PQ-4-Answer-331-2011-F

PQ-4-Answer-331-2011-F - 02 x t t p 00 x t>(d...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Answers to Practice Questions 4 – ACTSC 331, FALL 2011 1. (a) For t > 0, t p 00 x = t p 00 x = e - R t 0 ( μ 01 x + s + μ 02 x + s ) ds , t p 11 x = 1, t p 22 x = 1, t p 11 x = 1, and t p 22 x = 1. (b) Kolmogorov’s forward equation is d dt t p 01 x = μ 01 x + t t p 00 x , t > 0 , and the solution is t p 01 x = R t 0 s p 00 x μ 01 x + s ds for t > 0. (c) Kolmogorov’s forward equation is d dt t p 02 x = μ 02 x + t t p 00 x , t > 0 , and the solution is t p 02 x = R t 0 s p 00 x μ 02 x + s ds, t > 0. (d) i. The probability is 10 p 00 60 = 0 . 74082. ii. The probability is 10 p 01 60 = 0 . 172788. iii. The probability is 10 p 02 60 = 0 . 0863939. iv. π = 10000 ¯ A 02 60: 10 + 20000 ¯ A 01 60: 10 ¨ a 00 60: 10 = 480 . 52 . 2. (a) For t > 0, t p 00 x = t p 00 x = e - R t 0 ( μ 01 x + s + μ 02 x + s ) ds , t p 11 x = t p 11 x = e - R t 0 μ 13 x + s ds , t p 22 x = t p 33 x = t p 22 x = t p 33 x = 1 . (b) Kolmogorov’s forward equation is d dt t p 01 x = - μ 13 x + t t p 01 x + μ 01 x + t t p 00 x , t > 0 . (c) Kolmogorov’s forward equation is d dt t p 02 x = μ
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 02 x + t t p 00 x , t > . (d) Kolmogorov’s forward equation is d dt t p 13 x = μ 13 x + t t p 11 x , t > . (e) Kolmogorov’s forward equation is d dt t p 03 x = μ 13 x + t t p 01 x , t > . (f) i. The probability is 10 p 00 40 = 0 . 67032 . ii. The probability is 10 p 02 40 = 0 . 08242 . iii. The probability is 10 p 03 40 = 0 . 0187944 . iv. The probability is 10 p 13 40 = 0 . 139292 . v. π = 50000 ¯ A 02 40: 10 + 30000 ¯ A 03 40: 10 ¯ a 00 40: 10 = 562 . 20 . 1...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online