3a2 - 2p + ... 0.341 421 (0.049787) = 0. 127 487, 1 92 = - 7~5~(')0.341 421 (0. 127 487) + 7'5 (2) 0.0 426 78(0.049787) = 0.179161, 93 = - 7~5(1)0.341 421 (0.179161) + 7'5...

Ngày tải lên: 09/08/2014, 19:22

... (x) 1 2 3 4 5 6 7 8 9 10 0.150 0 .20 0 0 .25 0 0. 125 0.075 0.050 0.050 0.050 0. 025 0. 025 Furthermore, the frequency distribution is given in Table 6 .2. Table 6 .2 Frequency ... - 1 /2) h]"I h(k + 1) P3 + For k = 1 this reduces to 22 h 0 < 2 I-, 2 41 -90) - h("0 -go), PO - go) + ihpi + Pj h j-1 x2 - [(j - 1 /...

Ngày tải lên: 09/08/2014, 19:22

... distributions 25 2 MULTIVARIATE MODELS 0 02 04 0.6 OB 1 0 02 04 0.6 08 1 U U Fig. 8.15 BB6 copula density (6 = 2, 6 = 2) Fig. 8.16 BB6 copula pdf (6 = 2, 6 = 2) It ... = F2(X2). Because U and V are both Uniform (0,l) random variables with mean 1 /2 and variance 1/ 12, we can rewrite Spearman’s rho as E [~l(X1) ~2( X2)1 - E[Fl(Xl)IE...

Ngày tải lên: 09/08/2014, 19:22

... sample mean, 4 4 02 o2 Var(2X) = - var(x) = - = n 12n 3n For the adjusted maximum, the second moment is (n+ no2 (n+ 1 )20 2 ~- - E [ (FV,.)?] = - n2 n +2 (n +2) n for a variance ... (Inxj - p )2 a3 dl The second partial derivatives are d21 n - = 8/ 42 a2, n d2 1 lnzj -p - =-2c ar7dP j=1 a3 ’ 3c an2 a2 j=l r74 n d21 n (Inxj -...

Ngày tải lên: 09/08/2014, 19:22

... 2. 13~10-~’ 4 .22 ~ 5.63xlO-” 5 .29 ~ 1.89~ 2. 40~ lo-’‘ 6.16~10-~~ 1 .29 ~ 2. 26 x 9.33~10-~~ 3.64x 10 -21 7.57x10 -22 0.0000 0.0000 0.0003 0.0038 0. 023 6 0.0867 0.1718 0 .22 93 0 .21 56 ... 1 2 3 4 5 6 7f 20 ,5 92 2, 651 29 7 41 7 0 1 0 20 , 420 .9 2, 945.1 21 2.4 10 .2 0.4 0.0 0.0 0.0 20 ,596.8 2, 631.0 318.4 37.8 4.4 0.5 0.1 0.0...

Ngày tải lên: 09/08/2014, 19:22

... Exercise 12. 21 Model No. of parameters Negative loglikelihood Generalized Pareto 3 21 9.1 Burr 3 21 9 .2 Pareto 2 22 1 .2 Lognormal 2 221 .4 Inverse exponential 1 22 4.3 12. 20 From ... 14,075 2 1,766 3 25 5 4 45 5 6 6 2 7+ 0 1 02, 629 .6 15, 922 .0 1 ,23 5.1 63.9 2. 5 0.1 0.0 0.0 103, 723 .6 13,989.9 1,857.1 24 5 .2 32. 3 4 .2 0.6 0....

Ngày tải lên: 09/08/2014, 19:22