Lecture 12

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

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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...
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