tailieunhanh - Matematik simulation and monte carlo with applications in finance and mcmc phần 10

tầng lớp nhân dân' điểm 'ước tính giá = 100 = 1,919506766' ước tính tiêu chuẩn lỗi '= 0,00006567443424 13753? ¼ 0:01089879722 'K' = 45 ', sigma = 0,1,' n '= 16' bản sao '= 100, đường dẫn cho mỗi bản sao = 2500,' tầng lớp nhân dân 'điểm' ước tính giá = 100 = 6,055282128 ' ước tính tiêu chuẩn lỗi '= 0,0001914854216 | Appendices 301 Sample 100000 variates starting with initial value of 0 to produce a histogram showing their distribution together with sample mean and variance. seed randomize 59274 x mcmc -4 3 100000 res seq op 2 x t t histogram res title distribution of sampled x-values labels x density numbars 100 Mean describe mean res StdDev describe standarddeviation res seed 59274 Mean StdDev Reliability inference We are interested in determining survival probabilities for components. The age at failure of the ith component is X i 1 2 . These are . 302 Appendices Weibull random variables with parameters a and fl where the joint prior distribution is 4 . 1 1000 _ 1 r ã 7 1 a 1 5 4 1 1000 2 _ r õ 7 1 5 a and 2000 r 1 a 1 5 3000 r 1 a 1 . Then the joint posterior density given the data x i 1 . ng is a fl anfl na exp X1 X 1g o . Using an independence sampler with the proposal density equal to the joint prior and given that the current point is a fl the acceptance probability for a candidate point oc flc is 9 min 1 o 4c na exp 717 1 X1 Xn c 1 o exp 1 1 fl Mi 1 X1 . XnỴ-1 The candidate point is generated from the joint prior by noting that the marginal prior of ac is a symmetric triangular density on support 1 2 . Therefore given two uniform random numbers R1 and R2 we set Oc 1 R1 R2 . Given ac the conditional prior density of flc is i 2000 3000 A r 1 oc 1 r 1 oc 1 J so we set _ 1000 2 R3 r 1 oc 1 . The procedure fail performs k iterations of this independence sampler. It returns a matrix C having k rows and q 2 columns. The first q elements of the ith row give sampled survivor probabilities at specified ages y 1 . y q . The last two elements of the ith row give the sampled alpha and beta at this ith iteration point . The plots show that the burn-in time .