1 we get since derivative of a constant is zero 0 q

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: and letting t ! 1, we get (since, derivative of a constant is zero), 0 = Q with the requirement that probabilities sum up to 1 1 + 2 + ::: = 1: CS 522, v 0.94, d.medhi, W’99 8 if e is a column vector of the same dimension as  with each component being 1, we can write this in compact form as e = 1. Note that 0 = Q e = 1 6 is a linear system of equation which has a unique solution (under certain conditions). To illustrate, suppose that we have a system that takes three values (0, 1, 2). The probabilities are:  = 0 1 2 and Q = 0 is 2 ,q q 0 01 0 1 2 4 q10 ,q1 q20 and e = 1 is q21 3 q02 q12 5 = ,q2 000 0 + 1 + 2 = 1: If we know q ’s, then we can solve for  . 5. Birth-and-Death Process This is a special case of continuous-time Markov chain. You can only transition to a neighbor (one step away) or stay where you are in an infinitesimal time period, and the rate is state-dependent, one defined for arrival and the other for departure. = ,1 =  q =0 qi i+1 i qi i i ij i=0 1 2 ::: i = 1 2 ::: for j 6= i + 1 i , 1 7 i: Due to this situation and the requirement that the row sum to be zero (from (3)), we have qi = ,qii = i + i for i not in boundary state : Now, what d...
View Full Document

Ask a homework question - tutors are online