Writing f x for the density function of tq on 0 1 we

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5 0 0 1 2 3 1 0 0 0 2/5 0 3/5 0 2/5 0 1 0 Let g (i) = Ei V0 . g (0) = 0. Taking into account the rates at which jumps occur we have 1 2 /5 + (2) 5 g 13 2/5 g (2) = + g (1) + (3) 55 g 1 g (3) = + g (2) 3 g (1) = Inserting the last equation in the second one: g (2) = 13 2 2 + g (1) + + g (2) 55 15 5 or (3/5)g (2) = (1/3) + (3/5)g (1). Multiplying by 2/3’s and inserting this in the first equation we have 122 g (1) = + + g (1) 595 so (3/5)g (1) = 19/45 and g (1) = 19/27. To develop the analogue of (4.19) for exit times we note that if g (i) = Ei VA then g (i) = 0 for i 2 A, and for i 62 A g (i) = 1 i Multiplying each side by + X q (i, j ) g (j ) i j 6=i Q(i, i) we have X Q(i, i)g (i) = 1 + Q(i, j )g (j ) i = j 6=i which simplifies to X j Q(i, j )g (j ) = 1 for i 62 A. 136 CHAPTER 4. CONTINUOUS TIME MARKOV CHAINS Writing 1 for a vector of 1’s the solution, and writing R for the part of Q with i, j, 2 Ac : g = R 11 (4.21) In the barbershop example, the matrix R is 1 2 3 which has R 1 1 5 3 0 0 1/3 = @1/3 1/3 2 2 5 3 2/9 5/9 5/9 3 0 2 3 1 4/27 10/27A 19/27 Adding the entries on the first row we find g (1) = 1/...
View Full Document

Ask a homework question - tutors are online